"General mathematics"
AUG 1
2 9
1950
Southern Branch
of the
University of California
Los Angeles
Form L 1
This book is DUE on the last date stamped below
JUL 8 1*2?
JUL 1 8 W
I 2
2
7 1
8
'JUL 2 9 1953
MAN 2 7 1954
1962
- 2 3 1979
n L-9-15m-8,'26
GENERAL MATHEMATICS
BY
RALEIGH SCHORLING
IN CHARGE OF MATHEMATICS, THE LINCOLN SCHOOL OF TEACHERS
COLLEGE, NEW YORK CITY
AND
WILLIAM DAVID REEVE
TEACHERS' TRAINING COURSE IN MATHEMATICS IN THE COLLEGE
OF EDUCATION, ANI> HEAD OF THE MATHEMATICS DEPART-
MENT IN THE UNIVERSITY HIGH SCHOOL
THE UNIVERSITY OF MINNESOTA
GINN AND COMPANY
BOSTON NEW YORK CHICAGO LONDON
ATLANTA DALLAS COLUMBUS SAN FRANCISCO
2883
171-50
m
COPYRIGHT, 1919, BY
RALEIGH SCHORLING AND WILLIAM DAVID REEVE
ALL, RIGHTS RESERVED
819.11
7171
gtbenaum
GINN AND COMPANY PRO-
PRIETORS BOSTON U.S.A.
''A-
PREFACE
The purpose of this book, as implied in the introduction,
is as follows : to obtain a vital, modern scholarly course in
introductory mathematics that may serve to give such careful
training in quantitative thinking and expression as well-
informed citizens of a democracy should possess. It is, of
course, not asserted that this ideal has been attained. Our
achievements are not the measure of our desires to im-
prove the situation. There is still a very large " safety
factor of deud wood" in this text. The material purposes
to present such simple and significant principles of algebra,
geometry, trigonometry, practical drawing, and statistics,
along with a few elementary notions of other mathematical
subjects, the whole involving numerous and rigorous appli-
cations of arithmetic, as the average man (more accurately
the modal man) is likely to remember and to use. There
is here an attempt to teach pupils things worth knowing
and to discipline them rigorously in things worth doing.
The argument for a thorough reorganization need not
be stated here in great detail. But it will be helpful to
enumerate some of the major errors of secondary-mathe-
matics instruction in current practice and to indicate
briefly how this work attempts to improve the situation.
The following serve to illustrate its purpose and program :
1. The conventional first-year algebra course is charac-
terized by excessive formalism; and there is much drill
work largely on nonessentials. The excessive formalism is
iv GENERAL MATHEMATICS
greatly reduced in this text and the emphasis placed on
those topics concerning which there is general agreement,
namely, function, equation, graph, and formula. The time
thus gained permits more ample illustrations and applica-
tions of principles and the introduction of more significant
material.
2. Instead of crowding the many difficulties of the
traditional geometry course into one year, geometry in-
struction is spread over the years that precede the formal
course, and the relations are taught inductively by experi-
ment and by measurement. Many foreign schools and an
increasing number of American schools proceed on this
common-sense basis. This gives the pupil the vocabulary,
the symbolism, and the fundamental ideas of geometry.
If the pupil leaves school or drops mathematics, he never-
theless has an effective organization of geometric relations.
On the other hand, if he later pursues a formal geometry
course, he can work far more effectively because he can
concentrate on the logical organization of space relations
and the formal expression of these relations. The longer
" time exposure " minimizes the difficulties met in begin-
ning the traditional geometry courses and avoids the
serious mistake of forcing deductive logic and philosophic
criticism in these early years.
3. The traditional courses delay the consideration of
much interesting and valuable material that the field of
secondary mathematics has to offer, and which may well
be used to give the pupil very early an idea of what
mathematics means and something of the wonderful scope
of its application. The material of the seventh, eighth, and
ninth years is often indefensibly meaningless when com-
pared with that of many foreign curricula. Trigonometry,
containing many easy real problems, furnishes a good ex-
ample of this delay. Other examples are found in the
use of logarithms, the slide rule, standardized graphical
methods, the notion of function, the common construction
of practical drawing, the motivation of precise measure-
ment, a study of the importance of measurement in modern
life, and the introductory ideas of the calculus. It appears
that the mathematics student should be given an oppor-
tunity to use these important tools very early in his study.
They lend to the subject a power and interest that drills on
formal material cannot possibly give.
Particular emphasis is given to graphical representation
of statistics. The growing complexity of our social life
makes it necessary that the intelligent general reader
possess elementary notions of statistical methods. The
hundreds of articles in the current magazines so exten-
sively read demand an elementary knowledge of these
things in order that the pupil may not remain ignorant of
the common, everyday things of life. Brief chapters on
logarithms and the slide rule have been introduced in
order that a greater number of students may use these
practical labor-saving devices and in order that these devices
may function in the student's subsequent work, whether
in everyday life or in the classroom. Actual classroom
experience with these chapters has proved them to be
relatively simple and good material for eighth-grade and
ninth-grade students.
4. Mathematics needs to be reorganized on the side
of method. The information we now possess of individual
differences and effective devices in supervised study should
make the study of mathematics more nearly a laboratory
course, in which more effective work can be done.
vi ( i KN KJ{ A L M ATI 1 KM AT1CS
5. The teaching of algebra, geometry, and trigonometry
in separate fields is an artificial arrangement that does not
permit the easy solution of problems concerning projects
that correlate with problems met in the physical and bio-
logical sciences or the manual and fine arts. To reject the
formalism of algebra, to delay the demands of a logical
unit in geometry, and to present the simple principles of
the various branches of mathematics in the introductory
course opens the door to a greater variety of problems that
seem to be real applications. The pupil sees the usefulness
of the various modes of treatment of the facts of quantity.
Power is gained because the pupil is equipped with more
tools, in that the method of attack is not limited to one field.
6. One of the most curious characteristics of American
secondary-mathematics instruction is the obscurity in the
teaching of the function notion. It is generally agreed
that functional thinking (the dependence of one magnitude
upon another) constitutes one of the most fundamental
notions of mathematics. Because of the interrelations of
the equation, the formula, the function, the graph, and the
geometric relations inductively acquired, the material is
easily correlated around the function idea as the organizing
and unifying principle. The function concept (implicitly or
explicitly) dominant throughout helps to lend concreteness
and coherence to the subject. However, it would be false
to assume that this material is presented to establish the
principle of correlation. On the contrary, it happens that
correlation around the function notion, though incidental,
is a valuable instrument for accomplishing the larger aim,
which is to obtain a composite introductory course in mathe-
matics that all future citizens of our democracy should be
required to take as a matter of general scholarship.
PREFACE vii
7. The traditional reticence of texts has made mathe-
matics unnecessarily difficult for pupils in the early years.
The style of this book, though less rigidly mathematical, is
more nearly adapted to the pupils' mental age. The result
is a misleading length of the book. The book can easily be
taught in a school year of approximately one hundred and
sixty recitations. In the typical high school it will be taught
in the first year. (The Minnesota high schools taught it
in this grade, five recitations per week.) In schools which
control the seventh and eighth years the following are
also possibilities which have been tested by the authors
and cooperating teachers : (1) in the eighth year with daily
recitations ; (2) half of the book in the eighth year and the
remainder in the ninth year, with three recitations per week
(it was so used in the Lincoln School) ; (3) the course
may be started in the seventh year provided the class has
achieved good results in previous arithmetic work.
Specific references are given where material which is
not of the common stock has been taken consciously, the
purpose, however, being chiefly to stimulate pupils and
teachers to become familiar with these books for reasons
other than the obligations involved. Something of human
interest is added by relating some of the well-known stories
of great mathematicians. We are indebted to Professor
David Eugene Smith on questions relating to historical
material. In our thinking we are particularly indebted to
Professors Nunn, Smith, Breslich, and Myers. We shall
be obliged to all teachers who may think it worth while
to point out such errors as still exist.
THE AUTHORS
CONTENTS
CHAPTER PAGE
I. THE EQUATION 1
Solving an equation 6
Translation of an equation . . 12
Solution of verbal problems 16
Axioms 21
II. LINEAR MEASUREMENT. THE EQUATION APPLIED TO
LENGTH 26
Different units of length 28
Squared paper 32
Sum of two segments ; geometric addition 36
Polygons * 44
III. PROPERTIES OF ANGLES 47
Notation for reading angles 50
Measurement of angles ; the protractor 54
Measuring angles ; drawing angles 56
Comparison of angles - 59
Geometric addition and subtraction of angles 61
Parallel lines 68
How to construct a parallelogram 70
IV. THE EQUATION APPLIED TO AREA 74
Formula 78
Formula for the area of a parallelogram 79
Geometric interpretation of products 85
Algebraic multiplication 89
The accuracy of the result 93
V. THE EQUATION APPLIED TO VOLUME 98
Measurement of volume 99
Formula for the volume of a rectangular parallelepiped . 99
ix
x GE^'EKAL MATHEMATICS
CHAPTER PAGE
Formula for the volume of a cube 102
Exponents 102
Application of algebraic principles to geometric figures . 105
VJ. THE EQUATION APPLIED TO FUNDAMENTAL ANGLE
RELATIONS Ill
The sum of all the angles about a point on one side
of a straight line 112
Sum of all the angles about a point in a plane . . . 113
Supplementary angles ; supplement 116
Complementary angles 119
Vertical angles 122
Important theorems relating to parallel lines .... 126
VII. THE EQUATION APPLIED TO THE TRIANGLE .... 130
The sum of the interior angles 131
Right triangle -. . 135
Exterior angles of a triangle 139
The construction of triangles 142
VIII. POSITIVE AND NEGATIVE NUMBERS. ADDITION AND
SUBTRACTION , 150
Use of signs 151
Geometric representation of positive numbers. Origin . 152
Geometric representation of negative numbers . . . 153
Algebraic addition 162
Subtraction illustrated by the number scale .... 170
Algebraic subtraction 171
Subtraction of polynomials 173
IX. POSITIVE AND NEGATIVE NUMBERS. MULTIPLICATION
AND DIVISION. FACTORING 178
Geometric illustration of law of signs 178
Law of signs illustrated by a balanced bar 180
Multiplication of positive and negative numbers . . . 182
Special products 192
Law of signs in division 195
Factoring 198
CONTENTS xi
CHAPTER PAGE
Distinction between identity and equation 204
Use of factoring in identities for calculating areas . . 205
X. GRAPHICAL REPRESENTATION OF STATISTICS; THE
GRAPH OF A LINEAR EQUATION 214
Pictograms 214
Practice in interpreting the bar diagram 222
How to construct a bar diagram 224 '
Practice in interpreting graphic curves 231
How the. graphic curve is drawn 233
Normal distribution 257
Symmetry of a curve 259
Graph of constant cost relations 262
Graphs of linear equations 263
XI. GAINING CONTROL OF THE FORMULA; GRAPHICAL
INTERPRETATION OF FORMULAS 273
Solving a formula 276
Graphical illustration of a motion problem 283
Translating rules of procedure into formulas .... 287
Graph of the centigrade-Fahrenheit formula .... 288
Evaluating a formula 290
XII. FUNCTION 299
Graph of a function 301
Solving the function set equal to zero 304
Direct variation 305
Graphing direct variation 308
Graphing inverse variation . . . ... . . . . . 311
XIII. SIMILARITY; CONSTRUCTION OF SIMILAR TRIANGLES . 314
Summary of constructions for similar triangles . . . 317
Algebraic problems on similar figures ...... 319
Proportion 322
Construction of a mean proportional 332
Fourth proportional construction 334
Verbal problems solved by proportion 336
Proportionality of areas 341
xii GENERAL MATHEMATICS
CHAPTER PAGE
XIV. INDIRECT MEASUREMENT; SCALE DRAWINGS; TRIG-
ONOMETRY 345
Similar right triangles :;.",:,
Trigonometric ratios :;f>!t
Table of trigonometric ratios 361
Verbal trigonometry problems 362
XV. THEORY AND APPLICATION OF SIMULTANEOUS
LINEAR EQUATIONS 367
Graphic solution 369
Algebraic methods for solving simultaneous linear
equations 373
Summary of methods of elimination 379
Classified verbal problems 384
XVI. GEOMETRIC AND ALGEBRAIC INTERPRETATION OF
ROOTS AND POWERS 390
The theorem of Pythagoras 397
Constructing the square root of a number .... 404
Fractional exponents another means of indicating
roots and powers 412
XVII. LOGARITHMS * .... 424
Logarithms defined , 427
Exponential equations 443
Interest problems solved by logarithms 444
XVIII. THE SLIDE RULE . 449
Square roots found by means of the slide rule . . . 455
Verbal problems solved by the slide rule 458
XIX. QUADRATIC FUNCTIONS; QUADRATIC EQUATIONS . 462
How to solve a quadratic equation graphically . . . 465
The parabola 467
More powerful methods of solving quadratic equations 471
Maxima and minima algebraically determined . . . 479
INDEX 481
INTRODUCTION
The movement to provide an introductory course in general
mathematics is a part of an extensive movement toward mak-
ing the materials of study in secondary education more concrete
and serviceable. The trend of education expresses a determi-
nation that the seventh, eighth, and ninth school years should
be enriched by the introduction of such significant experiences
of science, civics, art, and other knowledge of human life as
all enlightened citizens of a democracy should possess. The
work of these grades cannot be liberalized by "shoving down"
the conventional material a year or so. The reorganization
must be more fundamental in order to revitalize and socialize
the mathematics of these grades.
Competent authorities in mathematics have from time to time
asserted, first, that American secondary-mathematics teaching
has been characterized by a futile attempt to induce all pupils
to become technical college mathematicians. Secondly, that
instead of giving pupils an idea of the real meaning of mathe-
matics and the wide range of its applications, they are forced
to waste a great deal of time on abstract work in difficult
problems in radicals, fractions, factoring, quadratics, and the
like, which do not lead to anything important in mathematics.
And, thirdly, that this meaningless juggling of symbolism fails
to meet the needs of the great number of pupils who go rather
early into their careers ; it also wastes time and effort on the
part of pupils with especial ability in the subject, who ought
to get an early insight into the scope and power of the real
science of mathematics.
xiv GENERAL MATHEMATICS
Quantitative thinking and expression play so large a part in
human experience that proper training in these matters will
always be important. The growing complexity of social and
industrial life is responsible for corresponding changes in the
use of quantitative relationships. Old applications' in many
instances are disappearing, but new ones growing out of present-
day relations are being introduced to take their places. These
changes require a new kind of introductory text in mathematics.
Action is forced by the demand that there shall be justification
of the time and effort given to each subject and each item in
the subject. New subjects which appear necessary in the proper-
training for citizenship are crowding the curriculum. Mathe-
matics too must justify its place " in the sun " by a thorough
reorganization that will meet modern needs. This is what is
meant by revitalizing mathematics.
The practical administrator will be impressed by the fact
that this program raises no administrative difficulties. The
pupil may be expected to develop greater power in algebra,
because the elimination of material which wastes time and
effort has made possible the emphasizing of the topics concern-
ing which there is general agreement. The supplementary
material which is drawn from the other subjects constitutes a
preparation for further study in these fields ; for example, the
text gives the pupil the vocabulary, the symbolism, and many
of the ideas of plane geometry.
This type of introductory course should appeal to the pro-
gressive educator because of a number of other features. The
" problem method" of teaching is followed throughout. Ration-
alized drills are provided in abundance. The course has been
used in mimeographed form by experienced teachers. Scores
of prospective teachers have found the treatment simple and
easy to present. Inexperienced teachers have gone out into
difficult situations and have taught the material with satis-
faction. Pupils following this course have made better progress
INTRODUCTION xv
than pupils following the traditional course, and both, pupils
and teachers manifest a degree of interest seldom seen in the
ordinary class in mathematics. The tests prepared by the
authors will save time for teachers and enable them, if they
desire, to diagnose their own situations and to compare their
results with those obtained in other institutions using the same
material. If the question is raised as to what students com-
pleting such a course will do when they get to college, it may
be replied that enough of them have already entered college
to convince the unbiased that they experience no handicap.
The more important point, however, is that such a course
enables one to understand and to deal with the quantitative
world in which he lives.
This course in reorganized introductory mathematics,
although but a part of a large movement in secondary educa-
tion which looks toward more concrete teaching and more
serviceable materials of study, has a further highly significant
aspect. It is a potent and encouraging evidence that high-
school teachers have become students of their own teaching,
and as a result are preparing their own textbooks in the midst
of real teaching situations, as the outcome of intelligent con-
'structive experimentation.
Probably very few books have been subjected previous to
publication to such thorough tests of teaching situations. The
authors have been shaping this course for many years. During
the last three years the manuscript as originally accepted by
the publishers has been taught in mimeograph form to more
than a thousand pupils distributed in a selection of typical
schools, among these being the following : Minneapolis Central
High School (large city high school), Bremer Junior High
School, Seward Junior High School, University of Minnesota
High School, Owatonna High School, Mabel High School
(small town), and the Lincoln School of Teachers' College.
Numerous consultations with the teachers in these schools
xvi GENERAL MATHEMATICS
resulted in many valuable suggestions which contributed directly
toward making the text easily teachable.
Each of the authors has taught secondary mathematics for
more than ten years in large public and private schools. They
have, supervised many teachers in training ; they have taught
teacher-training courses, and each during most of this time has
had unusual opportunities for free experimentation. The text
may be regarded by fellow teachers of mathematics as a report
which shows in organization and subject matter the things that
have seemed most useful.
LOTUS D. COFFMAN
OTIS W. CALDWELL
FIRST YEAR
CHAPTER I
THE EQUATION
1. A problem introducing the use of the equation. In
order to find the weight of a bag of candy, it was placed
on one pan of perfectly balanced scales (Fig. 1). The
candy, together with a 4-ounce weight, balanced 10 oz.
of weights on the
other pan. How
much did the bag
of candy weigh ?
It is a familial-
principle of bal-
anced scales that
if the same number
of ounces be taken
from each pan, the
balance is not disturbed. Hence, if we suppose that a
4-ounce weight could be removed from each pan, the
candy would be balanced by 6 oz.
This solves the problem, but let us analyze it a little
further. The important fact in the situation above is that
an unknown number of ounces of candy plus 4 oz. in
one pan balances 10 oz. in the other pan. If we agree
1
FIG. 1. THE BALANCED SCALES ILLUSTRATE
THE MEANING OF AN EQUATION
2 GENERAL MATHEMATICS
to let the letter w represent the number of ounces of
weight in the bag of candy and use the sign of equality
(=) to denote the perfect balance of the scales, the pre-
ceding mathematical fact may be conveniently translated
into the following expression : w + 4 = 10, where w + 4
denotes the weight in the left pan and 10 the weight in
the right pan. The abbreviated ("shorthand") statement,
w + 4=10, expresses equality and is called an equation.
The number to the left side of the equality sign is called
the left member of the equation, the number to the right
is the right member.
Just as the scales will balance if the same number of
ounces are taken from each pan, so ive may subtract the
same number from both aides of an equation and get another
f'/Hxtion. In the preceding problem the written work may
be. arranged thus:
f number of ounces of weight
Let to = J. . .,
^ in the bag of candy.
Then w + 4 = 10
4=4
Subtracting 4 from each "I ^
member of the equation, J
Thus, the- bag of candy weighs 6 oz.
The preceding problem illustrates the principle that if
the same number be subtracted from both members of an
equation, the remainders are equal ; that is, another equation
is obtained. [Subtraction Law]
EXERCISES
Find the value of the unknown numbers in the following
equations, doing all you can orally :
1. x + 2 = 6. 4. or + 11 = 18. 7. x + 10 = 27.
2. x + 6 = 10. 5. x + 13 = 23. 8. x + 14 = 21.
3. x + 7 = l3. 6. z+9 = 26. 9. x -f 33 = 44.
THE EQUATION
3
2. The importance of the equation. The equation is a
very important tool for solving problems in the mathe-
matical sciences. The equation gives us a new method of
attack on a problem, enabling us to solve many problems
which would be very difficult, if not impossible, without
its use.
3. Method of studying the nature of the equation. In
making a study of the equation we shall continue by con-
sidering some very simple problems in order that we may
clearly understand the laws which are involved. If these
laws are mastered in connection with the simple cases, it
will be easy to apply the equation as a tool for solving
the more complicated and difficult problems. In the next
article we shall continue to interpret the equation by
considering a problem in weighing.
4. Division Law. Two equal but unknown weights, to-
gether with a 1-pound weight, just balance a 16-pound and
a 1-pound weight to-
gether (Fig, 2). How
heavy is each un-
known weight?
Let p equal the
number of pounds in
one of the unknown
weights. Suppose that
1 Ib. be removed from
each pan, leaving *2p F IG . 2. THE BALANCED SCALES MAT UK USED
DOUnds in the left T(? ILLUSTRATE THE SUBTRACTION LAW AND
THE DIVISION LAW i
pan balancing the re-
maining 16 Ib. in the right pan. Then, if 2 p pounds
balances 16 Ib., p pounds (one half of the weight in the
left pan) must balance 8 Ib. (one half of the weight" in
4 GENERAL MATHEMATICS
the right pan). By the use of the equation the discussion
may be written in the following brief form:
f This is a translation of the tirst
" J \ sentence of the problem.
Subtracting 1 from "1 -
each member, / ^
Dividing each member of the equation by 2,
p=8.
This problem illustrates the principle that if both mem-
bers of an equation are divided by the name number (exclud-
ing division by zero, to be explained later), the quotients are
equal; that is, another equation is obtained. [Division Law~\
EXERCISES
Find the value of the unknown numbers, doing all you
can orally:
1. 2 x + 3 = 9. 12. 5 r + ^ 2 - = 13|.
2. 3x + 4 = 16. 13. 14 k + 7 = 79.
3. 2 a + 5 = 17. 14. 3 e + 4j = 9.
4. 3*4-7 = 28. 15. 15x4-0.5 = 26.
5. 5 r 4- 7 = 62. 16. 11 m + J = ^
6. 9s 4- 21 = 93. 17. 1.3y + 3 = 16.
7. 2y + l=S. 18. 11 y = 33.
8. 5 y 4- 3 = 15. 19. 1.1 x = 121.
9. 4x4-3.2=15.2. 20. 2.3x + 4 = 50.
10. 6^4-4 = 49. 21. 6.3 z 4- 2.4 = 15.
11. 9e + 8 = 116. 22. 5.3x4-0.34 = 2.99.
5. Addition Law. In Fig. 3 the apparatus is so arranged
that the 2-pound weight attached to the string which
passes over the pulley pulls upward on the bar at B. This
THE EQUATION
-Pulley
arrangement makes the problem different from the two
which we have considered. If there were no pulley attach-
ment, the weight pulling downward in the left pan would
be 5 a; pounds. Since there is a lifting force of 2 Ib. at B,
the downward pulling force in the left pan is 2 Ib. less
than 5 x pounds, or 5 x 2 pounds ; this balances the 18 Ib.
in the right pan. Hence the equation which describes the
situation in Fig. 3
is 5z-2 = 18. li-
the string be cut so
as to remove the up-
ward pull of 2 Ib.,
then a 2-pound
weight must be
added to the right
pan to keep the
scales balanced, for
removing the up-
ward pull of 2 Ib.
gave us a down-
ward pull in the left pan of 5 a? pounds. This is 2 Ib. more
than we had with the pulley attached, hence the necessity
of adding 2 Ib. to the right pan.
By the use of the equation the preceding discussion may
take the following brief form:
. . _ o _ i o / This expresses the
o \. original conditions.
M
Adding 2 to both members,
FIG. 3. IN THIS CASE THE SCALES ILLUSTRATE
THE ADDITION LAW
5 x
= 20
Dividing both members by 5,
This problem illustrates the principle that if the same
number is added to both members of an equation, the sums are
equal ; that is, another equation is obtained. [Addition Law\
6 GENERAL MATHEMATICS
EXERCISES
Find the value of the unknown number in each problem,
doing all you can orally :
1. x -5 = 10. 12. 12m -35 = 41.
2. 2 a; -15 = 13. 13. 9c- 3.2 = 14.8.
3. 3 a; -12 = 13. 14. 7 * - 2 = 5.7.
4. 3.r-8 = 17. 15. 14 A- - 5 = 21.
5. 12 y - 4 = 46. . 16. 2 y - 3.1 = 3.2.
6. 4 1 - 16 = 16. 17. 0.5 x - 3 = 4.5.
7. 19,--4i = 14f 18. 2 x -1=61
8. lly-9 = 79. 19. 3cc-9l = 17.r>.
9. 56- - 0.1 = 0.9. 20. 9.7- -7.5 = 73.5.
10. 4y-f = 7j. 21. 1.5 a; -3 = 1.5.
11. 7*- 4 = 26. 22. 1.6 x- 1.7 = 1.5.
6. Solving an equation ; check ; root. The process of
finding the value of the unknown number in an equation
is called solving the equation. To illustrate :
Let y + 3 = 8 be the equation.
"3 = 3
Then y 5, and the equation is said to be solved.
To test-, or check, the correctness of the result replace
the unknown number in the original equation by 5, ob-
taining 5 + 3 = 8. Since both members of the equation
reduce to the same number, the result y = 5 is correct.
When a number is put in place of a literal number it
is said to be substituted .for the literal number.
When both sides of an equation reduce to the same
number for certain values of the unknown number, the
equation is said to be satisfied. Thus, 3 satisfies the equa-
tion y + 2 = 5.
THE EQUATION 7
A number that satisfies an equation is a root of the
equation.
Thus, 5 is a root of the equation z + 3 = 8.
HISTORICAL NOTE. The word "root " first appears in the algebra
of Mohammed ibn Musa Abu Jafar Al-Khwarizmi (about A.D. 830).
The root of an equation (like the root of a plant) is hidden until
found. See Ball's "A Short History of Mathematics," p. 163.
EXERCISES
Solve the following equations and check the results :
1. 5y+3=18. 5. 26 + 2.7-1.3=11.4.
2. 7z-4=17. 6. 7x-3x + 3.1 = 7.1.
3. 2 a; -1.3 = 2.7. 7. 5* +14 -9 =15.
4. 3a + 4.5=7.5. 8. 7m 3f = 3j.
7. Terms; monomial; order of terms. The parts of an
expression separated by plus (+) and minus ( ) signs are
called the terms of a number. Thus, 2 a and 3 b are the
terms of the number 2 a-\- 3 b. A one-term number is called
a monomial.
EXERCISES
1.8-7+2 = ? 4. 8x7x + 2x = ?
2.8 + 2-7=? 5. 8x + 2x-7x = ?
3.2 + 8-7=? 6. 2z + 8z lx = ?
These problems illustrate the principle (to be discussed
more fully later) that the value of an expression is un-
changed if the order of its terms is changed, provided
each term carries with it the sign at its left. If no sign
is expressed at the left of the first term of an expression,
the plus sign is understood.
8. Similar and dissimilar terms. Terms which have a
common literal factor, as 2 r, 3 x, and 5 x, are similar terms.
Their sum is a one-term expression ; namely, 10 a;. When
8 GENERAL MATHEMATICS
terms do not have a common literal factor, as 2 x and
3 y, they are called dissimilar terms. Algebraic expressions
are simplified by combining similar terms. Combining
similar terms in either the right or the left member of an
equation gives us the same equation in simpler form.
HISTORICAL XOTE. The word "algebra" first appears about
A.D. 830 in an Arabian work called " Al-jebr wa'1-mukabala," written
by Al-Khwariznii. "Al-jebr," from which "algebra" is derived, may
be translated by the restoration and refers to the fact that the same
number may be added to or subtracted from both sides of the
equation; "al-mukabala" means the process of comparison, and some
writers say it was used in connection with the combination of
similar terms into one term.
The mathematical interest of the Arabs ran high. In the seventh
century religious enthusiasm had banded these nomadic tribes
into a conquering, nourishing nation. Enormous fortunes demanded
mathematical manipulation. Cantor cites a rumor of a merchant
whose annual income was about seven million dollars and a Christian
doctor of medicine whose annual income was about fifty thousand.
These fortunes gave them the necessary leisure time for culture
and learning. Among the many books translated was the Greek
geometry, Euclid's " Elements." See Ball's K A Short History of
Mathematics," p. 162, and Miller's "Historical Introduction to
Mathematical Literature," p. 83.
EXERCISES
Solve the following equations and check the results :
1. 2./- -7 =x + 3.
Subtract x from both members and proceed as usual.
2. 3x + 2 = x + S.
3. 5 i x 3x + 2-x 2 = 2x +
4. 16y-8y + 3y-2 = 5y
5. 20-4.r = 38 -10 a-.
6. 5.T -j-3 05 = .r +18.
THE EQUATION 9
7. 7 > +18 + 3 r = 32 + 2 r - 2.
8. 18 + 6 s + 30 + 6 s = 4 s + 8 + 12 + 3 * + 3 + s + 29.
9. 25 y + 20 - 7 y - 5 = 56 - 5 // + 5.
^J
9. Multiplication Law. Solve for x: - = 5.
A
This problem offers a new principle. If we translate it
into an English sentence, it reads as follows : One half of
what unknown number equals 5 ? If one half of a num-
ber equals 5, all of the number, or twice as much, equals
2 times 5, or 10. The problem may be solved as follows:
Multiplying both members by 2,
2 x = 2 -x 5.
x = 10.
The principle involved here may be further illustrated
by the scales, for if an object in one pan of a scales will
balance a 5-pound weight in the other, it will be readily
seen that three objects of the same kind would need 15 Ib.
to balance them. This may be expressed in equation form
as feitews : _ =
X U.
Mujtiplyijig both members by 3,
3 x = 15.
v.
From the preceding' discussion it is evident that if
both members of an equation are multiplied by the same
number, the products are equal: that is, another equation
is obtained. [Multiplication Law\
10 GENERAL MATHEMATICS
This multiplication principle is convenient when an
equation contains fractions. It enables us to obtain a
second equation containing no fraction but containing the
same unknown number. To illustrate this:
Let \ .- = 7.
Multiplying both members by 3,
3 x i- x = 3 x 7.
Reducing to simplest form, x = 21.
ORAL EXERCISES
Find the value of the unknown number in each of the
following equations :
tn /> & IT
The preceding list of problems shows that it is desirable
to multiply both members of the equation by some number
that will give us a new equation without fractions. The same
principle holds when the equation contains two or more
fractions whose denominators are different, as is illustrated
by the following problem :
Find x if 7-| = 2.
4 5
Solution. - - - = 2.
4 5
90 r 90 r
Multiplying by 20, = 40.
Simplifying, 5 x 4 x 40 ;
whence x = 40.
THE EQUATION 11
The fact that 4 and 5 will divide integrally (a whole
number of times) into the numerators gives us a new equa-
tion without fractions. Obviously there are an unlimited
number of numbers (for example, 40, 60, 80, etc.) which
we could have used, but it was advantageous to use the
smallest number in which 4 and 5 are contained integrally ;
namely, the least common multiple of 4 and 5, which is
20. The object of this multiplication is to obtain an equa-
tion in which the value of the unknown number is more
easily found than in the preceding one. This discussion
may be summarized by the following rule:
If the given equation contains fractions, multiply every term
in both members by the least common multiple {L.C.M.} of
the denominators in order to obtain a new equation which does
not contain fractions.
EXERCISES
Find the value of the unknown number in each problem,
and check :
2 ^ + ^-9 a 2x , x x , l
<S i . <? H. - -f- = - +
54 96 18 3
4. it + J y = 6. ' K
11 ^
5- y - I y = 7. 6
12 GENERAL MATHEMATICS
10. Definition of the equation; properties of the equation.
The foregoing problems were used to show that an equa-
tion iif a statement that two numbers are equal. It indicates
that two expressions stand for the same number. It may
be regarded as an expression of balance of values be-
tween the numbers on the two sides of the equality sign.
Some unknown number which enters into the discussion
of the problem is represented by a letter. An equation
is written which enables one to find the value of that
unknown number.
An equation is like a balance in that the balance of
value is not disturbed so long as like changes are made
on both sides. Thus, in equations we may add the same
number to both aides, or subtract the same number from
both sides, or we may multiply or divide both sides by the
same number (except division by zero); the equality is
maintained during all these changes.
On the other hand, the equality is destroyed if more is
added to or subtracted from one side than the other or if
one side is multiplied or divided by a larger number than
is the other side.
11. Translation of an equation. The list of problems in
the preceding exercises may appear abstract in the sense
that the equations do not appear to be connected in any
way either with a concrete situation of a verbal problem,
or with our past experience. However, such a list need
not be regarded as meaningless. Just as an English sen-
tence which expresses a number relation may be written
in the " shorthand " form of an equation, so, conversely,
an equation may be translated into a problem ; for ex-
ample, the equation 3x + 5 = 2x + 2Q may be interpreted
as follows : Find a number such that 3 times the num-
ber plus 5 equals 2 times the number plus 20. The
THE EQUATION 13
equation .c 3 = 5 may be regarded as raising the question
What number diminished by 3 ' equals 5 ? Or, again,
21 x + x + 2x + x = 140 may be considered as the trans-
lation of the following problem: What is the altitude of
a rectangle whose base is 21 times as long as the altitude
and whose perimeter is 140 ft. ?
EXERCISES
State each of the following in the form of a question or a
verbal problem :
1. 05-6 = 3. 7. 7r-2 = 6r + 8.
2. 2a- -1 = 10. 8. 5.2 x- 3 = 4.1 x+ 1.4.
3. 9 k -10 = 87. 9. 3* =12.
4. ly + 8 = 112. 10. 4* =16.
5. 7s- 3 = 81. 11. 2x + 3x + 4x = 18.
6. 3x + 2 = 2x + :;. 12. c = 2 T 2 rl.
12. Drill in the "shorthand" of algebra. The following
exercises give practice in translating number expressions
and relations from verbal into symbolical language:
1. Consecutive numbers are integral (whole) numbers which
follow each other in counting ; thus, 17 and 18, 45 and 46, are
examples of consecutive numbers. Begin at s -f 5 and count
forward. Begin at x + 3 and count backward. Give four con-
secutive integers beginning with 18 ; ending with 18 ; beginning
with x; ending with x. Give two consecutive even integers
beginning with 2 x. Give two consecutive even integers end-
ing with 2c.
2. The present age, in years, of a person is denoted by x.
Indicate in symbols the following : (a) the person's age fourteen
years ago ; (b) his age fourteen years hence ; (c) his age when
twice as old as now ; (d) 60 decreased by his age ; (e) his age
decreased by 60 ; (f) his age increased by one half his age.
14 GENERAL MATHEMATICS
3. A boy has a marbles and buys b more. How many has
he ? What is the sum of 'a and b ?
4. A boy having b marbles loses c marbles. How many
has he?
5. (a) The home team made 8 points in a basket-ball game
and the visiting team made 3 points. By how many points did
the home team win ? (b) If the home team scored h points
to the visitors' n points, by how many points did the home
team win ? (c) Substitute numbers for h in the last question
that will show the defeat of the home team, (d) If h = 5, what
must be the value of n when the game is a tie ?
6. What is the 5th part of n ? f of y ? f of t ?
7. Two numbers differ by 7. The smaller is s. Express
the larger number.
8. Divide 100 into two parts so that one part is .
9. Divide a into two parts so that one part is 5.
10. The difference between two numbers is d and the larger
one is I. Express the smaller one.
11. What number divided by 3 will give the quotient a?
12. A man's house, worth h dollars, was destroyed by fire.
He received i dollars insurance. What was his total loss ?
13. A is x years old and B lacks 5 yr. of being three times
as old. Express B's age.
14. A has m ties and B has n ties. If A sells B 5 ties, how
many will each then have ?
15. A man has d dollars and spends c cents. How many
cents does he have left ?
16. A room is I feet long and w feet wide. How many feet
of border does such a room require ?
17. The length of a rectangle exceeds its width by c feet.
It is w feet wide, (a) State the length of each side, (b) Find
the distance around the rectangle.
THE EQUATION 15
18. What is the cost of 7 pencils at c cents each ?
19. What is the cost of 1 sheet of paper if b sheets can be
bought for 100?
20. It takes 'two boys 5 da. to make an automobile trip.
What part can they do in 1 da. ? If it takes them d days,
what part of the trip do they travel in 1 da. ?
21. If a man drives a car at the rate of 31 mi. per hour,
how far can he drive in 3 hr. ? in 5 hr. ? in li hours ?
22. If a man drives n miles in 3 hr., how many miles does
he go per hour ?
23. A tank is filled by a pipe in m minutes. How much of
the tank is filled in 1 min. ?
24. The numerator of a fraction exceeds the denominator
by 3. (a) Write the numerator, (b) Write the fraction,
(c) Head the fraction.
25. A pair of gloves costs d dollars. What is the cost if the
price is raised 70? if lowered 70?
26. Write the sum of x and 17 ; of 17 and x. Write the
difference of x and 17; of 17 and x.
27. A class president was elected by a majority of 7 votes.
If the unsuccessful candidate received k votes, how many votes
were cast ?
13. Algebraic solution. Many problems may be solved
by either arithmetic or the use of the equation. When
the solution of a problem is obtained by the use of the
equation, 'it is commonly called an algebraic solution.
The following problems illustrate the important steps
in the algebraic solution of a problem. By way of contrast
an arithmetic solution is given for the first problem.
1. Divide a pole 20 ft. long into two parts so that one part
shall be four times as long as the other.
16 GEXKKAL .MATHEMATICS
ARITHMETICAL SOLUTION
The shorter part is a certain length.
The longer part is four times this length.
The whole pole is then five times as long as the shorter part.
The pole is 20 ft. long.
The shorter part is \ of 20 ft., or 4 ft.
The longer part is 4 x 4 ft., or 16 ft.
Hence the parts are 4 ft. and 16 ft. long respectively.
ALGEBRAIC SOLUTION
Let n = number of feet in the shorter part.
Then 4 n = number of feet in the longer part,
and n + 4 n, or 5 n length of the pole.
Then ."> n = '20.
n = 4.
4 n = 10.
Hence the parts are 4 ft. and 1(J it. long respectively.
/
2. A rectangular garden is three times as long as it is wide.
It takes 80 yd. of fence to inclose it. Find the width and
length.
ALGEBRAIC SOLUTION
Let x = number of feet in the width.
Then .3 x = number of feet in the length,
and ./ + :> x + x + 3 x = distance around the garden.
Then 8 x = 80.
x = 10.
3 x = 30.
Hence the width is 10 yd. and the length is 30 yd.
14. The important steps in the algebraic solution of verbal
(or story) problems. Before proceeding to the solution of
more difficult problems it is important that we organize the
steps that are involved. The preceding list of problems illus-
trates the following method for solving a verbal problem :
(a) In every problem certain facts are given as known
and one or more as unknown and to be determined. Read
the problem so as to get these facts clearly in mind.
THE EQUATION 17
(b) In solving the problem denote one of the unknown
numbers by some symbol, as y.
(c) Then express all the given facts in algebraic lan-
guage, using the number y as if it, too, were known.
(d) Find two different expressions which denote the
same number and equate them. (Join by the sign of
equality (=).)
(e) Solve the equation for the value of the unknown
number.
(f) Check the result by re-reading the problem, substi-
tuting the result in the conditions of. the problem to see
if these conditions are satisfied. Note that it is not suffi-
cient to check the equation, for you may have written the
wrong equation to represent the conditions of the problem.
EXERCISES
With the preceding outline of method in mind, solve the
following problems, and check :
PROBLEMS INVOLVING NUMBER RELATIONS
1. If six times a number is decreased by 4, the result is 26.
Find the number.
2. If four fifths of a number is decreased by 6, the result
is 10. Find the number.
3. The sum of three numbers is 12Q. The second is five
times the first, and the third is nine times the first. Find
the numbers.
4. The sum of three numbers is 360. The second is four-
teen times the first, and the third is the sum of the other two.
Find the numbers.
5. Seven times a number increased by one third of itself
equals 44. Find the number.
18 GENERAL MATHEMATICS
6. The following puzzle was proposed to a boy : " Think
of a number, multiply it by 4, add 12, subtract 6, and divide
by 2." The boy gave his final result as 13. What was his
original number ?
7. The sum of one half, one third, and one fourth of a
number is 52. What is the number ?
CONSECUTIVE-NUMBER PROBLEMS
8. Find two consecutive numbers whose sum is 223.
9. Find three consecutive numbers whose sum is 180.
10. Find two consecutive odd numbers whose sum is 204.
*
11. Find three consecutive even numbers whose sum is 156.
12. It is required to divide a board 70 in. long into five parts
such that the four longer parts shall be 1", 2", 3", and 4"
longer respectively than the shortest part. Find the lengths
of the different parts.
13. A boy in a manual-training school is making a bookcase.
The distance from the top board to the bottom is 4 ft. 7 in.,
inside measure. He wishes to put in three shelves, each 1 in.
thick, so that the four book spaces will diminish successively
by 2 in. from the bottom to the top. Find the spaces. .
PROBLEMS INVOLVING GEOMETRIC RELATIONS
14. The length of a field is three times its width, and the
distance around the field is 200 rd. If the field is rectangular,
what are the dimensions ?
15. A room is 15 ft. long, 14 ft. wide, and the walls contain
464 sq. ft. Find the height of the room.
16. The perimeter of (distance around) a square equals 64 ft.
Find a side.
NOTE. Such geometric terms as "triangle," "rectangle," "square,"
etc., as occur in this list of problems (14-24), are familiar from arith-
metic. However, they will later be defined more closely to meet
other needs.
THE EQUATION 19
17. Find the sides of a triangle if the second side is 3 ft.
longer than the first, the third side 5 ft. longer than the first,
and the perimeter is 29 ft.
18. Find the side of an equilateral (all sides equal) triangle
if the perimeter is 21f ft.
19. The perimeter of an equilateral pentagon (5-sided figure)
is 145 in. Find a side.
20. The perimeter of an equilateral hexagon (6-sided figure)
is 192 ft. Find a side.
21. Find the side of an equilateral decagon (10-sided figure)
if its perimeter is 173 in.
22. What is the side of an equilateral dodecagon (12-sided
figure) if its perimeter is 288 in. ?
23. A line 60 in. long is divided into two parts. Twice the
larger part exceeds five times the smaller part by 15 in. How
many inches are in each part ?
24. The perimeter of a quadrilateral A BCD (4-sided figure)
is 34 in. The side CD is twice as long, as the side AB; the
side AD is three times as long as CD; the side BC equals the
sum of the sides AD and CD. Find the length of each side.
MISCELLANEOUS PROBLEMS
25. Divide $48,000 among A, B, and C so that A's share
may be three times that of B, and C may have one half of
what A and B have together.
26. The perimeter of a rectangle is 132 in. The base is
double the altitude. Find the dimensions of the rectangle.
27. A and B own a house worth $16,100, and A has invested
twice as much capital as B. How much has each invested ?
28. A regulation football field is 56|^ yd. longer than it is
wide, and .the sum of its length and width is 163^ yd. Find
its dimensions.
20 GENERAL MATHEMATICS
29. A man has four times as many chickens as his neighbor.
After selling 14, he has 3^ times as many. How many had
each before the sale ?
30. In electing a president of the athletic board a certain
high school cast 1019 votes for three candidates. The first
received 143 more than the third, and the second 49 more
than the first. How many votes did each get ?
31. A boy has $5.20 and his brother has $32.50. The first
saves 200 each day and the second spends 100 each day. In
how many days will they have the same amount ?
32. One man has seven times as many acres as another.
After the first sold 9 A. to the second, he had 36 A. more
than the second then had. How many did each have before
the sale ?
33. To find the weight of a golf ball a man puts 20 golf
balls into the left scale pan of a balance and a 2-pound weight
into the right; he finds that too much, f but the balance is
restored if he puts 2. oz. into the left scale pan. What was
the weight of a golf ball ?
34. The number of representatives and senators together
in the United States Congress is 531. The number of repre-
sentatives is 51 more than four times the number of senators.
Find the number of each.
35. A boy, an apprentice, and a master workman have the
understanding that the apprentice shall receive twice as much
as the boy, and the master workman four times as much as
the boy. How much does each receive if the total amount
received for a piece of work is $105 ?
36. A father leaves $13.000 to be divided among his three
children, so that the eldest child receives $2000 more than
the second, and twice as much as the third. What is the
share of pa oh ?
THE EQUATION 21
37. A fence 5 ft. high is made out of 6-inch boards running
lengthwise. The number of boards necessary to build the
fence up to the required height is 5 and they are so placed
as to leave open spaces between them. If each of these open
spaces, counting from the bottom upwards, is half of the one
next above it, what must be the distances between the boards ;
that is, what will be the width of each of the open spaces ?
38. I paid $8 for an advertisement of 8 lines, as follows :
240 a line for the first insertion, 100 a line for each of the
next five insertions, and 20 a line after that. Firid the
number of insertions.
39. The annual income of a family is divided as follows :
One tenth is used for clothing, one third for groceries, and one
fifth for rent. This leaves $660 for other expenses and for
the savings account. How much is the income ?
15. Axioms. Thus far we have used the four following
laws in solving equations:
I. If the same number be added to equal numbers, the
sums are equal. [Addition Law]
II. If the same number be subtracted from equal numbers,
the remainders are equal. [Subtraction Law]
III. If equal numbers be multiplied by the same number,
the products are equal. [Multiplication Law]
IV. If equal numbers be divided by the same number
(excluding division by zero), the quotients are equal. [Division
Law]
Statements like the four laws above, when assumed to
be true, are called axioms. Usually axioms are statements
so simple that they seem evident. A simple illustration
is sufficient to make clear the validity of the axiom. For
example, if two boys have the same number of marbles
and 3 more are given to each, then our experience tells
22 GENERAL MATHEMATICS
us that again one boy would have just as many as the
other. This illustrates the validity of the addition axiom.
Hereafter the preceding laws will be called Axioms I, II,
III, and IV respectively.
16. Axiom V. In this chapter we have also made fre-
quent use of another axiom. In solving a verbal problem
we obtained the necessary equation by finding two expres-
sions which denoted the same number and then we equated
these two expressions. This step implies the following
axiom :
If two numbers are equal to the same number (or to equal
numbers'), the numbers are equal. [Equality Axiom]
Illustrate the truth of Axiom V by some familiar
experience.
The following exercises test and review the axioms.
EXERCISES
Solve the following equations, and check. Be able to state
at every step in the solution the axiom used.
1. 12 #-15 = 30.
Solution. 12 1 - 15 =30
15 = 15 (Axiom I)
Adding 15 to both members, 12 t = 45
Dividing both sides by 12, / = f f, or 3. (Axiom IV)
y y 1
2. - + - = -
24 3
Solution. Multiplying both sides of the equation by the least
common multiple of the denominators, that is, by 12,
12 y , 12 y 12
-2^ + -^ : = y (Axiom III)
Then 6 y + 3 y = 4.
THE EQUATION 23
By reducing the fractions of the first equation to lowest terms
we obtain the second equation, which does not contain fractions.
Combining similar terms, 9 y 4.
Dividing both sides of the equation by 9,
y = f. (Axiom IV)
3. 12a-+13 = 73. 8. 17s- 3s + 16s =105.
4. 18r-12r = 33. 9. 17x + 3x - 9x = 88.
5. 21^+15=120. 10. 16m + 2m 13m = 22%.
6. 28* -9 = 251. 11. 202y-152// + 6?/ = 280.
7. 20y+ 2y-18y = 22. 12. 3.4 x 1.2 x + 4.8 x = 70.
13. 3.5 y + 7.6 ?/- 8.6y=15.
14. 5.8 m 3.9 m + 12.6 m = 58.
15. 6 > - 3.5 > + 5.5 r = 68.
16. 3.41 x + 0.59 x - 1.77 x = 22.3.
17. 8 y 4.5 y + 5.2 y = 87.
18. 2s +7s - 3s- 6 = 24.
19. lx = 6. 27. ^ + |_| = 36.
20. J* = 2. 28. 15 = 3sc-3.
21. fa; = 6. Solution. Adding 3 to both
members,
22. fa; = 25. 18 = 3*.
^ jj. J)ividing both members by 3,
23- 3 + 2 =1 - 6=x.
Note that in the preceding
24. + = 3. problem the unknown appears
in the right member.
25. ^ + ^ = 8. 29. 17=2^-3.
30.
26.-^ = 6. 31.
24 GENERAL MATHEMATICS
32.^ = 4. 33.^ = 5.
x b
Solution. Multiplying both _ . 16 _
members by #, 3 x
1Q = ix. Multiply both members by 3 x.
Hence 4 = x. _ _ 3
<5o. - = 1.
4
Note that in the preceding
problem the unknown occurs in ,g 13 _ .
the denominator. 2
SUMMARY
17. This chapter has taught the meaning of the follow-
ing words and phrases : equation, members of an equation,
equation is satisfied, substituting, check, root of an equa-
tion, verbal problem, algebraic solution of a verbal problem,
term, monomial, literal number, similar terms, dissimilar
terms, order of terms, and axiom.
18. Axioms. In solving equations the following axioms
are used :
I. If the same number or equal numbers be added to equal
numbers, the sums are equaL [Addition Axiom]
II. If the same number or equal numbers be subtracted from
equal numbers, the remainders are equal. [Subtraction Axiom]
III. If equal tno/ihcrs be divided by equal numbers (exclud-
ing division by zero), the quotients are equal. [Division Axiom]
IV. If equal numbers be multiplied by the same number or
equal numbers, the products are equal. [Multiplication Axiom]
V. If two numbers are equal to the same number or to equal
numbers, the numbers are equal. [Equality Axiom]
19. If an equation contains fractions, a second equation
involving the same unknown may be obtained by multi-
plying every term of the given equation by the L.C.M.
THE EQUATION 25
of the denominators and then reducing all fractions to
integers. The solution of this equation is more readily
obtained than, that of the given equation.
20. The equation is a convenient tool for solving
problems. In solving a verbal problem algebraically ob-
serve the following steps:
(a) Use a letter for the unknown number called for in
the problem. Often the last sentence suggests the most
convenient choice.
(b) Express the given facts in terms of the unknown
(provided, of course, that these facts are not definite arith-
metical numbers).
(c) Obtain two expressions for the same number and
equate them. (This gives us the equation.)
(d) Solve the equation. Check the result by substitut-
ing in the conditions of the problem.
CHAPTER II
LINEAR MEASUREMENT. THE EQUATION APPLIED TO
LENGTH 1
21. Length, the important characteristic of lines. If in
drawing an object we lay the ruler on a sheet of paper and
pass a sharpened pencil as near the edge as possible, thus
obtaining what is familiar to us as a straight line, we are at
once concerned with the length of a part of the straight line
drawn. In fact, length is the important characteristic of a
line. In an exact sense a line has length only, not width nor
thickness. Thus, A ^
the edge of a table
FIG. 4. A LINE SEGMENT
has length only ;
the thickness and width of a crayon line are neglected ; the
wide chalk marks on a tennis court are not boundary
lines, but are made wide to help us see the real boundary
lines, which are the outside edges of the chalk marks.
The part of a line whose length we wish to determine
is a line segment or, briefly, a segment, as AB in Fig. 4.
A line segment has a definite beginning point and a
definite ending point. The word "point" is used to mean
merely position, not length, breadth, nor thickness. The
position of a point is shown by a short cross line ; that is,
a point is determined by two intersecting lines.
1 The pupil should now provide himself with a ruler one edge of which
is graduated to inches and fractional parts of an inch and the other to
units of the metric scale. He should also obtain a pair of compasses
and some squared paper ruled to the metric scale.
26
LINEAR MEASUREMENT 27
EXERCISE
Give examples of line segments that can be seen in the
classroom.
22. Measurement of length. In Fig. 5 the length of the
line segment AB is to be determined. One edge of your
ruler is graduated (divided) into inches and fractions of
an inch as is shown in Fig. 5. Place the division marked
zero on your ruler at A, with the edge of the ruler along
A B
i f
FIG. 5. How A LINE SEGMENT MAT BE MEASURED
the segment AB, and read the number of inches in the line
segment AB; that is, find what reading on the ruler is
opposite the point B.
In the preceding problem we compared the unknown
length of the line segment AB with the well-established
and well-known segment, the inch, and found the line
segment to be 21 times as long as the inch. Hence the
length of the line segment is 2^ in. When we determine
the length of a line segment we are measuring the line
segment. The segment with which we compare the given
line is called a unit segment or a unit of measurement. Hence
to measure a line segment is to apply a standard unit seg-
ment to it to find out how many times the unit segment is
contained in it.
EXERCISES
1. Draw a line segment and express its length in inches.
2. Measure the length of your desk in inches.
3. Measure the width of your desk in inches.
28 (rEXEKAL MATHEMATICS
23. Different units of length. The most familiar
inents for measuring are the foot rule, the yardstick, and
several kinds of tape lines. The fractional parts of the unit
are read by means of a graduated scale engraved or stamped
on the standard unit used. In Fig. 6, below, is shown a
part of a ruler. The upper edge is divided into inches
and fractional parts of an inch. What is the length of the
smallest line segment of the upper edge ?
The lower edge of the ruler is divided into units of
the metric (or French) scale. This system is based on the
I ' i ' I ' I ' ' t ' i ' i ' ' i ' i ' i ' i ' M i ' i ' M 1 1 1 1 1 ' M M i f m 1 1 1 1 ' 1 1 1
O INCH i 2 3
O 1 2 3 4
I .. CENTIMETERS | . , . ..| .... I.
u
' I->
iCm.
FIG. 6. PART OF A RULER, SHOWING DIFFERENT UNITS OF LENGTH
decimal system and is now very generally used in scien-
tific work in all countries. The standard unit is called the
meter (m.). It is divided into 1000 equal parts called milli-
meters (mm.). In the figure above, the smallest division
is a millimeter. Ten millimeters make a centimeter (cm.).
In the figure, AB is one centimeter in length, and you will
note that this is about two fifths of an inch. Ten centi-
meters make a decimeter (dm.) (about 4 in.), and ten deci-
meters make a meter (39.37 in., or about 1.1 yd.). We may
summarize these facts in the following reference table :
f 1 millimeter = 0.03937 inches
\linch = 2.54 centimeters
10 millimeters = 1 centimeter (0.3937 in., or nearly J in.)
10 centimeters = 1 decimeter (3.937 in., or nearly 4 in.)
10 decimeters = 1 meter (39.37 in., or nearly 3-^ ft.)
LINEAR MEASUREMENT 29
24. Advantages of the metric system. One of the advan-
tages of the system is that the value of the fractional part
of the meter is more apparent than a corresponding deci-
mal part of a yard. Thus, if we say a street is 12.386 yd.
wide, the decimal 0.386 tells us nothing about the smaller
divisions of a yardstick that enter into this number. At
best we would probably say that it is something over one
third of a yard. On the other hand, if we say that a road
is 12.386 m. wide, we know at once that the road is 12 m.
3 dm. 8 cm. 6 mm. wide. This last statement is far more
definite to one who has had .a little practice with the
metric system.
Obviously the advantages of the metric system lie in
the fact that ten line segments of any unit are equal to
one of the next larger. In contrast to this fact the multi-
pliers .of our system, though they may seem familiar, are
awkward. Thus, there are 1 2 in. in a foot, 3 ft. in a yard,
5^ yd. in a rod, 1760yd. in a mile, etc.
HISTORICAL NOTE. It is probable that most of the standard units
of length were derived from the lengths of parts of the human body
or other equally familiar objects used in measuring. Thus, we still
say that a horse is so many hands high. The yard is supposed to
have represented the length of the arm of King Henry I. Nearly all
nations have used a linear unit the name of which was derived from
their word for foot.
During the French Revolution the National Assembly appointed
a commission to devise a system that would eliminate the inconven-
ience of existing weights and measures. The present metric system
is the work of this commission.
This commission attempted to make the standard unit one ten-
millionth part of the distance from the equator to the north pole
measured on the meridian of Paris. Since later measurements have
raised some doubt as to the exactness of the commission's determina-
tion of this distance, we now define the meter not as a fraction of
the earth's quadrant, but as the distance, at the freezing temperature,
30 GENERAL MATHEMATICS
between two transverse parallel Jim-.* ruled on a bar of platinum-iridium
which is kept at the International Butqau of Weight.* and Measures, at
Sevres, near Paris.
25. Application of the metric scale. This article is in-
tended to give practice in the use of the metric system.
EXERCISES
1. With a ruler whose edge is graduated into centimeters
measure the segments AB and CD in Fig. 7.
FIG. 7
2. Measure the length and width of your desk with a centi-
meter ruler. Check the results with those of Exs. 2 and 3,
Art. 22.
3. Estimate the length of the room in meters and then
measure the room with a meter stick. If a meter stick is not
available, use a yardstick and translate into meters.
4. Turn to some standard text in physics (for example,
Millikan and Gale, pp. 2 and 3) and report to the class on the
metric system.
5. Refer to an encyclopedia and find out what you can
about the " standard yard " kept at Washington.
26. Practical difficulty of precise measurement. In spite
of the fact that measuring line segments is a familiar
process and seems very simple, it is very difficult to meas-
ure a line with a high degree of accuracy. The following
sources of error may enter into the result if we use a
yardstick : (1) the yardstick may not be exactly straight ;
(2) it may be a little too long or too short ; (3) it may
slip a little so that the second position does not begin at
LINEAR MEASUREMENT
31
the exact place where the first ended ; (4) the edge of
the. yardstick may not always be along the line segment;
(5) the graduated scale used for reading feet, inches, and
fractional parts of inches may not be correct. Nor do we
eliminate these errors by using other measuring devices.
For example, a tape line tends to stretch, but contracts if
wet, while a steel tape is affected by heat and cold.
From the preceding discussion it is apparent that a
measurement is always an approximation. The error can
be decreased but never wholly eliminated.
EXERCISES
1. Suppose you have measured a distance (say the edge of your
desk) with great care and have found it to be equal to 2 ft. 7f in.
Is this the exact length of the desk ? Justify your answer.
2. If you were to repeat the measurement with still greater
care, making use of a finer-graduated scale, is it likely that
you would find exactly the same result
as before ?
3. If you were asked to measure
the length of your classroom, what
would you use ? Why ?
27. The compasses. A pair of com-
passes (Fig. 8) is an instrument
that may be used in measuring line
segments. Since the use of com-
passes greatly decreases some of the
common errors in measuring that
have been pointed out, and is con-
sequently very useful in many forms of drawing which
require a high degree of accuracy, it will be helpful if the
student learns to use the compasses freely.
FIG. 8. A PAIR OF
COMPASSES
32 GENERAL MATHEMATICS
28. Measuring a line segment with the compasses. To
measure the line segment AB in Fig. 9 with the compasses,
place the sharp points of the compasses on A and B. Turn
FIG. 9
the screw which clamps the legs of the compasses. Then
place the points on the marks of the ruler and count the
number of inches or centimeters between them.
EXERCISES
1. With the compasses measure AB in Fig. 9, in inches.
2. With the compasses measure AB in Fig 9, in centimeters.
3. Multiply the result of Ex. 1 by 2.54. What do you observe ?
4. Estimate the number of centimeters in the length of this
page. Measure the length of this page with the compasses and
compare with your estimate.
29. Squared paper. Squared paper is another important
device which is often useful in measuring line segments.
Squared paper is ruled either A B
to inches and fractions of an
inch (used by the engineer) or
to the units of the metric scale.
A sample part of a sheet is
shown in Fig. 10. The method
of measuring with squared pa-
per is practically the same as
measuring with the compasses
and ruler. Thus, to measure
the line segment AB in Fig. 10
place the sharp points of the compasses on A and B.
Clamp the compasses. Place the sharp points on one of
FIG. 10. HOAV A LINE SEGMENT
MAY BE MEASURED BY THE USE
OF SQUARED PAPER
LINEAR MEASUREMENT
33
the heavy lines, as at E and D. Each side of a large
square being 1 cm., count the number of centimeters in
ED and estimate the remainder to tenths of a centimeter.
Thus, in Fig. 10 the segment ED equals 2.9 cm.
EXERCISES
1. Draw a line segment and measure its length in centi-
meters by the use of squared paper.
2. Why do you find it convenient to place one of the sharp
points of the compasses tyhere two heavy lines intersect ?
3. What are the advantages of measuring with squared
paper over measuring with a ruler ?
30. Measuring a length approximately to two decimal
places. By increasing the size of the unit of measurement
we may express the result with approximate accuracy to
two decimal places. In Fig. 11 let MN (equal to ten small
units, or 2 cm., of A %
the squared paper) '
be the unit. As
before, lay off AB
upon the squared
paper with the
compasses (CD in
the new position).
Now EF (a small
unit) equals 0.1 of
a unit, and 0.1 of EF equals 0.01 of the unit MN. Reading
the units in the line CD, we are sure the result is greater
than 2.7, for the crossing line is beyond that. But it is not
2.8. Why? Now imagine a small unit, as EF, divided
into tenths and estimate the number of tenths from the
end of the twenty-seventh small unit to D. This appears
N
Unit
FIG. 11
GENERAL MATHEMATICS
to be 0.4, but this is 0.04 of a unit;, hence CD equals
2.74 units. This means that CD is 2.74 times as long as
the line MN. Of course the 4 is only an approximation,
but it is reasonably close.
\
Kt-
EXERCISES
1. In Fig. 12 measure to two decimal places the segments
CD, DE, and EC. Compare the results of your work with
that of the other members of
the class.
2. Is the result obtained by
the method of Art. 30 more
accurate than the result ob-
tained by using 1 cm. as a unit
and claiming accuracy to only
one decimal place ?
FIG. 12
31. Equal line segments.
When the end points of one segment, as a in Fig. 13, coin-
cide with (exactly fit upon) the ends of another segment,
as 6, the segments a and b are said to be
equal. This fact may be expressed by
the equation a = b.
32. Unequal segments; inequality. If
the end points of two segments, as a and J,
cannot possibly be made to coincide, the segments are said to
be unequal. This is written a = b (read " a is not equal to 6").
The statement a 3= bis called . a |
an inequality. In Fig. 14 seg-
ment a is less than segment b
(written a < i), and seg-
ment c is greater than seg-
ment b (written c > 6).
I 1
I b - 1
FIG. 13. EQUAL LINE
SEGMENTS
V
FIG. 14. UNEQUAL LINE SEGMENTS
LINEAR MEASUREMENT 35
In the preceding equation and inequalities we need to
remember that the letters a, b; and c stand for the length
of the segments. They represent numbers which can be
determined by measuring the segments.
33. Ratio. This article will show that ratio is a funda-
mental notion in measurement.
INTRODUCTORY EXERCISES
(Exs. 1-4 refer to Fig. 15)
1. Measure the segment a accurately to two decimal places.
2. Measure the segment b
accurately to two decimal | 1
places.
3. What part of b is a ? \ I
4. Find the quotient of b FlG 15
divided by a.
The quotient of two numbers of the same kind is called
their ratio. The ratio is commonly expressed as a fraction.
Before forming the fraction the two quantities must be
expressed in terms of the same unit; for example, the
ratio of 2 ft. to 5 in. is the ratio of 24 in. to 5 in., that
is, - 2 ^ 4 -. The unit of measure is 1 in. Obviously there is
no ratio between quantities of different kinds ; for exam-
ple, there is no ratio between 7 gal. and 5 cm.
It should now be clear that every measurement is the
determination of a ratio either exact or approximate. Thus,
when we measure the length of the classroom and say it> is
10 m. long, We mean that it is ten times as long as the
standard unit, the meter ; that is, the ratio is. -L<>..
- - - - ' -. .- - - :
36
GENERAL MATHEMATICS
EXERCISES
1. The death rate in Chicago in a recent year was 16 to 1000
population. Express this ratio as a fraction.
2. An alloy consists of copper and tin in the ratio of 2 to 3.
What part of the alloy is copper ? What part is tin ?
3. A solution consists of alcohol and water in the ratio of
3 to 6. What part of the solution is water ?
4. The ratio of weights of equal volumes of water and cop-
per is given by the fraction How many times heavier is
copper than water ?
5. Water consists of hydrogen and oxygen in the ratio of
1 to 7.84. Express this ratio as a decimal fraction.
34. Sum of two segments ; geometric addition. It is
possible to add two line segments by the use of compasses.
Thus, in Fig. 16 if the segment a is laid off on the number
scale of squared paper from point A to point B and if in
turn b is laid off on the same line from B to C, then the
FIG. 16. GEOMETRIC ADDITION OF LINE SEGMENTS
sum of these lines can be read off at once. In Fig. 16 the
sum is 5.4 cm. The segment AC is the sum of a and b.
Very often in construction work we are not concerned
about either the length of the segments or their sum. In
that case lay off the segments as above on a working line
and indicate the sura of a and b as a +- b. Addition per-
formed by means of the compasses is a geometric addition.
LINEAR MEASUREMENT
37
FIG. 17
EXERCISES
1. In Fig. 17 find the sum of a, b, and c on a working
line. Indicate the sum.
. .
2. In Fig. 17 add the seg-
ments a, b, and c on the scale i - 1
line of a sheet of squared c
paper. Express the value of
a + b + c in centimeters.
3. On a working line draw
one line to indicate the num-
ber of yards of fencing
needed for the, lot in Fig. 18.
4. In Fig. 19 the whole
segment is denoted by c.
Show by measuring on
squared paper that c = a + b.
5. In Fig. 19 what rela-
tion exists between c and
either a or b? Why?
FIG. 18
FIG. 19
35. Axioms. Exs. 4 and 5, above, illustrate the following
two axioms :
VI. The whole is equal to the sum of all its parts.
VII. The whole is greater than any one of its parts.
EXERCISES
1. Draw the segments a = 2.3 cm.; b = 3.2 cm.; c = 1.3cm.
Draw the sum a + b + c.
2. Let a, b, and c denote three line segments. Draw a line
segment to represent 2a + 3b + c ; to represent 4 a -f b -f 2 c ;
to represent a -\- 3 b -j- 4 c.
38 GENERAL MATHEMATICS
3. In Fig. 20, if a, b, and c are three consecutive segments on a
straight line, such that a = c, show | , I |
by measuring that a + b = I + c.
FIG. 20
4. Show with out measuring that
a + b = b + c. What axiom does this fact illustrate ? Quote
the axiom.
36. Order of terms in addition. The fact that We get the
same sum when we lay off a segment a and then add b
that we do when we lay off b first and then add a is a
geometric illustration of the truth of the commutative law.
This law asserts that the value of a sum does not change
when the order of the addends is changed. In the first chap-
ter we illustrated this principle by a familiar experience
from arithmetic, as 2 + 5-1-4 = 2 + 4-}- 5.
EXERCISES
1. Illustrate the validity of the commutative law by a fact
from your everyday experience.
2. Add in the most advantageous way, using the commutative
law : .376 + 412 + 124 ; 2187 + 469 + 213 ; 36 + 142 + 164.
37. Difference of two line segments. The difference of
two line segments may also be found with the compasses.
To find out how much greater | . 1
the segment c is than the seg-
ment b we lay off the segment c
on a working line (Fig. 21) from
A to C, then lay the segment A. D b c
b backward from C toward A. FIG. 21. GEOMETRIC SUBTRAC-
Then the difference between
the segments b and c is expressed by the segment AD. In
equation form this may be written AD = c b. Illustrate
this method by comparing the lengths of two pencils.
c-b
LINEAR MEASUREMENT 39
EXERCISES
1. Transfer the line segments of Fig. 21 to squared paper
and express the difference between c and b in centimeters.
2. Subtract a line segment 3.5 cm. long from one 6 cm. long.
3. In Fig. 22 the line segment AB equals the line segment
MX. Show by measuring that M h N
AB - MB = MN - MB. \ h^-+ 1
A B
4. Ex. 3 is simpler if we ^ IG 22
write the fact in algebraic
form, using the small letters. Thus, a + c = b + c. How would
you show that a = b ?
5. What axiom of the first chapter is illustrated by Exs. 3
and 4 ? Quote the axiom.
6. If a = 3 cm., b = 2 cm., and c 1 cm., construct a line
segment representing 2 a + 3 b c ; representing 5a 2b + c.
7. If a, b, and c represent the length of three respective seg-
ments, construct 3 <z -f- 2 # c ; 4 a + 2 2 c ; 5 a 2 6.+ 3 c.
8. How long would each of the segments constructed in
Ex. 7 be if a = 5, b = 4, and c = 3 ?
38. Coefficient. The arithmetical factor in the term 2#
is called the coefficient of the literal factor #. When no
coefficient is written, as in #, we understand the coeffi-
cient to be 1. Thus, x means 1 x. The coefficient of a
literal number indicates how many times x
the literal number is to be used as an ~ ~
addend; thus, 5 a: means x+x+x+z+x
and can be expressed by the equation 5x = x+ x+x-\-x+x.
Written in this form we see that the use of a coefficient
is a convenient method of abbreviating. Geometrically a
coefficient may be interpreted as follows: Let x be the
length of the segment in Fig. 23. Then the 5 in 5x indi-
cates that the line segment x is to be laid off five times
40 GENERAL MATHEMATICS
consecutively on a working line. 5.r expresses the sum
obtained by this geometric addition. Find this sum. Usu-
ally the term "coefficient" means just the arithmetical factor
in a term, though in a more general sense the coefficient
of any factor, or number in a term, is the product of all the
other factors in that term. Thus, in 3 aby the coefficient
of y is 3 ab, of by is 3 a, of aby is 3.
EXERCISE
Give the coefficient in each of the following terms : 3 b ;
x 7x 8# 9 a-.
39. Polynomials. An algebraic number consisting of two
or more terms (each called a monomial), as bx + % y 2 2,
is a polynomial. The word "polynomial" is derived from
a phrase which means many termed. A polynomial of two
terms, as 5 x + 3 #, is a binomial. A polynomial of three
terms, as 2a-f3J4-4c, is a trinomial.
EXERCISE
Classify the following expressions on the basis of the
number of terms :
(a) 2 m + 3 n 5 x + r. (c) 6 x + 2 y.
(b) 6x. (d) a + 2& + 3.
40. Algebraic addition of similar terms. In simple prob-
lems we have frequently added similar terms. We shall
now review the process by means of the following example
in order to see clearly the law to be used in the more
complicated additions :
Add 4:r-!-3a: + 2:r.
Solution. 4 x can be considered as the sum of four segments each
x units long.
Therefore 4x = z + z + r + x.
LINEAR MEASUREMENT 41
Similarly, 3 x = x + x + x,
and 2x = z + x.
Adding,
4:X + 3x + 2x=:x + x + x + x + x + x + x + x + x, or 9 z.
Hence . 4z + 3z + 2z = 9z.
The preceding example illustrates the law that the sum
of two or more similar monomials is a monomial whose coeffi-
cient is the sum of the coefficients of the given monomials and
which has the same literal factor as the given monomials.
The advantages of adding numbers according to this
law may be seen by comparing the two solutions of the
following problem :
The tickets for a school entertainment are sold at 250 each
by a cominittee of five students, A, B, C, D, and E. A reports
38 sold; B, 42; C, 26; D, 39; and E, 57. At the door 173
tickets are sold. Find the total receipts.
Solution I.
A's receipts,
38
x
$0.25 =
$9.50
B's receipts,
42
x
0.25 =
10.50
C's receipts,
26
X
0.25 =
6.50
D's receipts,
39
X
0.25 =
9.75
E's receipts,
57
X
0.25 =
14.25
Door receipts,
173
X
0.25 =
43.25
Total receipts,
$93.75
Since the common factor in the above multiplication is
25, a simpler solution is obtained if the numbers of tickets
sold (coefficients) are first added and placed before the
common factor ; thus,
Solution II. A's receipts, 38 x $0.25
B's receipts,
42 x
0.25
C's receipts,
26 x
0.25
D's receipts,
39 x
0.25
E's receipts,
57 x
0.25
Door receipts,
173 x
0.25
Total receipts,
375 x
0.25
= $03.75
42 GENERAL MATHEMATICS
EXERCISES
1. Tickets were sold at c cents ; A sells 12; B, 15; C, 36;
and D, 14. There were 112 tickets sold at the gate. Find the
total receipts.
2. Express as one term 3-7 + 5-7-f4-7.
NOTE. A dot placed between two numbers and halfway up in-
dicates multiplication and is read "times." Do not confuse with
the decimal point.
3. Can you add 3 7 + 14 2 -f 5 4 by the short cut above?
4. Indicate which of the following sums can be written in
the form of monomials : 3x + 5a;; x + 7 x + 3; 13 + 5 -|- 3 ;
3 + 2 1 + 4.
5. Add as indicated : (a) 3 x + 20 x + 17 x + 9 x + ~x + 3 x ;
(b) 3y + y + 15y+ily + 2y; (c) 9* + 3s + 3s + 4 .s -j- 2s ;
6. The school's running track is / feet. While training, a
boy runs around it five times on Monday, six on Tuesday, ten
on Wednesday, seven 011 Thursday, six on Friday, and nine on
Saturday morning. How many feet does he run during the week ?
41. Subtraction of similar monomials. The law in sub-
traction is similar to the law in addition and may be
illustrated as follows:
Subtract 2 x from 5 x.
Solution. 5x = x + x + x + x + x.
2 x = x + x.
Subtracting equal numbers from equal numbers,
5 x 2 x = x + x + x, or 3 x.
Hence 5x 2x = 3 x.
The preceding example illustrates the law that tJie differ-
ence of two similar monomials is a monomial having a coefficient
equal to the difference of the coefficients of the given monomials
and having the same literal factor.
LINEAR MEASUREMENT
43
EXERCISES
1. Subtract 3 b from 146.
2. Write the differences of the following pairs of numbers as
monomials: lOce 3ic; 13x 5x; 12z 3z; VI k 5k; 2.68 r
-0.27/-; 1.03a-0.08a; fe-Je; f* i*.
3. The following exercises require both addition and sub-
traction. Write each result as a single term : <ix-}-6x 2x;
13x 2x + 3x; 11.5c + 2.3c - c; Ja + f a J a.
42. Triangle; perimeter. If three points, as A, B, and
(7 (Fig. 24), are connected by line segments, the figure
formed* is a triangle. The
three points are called ver-
tices (corners) of the triangle,
and the three sides a, 6, and A
c are the sides of the -tri-
angle. The sum of the three sides, as a -f- b + c (the
distance around), is the perimeter of the triangle.
EXERCISES
1. A yard has the form of an equal-sided (equilateral)
triangle, each side being x rods long. How many rods of
fence will be needed to inclose it ?
2. What is the sum of the sides
of a triangle (Fig. 25) whose sides
are 2 x feet, 2 x feet, and 3 x feet
long? Express the sum as a cer-
tain number of "times x.
3. What is the sum of the three sides 3b, 4b, and 6b of a
triangle ? Express the result as one term.
4. What is the perimeter of a triangle whose sides are 2x,
8 x } and 9 x ? Let p be the perimeter ; then write your answer
to the preceding question in the form of an equation.
2x
3x
FIG. 25
44
GENERAL MATHEMATICS
43. Polygons. A figure, as ABODE (see Fig. 26), formed
by connecting points, as A, J5, C, D, and E, by line
segments, is a polygon.
The Greek phrase from
which the word "polygon"
is derived means many cor-
nered. Polygons having 3,
4, 5, 6, 8, 10, ., n sides
are called triangle, quadri-
lateral, pentagon, hexagon,
octagon, decagon, - - ., n-gon FlG 2 6. A POLYGON
respectively. The sum of
the sides of a polygon is its perimeter. When all the
sides of a polygon are equal it is said to be equilateral.
EXERCISES.
1. What is the perimeter of each of the polygons in Fig. 27?
In each case express the result in the form of an equation ;
thus, for the first quadrilateral p = 12 x.
IX
FIG. 27
2. Show by equations the perimeter p of the polygons in
Fig- 28.
3. Show by equations the
perimeter of an equilateral
quadrilateral whose side is
11; 9; s; x; b; z\ 2e;
9 + 3; a + 5; a + d; x + 7 :
x-fy.
10 ,2
c
d
FIG. 28
LINEAR MEASUREMENT 45
4. Name the different figures whose perimeters might be
expressed by the following equations :
p = 3 s, p = 5 s, j) = 7 s, p = 9 s, p = 12 s, p = 20 s,
p = 4 s, p = 6 s, p = 8 s, p = 10 s, p = 15 s, p = ns.
5. Assume that all the figures in Ex. 4 are equilateral.
Find out how many of your classmates can give the name of
each polygon.
6. Assume that at least six of the polygons in Ex. 4 are
not equilateral. Sketch the figures of which the given equa-
tions may be the perimeters.
7. What is the perimeter of each figure of Ex. 4 if s = 2 in. ?
ifs = 3cm.? if s = 4yd.? ifs = 5ft.?
8. Determine the value of s in the equations p = 3 s, p = 4 s,
p = 5s, p = 6s, p = 10s, and p = 15s if in each case the
perimeter is 120 in.
9. What is the side of an equilateral hexagon made with a
string 144 in. long ? (Use all the string.)
10. Show by sketches polygons whose perimeters are ex-
pressed by p = 8 a + 6 ; by^? = 4 + 12; by^? = 66 + 6a;
by^> 4a + 25; by p = 3a -\- 2b.
11. Find the value of the perimeters in Ex.10 if a = 3
and b = 2 ; if a = 5 and b = 3 ; if a = 1 and b = 5.
12. If x = 2 and y 3, find the value of the following
expressions : 3x + ?/ ; 3x y; 3x 2 ?/ ; 2x 3|- ; 4 a; 2-^ ?/ ;
2.25 x-y; 2.27 x- 1.12 y.
SUMMARY
44. This chapter has taught the meaning of the follow-
ing words and phrases : line segment, point, measurement
of length, unit segment, standard unit, ratio, metric system,
coincide, intersect, equal segments, unequal segments, com-
mutative law, coefficient, polynomial, binomial, trinomial,
46 GEKEKAL MATHEMATICS
triangle, vertex of a polygon, vertices of a polygon, perim-
eter, sides of a triangle, polygon, quadrilateral, pentagon,
hexagon, octagon, decagon, n-gon, equilateral.
45. Axioms. The following axioms were illustrated:
VI. The whole is equal to the sum of all its parts.
VII. The whole is greater than any of its parts.
46. The following instruments have been used in measur-
ing line segments : the ruler, the compasses, and squared paper.
47. The following symbols were used :. = meaning does
not equal; < meaning is less than; > meaning is greater
than ; and a dot, as in 3 5, meaning times, or multiplied by.
48. A point is determined by two intersecting lines.
49. The metric system has certain advantages over our
English system.
50. The practical difficulty of precise measuring has
been pointed out. Five possible errors were enumerated.
Measurement implies the determination of a ratio.
51. The sum of two segments was found with the com-
passes. A law was discovered to serve as a short cut in
algebraic addition.
52. The difference of two segments was found with the
compasses and the law for algebraic subtraction was stated.
53. The Addition and Subtraction laws of Chapter I
were illustrated geometrically.
54. The perimeter of a figure may be expressed by an
equation.
55. The chapter has taught how to find the value of an
algebraic number when the value of the unknowns are
given for a particular cage; for example, how to find the
value of 3 x + 2 y when x- == 1 and #==2.
CHAPTER III
PROPERTIES OF ANGLES
56. Angle. If a straight line, as OX in either of the
drawings of Fig. 29, rotates in a plane about a fixed point,
as 0, in the direction indicated by the arrowheads (counter-
clockwise) until it reaches the position OT, it is said to
x
FIG. 29. ILLUSTRATING THE DEFINITION OF AN ANGLE
turn through the angle XOT. Thus, an angle is the amount
of turning made by a line rotating about a fixed point in a
plane (flat surface}. Note that as the rotation continues,
the angle increases.
57. Vertex. The fixed point (Fig. 29) is called the
vertex of the angle. (The plural of "vertex" is "vertices.")
47
48 GENERAL MATHEMATICS
58. Initial side ; terminal side. The line OX (Fig. 29)
is called the initial side of the angle. The line OT is called
the terminal side of the angle.
59. Symbols for " angle." The symbol for "angle" is Z;
for " angles," A Thus, " angle XOT" is written /^XOT.
60. Size of angles. From the definition of an angle given
in Art. 56 we see that it is possible for the line OX to stop
rotating (Fig. 29) so that the angle may contain any given
amount of rotation (turning).
EXERCISE
Draw freehand an angle made by a line OX which has
rotated one fourth of a complete turn ; one half of a complete
turn ; three fourths of a complete turn ; one complete turn ;
one and one-fourth complete turns.
61. Right angle ; straight angle ; perigon. If a line
rotates about a fixed point in a plane so as to make
T one fourth of a complete turn, the angle formed
is called -a right angle (rt. Z) (see Fig. 30, (a)).
-X
O
(a) Right Angle (b) Straight Angle (c) Perigon
FIG. 30. THREE SPECIAL ANGLES
If the line makes one half of a complete turn, the angle
formed is called a straight angle (st. Z) (see Fig. 30, (b)) ; if
the line makes a complete turn, the angle formed is called
& perigon (see Fig. 30, (c)).
EXERCISES
1. Draw freehand an angle equal to 1 right angle; 2 right
angles ; 3 right angles ; 4 right angles.
2. Draw freehand an angle equal to 1 straight angle;
2 straight angles ; 1|- straight angles ; 2^ straight angles.
PKOPEKTIES OF ANGLES
49
3. How many right angles are there in a half turn of a
rotating line ? in a whole turn ? in 5 turns ? in 3^ turns ? '
in x turns ? in - turns ? in - turns ?
2 4
4. How many straight angles are there in a half turn of a
rotating line ? in a whole turn ? in 1^ turns ? in 1| turns ?
5x
in x turns ? in 2x turns ? in - turns ?
4
5. Through how many right angles does the minute hand of
a watch turn in 3 hr. ? in 3^ hr. ? in 4 hr. ? in 1\ hr. ? in
x hours ? in 15 min. ? in 30 min. ? in 5 min. ? in 10 min. ?
6. How many straight angles are there in a perigon ? in
a right angle ? in 10 right angles ? in y right angles ?
62. Acute angle ; obtuse angle ; reflex angle. Angles
are further classified upon the basis of their relation to
T
FIG. 31. ACUTE ANGLE
FIG. 32. OBTUSE ANGLE
the right angle,' the straight angle, and the perigon. An
angle less than a right angle is called an acute angle,
Fig. 31. An angle
which is greater
than a right angle
and is less than- 'H
a straight angle
is called an obtuse
angle, Fig. 32. An
angle greater than
a straight angle
and less than a perigon is called a reflex angle, Fig. 33.
FIG. 33. REFLEX ANGLE
GENERAL MATHEMATICS
EXERCISES
4
1. Draw an acute angle ; an obtuse angle ; a reflex angle.
2. Point out, in the classroom, examples of right angles ;
of obtuse angles.
C\
C
D
(a) (b) (c)
FIG. 34. ILLUSTRATING THE VARIOUS KINDS OF ANGLES
3. In the drawings of Fig. 34 determine the number of acute
angles ; of right angles ; of obtuse angles ; of reflex angles.
63. Notation for reading angles. There are three common
methods by which one may denote angles: (1) Designate
the angle formed by two lines OX
and OT as the " angle XOT" or the
" angle TOX" (Fig. 35). Here the
first and last letters denote points
on the lines forming the angle, and
the middle letter denotes the point
of intersection (the vertex). In read-
ing " angle XOT" we regard OX as
the initial side and OT as the terminal side. (2) Denote the
angle by a small letter placed as # in Fig. 36. In writing
equations this method is the B
most convenient. (3) Denote
the angle by the letter which
is written at the point of in-
tersection of the two sides of
the angle, as " angle A" (Fig. 36). This last method is used
only when there is no doubt as to what angle is meant.
FIG. 35
FIG. 36
PKOPERTIES OF ANGLES
EXERCISE
51
In the drawings of Fig. 37, below, select three angles and
illustrate the three methods of notation described above.
(b)
FIG. 37
64. Circle. If a line OX be taken as the initial side
of an angle (see Fig. 38) and the line be rotated one
complete turn (a peri-
gon), any point, as P,
on the line OX will
trace a curved line
which we call a circle.
Thus, a circle is a
closed curve, all points
of which lie in the
same plane and are
equally distant from a
fixed point. FIG. 38. THE CIRCLE
52 GENERAL MATHEMATICS
65. Center ; circumference. The fixed point is the
center of the circle. The length of the curve (circle) is
called the circumference (distance around) of the circle.
66. Radius ; diameter. A line drawn from the center
of a circle to any point on the circle is a radius. Thus,
OP is a radius of the circle in Fig. 38. A line connecting
two points on the circle and passing through the center of
the circle is called a diameter.
From the definition of " radius " given above it is clear
that in a given circle or in equal circles one radius has
the same length as any other. Thus we obtain the follow-
ing important geometric relation, Radii of the same circle
or of equal circles are equal. ("Radii" is the plural of
" radius.")
67. Arc; to intercept; central angle. An arc is a part
of a circle. If two radii are drawn from the center of the
circle to two different points on the circle, they cut off an
arc on the circle. The symbol for." arc " is ""^ Thus, AB is
read " the arc AB" The angle formed at the center of the
circle is said to intercept the arc. The angle at the center
is called a central angle.
68. Quadrant; semicircle. An arc equal to one fourth
of a circle is called a quadrant. An arc equal to one half
of a circle is called a semicircle.
EXERCISES
1 . How does a diameter compare in length with a radius ?
2. How many quadrants in a semicircle ? in a circle ? In
what connection have we mentioned the idea expressed by the
word " quadrant " ?
PROPERTIES OF ANGLES
53
69. Degrees of latitude and longitude. The use that is
made of the circle in geography is no doubt familiar to
all of us. The prime meridian, that passes through Green-
wich, England (see Fig. 39), is the zero meridian. We
speak of places lying to the west of Greenwich as being
in west longitude and of those lying to the east of
Greenwich as being in east longitude (see Fig. 39). Since
it takes the earth twenty-four hours to make one complete
rotation, the sun apparently passes over one twenty-fourth
N.P.
N.P.
S.P.
LATITUDE AND LONGITUDE
of the entire distance around the earth every hour. Thus,
points lying a distance of one twenty -fourth of a complete
turn apart differ in time by an hour.
In order to carry the computations further the entire
circle is divided into three hundred and sixty equal parts,
each of which is called a degree (1) of longitude. In
order to express fractional parts of the unit each degree
is divided into sixty minutes (60') and each minute into
sixty seconds (60"). With this agreement the longitude
of a place is determined.
The position of a place north or south of the equator is in-
dicated by the number of degrees of north or south latitude.
54 GENERAL MATHEMATICS
EXERCISES
1. What is the greatest longitude a place can have ? the
greatest latitude ?
2. How many seconds are there in a degree of longitude ?
in a degree of latitude ? *
( $7 What is the length of a degree arc on the earth's
surface ? of a minute arc ? of a second arc ? Try to find out
how accurately the officers of a ship know the location of
the ship out in mid-ocean.
4. How would you read 25 14' west longitude ? 33 5' 17"
north latitude ?
5. Compare the method of locating by latitude and longitude
to the method used in locating a house in a large city.
6. Find out in what longitude you live ? in what latitude ?
70. Amount of rotation determines the size of an angle.
If we remember that an angle is formed by rotating a line
in a plane about a fixed point, it will be clear that the
Fio. 40
size of the angle depends only on the amount of turning,
not upon the length of the sides. Since the sides may
be extended indefinitely, an angle may -have short or long
sides. In Fig. 40 the angle A is greater than angle B, but
the sides of angle B are longer than the sides of angle A.
71. Measurement of angles; the protractor. In many
instances the process of measuring angles is as important
as that of measuring distances. An angle is measured
when we find how many times it contains another angle
selected as a unit of measure.
PROPERTIES OF ANGLES
55
FIG. 41. THE PROTRACTOR
The protractor (Fig. 41) is an instrument devised for
measuring and constructing angles. The protractor com-
monly consists of a semicircle divided into one hundred
and eighty equal parts. Each of these equal parts is called
a degree of arc (1). In the geography work referred to in
Art. 69, the unit for longitude and latitude was the degree
of arc. In the measure-
ment of angles we shall
consider a unit corre-
sponding to a unit of
arc and called a degree
of angle.
If straight lines are
drawn from each of
these points of division
on the semicircle to the
center 0, one hundred
and eighty equal angles are formed, each of which is a
degree of angle (I )-. Thus, the unit of angular measure
is the degree. A degree is divided into sixty equal parts,
each of which is called a minute (!').
Each minute is divided into sixty equal parts, eack
of which is called a second (1"). Of course the minute
and the second graduations are not shown on the pro-
tractor. Why not ?
EXERCISES
1 . How many degrees in a right angle ? in a straight angle ?
in a perigon ?
2. A degree is what part of a right angle ? of a straight
angle ? of a perigon ?
3. What angle is formed by the hands of a clock at 3 o'clock ?
at 6 o'clock ? at 9 o'clock ? at 12 o'clock ? at 4 o'clock ? at
7 o'clock ? at 11 o'clock ?
56 GENERAL MATHEMATICS
4. Give a time of day when the hands of a clock form a
right angle; a straight angle.
5. What is the correct way to read the following angles :
15 17' 2"? 5 0' 10"?
6. How many degrees are there in three right angles ? in
four straight angles ? in one third of a right angle ? in two
thirds of a straight angle ? in one fifth of a right angle ? in
x straight angles ? in y right angles ? in 2 x right angles ?
7. Ordinary scales for weighing small objects are sometimes
made with a circular face like a clock face. The divisions of
the scale indicate pounds. If the entire face represents 12 lb.,
what is the angle between two successive pound marks ?
8. What is the angle between two successive ounce marks
on the face of the scale in Ex. 7 ?
72. Measuring angles ; drawing angles. The protractor
may be used to measure a given angle. Thus, to measure
a given angle x place the protractor so that the center of
the protractor (point in Fig. 42) falls upon the vertex
and make the straight edge of the protractor coincide
with (fall upon) the initial side of the given angle x. Now,
observe where the terminal side of the given angle intersects
(crosses) the rim of the protractor. Read the number of
degrees in the angle from the scale on the protractor.
EXERCISES
1. Draw three different angles, one acute, one obtuse, and
one reflex. Before measuring, estimate the number of degrees in
each angle. Find the number of degrees in each angle by means
of the protractor. Compare the results with the estimates.
2. Draw freehand an angle of 30; of 45; of 60; of 90;
of 180 ; of 204. Test the accuracy of the first four angles by
means of the protractor.
PEOPERTIES OF ANGLES
57
The protractor is also useful in constructing angles
of a required size ; for example, to construct an angle
of 42 draw
a straight line
OX (Fig. 42)
and place the
straight edge
of the pro-
tractor on the
FIG. 42. MEASURING ANGLES WITH A PROTRACTOR
that the center
rests at 0. Count 42 from the point on OX where the
curved edge touches OX and mark the point A. Connect
A and 0, and the angle thus formed will contain 42.
EXERCISES
1. Construct an angle of 25; of 37; of 95; of 68; of
112 ; of 170. Continue this exercise until you are convinced
that you can draw any required angle.
2. Construct an angle equal to a given angle ABC by means
of the protractor.
HINT. Draw a working line MN. Measure the angle ABC.
Choose a point P on the line MN as a vertex and construct an
angle at P containing the same number of degrees as the given
angle ABC. The angle is then constructed as required.
3. Draw a triangle. How many angles does it contain ?
Measure each with the protractor.
73. How to measure angles out of doors. It is possible
to measure angles out of doors by means of a simple field
(out-of-door) protractor, so that some simple problems in sur-
veying can be solved. Such a field protractor may be made
by a member of the class, as shown on the following page.
58
GENERAL MATHEMATICS
Secure as large a protractor as possible and fasten it on
an ordinary drawing board. Attach the board to a camera
tripod (if this is not to be had, a rough tripod can be
made). Make a slender pointer which may be attached at
the center of the circle with a pin so that it may swing freely
about the pin as a pivot. Place two inexpensive carpenter's
levels on the board, and the instrument is ready for use.
Thus, to measure an angle ABC (suppose it to be an
angle formed by the intersection of an avenue, BA, and a
street, -BC), first put the board in a horizontal position
(make it stand level). Then place the center of the circle
over the vertex of ;
the angle to be meas-
ured and sight in
the direction of each
side of the angle,
noting carefully the
reading on the pro-
tractor. The number
of degrees through
which the pointer
is turned in passing
from the position of
BA to that of BC
is the measure of
angle ABC.
74. Transit. When
it is important to
secure a higher de-
gree of accuracy than is possible with the instrument
described in Art. 73, we use an instrument called a transit
(Fig. 43). This instrument is necessary in surveying.
Three essential parts of the transit are (1) a horizontal
FIG. 43. THE TRANSIT
59
graduated circle for measuring angles in the horizontal
plane (see D in Fig. 43) ; (2) a graduated circle, C, for
measuring angles in the vertical (up-and-down) plane ; and
(3) a telescope, AB, for sighting in the direction of the
sides of the angle. For a fuller description of the transit
see a textbook in trigonometry or surveying.
HISTORICAL NOTE. The division of the circle into three hundred
and sixty degrees and each degree into sixty minutes and each
minute into sixty seconds is due to the Babylonians. Cajori cites
Cantor and others somewhat as follows : At first the Babylonians
reckoned the year as three hundred and sixty days. This led them
to divide the circle into three hundred and sixty degrees, each degree
representing the daily part of the supposed yearly revolution of the
sun around the earth. Probably they were familiar with the fact that
the radius could be applied to the circle exactly six times and that
as a result each arc cut off contained sixty degrees, and in this way
the division into sixty equal parts may have been suggested. The
division of the degree into sixty equal parts called minutes may have
been the natural result of a necessity for greater precision. Thus the
sexagesimal system may have originated. " The Babylonian sign * is
believed to be associated with the division of the circle into six equal
parts," and that this division was known to the Babylonians seems
certain " from the inspection of the six spokes in the wheel of a royal
carriage represented in a drawing found in the remains of Nineveh."
Henry Briggs attempted to reform the system by dividing the
degree into one hundred minutes instead of into sixty, and although
the inventors of the metric system are said to have proposed the
division of the right angle into one hundred equal parts and to
subdivide decimally, instead of the division into ninety parts, we
have actually clung to the old system. However, there is a tend-
ency among writers to divide each minute decimally ; for example,
52 10.2' instead of 52 10' 12". See Cajori, " History of Elementary
Mathematics," 1917 Edition, pp. 10, 43, and 163.
75. Comparison of angles. In order to make a comparison
between two angles, we place one over the other so that the
vertex and the initial side of one coincide with the vertex and
the initial side of the other. If the terminal sides coincide,
60
GENERAL MATHEMATICS
the angles are equal ; if the terminal sides do not coincide,
the angles are unequal assuming, of course, in both cases,
that each of the two angles compared is less than 360. In
the exercises and articles that follow we consider no angle
greater than 360.
EXERCISES
1. Compare angles x, y, and 2, in Fig. 44, and arrange them
in order as to size.
HIXT. Make a tracing of each on thin paper and try to fit each
on the other.
FIG. 44
2. Construct an angle equal to a given angle ABC. Lay a thin
sheet of paper over the angle ABC and make a tracing of it. Cut
out the tracing and paste it to another part of the paper. The
angle thus shown is equal to the angle ABC.
3. Try to draw freehand two equal angles.
Test your drawings by the method of Ex. 1.
4. Draw freehand one angle twice as
large as another. Test your drawings with
the protractor.
FIG. 45
76. Adjacent angles ; exterior sides.
Angles x and y in Fig. 45 are two angles
which have a common vertex and a com-
mon side between them. The angles x and y are said to be
adjacent angles. Thus, adjacent angles are angles that have
the same vertex and have a common side between them.
The sides OT and OR are called the exterior sides.
PKOPERTIES OF ANGLES
61
E
EXERCISES
1. Indicate which angles in Fig. 46 are adjacent. Point out
the common vertex and the common side in each pair of
adjacent angles.
2. Draw an angle of 45 adjacent
to an angle of 45; an angle of 30
adjacent to an angle of 150; an
angle of 35 adjacent to an angle
of 80.
3. Do you notice anything partic-
ularly significant in any of the parts
of Ex. 2 ?
4. Draw an angle of 30 adjacent
to an angle of 60. What seems to
be the relation between their exterior
sides ? Does this relation need to
exist in order that the angles shall
be adjacent ? What total amount of turning is represented ?
77. Geometric addition and subtraction of angles. Exs. 2
and 4, above, suggest a method for adding any two given
angles. Thus, to add a given angle y to a given angle x,
B
FIG. 46
FIG. 47. GEOMETRIC ADDITION OF ANGLES
Fig. 47, angle y is placed adjacent to angle z, and the re-
sulting angle is called the sum of x and y. The angles may
be transferred to the new position either by means of tracing
paper or, more conveniently, by means of the protractor.
62 GENERAL MATHEMATICS
EXERCISE
Add two angles by placing them adjacent to each other.
We may also find the difference between two angles.
In Fig. 48 the two given angles are x and y. Place tin-
smaller angle, y, on the larger, a:, so that the vertices and
o
FIG. 48. GEOMETRIC SUBTRACTION OF ANGLES
one pair of sides coincide. The part remaining between the
other two sides of x and y will be the difference between
x and y. Thus, in Fig. 48 we obtain Zz /.y=/.AOC.
EXERCISES
1. Draw three unequal angles x, y, and 2, so that y~>x and
x > z. Draw an angle equal to x + y + z ; equal to y x + z ;
equal to y -\- x - z.
2. Draw an angle of 60 and draw another of 20 adjacent
to it. What is their sum ? Fold the 20-degree angle over the
60-degree angle (subtraction) and call the difference x. What
is the equation which gives the value of x ?
78. Construction problem. At a given point on a given
line to construct by means of ruler and compasses an
angle equal to a given angle. In this construction we
make use of the following simple geometric relation be-
tween central angles and their intercepted arcs : In the
same circle or in equal circles equal central angles intercept
PROPERTIES OF ANGLES 63
equal arcs on the circle. For example, if the central angle
contains nineteen angle degrees, then the intercepted arc
contains nineteen arc degrees.
The student may possibly see that this geometric rela-
tion is implied in our definitions of Art. 71. However,
the two following paragraphs will assist him in under-
standing its application.
Make a tracing of the circle and the angle ABC .in
Fig. 49, (a), and place B upon E in Fig. 49, (b). The
angles must coincide because they are given equal. Then
the circle whose center
is B (circle B) must
coincide with the cir-
cle whose center is E
(circle -E 1 ), because the
radii of equal circles
are equal. Then A will
FIG. 49
fall on Z>, and C on F;
that is, the arc CA will fall on the arc FD, and the arcs
are therefore equal.
It is easy to see that the following statement is also
true : In the same circle or in equal circles equal arcs on the
circle are intercepted by equal central angles. For circle M
can be placed on circle E so that arc CA coincides with
arc FD, since these arcs are given equal, and ''so .that B
falls on E. A will fall on Z>, and C on F. Then the angles
must. coincide and are therefore equal.
The two preceding geometric relations make clear why
the protractor may be used to measure angles as we did
in Art. 71. The method used there is based upon the idea
that every central angle of one degree intercepts an arc
of one degree on the rim of the protractor ; that is, when
we know the number of degrees in an angle at the center
64
GENEKAL MATHEMATICS
of a circle we know the number of degrees in the arc
intercepted by its sides, and vice versa.
The idea can be expressed thus: A central angle is
measured by the arc intercepted by its sides (when angular
degrees and arc degrees are used as the respective units
of measure).
How many degrees of an arc are intercepted by a central
angle of 30 ? of 40 ? of 60.5 ? of n ?
We are now ready to proceed with our problem: At
a given point on a given line to construct by means of
ruler and compasses an angle equal to a given angle.
FIG. 60. CONSTRUCTING AN ANGLE EQUAL TO A GIVEN ANGLE
Construction. Let DEF in Fig. 50 be the given angle and let P
be the given point on the given line AB.
With E as a center and ER as a radius draw a circle. With P as
a center and with the same radius {ER^ draw another circle. Place
the sharp point of the compasses at R and cut an arc through M.
With S as a center and the same radius cut an arc at N.
The Z.BPC is the required angle. Why?
EXERCISES
1. Check the correctness of your construction for the pre-
ceding directions by measuring with a protractor.
2. How many ways have we for constructing an angle equal
to a given angle ?
PROPERTIES OF ANGLES 65
3. Construct two angles, designating one of them as contain-
ing a degrees and the other as containing b degrees ; using the
angles a and b, construct an angle containing a + b degrees ;
construct an angle containing 2 a -f- b degrees.
4. Choose a and b in Ex. 3 so that a >b, then construct an
angle equal to the difference of the two given angles.
5. Construct an angle equal to the sum of three given angles.
79. Perpendicular. We have seen in Art. 61 how right
angles are formed by rotation. If two lines form right
angles with each other, they are said to be perpendicular
to each other. The symbol
for " perpendicular " is _L. ^
80. Construction problem.
At a given point on a given
line to erect a perpendic-
ular to that line by using A
ruler and compasses. c \ I D
FlG. 51. HOW TO ERECT A
Construction. Let AB be the PERPENDICULAR
given line and P the given
point (Fig. 51). With P as a center and with a convenient radius
draw arcs intersecting AB~&t C and D.
With C and D as centers and with a radius greater than \ CD
draw arcs intersecting at E. Draw EP. Then EP^s the required
perpendicular.
EXERCISES
1. Test the accuracy of your construction in the problem of
Art. 80 by using a protractor.
2. Why must the radius CE in Fig. 51 be greater than ^ CD?
3. In Fig. 52 how < =-
A
would you draw a ^ IG 2
perpendicular to the
line AB at the point A? Do the construction work on paper.
06 GENERAL MATHEMATICS
81. Construction problem. How to bisect a given line
segment AB.
Construction. Let AB be the given
line segment (Fig. 53). With A as
a center and with a radius greater
A I I 171
than ^ AB describe arcs above and *
below AB. With B as a center and
with the same radius as before de-
scribe arcs above and below and \ /~
intersecting the first arcs at C and D.
Draw CD. Then E is the point of FIG. 53. How TO BISECT
bisection for AB. * LINE SEGMENT
82. Perpendicular bisector. The line CD in. Fig. 53 is
called the perpendicular bisector of AB.
EXERCISES
1. How may a line be divided into four equal parts ? into
eight equal parts ?
2. Draw a triangle all of whose angles are acute (acute-
angled triangle). Construct the perpendicular bisectors of
each of the three sides of the triangle.
3. Cut out a paper triangle and fold it so as to bisect
each side.
4. Draw a trianglfe in which one angle is obtuse (obtuse-
angled triangle) and draw the perpendicular bisectors of the
three sides.
5. Draw a triangle in which one angle is a right angle and
construct the perpendicular bisectors of the sides.
6. Draw a triangle ABC. Bisect each side and connect each
point of bisection with the opposite vertex.
83. Median. A line joining the vertex of a triangle to
the mid-point of the opposite side is called a median.
PKOPEKTIES OF ANGLES
EXERCISE
Draw a triangle ; construct its medians.
84. Construction problem. From a given point out-
side a given line to drop a perpendicular to that line.
Construction. Let AB be the given line and P the given point
(Fig. 54). With P as a center and p
with a radius greater than the dis-
tance from P to AB describe an arc
cutting AB at M and R. With M
and R as centers and with a radius
greater than \ MR describe arcs
either above or below (preferably be-
low) the line AB. Connect the point
of intersection E with P. Then the
line PD is perpendicular to AB, as
required. Test the accuracy of your
work by measuring an angle at D.
\/E
FIG. 54. How TO DROP A.
PERPENDICULAR
EXERCISES
1 . Why is it preferable to describe the arcs in Fig. 54 below
the line AB ?
2. Draw a triangle ABC all of whose angles are acute and
draw perpendiculars from each vertex to the opposite sides.
85. Altitude. An altitude of a triangle is a line drawn
from a vertex perpendicular to the opposite side.
EXERCISES
1. Draw a triangle in which one angle is obtuse and draw
the three altitudes.
2. Draw a triangle in which one angle is a right angle and
draw the three altitudes.
3. When d.o the altitudes fall inside a triangle ? outside?
68
GENERAL MATHEMATICS
86. To bisect a given angle. Suppose angle ABC to
be the given angle (Fig. 55). With the vertex B as a
center and with a convenient radius draw an arc cutting
HA and BC at X and Y re-
spectively. With X and 1'
as centers and with a radius
greater than ^ XY draw arcs
meeting at D. Join B and D.
Then BD is the bisector of
/.ABC.
FIG. 55. How TO BISECT
AN ANGLK
EXERCISES
1. Bisect an angle and check by folding the paper so that
the crease will bisect the angle.
2. Bisect an angle of 30 ; of 45 ; of 60 ; of 5)0.
3. Divide a given angle into four equal parts.
4. Draw a triangle whose angles are all acute and bisect
each of the angles.
5. Draw a triangle in which one angle is obtuse and bisect
each of the angles ; do the same for a triangle in which one
angle is a right angle.
87. Parallel lines. AB and CD in Fig. 5(3 have had
the same amount of angular rotation from the initial line
EF. Thus, they have the
same direction and are said
to be parallel. The symbol
for ''parallel" is II. Thus,
AB II CD is read "AB is par-
allel to CD"
A C
FIG. 56. PARALLEL LINES
88. Corresponding angles ;
transversal. Angles x and
// in Fig. 56 are called corresponding angles. The line EF
69
is called a transversal. It is clear that the lines are par-
allel only when the corresponding angles are equal and
that the corresponding angles are equal only when the
lines are parallel.
EXERCISES
1. Draw figures to illustrate the importance of the last
statement in Art. 88, above.
2. Point out the parallel lines you can find in the classroom.
89. Construction problem. How to draw a line parallel
to a given line.
Construction. Choose a point P
outside the given line A B in Fig. 57.
Draw a line through P so as to form
a convenient angle x with AR. Call
the point of intersection D. At P,
using DP as initial line, construct an
angle y equal to angle x (as shown)
by the method of Art. 78. Then PR j[
and AB are parallel because they
have had the same amount of rota- FIG. 57. How TO DKAW
tion from the initial line PD. PARALLEL LISKS
EXERCISES
1. Construct a line parallel to a given line through a given
point outside the line.
2. A carpenter wants a straight-edge board to have parallel
ends. He makes a mark across each end with his square. Why
will the ends be parallel ?
3. In Fig. 57 if angle x = 60, what is the number of degrees
in Z. y ? Give a reason for your answer.
4. Two parallel lines are cut by a transversal so as to form
two corresponding angles (x + 125) and (3 x + 50). Find x and
the size of each angle. Make a drawing to illustrate your work.
GENERAL MATHEMATICS
FIG. 58
5. In Fig. 58 if AB II CD, what other angles besides x and y
are equal corresponding angles ?
6. In Fig. 58, /.x = Z.y. Bisect
Z x and Z y and show that these
bisectors are parallel to each other.
90. Parallelogram. If onepair
of parallel lines cross (intersect)
another pair, the four-sided
figure thus formed is called a
parallelogram ; that is, a paral-
lelogram is a quadrilateral whose opposite sides are parallel.
91. How to construct a parallelogram. If we remember
the method used in Art. 89 for constructing one line parallel
to another, it will be
easy to construct a
parallelogram. Thus,
draw a working line
AB (Fig. 59). Draw
AR making a conven-
ient angle with AB.
Through any point, as
P, on AR draw a line
P V parallel to AB. Through any point M on A B draw a line
MT parallel to AR. The figure AMSP is a parallelogram,
for its opposite sides are parallel.
92. Rectangle. If one of the in-
terior angles of a parallelogram is a
right angle, the figure is a rectangle
(Fig. 60). Thus, a rectangle is a
parallelogram in which one interior angle is a right angle.
FIG. 59. How TO CONSTRUCT A
PARALLELOGRAM
FIG. 60
PROPERTIES OF ANGLES 71
EXERCISE
Show that all the angles of a rectangle are right angles.
HINT. Extend the sides of the rectangle.
93. Square. If all the sides of a rec-
tangle are equal, the figure is called a
square (Fig. 61).
FIG. 61
EXERCISES
1. Give examples of rectangles ; of squares.
2. Construct a rectangle having two adjacent sides equal to
5 cm. and 8 cm. respectively (use compasses ' and straight-
edge only). , a ,
3. Construct a rectangle hav- , .
ing the two adjacent sides equal
to the line segments a and b in FlG - 62
K- 62..
-
4. Construct a square whose
side is 7 cm. long. FlG> 63
5. Construct a square a side of which is a units long (use
line a in Fig. 63).
SUMMARY
94. This chapter has taught the meaning of the follow-
ing words and phrases : angle, vertex, vertices, initial side
of an angle, terminal side of an angle, right angle, straight
angle, perigon, acute angle, obtuse angle, reflex angle,
circle, center, circumference, radius, diameter, radii, arc,
intercept, central angle, quadrant, semicircle, latitude, lon-
gitude, degree of latitude, degree of longitude, minute,
second, size of an angle, protractor, degree of, arc, degree
of angle, adjacent angles, exterior sides of an angle, field
7^ (JKNERAL MATHEMATICS
protractor, transit, perpendicular bisector, perpendicular to
a line, altitude of a triangle, median of a triangle, bisector
of an angle, parallel lines, corresponding angles, transversal,
parallelogram, rectangle, and square.
95. The following symbols have been introduced: Z for
angle ; rt. /. for right angle ; A for angles ; v for arc ;
_L for is perpendicular to ; II for is parallel to ; for dei/n-f
or degrees ; ' for minute or minutes ; " for second or seconds.
96. The following notations have been discussed : (1) no-
tation for denoting and reading angles; (2) notation for
denoting a circle by its center.
97. This chapter has presented the important methods of
1. Classifying angles.
' 2. Measuring angles.
3. Comparing angles.
4. Drawing angles containing any amount of turning or
any number of degrees.
5. Adding and subtracting angles.
6. Measuring angles out of doors.
98. In this chapter the pupil has been taught the follow-
ing fundamental constructions :
1. To draw a circle.
2. To draw an angle equal to a given angle.
3. To draw a line perpendicular to a given line at a
given point.
4. To draw the perpendicular bisectors of the sides of
a triangle.
5. To draw the medians of a triangle.
6. To draw a line perpendicular to a given line from a
given point outside the line.
7. To draw the altitudes of a triangle.
PBOPEKTIES OF ANGLES 73
8. To bisect a given angle.
9. To draw the bisectors of the angles of a triangle.
10. To draw a line through a given point parallel to a
given line.
11. To construct a parallelogram.
12. To construct a rectangle.
18. To construct a square.
99. This chapter has taught the pupil to use the follow-
ing instruments and devices : tracing paper, the protractor,
and the field protractor.
IMPORTANT GEOMETRIC RELATIONS
100. Radii of the same circle or of equal circles are equal.
101. In the same circle or in equal circles equal central
angles intercept equal arcs on the circle.
102. In the same circle or in equal circles equal arcs on
the circle are intercepted by equal central angles.
103. A central angle is measured by the arc intercepted
by its sides.
IMPORTANT DEFINITIONS
. 104. An angle is the amount of turning made by a line
rotating about a fixed point in a plane.
105. A circle is a closed curve all points of which lie in
the same plane and are equidistant from a fixed point.
106. (1) A quadrilateral whose opposite sides are parallel
is a parallelogram. (2) A rectangle is a parallelogram in
which one interior angle is a right angle. (3) A square is a
rectangle with all sides equal.
CHAPTER IV
FIG. 64.
THE EQUATION APPLIED TO AREA
107. Measuring areas. If we determine the amount of
area inclosed within a polygon, as in the triangle ABC
in Fig. 64, we are measuring the area of the triangle. As in
measuring length, the process is
one of comparison. We compare
the area of the given polygon
with some standardized (defined
and accepted) unit of area and
determine how many units are contained in the polygon ;
that is, we determine the ratio between the area of the
given polygon and a standard unit of area.
108. Unit of area. The unit of area is a square each
of whose sides is a standard unit of length. Such a unit
involves length and width. Thus, we may
measure area and express the result in
square feet, square inches, square meters,
square centimeters, etc.
109. Practical method of estimating area.
A practical way to estimate the area of a
polygon is to transfer it to squared paper
by means of tracing paper and then count the num-
ber of square units inclosed within the figure. If the
bounding lines cut the squares, it becomes necessary to
approximate. In such approximations we should be careful,
but we should not go beyond reasonable limits of accuracy.
74
1cm.
1 cm.
FIG. 65. UNIT OF
AREA ix THE MET-
RIC SYSTEM
THE EQUATION APPLIED TO AREA
75
EXERCISES
1. The six figures in Fig. 66 were transferred by means of
tracing paper. Estimate the areas of each of them by counting
the squares. Express the areas either as square centimeters
or as square millimeters.
HINT. One small square equals 4 sq. mm.
FIG. 66. ESTIMATING AREAS BY MEANS OF SQUARED PAPER
2. If the paper were ruled much finer, would you get a
more accurate estimate ? Give an argument for your answer.
3. Do you think that any of your results are accurate?
4. Write a paragraph in
precise terms, supporting your
answer to Ex. 3.
C-
110. Area of a rectangle.
CASE I. The sides of the rec-
tangle that has been trans-
ferred to the squared paper of
Fig. 67 are integral multiples
of 1 cm. Using 1 sq. cm. as a unit, there are two rows of
FIG. 67. How TO FIND THE AREA
OF A RECTANGLE
76
GENERAL MATHEMATICS
units, and four units in a row. Counting, we see that the
area equals 8, or 2 x 4. The law in this case is : The area
equals the base times the altitude. In equation form this
law may be written A = b x a.
111. Area of a rectangle. CASE II. Let us suppose that
we are given a square whose sides are not integral multiples
of 1 cm. : for example, a rectangle whose base (length)
is 2.3 cm. and whose altitude (width) is 1.3 cm. If we
assume that the preceding law holds, then we ought to
get 2.3 x 1.3 = 2.99 sq. cm. Instead
of putting the rectangle on the kind
of squared paper used in Case I,
let us draw it again, by means
of tracing paper, on squared paper
that is ruled to a smaller unit,
the millimeter, as in Fig. 68. Since
there are 23 mm. in 2.3 cm. and
13 mm. in 1.3 cm., if we temporarily adopt the square
millimeter as a unit of area, then the sides of the rectangle
are, as in Case I, integral multiples of the unit of length
(in this case the millimeter). Hence there are 13 rows
of units with 23 in a row, or 299 sq. mm. But there are
100 sq. mm. in 1 sq. cm. ; hence, dividing 299 by 100, the
result is 2.99 sq. cm., which is precisely the same number
as that obtained by assuming the law of Case I.
112. Area of a rectangle. CASE III. This process of
temporarily adopting a smaller square can be continued.
If, for example, the base of a rectangle is 2.13 cm. and
the altitude 1.46 cm., we may imagine the rectangle to be
drawn upon squared paper still finer ruled, that is, ruled
to 0.1 of a millimeter. From here the reasoning is the
same as in Cases I and II.
FIG.
THE EQUATION APPLIED TO AREA
77
EXERCISES
1. Finish the reasoning of the foregoing paragraph.
2. The base of a rectangle is 3^ cm. and its altitude is 5-| cm.
Show that the area may be found by counting squares.
3. The base of a rectangle is 5| cm. and its altitude is-2| cm.
What unit would you temporarily adopt to find the area?
Express the area in square centimeters.
The preceding exercises show that if the sides of a
rectangle involve fractions which may be expressed as
exact decimal parts of a unit, the problem is the same
as in Case II.
113. Second method for finding area of a rectangle. It is
possible to show that the transfer of a rectangle to the
squared paper by means of tracing paper was unnecessary.
Suppose we are given a rec-
tangle AS CD (see Fig. 69)
whose base is 4 cm. and
whose altitude is 3 cm. We
wish to find the area. Draw
a perpendicular line to the
line AB at the end of each
unit segment ; that is, at the
points E, F, and G (review
the method of Art. 80).
Also construct perpendiculars to the line AD at the points
H and L Then each small square is a unit of measure
(by definition), and the figure is divided into three rows
of units with four in a row. By counting, the area equals
12 (that is, base times altitude').
II
I
A
E F G
FIG. 69
78 GENERAL MATHEMATICS
EXERCISES
1. The base of a rectangle is 4.3cm. long, and its altitude
is 1.7 cm. Show how to find the area of this rectangle by count-
ing, but without the use of squared paper.
2. Apply the law A = b x . What advantage has this law
over the method of Ex. 1 ?
114. Formula. An equation which expresses some prac-
tical rule from arithmetic, the shop, the trades, the sciences,
the business world, etc. is called a, formula. Thus, Aaxb
is a practical formula for finding the area of a rectangle.
The plural of "formula" is "formulas" or "formulae."
115. Formula for the area of a square. The square is
a special case of a rectangle ; that is, it is a rectangle in
which a = b. The formula can be developed by the same
method as for a rectangle. The only difference in the
reasoning is that in every case there are as many rows of
square units as there are square units in a row. (Why ?)
Hence the formula for the area of a square is A = b x b.
This formula may be written A = J 2 , where i 2 means b x b,
and the formula is read "A equals b square."
EXERCISES
1. By the method of counting squares find the area of a
square whose side is 2|- cm. long.
2. Apply the formula A = Z/ 2 to the square in Ex. 1. .Com-
pare results.
3. Find the area of a square whose side is oft.; a feet;
x inches ; y yards ; m meters ; 0.07 mm. ; 2.41 m.
4. How many feet of wire fencing are needed to inclose a
square lot whose area is 4900 sq. ft. ? b 2 sq. ft. ? 4 r 2 sq. yd. '.'
THE EQUATION APPLIED TO AREA 79
5. Express by an equation the area A of a rectangle that
is 8 in. long and 5 in. wide ; 8 in. long and 4 in. wide ; 8 in.
long and 3 in. wide ; 8 in. long and 6^ in. wide.
6. Express by an equation the area A of a rectangle 12 in.
long and of the following widths: 6 in.; 8^ in.; 9^ in.; 10| in.;
x inches ; y inches.
7. A mantel is 54 in. high and 48 in. wide. The grate is
32 in. high and 28 in. wide. Find the area of the mantel and
the number of square tiles contained in it if each tile is 3 in.
on a side.
8. How many tiles 8 in. square are needed to make a walk
60 ft. long and 4 ft. wide ?
9. Express by equations the areas of rectangles 1 in. long
and of the following widths : 12 in. ; 9 in. ; h inches ; n inches ;
x inches ; a inches.
10. Express by equations the areas of rectangles of width w
and of the following lengths : 8 ; 10 ; 12^ ; x ; a ; I ; b ; z.
11. In each case write an equation for the other dimension
of the rectangle, having given (a) altitude 8 in. and area
32 sq. in. ; (b) altitude 5 ft. and area 7^ sq. ft. ; (c) base 9 ft.
and area 30 sq. ft. ; (d) base 6 in. and area 27 sq. in. ; (e) base
3 in. and area A square inches ; (f) base 5 in. and area A square
inches ; (g) altitude a inches and area A square inches ; (h) base
b inches and area A square inches.
116. Formula for the area of a parallelogram. Fig. 70
shows a parallelogram that has been transferred to that
position by means of tracing paper. We wish to find its
area. The line AB is produced (extended), and perpendic-
ulars are dropped from D and C to the line AB (see Art. 84
for method of constructions), thus forming the triangles
AED inside and BFC adjoining the given parallelogram.
80
GEXKi; A L MATHEMATICS
EXERCISES
(Exs. 1-7 refer to Fig. 70)
1. Estimate by count the number of square units in the
triangle A ED.
2. Estimate the number of square units in the triangle BFC.
3. Compare the results of Exs. 1 and 2.
4. If the area of the triangle .BFC equals the area of the
triangle A ED, what is the relation between the area of the
rectangle CDEF and the area of the parallelogram AB CD?
-
FlG. 70. HOW TO FIND THE AREA OF A PARALLELOGRAM BY
MEANS OF SQUARED PAPER
5. What is the formula for the area of the rectangle CDEF?
Write the formula.
6. What seems to be the relation between the base of the
parallelogram and the base of the rectangle ? What evidence
have you to support your answer ?
7. What is the relation between the altitude of the paral-
lelogram and the altitude of the rectangle ? Give the evidence.
8. What seems to be the formula expressing the area of
a parallelogram ?
9. Without using squared paper construct a parallelogram
(use ruler and compasses and follow the method of Art. 91).
Divide the parallelogram into two parts a triangle and a
THE EQUATION APPLIED TO AREA 81
quadrilateral (as in Fig. 71). Now shift the triangle to the
other side so as to form a rectangle. Show that the rectangle
is equal to the parallelogram.
The preceding exercises furnish evi-
dence to support the following law :
The area of a parallelogram equals the
product of its base and altitude. This law may be written
in the form of the following well-known formula:
A = a x b.
EXERCISE
Find the area of a parallelogram if b ' = 17 in. and a = 5.3 in. ;
if b = 15.4 in. and a 9.2 in.
117. Area of a rhombus. The rhombus (Fig. 72) is a
special case of the parallelogram, as it is a parallelogram
with all its sides equal. Hence its area
equals its base times its altitude.
118. The area of a triangle. The ex-
ercises that follow will help the pupil
to understand the formula for the area of a triangle.
EXERCISES
(Exs. 1-12 refer to Fig. 73)
1. Draw a triangle ABC as shown in Fig. 73.
2. Through C draw a line CD parallel to AR (review the
method of Art. 89).
3. Through B draw a line parallel to AC, meeting CD at D.
4. What kind of a quadrilateral is the figure ABDC? Why ?
5. By means of tracing paper transfer the parallelogram
to squared paper.
82 GENERAL MATHEMATICS
6. Estimate the number of square units in triangle ABC.
7. Estimate the number of square units in triangle CBD.
8. Compare the results of Exs. 6 and 7. What relation
does the triangle bear to t^e parallelogram ?
9. What seems to be the relation between the base of the
triangle and the base of the parallelogram ? Why ?
10. What is the relation between the altitude of the tri-
angle and the altitude of the parallelogram ? Explain' why.
11. What is the formula
then for the area of any
parallelogram ?
12. What appears to be
the formula for the area
of a triangle?
FIG. 73. How TO FIND THE AREA
13. Construct a paral- OF A TRIANGLE
lelogram ABCD. Construct
the diagonal AC (a line joining opposite vertices). With a
sharp knife cut out the parallelogram and cut along the
diagonal so as to form two triangles. Try to make one triangle
coincide with the other.
14. What conclusion does the evidence of Ex. 13 support ?
The preceding exercises furnish evidence to show that
the area of a triangle is equal to one half the product of
its base and altitude. This law may be written in the
form of the following formula:
ab
I
119. Area of a trapezoid. A quadrilateral having only
two sides parallel is called a trapezoid (Fig. 74). The
parallel sides are said to be its bases. In Fig. 74 the upper
THE EQUATION APPLIED TO AREA 83
base of the trapezoid is 6, the lower base is a, and the
altitude is h. To find the area draw the diagonal BD.
The area of the triangle ABD = -.h. Why ?
The area of the triangle BCD = --h. Why ?
2,7
Therefore the area of the trapezoid = a '- + b - Why ?
77 2t 2
Note that a and b - are similar terms. Why ? In the
h
first term a is the coefficient of and in the second term
b is the coefficient; hence, adding coefficients, as we may
always do in adding
similar terms, the area
of trapezoid is (a + b) -
We can only indicate A
the sum of the two FlG - 74 ' How J FIND THE AREA OF
A TRAPEZOID
bases until we meet an
actual problem. The parenthesis means that a + b is to
be thought of as one number. The law is: The area of
a trapezoid is equal to one half the product of its altitude by
the sum of its bases. This law may be written in the form
of the following formula:
EXERCISES
1. Find the area of the trapezoid whose altitude is 12.6 in.
and whose bases are 8 in. and 4.6 in. respectively.
2. The altitude of a parallelogram is 3 x + 2, and its base
is 4 in. Write an algebraic expression representing its area.
Find the value of x when the area is 28 sq. in.
84
GENERAL MATHEMATICS
D
70
FIG. 75
3. The altitude of a triangle is 10 in., and the hast- is
3 x + 2 in. Write an algebraic number representing the area.
Find the value of ./ when the area is 55 sq. in.
4. A man owns a city lot with the form
and dimensions shown in Fig. 75. ^He wishes
to sell his neighbor a strip AEFD having a
frontage DF equal to 10 ft. If the property
is worth $5600, how much should he receive
for the strip '.'
5. Of what kinds of polygons may the
following equations express the areas '.'
(a) A=x\ (h) A=l
(b) A = 3.*-. _3
(d) -i =5 (a: + 3). 4
(e) .4 = (. + 2).
(f) .4 = a(5 + 4),
(g) A = *(y + 2).
6. Find the value of ,1 in Ex. 5 when x = 3, y = 2, a = 4,
b = 1, and c = 5.
7. What quadrilaterals contain right angles ?
8. In what respect does the square differ from the rectangle ?
9. Having given a side, construct a square, using only ruler
and compasses.
HINT. Review the method for constructing a perpendicular to a
line segment (Art. 80).
10. Hg>w does a square differ from a rhombus ?
11. Is a rhombus a parallelogram ? Is a parallelogram a
rhombus ?
12. Construct a rhombus with ruler and compasses, given a
side equal to 5 cm. and given the included angle between two
adjacent sides as 41.
HINT. Use the construction for parallel lines (Art. 89).
THE EQUATION APPLIED TO AREA
85
GEOMETRIC INTERPRETATION OF PRODUCTS
120. A monomial product. The formulas for the area of
the rectangle, the square, the triangle, the trapezoid, etc.
show that the product of numbers
may be represented geometrically ;
for example, the product of any
two numbers may be represented
by a rectangle whose dimensions FIG. ~ 6 - ILLUSTRATING A
are equal to the given numbers. .
Thus the rectangle in Fig. 76 represents the product ab.
ab
EXERCISES
1 . Sketch a rectangle to represent the product 6 x.
2. Sketch an area to represent kry.
3. Show from Fig. 77 that the area
4 x 2 . What is the product of 2 a- x 2x?
4. Show by means of a figure the area
of a rectangle 3 a by 5 a.
5. Draw a figure to represent the prod- x
uct of 5 a- and 4 x.
6. On squared paper draw an area repre-
lio. 77. ILT.USTRAT-
senting the product ab. To the same scale 1NG THE SQL-ARE OF
draw the area ba. Compare the areas. A MONOMIAL
7. Show by a drawing on squared paper that 4-5 = 5-4.
121. Law of order. The last two exercises illustrate
that in algebra, as in arithmetic, the fen 'torn of a product
may be changed in order without changing the value of the
product. Thus, just as 2x3x5 = 5x3x2, so xyz zyx.
This is called the Commutative Law of Multiplication.
i.
x
x
3 expressed
x x
by
X
X
x*
X X
GENERAL MATHEMATICS
EXERCISE
Simplify the following : (a) 2 x 3 y . 4 z ; (b) (2 x y) (3 x y} ;
(c) 4 x 2/ 3 x my.
122. Product of a polynomial and a monomial. The
formula for the trapezoid suggests the possibility of draw-
ing areas to represent the product of a sum binomial by
a monomial. The process is illustrated by the following
exercises.
EXERCISES
1. Express by means of an equation the area of a rectangle
of dimensions 5 and x + 3 (see Fig. 78). The area of the
whole rectangle equals 5 (x -\- 3). D A F
Why? If a perpendicular be erected
at B (see Art. 80 for method), the 5
rectangle is divided into two rec-
tangles. The area of DCBA equals
5x. Why? The area of ABEF
equals 15. Why? It is now easy PRODUCT OF A POLYNOMIAL
f. T ,, , AND A MONOMIAL
to hnd the entire area DCEF.
5x
15
C X B 3 E
FIG. 78. ILLUSTRATING THE
Since DCEF = DCBA + ABE1*,
T> (x + 3) =5 x + 15.
2. Show from Fig. 79 that a(x + ?/) = ax + "//.
3. Show by Fig. 80 that a (x + T/ + z) = ax + a?/ + az.
Why ?
Why?
x y
FIG. 79
X y Z
FiG. 80
4. Draw an area to represent bm + In + be.
5. Draw an area to represent 2 ax + 2 ay + 2 az.
6. Kepresent 2cc + 4?/ + 6byan area.
7. Sketch a rectangle whose area equals 2 ax + 2 ay + 6 az.
EUCLID
88 GENERAL MATHEMATICS
HISTORICAL NOTE. The word "geometry" comes from a Greek
phrase which means to measure the earth. The early Egyptians had
serious need for a reliable method of measuring the land after each
overflow of the Nile. The early history of geometry appears to rest
on this practical basis.
The oldest collection of geometry problems is a hieratic papyrus
written by an Egyptian priest named Ahmes at a date considerably
earlier than 1000 H.C., and this is believed to be itself a copy of some ,
other collection a thousand years older.
Ahmes commences that part of his papyrus which deals with
geometry by giving .some numerical instances of the contents of
barns. Since we do not know the shape of the barns, we cannot
check the accuracy of his work. However, he gave problems on
pyramids. The data and results given agree closely with the dimen-
sions of the existing pyramids.
Geometry took definite form as a science when Euclid (about
300 B.C.) wrote his "Elements of Geometry." The proofs of his
text were so excellent that the book replaced all other texts of
the time and has held an influential position to this day. The
form of Euclid is practically the same as most American geom-
etry texts, and in England boys still say they are studying Euclid
(meaning geometry).
We know little of Euclid's early life. He may have studied in
the schools founded by the great philosophers Plato and Aristotle at
Athens, in Greece. He became head of the mathematics school
at Alexandria, Egypt, and proceeded to collect and organize into a
set form the known geometric principles. He is said to have insisted
on the knowledge of geometry for its own sake. Thus, we read of
his telling the youthful Prince Ptolemy, " There is no royal road to
geometry." At another time, so the story goes, when a lad who had
just begun geometry asked, "What do I gain by learning all this
stuff ? " Euclid made his slave give the boy some coppers, " since,"
said he, " he must make a profit out of what he learns."
Euclid organized his text so as to form a chain of reasoning, begin-
ning with obvious assumptions and proceeding step by step to results
of considerable difficulty. The student should read about his work in
Ball's "A Short Historyof Mathematics." Cajori's" History of Elemen-
tary Mathematics " and Miller's " Historical Introduction to Mathe-
matical Literature " are further sources of information about Euclid.
THE EQUATION APPLIED TO AREA 89
123. Partial products. The products ax, ay, and az in
the polynomial ax + ay 4- az are called partial products.
Each term of such an expression may be used to represent
the area of some part of a rectangle. For example, Fig. 80
shows a rectangle divided into the parts ax, ay, and az
respectively. Here the polynomial ax + ay + az may be
said to represent the area of the whole rectangle.
124. Algebraic multiplication. The list of exercises in
Art. 122 also shows that the product of a polynomial and a
monomial is found by multiplying each term of the polynomial
by the monomial and then adding the partial products.
EXERCISES
1. Perform the following indicated multiplications :
(a) 3(2 x + 3 ij). (c) 3c(2 + 36).
(b) (5x + 2z)4:a. (d)3e(2c + 3).
2. Letting x = 3, y = 1, and z = 2, find the value of the
following numbers : (x + y)z; 2 (x -j- ?/) 2 ; 3 x + 2 (y + z) ;
2z); 2s + 5(x + ,/).
125. Geometric product of two polynomials. The product
of a + c and x + y may be indicated as (a + c) (x + y}.
This product may be represented
cy
, . ,, , <-,.,.. , ~ ay
geometrically (rig. ol) by a rec
tangle whose base is x + y and
whose altitude is a + c. FIG. 81. ILLUSTRATING
The rectangle is composed of. THE PRODUCT OF TWO
four rectangles: ax, ay, ex, and cy.
By the axiom that the whole is equal to the sum of its parts,
the whole rectangle, ( + <?)(#+?/), equals ax+ay + cx + cy,
the sum of the parts.
90 GENERAL MATHEMATICS
EXERCISES
1: Sketch a rectangle whose area will be the product of
(a + ft) (e + d).
2. Find the geometric product of (c + t) (m + n).
3. Perform the multiplication (2 -f x)(m -f w) by means of
a geometric figure.
4. Find the product (3 x + 2 ?/) (a + b). Sketch the area
represented by this product.
5. Find the product (a + b~)(x + ;/ + z), using a geometric
figure.
126. Algebraic product of two polynomials. The figures
drawn in the preceding exercises indicate a short cut in
the multiplication of two polynomials. Thus, a polynomial
is multiplied by a polynomial by multiplying each term of
one polynomial by every term of the other and adding the
partial products.
EXERCISES
1. Using the principle of Art. 126, express the following
indicated products as polynomials :
(a) (m + n) (a + 5). (g) 3(2 a 2 + a + 5).
(f) 5(4 +7 + 3). (j) (3x
(k) (5 b + 2 c + 3 d) (2 x + 3 y + 4 z).
(1) (2 m + 3 n -f- p] (3 a + 7 J + 5*).
2. One side of a rectangle is 4yd. and the other is 6yd.
How much wider must it be made so as to be l times as
large as before? /
THE EQUATION APPLIED TO AREA 91
3. Multiply the following as indicated and check the result :
Solution. 2 x (x + 2 y + y-) = 2 x 2 + 4 xy +
Check. Let x = 2 and y 3.
2 x 2 + 4 xy + 2 xy" = 8 + 24 + 36 = 68.
NOTE. Avoid letting x = 1, for in this case 2 x, 2 x 2 , 2 x 8 , etc. are
each equal to 2. Why?
4 . Multiply the following as indicated and check the results :
(a)
(b)
(c) (m + n + a) (m + n + &).
(d) (m + n + 2
(e) (0.4 x + 0.3 y + 0.6 s) (10 x + 20 y + 30 z).
127. Geometric square of a binomial. The product of
(x + y*)(x + yy, or (z + y) 2 , is an interesting special case
of the preceding laws. The prod-
uct may be represented by a square
each of whose sides is x + y (see
Fig. 82). The square is composed of
four parts, of which two parts are
equal. Since these two parts are repre-
sented by similar algebraic terms, they
may be added ; thus, xy -\-xy-2 xy. x V
Hence the area of a square whose F IG. 82. ILLUSTRATING
j o . r> o rru THE SQUARE OF A Bi-
side isaj-t-yisar+Jsa^ + ^r. 1 he
same product is obtained by apply-
ing the law for the product of two polynomials ; thus,
x +y
x +y
xy
xy
+
xy
xy +
2 xy
92 GENERAL MATHEMATICS
In algebraic terms we may say that the square of the
sum of tivo numbers equals the square of the first, plus twice
the product of the two numbers, plus the square of the second.
Use Fig. 82 to show what this law means.
EXERCISES
1. By means of figures express the following squares as
polynomials :
(b) (m + n) 2 . (e) (x + 2) 2 . (h) (2 x + y) 2 .
(c) (c + d)*. (f ) (m + 3f. (i) (2 x + 3 y) 2 .
2. Sketch squares that are suggested by the following
trinomials :
(a) a 2 + 2 ah + lr. (e) m z + 8 MI + 16.
(b) x 2 + 2 ax + a 2 . (f) .r 2 + 10 a- + 25.
(c) A- 2 + 2 A->- -(- r. (g) 49 + 14 x + z 2 .
(d) x 2 + 4* + 4. (h) c 2 + c + ],
3. Indicate what number lias been multiplied by itself to
produce
(a) or 2 + 2 vy + /. (c) z 2 + 6 a- + 9.
(b) r 2 + 4 r + 4. (d) & a + 10 // + 25.
4. What are the factors in the trinomials of Ex. .3 '.'
5. The 'following list of equations review the fundamental
axioms as taught in Chapter I. Solve each equation and check
by the methods of Chapter I.
(a) 30 + 4) =22 + ,-.
(b) 9 + 35) =5 (2 a + 45).
(c) 3(x + 15)+ 5 = 2(2* + 9) + 4(.r + 3).
(d) ?fe2-8. W |i-| = 1 . (h)f-f=a
2 . fc, 5 + 5-2. (i)^^ = 8.
.
93
128. Evaluation. The area of each of the geometrical
figures considered in this chapter has been found to depend
upon the dimensions of the figure. This dependence has
been expressed by means of formulas, as A = ab in the case
of the rectangle. Whenever definite numbers are substi-
tuted in the expression ab in order to find the area, A,
for a particular rectangle, the expression ab is said to be
evaluated. This process implies getting practical control
of the formulas.
EXERCISES
1. Find the value of A in the formula .1 = ab when
a = 22.41 ft. and b = 23.42 ft.
2. Find the value of J in the formula A = when
a = 12.41 ft. and I = 2.144 ft.
3. Find the value of .1 in the formula A = (<i -)- fi) when
a = 12.42 ft, b = 6.43 ft., and h = 20.12 ft.
129. The accuracy of the result. In finding the area in
Ex. 1 above we get A = (22.41) (23.42)= 524.8422 sq. ft.
This is a number with four decimal places. As it stands
it claims accuracy to the ten-thousandth of a square foot.
The question arises whether this result tells the truth.
Suppose the numbers above represent the length and width respec-
tively of your classroom. Does the product 524.8422 sq. ft. indicate
that we actually know the area of the floor accurate to one ten-
thousandth of a square foot? Shall we discard some of the deci-
mal places ? If so, how many are meaningless ? How much of the
multiplication was a waste of time and energy ? These questions are
all involved in the fundamental question How many decimal places
shall we regard as significant in the process of multiplication ? .
It is important that we have a clear understanding of
the question. For if we carry along in the process mean-
ingless decimals we are wasting time and energy, and,
94
what is more serious, we are dishonestly claiming for
the result an accuracy which it does not have. On the
other hand, we are not doing scientific work when we
carelessly reject figures that convey information.
The following facts are among those which bear on our
problem :
(a) In Art. 26 we pointed out that any number obtained
by measurement is an approximation.' The application of
the area formulas involves the measurement of line seg-
ments. Hence an area is an approximation. This fact
alone is sufficient to make us exceedingly critical of the
result 524.8422 sq. ft. as an absolutely accurate measure
of the area of the classroom floor.
(b) If we measure the length of a room with a reliable
tape measure and record the result as 23.42 ft., this does
not mean that we regard the result as absolutely accurate.
If the scale is graduated to hundredths of a foot, it means
that 23.42 ft. is the result nearest to the true value. The
eye tells us that 23.425 ft. is too high and 23.415 ft. is
too low, but that the result may be anywhere between
these. Thus, the length of the room lies anywhere between
23.415 ft. and 23.425 ft. Similarly, the width may be
anywhere between 22.405 ft. and 22.415 ft. The student
should practice measuring objects with a yardstick or a
meter stick till the point of this paragraph is clear to him.
Test question : How does 2.4 ft. differ from 2.40 ft. ?
Multiplying the smallest possible length (23.415 ft.)
of the classroom by the smallest width (22.405 ft.) we
get a possible area of 524.613075 sq. ft. By multiplying
the greatest length (23.425 ft.) by the greatest width
(22.415 ft.) we get 525.071375 sq. ft. Subtracting the
smallest possible area from the largest possible area gives
us a range of over 0.45 of a square foot. In short, the
THE EQUATION APPLIED TO AREA 95
result might be wrong by practically one half of a square
foot. We are not actually sure of the third figure from the
left. It mayfce a 4 or a 5. We shall be reasonably near
the truth if w% record the result simply as 524.? sq. ft.,
a number chosen roughly halfway between the largest and
smallest possible areas.
It can thus be shown that the product of two approxi-
mate four-place numbers is not to be regarded accurate
to more than four places.
*130. Abbreviated multiplication. It is apparent in the
preceding discussion that it is a waste of time to work
out all the partial products in multiplication. It is easier
(when the habit is once established) to work out only
the partial products which go to make up the significant
part of the answer.
Thus, 47.56 x 34.23 may take the following forms :
ABBREVIATED FORM USUAL METHOD
47.56 By multiplication we get
34.23 47.56
1427 34.23
190
10
1
1628.
The difference is accidentally only a little more than
0.02 sq. ft. It can be shown by the method used in the
classroom problem (Art. 129) that 1628 is easily in the
range of probable areas ; that is, we are not actually sure
about the fourth figure from the left.
* Hereafter all articles and exercises marked with an asterisk may be
omitted without destroying the sequence of the work.
1
4268
9
512
190
24
1426
8
1627
9788
96 GENERAL MATHEMATICS
The abbreviated method consists of writing only the
significant parts of the usual method (see numbers to left
of the line). Add 1 unit when the number^o the right is
the figure 5 or larger. The method will pppear awkward
until sufficiently practiced.
A similar discussion concerning accuracy could be given
for division. In addition or subtraction it is easy to
see that the sum or difference of two numbers cannot be
regarded as more accurate than the less accurate of the
two numbers. Illustrate the truth of the last statement.
While the discussion of this very important topic has
been by no means complete, perhaps enough has been
said to fulfill our purpose, which is to make the student
exceedingly critical of results involving the significance of
decimal places.
EXERCISES
*1. Assuming that the dimensions of a hall are measured
with a reliable steel tape and that the dimensions are recorded
as 47.56 ft. and 34.23 ft. respectively, show by the method used
in Art. 129 that the difference between the smallest and the
largest possible area of the hall is actually over four fifths of
a square foot.
*2. By means of the abbreviated multiplication method
write the product of 46.54 and 32.78 ; of 23.465 and 34.273.
*3. Multiply by the usual method and compare the short-
cut result with this result.
*4. Which result is the more accurate ?
SUMMARY
131. This chapter has taught the meaning of the follow-
ing words and phrases : area, measuring area, unit of area,
rhombus, trapezoid. Commutative Law of Multiplication,
partial products, parenthesis, formula, formulas.
THE EQUATION APPLIED TO AREA 97
132. The following formulas have been taught:
(a) A = ba. (For the area of a rectangle.)
(b) A = b 2 . (For the area of a square.)
(c) A=bh. (For the area of a parallelogram.)
(d) A = (For the area of a triangle.)
7
(e) A = (a + >) - (For the area of a trapezoid.)
a
(f ) (x + #) 2 = y?+ 2 xy + 3/ 2 . (For the area of a square
whose side is x
133. The product of two numbers may be represented
geometrically as an area.
134. The algebraic product of a monomial and a poly-
nomial is found by multiplying each term of the polynomial
by the monomial and then adding the partial products.
135. The product of two polynomials is the sum of all
the partial products obtained by multiplying each term
of one polynomial by every term of the other.
*136. We need to be very critical of the number of
decimal places that we submit in a result. The product of
two approximate four-digit numbers is only approximately
correct for four digits.
CHAPTER V
THE EQUATION APPLIED TO VOLUME
137. Solids. The drawings in Fig. 83 represent geo-
metric solids. A solid is commonly thought of as an object
that occupies a portion of space. It is separated from
Oblique Paral-
lelepiped
Cube
Rectangular Paral-
lelepiped
Triangular
Pyramid
Sphere
FIG. 83. FAMILIAR SOLIDS
Frustum of a
Pyramid
the surrounding space by its surface. In geometry we
study only the form of the solid and its size. We are
not interested in color, weight, etc. A solid differs from
the figures we have been studying in that it does not lie
altogether in a plane, but involves a third dimension. What
figures in two dimensions are suggested by the solids in
Fig. 83 ? For example, the square is suggested by the cube.
98
THE EQUATION APPLIED TO VOLUME
99
138. Cube. The cube has six faces all of which are
squares. Two faces intersect in an edge. How many edges
has a cube ? How many corners ? How is a corner formed ?
139. Oblique parallelepiped. The faces of an oblique
parallelepiped are all parallelograms. How many faces
has it? How many vertices? How many edges?
140. Rectangular parallelepiped. The faces of a rec-
tangular parallelepiped are rectangles.
141. Measurement of volume ; unit of volume. When we
determine the amount of space inclosed within the surface
of a solid we are measuring the volume of the solid. To
measure the volume of a solid we compare the solid with
a cube each of whose edges equals a unit of length. The
volume is expressed numerically by the number of times
the unit cube goes into the solid. The unit cube is called
the unit of volume.
142. Formula for the volume of a rectangular parallele-
piped. In Fig. 84 a rectangular parallelepiped is shown
which is 5 cm. long, 3 cm. wide,
and 4 cm. high. The unit cube
is represented by K. Since the
base of the solid (the face
on which it stands) is 5 cm.
long and 3 cm. wide, a layer of
3 x 5 unit cubes could be placed
upon it. Since the solid is 4 cm.
high, it contains 4 layers of unit
cubes ; that is, 4x3x5, or 60,
unit cubes. Thus the volume
of a rectangular parallelepiped
is obtained by multiplying the length by the width by the
height. This law may be expressed by the formula V= Iwh.
B
/ /
7
N
R
FIG. 84. How TO FIND THE
VOLUME OF A RECTANGULAR
PARALLELEPIPED
100
EXERCISES
1. Find the volume of a rectangular parallelepiped if its
dimensions are I = 63 in., h = 42 in., and w 56 in.
*2. If in the preceding discussion the edges of the rec-
tangular parallelepiped had not been given as integral multiples
of the unit cube, it would have been necessary temporarily to
adopt a smaller unit cube. Show that the formula V = Iwh
holds when I = 2.3 cm., h 3.4 cm., and w = 1.7 cm.
HINT. Follow the method suggested in Art. 111.
*3. Show by means of a general discussion that the formula
would be true if I = 3j, h = 2j, w = 3f .
HINT. See Ex. 3, Art. 112.
* 143. Volume of an oblique parallelepiped. Fig. 85 shows
in a general way the method used in a more advanced
mathematics course to show that the formula V= Iwh holds
in
}h
R
FIG. 85. MODEL ILLUSTRATING HOW TO FIND THE VOLUME OF AN
OBLIQUE PARALLELEPIPED
even for aji oblique parallelepiped. Parallelepiped III is a
rectangular parallelepiped, and we know the formula holds
for it. Parallelepiped II is a right parallelepiped (it has
THE EQUATION APPLIED TO VOLUME 101
four rectangular faces, and two are parallelograms) and
by advanced methods is shown equal to parallelepiped III.
Parallelepiped I is oblique and is shown equal to parallele-
piped II. Since parallelepiped I equals parallelepiped II,
and parallelepiped II in turn equals parallelepiped III, the
formula holds for parallelepiped I. The student should not
be concerned if he cannot fully understand this discussion.
He should be ready to apply the formula for an oblique
parallelepiped when the need for it arises in shop or factory
just as he does many principles of arithmetic.
EXERCISES
*1. It will be easy for some student to make models of the
preceding figures in the shop. Thus, to show parallelepiped II
equal to parallelepiped III construct parallelepiped II and
drop a perpendicular from D to the base. Then saw along the
edges MD and DF. Place the slab obtained on the right side,
and parallelepiped II will look exactly like parallelepiped III.
This will be helpful to your classmates, and you will find the
exercise easy and interesting.
*2. A much more difficult and interesting exercise is to make
parallelepiped I look like parallelepiped III.
HINT. Construct RK to AC and KI to AC. Saw along the
edges RK and KI and place the slab obtained on the left side. Now
the figure will be transformed to parallelepiped II. Continue as in
Ex. 1 to make parallelepiped I look like parallelepiped III.
3. A rectangular reservoir 120 ft. long and 20 ft. wide con-
tarns water to a depth of 10.5 ft. A second reservoir 125 ft. long
and 18 ft. wide contains water to a depth of 36.5 ft. How much
more water is there in the. second reservoir than in the first ?
4. A rectangular tank 6 ft. long, 4 ft. wide, and 5 ft. deep
is to be lined with zinc ^ in. thick. How many cubic feet of
zinc will be required if 4 sq. ft. are allowed for overlapping ?
102 GENERAL MATHEMATICS
5. If 1 cu. in. of pure gold beaten into gold leaf will cover
30,000 sq. ft. of surface, what is the thickness of the gold leaf ?
6. An open tank is made of iron ^ in. thick. The outer
dimensions are as follows : length, 3 ft. ; width, 1 ft. 9 in. ;
height, 2 ft. If 1 cu. ft. of iron weighs 460 lb., find the weight
of the tank.
7. In a rainfall of 1 in. how many tons of water fall upon
an acre of ground if 1 cu. ft. of water weighs 62.5 lb. ?
144. Formula for the volume of a cube. The volume of
a cube is computed in the same way as that of a parallele-
piped. The cube is a special case in the sense that the
length, width, and height are all equal. Hence, if s equals an
edge of a cube, the volume may be expressed by the formula
V = s x s x s. The formula V= s x s x s may be written
more briefly V= s 3 (read "V equals s cube"); s 3 being an
abbreviated form of s x s x s. The formula may be trans-
lated into the following law : The volume of a cube equals
the mbe of an edge.
EXERCISES
1. Find the volume of a cube whose edge is ^ in. ; ^ in. ; ^ in.
2. Find the volume of a cube whose edge is 1^ in.; 2.2 cm.;
3 in.; 1 m. ; 0.01 m.
145. Equal factors ; exponents ; base ; power. The prod-
ucts of two equal dimensions and three equal dimen-
sions have been represented by the area of a square and
the volume of a cube respectively. Hence the notation
" a square " and "s cube." The product of four equal fac-
tors cannot be represented geometrically, though you may
already have heard people talk vaguely about the fourth
dimension. However, the product of four equal algebraic
factors, say s x s x s x s, is as definite as 2x2x2x2
. THE EQUATION APPLIED TO VOLUME 103
in arithmetic. Thus, we extend the process indefinitely in
algebra and write sxsxsxs = s* (read "s fourth") or
bxbxbxbxb = b* (read " b fifth "), etc.
The term b 5 is obviously much more convenient than
bxbxbxbxb. The 5 hi b 6 is called an exponent. It
is a small number written to the right and a little above
another number to show how many times that number is to
be used as a factor. In 5 3 (meaning 5x5x5) the 3 is
the exponent, the number 5 is the base, and the product
of 5 x 5 x 5 is the power. Thus, 125, or 5 3 , is the third
power of 5. When no exponent is written, as in x, the
exponent is understood to be 1. Thus, in 2 xy, both x and
y are each to be used only once as t a factor. The mean-
ing is the same as if the term were Mjritten 2x l y l .
EXERCISES
1. State clearly the difference between a coefficient and an
exponent. Illustrate with arithmetical numbers.
2. Letting a = 5, give the meaning and the value of each
of the following numbers :
(a) 3 a. (c) 2 a. (e) 4 a. (g) 5 a. (i) 2 a 2 . (k) 4 a 2 .
(b) a 8 . (d) a?. (f) a 4 . (h) a 5 . (j) 3 a 2 . (1) 2 a 4 .
3. Write the following products in briefest form :
1111 333 a a a
yyyyy, 5-5-5-5; g-^ji m'>
2J . 5J 2J ; 5 . 4 V4 f 6 3 ^ 3 3 * . y . y . y.
4. Find the value of 2 s ; 6 8 ; (i) 4 ; 3 8 ; (1.3) a ; 9 8 ; (0.03) 8 ;
(1.1) 8 .
5. Letting m = 2 and n = 3, find the value of the follow-
ing polynomials : m 2 + 2 mn + w 2 ; m" + 3 ra 2 n, + 3 wp 2 + w 8 ;
); 5(m + n); 6(2 w s + 3m 2 -f- 4mn + n + 3).
104 GENERAL MATHEMATICS
6. Find the value of the following numbers, where z 3 :
(a) 2z; (b) *; (c) (2*) 2 ; (d) 2* 2 ; (e) 3z 8 ; (f) (3z) 3 ;
(g) (3*) 2 -
7. Letting a; = 1, y = 2, z = 3, and u = 4, find the value of
the following :
xy + xz + yu + zu x* + 4 x 3 # -f- 6 x 2 y 2 4- 4 x^ 3 + 2/ 4
# -f y + + w xy
146. Exponents important. Since the subject of expo-
nents is fundamental to a clear understanding of two very
important labor-saving devices, namely the slide rule and
logarithms, which we shall presently study, it is necessary
to study the laws of exponents very carefully.
147. Product of powers having the same exponents. The
law to be used in this type may be illustrated by the
problem, " Multiply a 2 by a 3 ." The expression a 2 means
a a, or aa, and the expression a? means a a a, or aaa.
Hence a 2 a 3 means aa aaa, or, hi short, a 6 .
EXERCISE
In each case give orally the product in briefest form :
(a) 3 2 -3 3 . (i) x s - x 6 . (q) x 2 m.
(b) 6-6 8 . (j) ax-x. (r)6-6.*>fl
(c) 5 2 5*. (k) b e b. ( s ) 4 tfc 5 iV.
(d) 10 10 8 . (1) b-b. ( ; t ) xy- x 2 yz 2 2 xifz s .
(e) x x 2 . (m) e e 2 2. (u) (2 xyf.
(f) 12 2 .12 S . (n) c-c 3 . (v) (2xV) 3 .
(g) x 2 x s . ( o ) x x 5 . (w) ' (3 x 2 ?/) 2 .
(h) x x 4 . (p) m a- 2 . (x) 3 5 2 5 2 3 8 .
The exercise above shows that the product of two or
more factors having the same base is a number ivJwse base
is the same as that of the factors and whose exponent is the
mm of the, exponent* of the factors ; thus, b 2 & . / = 2> 10 .
THE EQUATION APPLIED TO VOLUME 105
148. Quotient of powers having the same base. The quo-
tient of two factors having the same base may be simplified
by the method used in the following problem:
Divide b 6 by b\
HINT. Since a fraction indicates a division, this quotient may
6 5
be indicated in the form
cr
b 5 b-b'b-b'b
Solution. - = - - --
&* b b
Hence, dividing numerator and denominator by b b, or b 2 ,
EXERCISE
In each case give orally the quotient in briefest form :
2 (c)ft'-*-P. (g)* 6 ^* 2 .
w 2" ' (d) b 6 -- b\ (h) x 6 -4- x s .
5_ (e)b" + b*. (i)x + x*.
W 5 2 ' (f) b + bv (j) x 21 + x".
The preceding exercise shows that the quotient obtained
by dividing a power by another power having the same base
is a number whose base is the common base of the given
powers and whose exponent is obtained by subtracting the
exponent of the divisor from the exponent of the dividend.
Thus, a: 10 * x 3 = x 1 . Here x 1 has the same base as the dividend x w
and the divisor X s ; and its exponent, 7, is obtained by subtracting 3
from 10.
149. Review list of problems involving application of
algebraic principles to geometric figures. The following
exercises are intended to help the student to see how to
apply algebraic principles to geometric figures.
106
2X+3
FIG.
EXERCISES
1. Find an algebraic number which will express the sum
of the edges of the solid in Fig. 86.
2. If the sum of the edges of the solid in Fig. 86 is 172, what
are the actual dimensions of the solid ?
3. Find an algebraic expression
for the total surface of the solid in
Fig. 86. Also for the volume.
4. What is the total surface and vol-
ume of the solid in Fig. 86 if x equals 10 ?
5. Express algebraically the sum of the edges of the cube
in Fig. 87.
6. If the sum of the edges of the culx-
in Fig. 87 is 112, what is the length of
one edge ?
7. Express algebraically the total surface
and volume of the cube in Fig. 87.
8. What is the total surface of the cube in Fig. 87 if x = 2 ?
9. The edge of a tetrahedron (Fig. 88) is denoted by
2 x + 1. Express algebraically the sum of all
the edges of the tetrahedron.
NOTE. A tetrahedron is a figure all of whose
edges are equal and whose faces are equal equilateral
triangles.
10. Find the length of an edge of the tetra-
hedron in Fig. 88 if the sum of the edges is 40.5 cm.
11. Fig. 89 shows a frustum of a pyra-
mid. The upper and lower bases are equi-
lateral pentagons ; the sides are trapezoids
with the edges denoted as in the figure.
Find the sum of all the edges. If e
equals 2. what is the sum of the edges ? FIG. 89
DESCARTES
108 GENERAL MATHEMATICS
HISTORICAL NOTE. The idea of using exponents to mark the
power to which a quantity was raised was due to Rene 1 Descartes,
the French philosopher (1596-1650). It is interesting to read of the
struggle for centuries on the part of mathematicians to obtain
some simple method of writing a power of a number. Thus, we read
of the Hindu mathematician Bhaskara (1114- )*using the initials
of the Hindu words " square " and " solid " as denoting the second and
third power of the unknown numbers in problems, which he gave a
practical setting with many references to fair damsels and gallant
warriors. In the following centuries a great variety of symbols for
powers are used ; for example, arcs, circles, etc., until we come to a
French lawyer, Frangois Vieta (1540-1603), who wrote on mathe-
matics as a pastime. Vieta did much to standardize the notation of
algebra. Thus, in the matter of exponents he employed " A quad-
ratus x " " A cubus," to represent z 2 and x 8 , instead of introducing a
new letter for each power. From this point it is only a step to
Descartes's method.
The biographies of the three mathematicians Bhaskara, Viet;a, and
Descartes are exceedingly interesting. Thus, you may enjoy reading
of Bhaskara's syncopated algebra in verse, in which many of the
problems are addressed to " lovely and dear Lilavati " (his daughter)
by way of consolation when he forbade her marriage.
You may read of Vieta's being summoned to the court of Henry IV
of France to solve a problem which involved the 45th power of x.
The problem had been sent as a challenge to all mathematicians
in the empire. Vieta appeared in a few moments and gave the king
two correct solutions. Next King Henry asked Vieta to decipher
the Spanish military code, containing over six hundred unknown
characters, which was periodically changed. King Henry gave the
cipher to Vieta, who succeeded in finding the solution to the system,
which the French held greatly to their profit during the war.
Or you may read of Descartes, a member of the nobility, who
found the years of his army life exceedingly irksome, for he craved
leisure for mathematical studies. He resigned his commission in
1621 and gave his time to travel and study. In 1637 he wrote a book,
" Discourse on Methods." In this text he made considerable advance
toward the system of exponents now used. The text shows that he
realized the close relation existing between geometry and algebra.
He is often called "the father of modern algebra."
THE EQUATION APPLIED TO VOLUME 109
12. A tetrahedron may be constructed from a figure like
Fig. 90. Draw the figure on cardboard, using a larger scale.
Cut out the figure along the heavy lines ; then fold along
the dotted lines. Join the edges by means of gummed paper.
FIG. 90. How TO CON-
STRUCT A TETRAHEDRON
FIG. 91. How TO CONSTRUCT
A CUBE
13. The cube may be constructed from a figure like Fig. 91.
Draw the figure on cardboard, using a larger scale ; for example,
let x = 3 cm. Cut out the figure
along the heavy lines, then fold
along the dotted lines. Join the
edges by means of gummed paper.
This will form a model of a cube.
14. Measure the edge of the
cube constructed for Ex. 13 and
compute the area of the whole
surface. Find the volume also.
15. A rectangular parallele-
piped may be constructed from
a figure like Fig. 92. Compute the volume of the solid and the
area of the surface.
SUMMARY
150. This chapter has taught the meaning of the follow-
ing words and phrases : a solid, surface of a solid, volume
of a solid, unit of volume, cube, parallelepiped, rectangular
parallelepiped, right parallelepiped, oblique parallelepiped,
triangular pyramid, exponent, base, power, tetrahedron.
FIG. 92. How TO CONSTRUCT A
RECTANGULAR PARALLELEPIPED
110 GENERAL MATHEMATICS
151. The volume of a solid is determined by applying
the unit cube to see how many times it is contained in the
solid. The process is essentially comparison. The unit cube
is a cube each of whose edges is one unit long.
152. The following formulas have been used :
v = Iwh,
v=s 3 .
153. The product of factors having a common base
equals a number whose base is the same as the factors
and whose exponent is the sum of the exponents of the
factors.
154. The quotient obtained by dividing a power by
another power having the same base is a number whose
base is the common base of the given powers and whose
exponent is obtained by subtracting the exponent of the
divisor from the exponent of the dividend.
CHAPTER VI
THE EQUATION APPLIED TO FUNDAMENTAL
ANGLE RELATIONS
155. Fundamental angle relations. In Chapter III we
discussed the different kinds of angles and the methods of
constructing them. In this chapter we shall study some
of the fundamental relations between angles and see how
the equation is applied to them.
156. Relation of exterior sides of supplementary adjacent
angles. Draw two adjacent angles of 64 and 116, of 75
and 105, of 157 and 23. What is the sum of each pair?
What is the relation of the exterior sides of each pair ?
E
FIG. 93
157. Important geometric relation. The preceding article
illustrates the following geometric relation : If the sum of
two adjacent angles is a straight angle, their exterior sides
form a straight line.
ill
112 GENEKAL MATHEMATICS
EXERCISES
1. Show that the geometric relation stated in Art. 157 agrees
with the definition of a straight angle (Art. 61).
2. In Fig. 93 read the number of degrees in angles XOA,
XOB, XOC, XOY, XOD, XOE.
3. What is the sum of /.XOA and Z.AOB?
4. Express /.XOD as the sum of two angles.
5. Express Z.AOB as the difference of two angles.
6. Express /.XOE as the sum of three angles.
158. Sum of all the angles about a point on one side
of a straight line. Draw a line AB and choose a point
P on it. Draw lines Pit,
PS, and PT as shown in
Fig. 94 and find the sum
of the four angles formed.
Estimate first and then
measure with the pro-
tractor. What seems to
be the correct sum ? Ex-
press the sum of the angles x, y, z, and w by means of
an equation. Give a word statement for the equation.
159. Important geometric relation. Art. 158 illustrates
the truth of the geometric relation that the sum of all the
angles about a point on one side of a straight line is a straight
angle (180).
EXERCISES
1. Find the value of x and the size of each angle in Fig. 95.
2. In the following examples each expression represents one
of the angles into which all the angular space about a point
on one side of a straight line has been divided. Write an
equation expressing the sum of all the angles, solve for x,
FUNDAMENTAL ANGLE RELATIONS
113
and find the size of each angle in degrees. Draw figures to
illustrate your results in the first three examples.
(a) x, 3 a-, 5 x + 9.
(b) 3x, 4z-10, 33-2*, Wx + 7.
(c) 2 x + 18}, 5 x + 91, 8J- + a-.
(d) 2 (x + 5), 3 x + 24, 2 (35 + x).
(e) 5.83 x, 3,94 a-, 1.27 a- + 11.55, 138.45 - 8.04 x.
FIG. 95
FIG. 96
160. Sum of all the angles about a point in a plane. If
we choose a point in a plane, as P in Fig. 96, and from this
point draw four lines so as to make four angles, we can
measure these angles and thus determine the sum of all
the angles in a plane about a point.
EXERCISES
1. Measure the angles in Fig. 96 and write an equation
expressing their sum.
2. What is another way
to show that the sum of the
angles that exactly fill a plane
about a point is two straight
angles (360)? See the defi-
nition of a perigon (Art. 61).
3. Find the value of x and the size of each angle in Fig. 97.
3Z+80
FIG. 97
114 GENERAL MATHKMATICS
4. The expressions in the following examples represent the:
angles into which the angular space about a point in a plane
has been divided. Find the size of each angle.
(a) 3 x, x, 2 x + 35, 125 - at.
(b) 2x, 72 + 3x, 4* - 10, 118.
(c) 10 x + 20-J-, 35| - x, 8 x + 49.
(d) 5 a-, 3x + 27f,*7z - 20, 9a-. + 112j.
(e) x + 1, 7 (a; + 1), 3 (35 + x), 2 x + 169.
(f) 3x, 117 + 15 a-, 9 a; -27.
(g) 14 x + 48, 28 x + 106f , 133^ - 6 x.
The first two exercises in this article show that the sum
of all the angles about a point in a j)lafie is 360.
161. Left side of an angle; right side of an angle. If in
Fig. 98 we imagine ourselves standing at the vertex of
/.ABC and looking off over the angular space, say in the
direction BD, then the side BC Q
is called the left side of the angle
(because it lies on our left), and - a* -
the side BA is called the right s\f&^~~'
side of the angle. ^-''"
B Right A
162. Notation. In lettering an-
, , fi . . . f . FIG. 98
gles and figures it is often desir-
able to denote angles or lines that have certain characteristic
likenesses by the same letter so as to identify them more
easily. It is clear that to use I for the left side of one
angle and the same I for the left side of another angle
in the same discussion might be misleading. In order to
be clear, therefore, we let ^ stand for the left side of one
angle, Z 2 stand for the left side of a second angle, and
1 3 stand for the left side of a third angle, etc. Then the
three sides would be read " I sub-one," " I sub-two," " I sub-
three," etc.
FUNDAMENTAL ANGLE RELATIONS
115
li
FIG.
163. Important geometric relation. Two angles, a; 1 and
x v in Fig. 99, are drawn so that their sides are parallel left
to left and right to
right. How do they
seem to compare in
size ? Check your
estimate by measur-
ing with a protractor.
Give an argument
showing that x-^ = x
This article shows
that if two angles have their sides parallel left to left and right
to right, the angles are equal.
EXERCISE
Draw freehand two obtuse angles so that their sides will
look parallel left to left and right to right. (The angles should
be approximately equal. Are they ?)
HINT. Take -two points for vertices and in each case imagine
yourself standing at the point. Draw the left sides to your left and
the right sides to your right. Assume the drawing correct and prove
the angles equal.
164. Important geometric relation. Two angles, x and
y, in Fig. 100, have been drawn so that their sides are par-
allel left to right and right to left. What relation seems
to exist between them ? Meas-
ure each with a protractor.
Give an argument showing
that z + # = 180.
This article shows that if
two angles have their sides par-
allel left to right and right to
left, their sum is a straight angle. F IG . 100
1 1
116 GENERAL MATHEMATICS
EXERCISE
Practice drawing freehand a pair of angles whose sides are
parallel according to the conditions in the theorem of Art. 164.
Is the sum approximately 180 ?
165. Supplementary angles ; supplement. Two angles
whose sum is equal to a straight angle (180) are said
to be supplementary angles. Each
angle is called the supplement of
the other.
166. Supplementary adjacent F 101
angles. Place two supplementary
angles adjacent to each other as in Fig. 101. Angles so
placed are called supplementary adjacent angles.
EXERCISES
1. In Fig. 101 what is the angle whose supplement is Zee?
2. In Fig. 102 are several angles, some pairs of which are
supplementary. Make tracings of these angles on paper and
by placing them adjacent decide which pairs are supplementary.
FIG. 102
3. State whether the following pairs of angles are sup-
plementary : 40 and 140 ; 30 and 150 ; 35 and 135 ; 55
and 135.
4. How many degrees are there in the supplement of an
angle of 30 ? of 90 ? of 150 ? of x ?
2 s
5. What is the supplement of y ? of z ? of 3 w ? of ?
o
117
6. Write the equation which expresses the fact that y and
130 are supplementary and solve for the value of y.
7. Write equations that will show that each of the follow-
ing pairs of angles are supplementary :
(a) y and 80. (d) 30 and y + 40.
(.b) 90 and z. (e) 3 y + 5 and 12 y - 4.
(c) x and y. (f) f x and lx + 75 J.
8. Two supplementary angles have the values 2x + 25
and x + 4. Find x and the size of each angle.
9. What is the size of each of two supplementary angles
if one is 76 larger than the other ?
10. One of two supplementary angles is 33 smaller than the
other. Find the number of degrees in each.
11.. What is the number of degrees in each of two supple-
mentary angles whose difference is 95 ?
12. Find the value of x and the angles in the following
supplementary pairs :
(a) x and 6 x.
(b) 2 x and 3 x + 2.
(c) 4 x and 6 x.
(d) 2 aj + 5 and 7 x - 8.
13. Write the following expressions in algebraic language :
(a) Twice an angle y.
(b) Four times an angle, plus 17.
(c) 23 added to double an angle.
(d) Seven times an angle, minus 14.
(e) 45 less than an angle.
(f ) 52 subtracted from four times an angle.
(g) Twice the sum. of an angle and 10.
(h) One half the difference of 22 and x.
14. If an angle is added to one half its supplement, the
sum is 100. Find the supplementary angles.
118 GENERAL MATHEMATICS
15. If an angle is increased by 5 and if one fourth of its
supplement is increased by 25, the sum of the angles thus
obtained is 90. Find the supplementary angles.
16. Construct the supplement of a given angle.
17. Find the size of each of the following adjacent pairs of
supplementary angles :
(a) 0llft : +i (d) -60, 130- *.
(b) | x + 32, 88 - \x. (e) 2 (* + 10), ^
(c) | + 150, I - 10. (f) 65 + 2 -f, 92 + ^.
167. Construction problem. To construct the supplements
of two equal angles.
Construction. Let a; and
y be the given angles.
Construct Z z, the supple-
ment of Zz, adjacent to z /x w /y
it (Fig. 103). In the same
manner construct Zw-, the Fm - 103 ' HOWTO CONSTRUCT THE SCPPLK-
, . MENTS OF T\VO GlVEX AHOLES
supplement of Z.y.
Compare the supplements of Z x and Z y and show that
/-Z=/-W.
This article shows that the Supplements of equal angles
are equal.
EXERCISES
1. Prove the preceding fact by an algebraic method.
HINT. In Fig.103 prove that if Zx + Zz =180 and Z.y + Zro =180,
then Zz = Zw.
2. Are supplements of the same angle equal ? Why ?
FUNDAMENTAL ANGLE RELATIONS
119
3. Show that the bisectors of two supplementary adjacent
angles are perpendicular to each other ; for example, in Fig. 104
show that Z x + Z. y = 90.
4. In Fig. 104, if Z. BOD = 60
and Z. AOD = 120, find the size
of Z. x and Z. y.
*5. The following examples
furnish a review. In each case
solve for the value of the un-
known, and check.
() Y
(b) \y
OD
(e)
t ^
(g)
. " 19
h < = 12.
16 + -* = 3.
o
168. Complementary angles. If the
sum of two angles is a right angle, the
two angles are called complementary
angles. Each angle is called the com-
plement of the other. Thus, in Fig. 105
Z.x is the complement of Z y.
FIG. 105
EXERCISES
1. What is the complement of 30 ? of 60 ?
2. Are 23 and 57 complementary ? 32 and 58 ?
3. Draw two complementary angles of 40 and 50 and
place them adjacent. Check the construction.
4. What is the relation existing between the exterior sides
of two adjacent complementary angles ?
120
GENERAL MATHEMATICS
5. In Fig. 106 decide by means of tracing paper which
pairs of angles seem to be complementary.
6. What are the complements of the following angles :
20? 50? 12? 48|? x? 3y? ^?
7. 40 is the complement of y. How many degrees does
y represent ?
8. Write the equation which says in algebraic language
that x and 50 are complementary and solve for the value of x.
FIG. 106
9. In the equation x + y = 90 is there more than one
possible pair of values of x and y ? Explain.
10. Write equations that will express the fact that the
following pairs of angles are complementary :
(a) x and 40. (c) x + 25 and x - 30.
(b) 35 and y. (d) 2 x - 3 and 3 x + 8.
11. Write the following expressions in algebraic language:
(a) The sum of angle x and angle y.
(b) Four times an angle, increased by 15.
(c) 85 diminished by two times an angle.
(d) Five times the sum of an angle and 13.
(e) Three times the difference between an angle and 12.
(f) Four times an angle, minus 6.
FUNDAMENTAL ANGLE RELATIONS 121
12. Find two complementary angles such that one is 32
larger than the other.
13. Find two complementary angles such that one is 41
smaller than the other.
14. Find the number of degrees in the angle x if it is the
complement of 3 x ; of 5 x ; of 8^ x -
15. Find the number of degrees in the angle x if it is
the complement of twice itself; of five times itself; of one
third itself.
16. Construct the complement of a given acute angle.
169. Construction problem. To construct the comple-
ments of two equal acute angles.
Construction. Let Zx and /.y be the given angles. Draw the
complement of x adjacent to it (Fig. 107). Do the same for y.
FlG. 107. HOW TO CONSTRUCT THE COMPLEMENTS OF. TWO GlVEN ANGLES
Compare the complements of Ax and y and show that
Z0 = /.w.
This construction problem shows that complements of
equal angles are equal.
EXERCISES
1. Prove the preceding relation by an algebraic method.
2. Does it follow that complements of the same angle are
equal? Why?
122
GENERAL MATHEMATICS
170. Vertical angles. Dra\v two in-
tersecting straight lines AB and CD as
in Fig. 108. The angles x and z are
called vertical, or opposite, angles. Note
that vertical angles have a common
vertex and that their sides lie in the
same straight line but in opposite direc-
tions. Thus, vertical angles are angles
which have a common vertex and their
sides Jying in the same straight line -A D
but in opposite directions. Are w and FIG. 108. VERTICAL
y in Fig. 109 vertical angles?
ANGLES
EXERCISES
(Exs. 1-6 refer to Fig. 108)
1. Make a tracing of /- x and /- z and compare them as to size.
2. Check your estimate in Ex. 1 by measuring the two angles
with a protractor.
3. What is the sum of /- x and /. y ? of Z. z and /- y ?
4. Show that x + y = z -j- y.
5. How does Ex. 4 help in obtaining the relation between
x and z ? What is this relation ?
6. Show that y + x = x -+- w and from this that y = w.
The six exercises above show that if two lines intersect,
the vertical angles are equal.
171. Value of mathematical thinking. The preceding
relation between vertical angles is of course so easily seen
that in most cases the truth would be granted even with-
out measuring the angles involved. However, the discus-
sion in Exs. 3-6 above is another simple illustration of
123
the power of mathematical thinking which makes the dis-
covery of new truths rest finally on nonmeasurement, that
is to say, on an intellectual basis. This type of thinking
will be used to an increasing extent in subsequent work.
EXERCISES
1. Upon what does the proof (Exs. 3-6, Art. 170) of
the geometric relation concerning vertical angles rest ?
2. Find x and the size of
each angle in Fig. 109. ~^_^__ o
First method. Since vertical
angles are equal,
Then 3r + 4 = 2ar + 10.
Subtracting 4 from each member,
3x = 2x + 6.
Subtracting 2 x from each member,
* = .
Substituting 6 for x, 3 x + 4 = 3 6 + 4 = 22,
2 x + 10 = 2 . 6 + 10 = 22,
9 x + 104 = 9 . 6 + 104 = 158 (for Z. BOC),
and since vertical angles are equal,
Check. 22 + 22 + 158 + 158 = 360.
Second method. By definition of supplementary angles,
8* + 4 + 9* + 104 =180.
Solving, . r = 6.
The remainder of the work is the same as that of the first method.
124 GENERAL MATHEMATICS
3. Find the values of the unknowns and each of the follow-
ing vertical angles made by two intersecting straight lines :
(a) 3 x + 15 and 5x 5. (f) -J x + $ x and f x + 55.
(b) , + 105 and IS* + 15. * 5,
(c) cc. 10 and 2 a; 160. 4 2
(d) + 181 and + 21. (h) + and + 18.
Q ~ 8 T ^ -> 9 T
(e)|*-8andj + 12. (i) ^ - - ^ and ^ + 11 ].
172. Alternate-interior angles. In Fig. 110 the angles
x and y, formed by the lines AB, CD, and the transversal
EF, are called alternate-interior angles (on alternate sides
of EF and interior with respect to AB and CD~).
FIG. 110 E FIG. Ill
EXERCISES
(Exs. 1-4 refer to Fig. Ill)
1 . Measure and compare Z x and Z //.
2. The lines AB and CD and FE are drawn so that Z.x=Z.y.
What seems to be the relation between the lines AB and CD?
3. Show that if Z x = Z y, then Z y = Z z.
4. Show that 4-^ is parallel to CD (see the definition for
parallel lines, Art. 87).
FUNDAMENTAL ANGLE RELATIONS
125
Exercises 1-4 show that if the alternate-interior angles
formed by two lines and a transversal are equal, the lines are
parallel.
The proof may take the following brief form:
Proof. In Fig. Ill, Z x = Z y (given). Z x = Z z (vertical angles are
equal). Then Z?/ = Z.z (things equal to the same thing are equal to
each other). Therefore AB II CD (by definition of II lines, Art. 87).
EXERCISE
In Fig. 112 construct a line through P parallel to the line
AB by making an alternate-interior angle equal to /.x. Show
why the lines are parallel. F
A \ B
FIG. 112
173. Interior angles on the same side of the transversal.
In Fig. 113 angles x and y are called interior angles on the
same side of the transversal.
EXERCISES
1. Measure angles x and y in Fig. 113 and find their sum.
2. In Fig. 114 the lines are drawn so that /Lx + Z.y = 180.
What seems to be the relation
between AB and CD?
3. Prove that if the interior
angles on the same side of a,
transversal between two par-
allel lines are supplementary,
the lines are parallel.
'\-
-D
FIG. 114
126 GENERAL MATHEMATICS
4. In Fig. 115 select all the pairs of corresponding angles,
alternate-interior angles, and in- F
terior angles on the same side of /
x /?/
the transversal. ./' B
174. Important theorems relat-
ing to parallel lines. The follow- c y^ D
ing exercises include theorems /
which supplement the work of 'E
Arts. 172 and 173. FIG. 115
EXERCISES
1. Show by reference to the definition of parallel lines in
Art. 87 that if two parallel lines are cut by a transversal, the
corresponding angles are equal.
2. Show that if two parallel lines are cut by a transversal,
the alternate-interior angles are equal.
3. Show that if two parallel lines are cut by a transversal,
the interior angles on the same side of the transversal are
supplementary.
4. Two parallel lines are cut by a transversal so as to
form angles as shown in Fig. 116. Find x and the size of
all the eight angles in the figure.
V Ab <K'^
\ X
FIG. 116 FIG. 117
5. Find x and all the eight angles in Fig. 117.
6. Draw two parallel lines and a transversal. Select all the
equal pairs of angles ; all the supplementary pairs.
FUNDAMENTAL ANGLE RELATIONS 1.27.
175. Outline of angle pairs formed by two lines cut by
a transversal. When two lines are cut by a transversal,
as in Fig. 118,
"a and e~]
the angles of the b and /
7 IT?- are called corresponding angles ;
angle pairs a and h
c and g\
angles c, d, e, f are called interior angles ;
angles a, 5, #, h are called exterior angles ;
the angles of the f d and e 1 are called interior angles on the
angle parrs j c and /j same side of. the transversal ;
*:',, f , , J on opposite sides of the trans-
the angles of the \ a and /
, > versal are called alternate-
angle pairs c and e\
J interior angles ;
, j "] on opposite sides of the trans-
the angles ot the b and h\ , . ,
^ , Y versal are called alternate-
angle pairs a and a \
[ J exterior angles.
FIG. 119
The student should remember
(a) that corresponding angles are equal,
(b) that alternate-interior angles are equal,
(c) that alternate-exterior angles are equal,
(d) that interior angles on the same side of
the transversal are supplementary,
only when the lines cut by the transversal are parallel (Fig. 119).
128 GENERAL MATHEMATICS
SUMMARY
176. This chapter has taught the meaning of the follow-
ing words and phrases: left side of an angle, right side
of an angle^ parallel right to right and left to left, par-
allel right to left and left to right, supplementary angles,
supplement, supplementary-adjacent angles, complementary
angles, complement, vertical angles, alternate-interior angles,
interior angles on the same side of the transversal.
177. The following fundamental constructions have been
presented :
1. How to construct the supplement of a given angle.
2. How to construct the supplements of two equal given
angles.
3. How to construct the complement of -a given angle.
4. How to construct the complements of two equal
angles.
5. A new method of drawing parallel lines.
6. How to form vertical angles.
178. This chapter has discussed the following funda-
mental geometric relations:
1. If the sum of two adjacent angles is a straight angle,
their exterior sides form a straight line.
2. The sum of all the angles about a point on one side
of a straight line is a straight angle (180).
3. The sum of all the angles in a plane about a point
is two straight angles (360).
4. If two angles have their sides parallel left to left and
right to right, the angles are equal.
5. If two angles have their sides parallel left to right
and right to left, the angles are supplementary.
6. Supplements of the same angle or of equal angles
are equal.
FUNDAMENTAL ANGLE RELATIONS 129
7. Complements of the same angle or of equal angles
are equal.
8. If two lines intersect, the vertical angles formed
are equal.
9. Two lines cut by a transversal are parallel
(a) if the corresponding angles are equal ;
(b) if the alternate-interior angles are equal;
(c) if the interior angles on the same side of the trans-
versal are supplementary.
10. If two parallel lines are cut by a transversal, then
(a) the corresponding angles are equal ;
(b) the alternate-interior angles are equal ;
(c) the interior angles on the same side of the trans-
versal are supplementary.
CHAPTER VII
THE EQUATION APPLIED TO THE TRIANGLE
179. Notation for triangles. It is customary to denote
the three points of intersection of the sides of a triangle
by capital letters and the three sides which He opposite
these respective sides with the corresponding small letters.
Thus, in Fig. 120 we denote
the points of intersection
of the sides (the vertices)
by A, B, and C, and the
sides opposite by a, 6, and c.
The sides may also be
read BC, AC, arid AB. The symbol for " triangle" is a small
triangle (A). The expression A ABC is read " triangle ABC."
The three angles shown in Fig. 120 are called interior angles.
180. Measuring the interior angles of a triangle. We shall
now consider some of the methods of measuring the interior
angles of a triangle.
EXERCISES
ANGLE
TRIANGLE ABC
No. OF DEGREES
TRIANGLE DEF
No. OF DEGREES
TRIANGLE GHI
No. OF DEGREES
Estimated
Measured
Estimated
Measured
Estimated
Measured
X
y
z
Sum
130
131
1. Fill in the table on the preceding page with reference
to the triangles ABC, DEF, and GHI (Fig. 121).
FIG. 121
2. Draw a triangle on paper (Fig. 122). Cut it out and
tear off the corners as shown in Fig. 123. Then place the
three angles adjacent as shown. What
seems to be the sum of the three angles
of the triangle ? Test your answer with
a straightedge.
181. Theorem. The results of Exs. 1 FIG. 122
and 2, above, illustrate the geometric
relation that the sum of the interior
angles of a triangle is a straight angle
(180). The statement "The sum
of the interior angles of a triangle is
a straight angle" can be proved by more advanced geo-
metric methods. Such a statement of a geometric relation
to be proved is called a theorem.
182. More advanced methods of
proof for the preceding theorem. The
truth of the theorem, that the sum of
the interior angles of a triangle is Ji
180 may be illustrated as follows:
Draw a triangle as in Fig. 124. Place
a pencil at A as indicated in the figure, noting the direction
in which it points. Rotate the pencil through angle A as
FIG. 123
FIG. 124
132
GENERAL .M ATM K.MATK 'S
sliown by the arrowhead. Then slide it along AB to the
position indicated in the figure. Rotate the pencil next
through angle B as indicated and slide it along BC to
the position sliown. Then rotate the pencil through angle
C to the last position shown. This rotation through
angles .1. /;, and C leaves the point of the pencil in what
position in respect to its original position ? What part of
a complete turn has it made ? Through how many right
angles has it turned ? Through
how many straight angles ?
Through how many degrees?
The theorem that ' : the sum
of the interior angles of a tri-
angle is 180 " may be proved
as follows :
(oven triangle AB<'(Y\%. 12o), to prove that Z.I + Z/> + ZC' = 180.
Proof
STATE.MK.NT>
REASONS
Draw
Then
And
Z if - Z B.
But Z ./.; 4- Z // + A : = 180
.-. z.i
Because corresponding angles
formed by two parallel lines cut
by a transversal are equal.
Because alternate-interior an-
gles formed by two parallel lines
cut by a transversal are equal.
Because the sum of all the
angles about a ] >oiut in a plane on
one side of a straight line is 180.
By substituting Z.A for Zr.
i Z B for Z if. and Z C for Z r.
This is n more formal proof of the theorem, inasmuch as it
is independent of measurement. Write an equation which will
express the number of degrees in the sum of the angles of
a triangle.
. EQUATION APPLIED TO THE TRIANGLE 133
This equation is a very useful one, as it enables us to
find one angle of a triangle when the other two are known.
Thus, if we know that two angles of a triangle are 50
and 70, we know that 60 is the third angle. This is of
great practical value to the surveyor, who is thus enabled
to know the size of all three angles of a triangle by
measuring only two directly.
HISTORICAL XOTK. Thales (040 B.C. - about o50 B.C.), the founder
of the earliest Greek school of mathematics, is supposed to have known
that the sum of the angles of a triangle is two right triangles.
Someone has suggested that this knowledge concerning the sum
of the angles of a triangle may have been experimentally demon-
strated by the shape of the tiles used in paving floors in Th-ales'
day. What has been regarded as the most remarkable geometrical
advancement of Thales was the proof of a theorem which depended
upon the knowledge that the sum of the angles of a triangle is two
right angles. It is related that when Thales had succeeded in proving
the theorem, he sacrificed an ox to the immortal gods. The large
number of stories told about Thales indicates that he must have
been a man of remarkable influence and shrewdness both in science
and in business. Thus, we read that at one time he cornered the olive
market and that at another time he was employed as engineer to
direct a river so that a ford might be constructed. The following
story is told illustrative of his shrewdness :
It is said that once when transporting some salt which was loaded
on mules one of the animals, slipping in a stream, got its load wet and so
caused some of the salt to be dissolved. Finding its burden thus light-
ened, it rolled over at the next ford to which it came ; to break it of
this trick, Thales loaded it with rags and sponges, which, by absorbing
the water, made the load heavier and soon effectually cured it of its
troublesome habit. 1
183. Problems involving the theorem " The sum of the
interior angles of a triangle is a straight angle." In
the problems that follow the pupil will need to apply the
theorem proved in the preceding article.
1 Ball, "A Short Account of the History of Mathematics," p. 14.
134 GENERAL MATHEMATICS
EXERCISES
In the following problems
(a) Draw freehand the triangle.
(b) Denote the angles properly as given.
(c) Using the theorem of Art. 182, write down the equation
representing the conditions of the problem.
(d) Solve the equation and find the value of each angle.
(e) Check your solution" by the conditions of the problem.
1. The angles of a triangle are x, 2x, and 3x. Find x and
the number of degrees in each angle.
2. The first angle of a triangle is twice the second, and the
third is three times the first. Find the number of degrees in
each angle.
3. If the three angles of a triangle are equal, what is the
size of each ?
4. If two angles of a triangle are each equal to 30, what
is the value of the third angle ?
5. One angle of a triangle is 25. The second angle is 55
larger than the third. How large is each angle ?
6. The first angle of a triangle is four times the second,
and the third is one half the first. Find e.ach angle.
7. Find the angles of a triangle if the first is one half
of the second, and the third is one third of the first.
8. The first angle of a triangle is two fifths as large as
another. The third is four times as large as the first. How
large is each angle ?
9. Find the angles of a triangle if the first angle is 16 more
than the second, and the third is 14 more than the second.
10. Find the angles of a triangle if the difference between
two angles is 15, and the third angle is 43..
11. The first angle of a triangle is 30 more than the second,
and the third is two times the first. Find the angles.
EQUATION APPLIED TO THE TRIANGLE 135
12. Find the angles of a triangle if the first angle is twice
the second, and the third is 15 less than two times the first.
13. The angles of a triangle are to each other as 1, 2, 3.
What is the size of each ?
HINT. Let x = the first, 2 x the second, and 3 x the third.
14. Find the angles of a triangle if the first is 2^ times the
second increased by 10, and the third is one fourth of the
second.
15. In a triangle one angle is a right angle; the other two
%K>
angles are represented by x and - respectively. Find each angle.
16. How many right angles may a triangle have ? How
many obtuse angles ? How many acute angles at most ? How
many acute angles at least ?
17. Two angles x and y of one triangle are equal respec-
tively to two angles m and n of another triangle. Show that
the third angle of the first triangle equals the third angle of
the second triangle.
184. Theorem. By solving Ex. 17 we obtain the theorem
If two angles of one triangle are equal respectively to two
angles* of another triangle, the third angle of the first is equal
to the third angle of the second.
185. Right triangle. If one angle of a triangle is a right
angle, the triangle is called a right triangle. The symbol
for " right triangle " is rt. A.
EXERCISES
1. Show that the sum of the acute angles of a right triangle
is equal to a right angle.
2. Find the values of the acute angles of a right triangle if
one angle is two times the other ; if one is 5 more than three
times the other.
130
GENEKAL MATHEMATICS
/' /'
3. The acute angles of a right triangle are ^ and r
Find
and the number of degrees in each angle.
4. Draw a right triangle on cardboard so that the two acute
angles of the triangle will contain 30 and 60 respectively.
Use the protractor.
HINT. First draw a right angle. Then at any convenient point
in one side of the right angle construct an angle of 60 and produce
its side till a triangle is formed. Why does the third angle equal.30?
*5. Cut out the cardboard triangle made in Ex. 4 and tell
what angles may be constructed by its use without a protractor
or tracing paper.
6. Draw on cardboard a right triangle whose acute angles
are each equal to 45, cut it out, and show how it may be used
to draw angles of 45 and 90 respectively.
186. Wooden triangles. A wooden triangle is a triangle
(usually a right one) made for convenience in drawing
triangles on paper or on the blackboard (see Fig. 126).
The acute angles are usually 60 and 30
or 45 and 45. These
wooden right triangles fur-
nish a practical method of
drawing a perpendicular to
a line at a given point on
that line. If no triangles
of this kind can be had,
a cardboard with two per-
pendicular edges or a card-
board right triangle will serve the purpose just as well.
187. Set square. A set square is made up of a wooden
triangle fastened to a straightedge so that it will slide
along the straightedge (see Fig. 127).
FIG. 126. WOODEN
TRIANGLE
FIG. 127. SET
SQUARE
EQUATION APPLIED TO THE TRIANGLE 137
EXERCISES
1. Construct a right triangle, using the method of Art. 80.
2. The set square (Fig. 127) is a mechanical device for drawing
parallel lines. Show how it may be used to draw parallel lines.
3. Show that two lines perpendicular to the same line are
parallel (see the definition of parallel lines, Art. 87).
4. Show how a wooden triangle may be used to draw
parallel lines.
5. What are three ways of drawing parallel lines?
188. Problems concerning the acute angles of a right
triangle. The following problems will help the student to
understand the relations concerning the acute angles of
a right triangle.
EXERCISES
1. Draw a right triangle as in Fig. 128 and show that angles
. I and B are complementary ; that is, show that /. A + /.B = 90.
NOTE. In lettering a right triangle
ABC the letter C is usually put at the
vertex of the right angle.
2. State a theorem concerning
the acute angles of a right-angled
triangle.
3. If Z.I =Z B (Fig. 128), what
is the size of each ? How do the sides about the right angle
compare in length ?
4. If /.A is twice as large as /.B (Fig. 128), what is the size
of each ?
*5. Using the method of Ex. 4, Art. 185, draw a right triangle
whose acute angles are 30 and 60 respectively. Measure
the side opposite the 30-degree angle. Measure the side oppo-
site the 90-degree angle. (This side is called the hypotenuse.)
Compare the two results obtained.
138
GENERAL MATHEMATICS
6. To measure the distance AB across a swamp (Fig. 129), a
man walks in the direction AD, so that /.BAD= 60, to a point
C, where ^BCA = 90. If AC = 300 yd.,
what is the length of AB ?
7. Find the number of degrees in each
acute angle of a right triangle if one angle is
(a) four times the other ;
(b) three fourths of the other ;
(c) two and a half times the other ;
(d) 5" more than three times the other ;
(e) 5 less than four times the other.
*8. Practice drawing angles of 30, 45, 60, and 90 by
using wooden or cardboard triangles.
Ex. 5 illustrates the truth of the theorem In a right
triangle whose acute angles are 30 and 60 the side opposite
the 30 angle is one half the hypotenuse. This theorem will
be proved formally later. It is very important because of
its many practical applications in construction work and
elsewhere.
189. Isosceles triangle ; base angles. A triangle which
has two equal sides is called an isosceles triangle. The
angles opposite the equal sides are called the base angles of
the isosceles triangle.
EXERCISES
1. Two equal acute angles of a right triangle are repre-
sented by 2 x + 5 and 3 x 15. Find the size of each angle.
2. Draw a right triangle ABC
(Fig. 130). Draw a line from
CAB; call the foot of the per-
pendicular P. Show that the
perpendicular (CP) divides the
A ABC into two right triangles. FIG. 130
EQUATION APPLIED TO THE TRIANGLE 139
3. In Ex. 2 the angle x is the complement of what two
angles ? What is the relation between these two angles ?
\4\ In Ex. 2, /.y is the complement of two angles. Indicate
them. Show that /.
5. Draw freehand three isosceles triangles.
190. Scalene triangle. A scalene triangle is a triangle
no two of whose sides are equal.
EXERCISES
1. Draw freehand a scalene triangle.
2. Do you think that a right triangle whose acute angles
are 30 and 60 is a scalene triangle ? Support your answer.
3. Draw an obtuse-
angled scalene triangle.
191. Exterior angles
of a triangle. If the
three sides of a tri-
angle are extended, one
at each vertex, as in ''
Fig. 131, the angles thus FlG - 13L I^STRATING THE EXTERIOR
ANGLES OF A TRIANGLE
formed (x, y, and z)
are called exterior angles of the triangle ABC.
EXERCISES
1. How many exterior angles can be drawn at each vertex
of a triangle ?
2. How many interior angles has a triangle ? How many
exterior angles ?
3. Draw a triangle and extend the sides as in Fig. 131.
Measure the three exterior angles with a protractor. What is
their sum ?
14<
GENERAL MATHEMATICS
FIG. 132
4. I>ra\\ another triangle and extend the sides as in. Fig. 131.
Cut out the exterior angles (taking one at each vertex) with a
pair of scissors and place them next to
each other with their vert ices together.
What does their sum seem to be'/
5. Find the sum of the three ex-
terior angles x, //. and .-.- in Fig. 132 In-
rotating a pencil as indicated by the
arrowheads.
6. Show that tin- mini nf flic e;rti'ri<n-
niKjli-x i >f a triangle (taking one at each vertex) /\ -><>0 (two
straight angles).
HINT. How many degrees are in the sum x + w/? // + n ? z + r?
(S,-,. Fig. l:;i.)
Show that tlie sum
(.' + m) + (// + n) + (z + ?) = 3 X 180 = 540.
Then this fact may be expressed as follows :
(x + ?/ + r) + (HI + n + r) = 540.
But ( x + y + z -) = 180. ^ Why ?
Therefore (m + n + r) = :5i;o . \Vliy?
7. The three interior angles of a triangle are equal. Find
the size of each interior and each exterior angle.
FIG. 133
8. Find the value of the interior and exterior angles in the
triangle of Fig. 133.
EQUATION APPLIED TO THE TRIANGLE 141
FICJ. 134
9. Show that the exterior angle x of the triangle AB(.'
in Fig. 134 is equal to the sum of the two nonadjacent interior
angles A and C.
10. Using Fig. 135, in which
BD II ,1 (7, prove that an ex-
terior (inrjle of a triangle is / -\ \x
('ijiinl f'> flif sifin of the two A
nonadjficent interior angles.
Note that two different methods are suggested by this figure.
11. Prove Ex. 10 by drawing a line through C parallel to AB.
HINT. Extend line A C.
12. Draw a quadrilateral. Tear off the corners and place
the interior angles
next to each other
by the method of
Ex. 2, Art. 180.
What does the sum
of the interior an-
gles seem to be ?
13. Draw a quadrilateral as in Fig. 136. Draw the diagonal
.4 C. This divides the quadrilateral into two triangles. What is
the sum of the interior angles
in each triangle ? What, then,
is the sum of the interior
angles of a quadrilateral ?
14. Draw a quadrilateral as
in Fig. 136. Produce each side
(one at each vertex). What
do you think is the sum of
the exterior angles of the
quadrilateral ? Check your estimate by measuring the angles.
15. Find the angles of a quadrilateral in which each angle
is 25 smaller than the consecutive angle.
FIG. 136
142 GENEEAL MATHEMATICS
16. Prove that the consecutive angles of a parallelogram
are supplementary ; that is, prove x + y = 180, in Fig. 137.
'17. Prove that the
opposite angles of a par- _Dl /<?___
allelogram are equal.
HINT. In Fig. 137 show
that Zx = Za = Zz.
18. If one angle of
a parallelogram is twice ' FIG. 137
as large as a consecutive
angle, what is the size of each angle in the parallelogram?
19. The difference between two consecutive angles of a
parallelogram is 30. Find the size of all four angles in the
parallelogram.
20. Show that the sum of the interior angles of a trapezoid
is two straight angles (180).
21. Prove that two pairs of
consecutive angles of a trape-
zoid are supplementary. (Use
Fig. 138.)
22. In Fig. 138, Z D is 40 Fl0 ' 138
more than Z. A, and /.B is 96 less than Z. C. Find the number
of degrees in each angle.
192. The construction of triangles. We shall now proceed
to study three constructions which require the putting
together of angles and line segments into some required
combination.. With a little practice the student will see
that the processes are even simpler than the thinking
involved in certain games for children which require the
various combinations of geometric forms.
These constructions are very important in all kinds of
construction work ; for example, in shop work, mechanical
EQUATION APPLIED TO THE TRIANGLE 143
drawing, engineering, and surveying. The student should
therefore master them.
193. Construction problem. To construct a triangle when
the three sides are given.
Construction. Let the given sides be a, b, and c, as shown in Fig. 139.
Draw a working line X Y, and lay off side c, lettering it AB. With
A as a center and witk a radius equal to b construct an arc as shown.
c
FIG. 139. How TO CONSTRUCT A TRIANGLE WHEN THREE SIDES
ARE GIVEN
With B as a center and with a radius equal to a construct an arc
intersecting the first. Call the point of intersection C. Then the
triangle is constructed as required.
EXERCISES
v 1. Construct triangles with the following sides :
- (a) a = 5 cm., b = 5 cm., c = 8 cm.
(b) a = 1 cm., b = 8 cm., c = 4 cm.
(c) a = 7 cm., b = 9 cm., c = 3 cm.
2. Is it always possible to construct a triangle when three
sides are given ? a
3. Construct a triangle,
using the sides given in | 1
Fig. 140.
. c
4. Compare as to size and
, , . , , FIG. 140
shape the triangle drawn by
you for Ex. 3 with those drawn by other pupils. (See if the
triangles will fold over each other.)
1 (JKNKKAL MATHEMATICS
5. Make a wn.id.-n triangle by nailing three sticks together.
Is it possible to change the shape of the triangle without break-
ing a stick or removing the corner nails '/
6. A great deal of practical use is made of the fact that
a triangle is a rigid figure ; for example, a rectangular wooden
gate is usually divided into two triangles by means of a wooden
diagonal so as to make the gate more stable (less apt to sag).
Try to give other examples of the practical use that is made
^of the stability of the triangular figure.
7. Construct an isosceles triangle having given the base
and one of the two equal sides.
HINT. Use c in Fig. 140 as the base and use // twice: that is, in
Uais case take a = />.
8. Measure the base angles of the isosceles triangle drawn
for Ex. 7. What appears to be the relation between the base
.angles of an isosceles triangle '.'
9. Make tracings of the base angles drawn for Ex 7 and
attempt to fold one angle over the other. Do the two angles
appear to represent the same amount of rotation ?
10. Compare your results witli those obtained by your
classmates.
NOTE. Results obtained from Exs. 7-10 support the following
theorem : Tin- buxe anglt-x of an isosceles triangle fire ef/>inl.
11. Construct an equilateral triangle having given a side.
12. Study the angles of an equilateral triangle by pairs in
the manner suggested by Exs. 8-9. State
the theorem discovered.
13. To measure the distance AB
(Fig. 141) we walk from B toward M
so that Z/J = 50, until we reach C, a /
point at which /.ACB = 50. What line
must we measure to obtain A B ? Why ? FIG. 141
EQUATION APPLIED TO THE T1UANULE 145
14. At a point C (Fig. 142), 70 ft. from the foot of a
pole AB, the angle ACB was found to be 45. How high is
the pole ?
*15. Walking along the bank of a river from A to C
(Fig. 143), a surveyor measures at B the angle ABD, and
walking 500 ft. further to C, he finds angle BCD to be one
half of angle A BD. What is the distance BD ?
D
C A
FIG. 142
FIG. 143
194. Construction problem. To construct a triangle, hav-
ing given two sides and the angle included between the
sides.
Construction. Let the given sides be a and b and the given angle
be Z C, as shown in Fig. 144.
B
FIG. 144. How TO CONSTRUCT A TRIANGLE WHEN Two SIDES AND THE
INCLUDED ANGLE ARE GIVEN
Draw a working line XY, and lay off CA equal to b. At C draw
an angle equal to angle C by the method of 78. With C as a
center and a radius equal to a lay off CB equal to a. Join B and A,
and the triangle is constructed as required.
146
GENERAL MATHEMATICS
EXERCISES
1. Is the construction given in Art. 194 always possible ?
2. Draw triangles with the following parts given :'
(a) a = 3 cm., b = 4 cm., Z C = 47.
(b) c = lin., 6 = 2i
(c) J = ljin., c = lf
3. Construct a triangle with the parts given in Fig. 145.
Z.I = 112.
= 87.
FIG. 145
4. Compare as to size and shape the triangle drawn for Ex. 3
with those drawn by other students in your class. (Place one
triangle over the other and see if they fit.)
195. Construction problem. To construct a triangle when
two angles and the side included between them are given.
FIG. 146. How TO CONSTRUCT A TRIANGLE WHEN Two ANGLES AND THE
SlDE INCLUDED BETWEEN THEM ARE GIVEN
Construction. Let ZA and ZB be the given angles and line c be
the given included side (Fig. 146).
EQUATION APPLIED TO THE TRIANGLE 147
Lay down a working line A'Fand lay off AB equal to linec on^it.
At A constrict an angle equal to the given angled; at B construct
an angle equal to the given angle B and produce the sides of those
angles till they meet at C, as shown. Then the t\ABC is the
required triangle.
EXERCISES
1. Draw triangles with the following parts given :
(a) Z A = 30, Z B = 80, e = 4 cm.
(b) Z C = 110, Z = 20, a = 2 in.
2. Draw a triangle with the parts as given in Fig. 147.
3. Is the construction of Ex. 2 always possible ?
FIG. 147
4. Compare as to size and shape the triangle drawn for Ex. 2
with those drawn by other members of your class. (Fold them
over each other and see if they fit.)
SUMMARY
196. This chapter has taught the meaning of the follow-
ing words and phrases : right triangle, cardboard triangle,
wooden triangle, set square, isosceles triangle, scalene
triangle, interior angles of a triangle, exterior angles of
a triangle, base angles.
The following notations have been given : notation
for the angles and sides of triangles, notation for right
triangle (rt. A).
148
197. This chapter has presented methods of
1. Finding the sum of the interior angles of a triangle.
Finding the sum of the exterior angles of a triangle.
3. Drawing right triangles by means of wooden or card-
board triangles.
4. Drawing parallel lines by means of the wooden tri-
angle or the set square.
198. This chapter has taught the pupil the following
constructions :
1. Given three sides of a triangle, to construct the triangle.
2. Griven two sides and the included angle of a triangle,
to construct the triangle.
3. Given two angles and the included side of a triangle,
to construct the triangle.
199. The following theorems have been presented in
this chapter:
1. The sum of the interior angles of a triangle is a
straight angle (180).
2. The sum of the exterior angles of a triangle is two
straight angles (360).
3. If two angles of one triangle are equal respectively
to two angles of another triangle, the third angle of the
first triangle is equal to the third angle of the second.
4. The acute angles of a right triangle are comple-
mentary.
5. In a right triangle whose acute angles are 30 and
60 the side opposite the 30-degree angle is one half the
hypotenuse.
6. An exterior angle of a triangle is equal to the sum
of the two nonadjacent interior angles.
7. The sum of the interior angles of a quadrilateral is
four right angles (360).
8. The sum of the exterior angles of a quadrilateral
is four right angles (360).
9. The opposite angles of a parallelogram are equal.
10. The consecutive angles of a parallelogram are
supplementary.
11. Two pairs of consecutive angles of a trapezoid are
supplementary .
12. The base angles of an isosceles triangle are equal.
13. An equilateral triangle is equiangular (all angles
equal).
CHAPTER VIII
POSITIVE AND NEGATIVE NUMBERS. ADDITION AND
SUBTRACTION
200. Clock game. Mary and Edith were playing with a
toy clock. Each took her turn at spinning the hand. The
object of the game was to guess the number on which the
hand of the clock stopped. A correct guess counted five
points. If a player missed a guess by more than three,
she lost three points. If she came within three she either
won or lo'st the number of points missed, according to
whether she had guessed under or over the correct number.
After five guesses they had the following scores :
MARY Solution. The score as kept by the
Won 2 players (the words are inserted) appeared
Lost 1 as follows :
Lost 3
Won 2
Won 1
EDITH
Lost 1
Lost 2
Won 5
Won 2
Lost 2
Who won the
game? Edith won the game, 2 to 1.
150
MARY
EDITH
First score
First score
CD
Lost
1
Lost
2
Second score
1
Second score
(D
Lost
3
Won
5
Third score
d)
Third score
2
Won
2
Won
2
Fourth score
Fourth score
4
Won
1
Lost
2
Final score
1
Final score
2
POSITIVE AND NEGATIVE NUMBERS 151
201. Positive and negative numbers. Algebraic numbers.
The adding of scores in many familiar games like the one
cited above illustrates an extension of our idea of count-
ing that will be found very useful in our further study of
mathematics. It is important to notice that the players
began at zero and counted their scores in both directions.
Thus, Mary began with zero and won two points, and the
2 she wrote meant to her that her score was two above
zero. On the other hand, Edith had zero and lost one.
In order to remember that she was " one in the hole "
she wrote 1 within a circle. On the next turn she lost
two more, and she continued to count away from zero
by writing 3 within a circle. The same idea is shown in
Mary's score. Her second score was 1. On the next turn
she lost three. She subtracted 3 from 1. In doing so she
counted backward over the zero point to two less than zero.
In writing the scores it was necessary to indicate whether
the number was above or below zero.
We shall presently have numerous problems which
involve pairs of numbers which possess opposite qualities,
like those above. It is generally agreed to call num-
bers greater than zero positive and those less than zero
negative. Such numbers are called algebraic numbers.
The opposite qualities involved are designated by the
words " positive " and " negative." In the preceding game,
numbers above zero are positive, whereas numbers below
zero are negative.
202. Use of signs. To designate whether a number is
positive or negative we use the plus or the minus sign.
Thus, 4- 4 means a positive 4 and 4 means a negative 4.
The positive sign is not always written. When no sign pre-
cedes a number the number is understood to be a positive
number. Thus, 3 means + 3.
152
The following stock quotations from the Chicago Daily
Tribune (March 24, 1917) illustrate a use of the plus and
minus signs :
CHICAGO STOCK EXCHANGE
SALES
HIGH
Low
CLOSE
NET
American Radiator . . .
20
295J
295
296
-2
Commonwealth Edison . .
79
136i
136 J
136 J:
-1
Diamond Match
40
81
81
81
Swift & Co
515
114^
142f
1441
+ ~'-
Peoples Gas
397
95"
90^
91
- 5
Prest-o-Lite
40
1292
129
129
-3
Sears Roebuck
325
192
190*
19H
~ <>
Booth Fisheries
20
79
79
79
+ T
The last column shows the net gain or loss during the
day; for example, American Radiator stock closed two
points lower than on the preceding day, Swift & Co.
gained 2|, Peoples Gas lost 5, etc. The man familiar
with stock markets glances at the first column to see
the extent of the sales and at the last column to see
the specific gain or loss. To check the last column one
would need the quotations for the preceding day.
In order that we may see something of the importance
of the extension of our number system by the preceding
definitions, familiar examples of positive and negative
numbers will be discussed.
203. Geometric representation of positive numbers. Origin.
We have learned in measuring a line segment that when a
unit (say^[_^) is contained five times 'in another segment,
the latter is five units long. (In general, if it is contained
a times, the measured segment is a units long.) We may
also say that the two segments represent the numbers 1 and
o respectively. This suggests the following representation
POSITIVE AND NEGATIVE NUMBERS 153
of positive integers : On any line OX (Fig. 148) beginning
at (called the origin of the scale) but unlimited in the
direction toward A', we lay off line segments OA, AB, BC,
CD, DE, etc., each one unit in length. We thus obtain a
series of points on the o A B c D E F
line, each of which I 1 1 \ 1 1 1 1 1 x
corresponds to some 012345678
positive integer, and FlG - 148 - GEOMETRIC REPRESENTATION OF
T , r , POSITIVE NUMBERS
vice versa. We also
note that the line segment which connects the origin with
some point (say JE") is the corresponding number of units
in length. (Thus, OE = 5.) Hence the positive integers
+ 1, -f 2, -f 3, + 4, 4- 5, etc. are represented by OA, OB,
OC, OD, OE, etc.
204. Geometric representation of negative numbers. Num-
ber scale. If we prolong OX to the left from in the
direction OX' (Fig. 149), we may lay off line segments
in either of two opposite directions. Segments OC and
OD' differ not only hi length but also in direction. This
difference in direction is indicated by the use of the
D' C' B 1 A' A B C D
X r 1 1 1 1 1 1 1 1 1-
-4-3-2-1 1 2 3 4
FIG. 149. GEOMETRIC REPRESENTATION OF NEGATIVE NUMBERS
signs + and . Thus, OC represents + 3, whereas OD' rep-
resents 4. The integers of algebra may now be arranged
in a series on this line called the number scale, beginning
at and extending both to the right and to the left.
In order to locate the points of the number scale (Fig. 149)
we need not only an integer of arithmetic to determine
how far the given point is from zero but also a sign of
quality to indicate on which side of it is found.
154 GENEKAL MATHEMATICS
EXERCISES
1. Locate the following points on the number scale : + 3,
-3, -7, + 12, -8, +6.
2. If we imagine that each of the players in the clock
game (Art. 200) has a string or tape measure graduated to
the number scale, with a ring on it, they could add their
scores from time to time by sliding the ring back and forth.
Indicate on the number scale how the sum of the following
consecutive scores could be found : Began at 0, lost 1, lost 2,
won 5, won 2, lost 1.
3. Starting from the middle of the field in a certain football
game, the ball shifted its position in yards during the first
fifteen minutes of play as follows: .+ 45, 15, +11, 10,
2, + 23. Where was the ball at the end of the last play ?
205. Addition and further use of positive and negative
numbers. The preceding exercises show that positive and
negative numbers may be added by counting, the direc-
tion (forward or backward) in which we count being
determined by the sign (+ or ) of the numbers which
we are adding. Thus,
To add + 4 to + 5 on the number scale of Fig. 149, begin at + 5
and count 4 to the right.
To add 4 to + 5 begin at + 5 and count 4 to the left.
To add 4 to 5 begin at 5 and count 4 to the left.
To add + 4 to 5 begin at 5 and count 4 to the right.
The results are as follows :
+ 5 +(+4) = + 9; +5+(-4) = + l; -5 + (-4) = -9;
+ 5 + ( 4) = -f 1 is read "positive 5 plus negative 4 equals
positive 1."
5 +(4) = 9 is read "negative 5 plus negative 4 equals
negative 9."
7
POSITIVE AND NEGATIVE NUMBERS 155
EXERCISES
1. Give the sum in each of the following. Be prepared to
interpret the result on the number scale.
- (a) 3 +(+2). (g) 6+(-l). - (m) -2+(-5).
(b) 4 +(- 3). (h) 6 +(- 3). ^(n) - 4 + 6.
2. On a horizontal straight line, as X'OX in Fig. 150, con-
sider the part OX as positive, and the part OX' as negative.
Construct line segments correspond-
ing to the following numbers : 2,
+ 6, - 3, - 5, 0.
3. Consider OF as positive and OY' x r
as negative and construct segments on
the line YOY' corresponding to 4,
- 2, + 3, + 4, - 3, 0.
4. A bicyclist starts from a certain
point and rides 18 mi. due northward -p IG JCQ
(4- 18 mi.), then 12 mi. due south-
ward ( 12 mi.). How far is he from the starting point?
5. How far and in what direction from the starting point is
a traveler after going eastward (+) or westward ( ) as shown
by these pairs of numbers : + 16 mi., then 3 mi.? 2 mi., then
-f- 27 mi.? 16 mi., then + 16 mi.? + 100 mi., then 4- 52 mi.?
6. Denoting latitude north of the equator by the plus sign
and latitude south by the minus sign, give the meaning of the
following latitudes : 4- 28, + 12, - 18, + 22, - 11.
7. Would it be definite to say that longitude east of Green-
wich is positive and west is negative ? What is the meaning
of the following longitudes': 4- 42? + 142? - 75? - 3 ?
156 GENERAL MATHEMATICS
8. A vessel starting in latitude + 20 sails + 17, then G3,
then + 42, then 1G. What is its latitude after all the
sailings ?
9. What) is the latitude of a ship starting in latitude 53
after the following changes of latitude : + 12, - 15, + 28,
- 7, + 18, - 22, + 61 ?
206. Double meaning of plus and minus signs. Absolute
value. The many illustrations of negative number show
that there is a real need in the actual conditions of life
for the extension of our number scale so as to include
this kind of number. It must be clear-ly understood that
a minus sign may now mean two entirely different things :
(1) It may mean the process of subtraction, or (2) it may
mean that the number is a negative number. In the latter
case it denotes quality. Show that a similar statement is
true for the plus sign.
This double meaning does not often confuse us, since it is
usually possible to decide from the context of the sentence
which of these two meanings is intended. Sometimes a
parenthesis is used to help make the meaning clear, thus
4 + ( 3) means add a negative 3 to a positive 4.
Sometimes we wish to focus attention merely on the
number of units in a member regardless of sign. In that
case we speak of the absolute value, or numerical value.
Thus, the absolute value of either -f 4 or 4 is 4.
207. Forces. In mechanics we speak of forces acting
in opposite directions as positive and negative. Thus, a
force acting upward is positive, one acting downward is
negative.
POSITIVE AND NEGATIVE NUMBERS 157
EXERCISES
1. Three boys are pulling a load on a sled, one with a force
of 27 lb., another with a force of 56 lb., and the third with a
force of 90 lb. With what force is the load
being pulled ?
2. Two small boys are pulling a small
wagon along; one pulls with a force of
23 lb., and the other pulls with a force of
36 lb. A boy comes up behind and pulls
with a force of 47 lb. in the opposite direc-
tion from the others. What is the result ?
FIG. 151
3. An aeroplane that
can fly 48.3 rni. an hour
in still air is flying
against a wind that
retards it 19.6 mi. an hour. At what rate
does the aeroplane fly ?
4. A balloon which exerts an upward
pull of 512 lb. has a 453-pound weight
attached to it. What is the net upward
or downward pull ?
5. A toy balloon (Fig. 151) tends to pull
upward with a force of 8 oz. What happens
if we tie a 5-ounce weight to it ? an 8-ounce ?
6. A boy can row a boat at the rate of
4 mi. per hour. How fast can he go up
a river flowing at the rate of 2^ mi. per
hour ? How fast can he ride down the
river ? How fast could he go up a stream
flowing 5 mi. per hour ?
208. The thermometer. The ther-
mometer (Fig. 152) illustrates the idea
of positive and negative numbers in two
Z12-
100-
70-
32-
FIG. 152. THE TIIKK-
SIOMETER ILLUSTRATES
THE IDEA OF POSI-
TIVE AND NEGATIVE
NUMBEBS
158 GENERAL MATHEMATICS
ways. In the first place, the number scale is actually pro-
duced through the zero, and degrees of temperature are
read as positive (above zero) or negative (below zero).
In the second place, the thermometer illustrates positive
and negative motion discussed in the preceding article.
Thus, when the mercury column rises, its change may be
considered as positive, + 5 indicating in this case not a
reading on the thermometer, as before, but a change (rise)
in the reading ; similarly, 3 in this sense indicates a
drop of 3 in the temperature from the previous reading.
EXERCISES
1. What is the lowest temperature you have ever seen
recorded ?
2. The top of the mercury column of a thermometer stands
at at the beginning of an hour. The next hour it rises 5,
and the next 3; what does the thermometer then read ?
3. If the mercury stands at 0, rises 8, and then falls 5,
what does the thermometer read ?
4. Give the final reading in each case :
A first reading of 10 followed by a rise of 2.
A first reading of 10 followed by a fall of 12.
A first reading of 20 followed by a fall of 18.
A first reading of x followed by a rise of y.
A first reading of x followed by a fall of y.
A first reading of a followed by a rise of a.
A first reading of a followed by a fall of a.
A first reading of a followed by a fall of a.
5. The reading at 6 P.M. was 7. What was the final read-
ing if the following numbers express the hourly rise or
fall : + 2, + 1, 0, - 3, - 3, - 2, - 2, - 1, - 1, - 3,
1 4- 2 ?
POSITIVE AND NEGATIVE NUMBERS 159
6. Add the following changes to find the final reading, the
first reading being : + 3, + 2, - 4, - 3, - 2, -f 3.
7. The differences in readings of a thermometer that was
read hourly from 5 A. M. until 5 P. M. were as follows : 3, 4,
7, 10, 12, 9, 8, 5, 0, -22, -17, -12. How did
the temperature at 5 P.M. compare with 'that at 5 A.M.?
If the. temperature at 5 A.M. was -f- 20, make a table showing
the temperature at each hour of the day.
209. Positive and negative angles. By rotating line AB
in a plane around A until it takes the position AC the
angle BAG is formed (Fig. 153).
By rotating AB in the opposite direc-
tion angle BAC is formed. To dis-
tinguish between these directions
one angle may be denoted by the
plus sign, and the other by the FlG 153
minus sign. We agree to consider
an angle positive when it is formed by rotating a line
counterclockwise and negative when it is formed by clock-
wise rotation. This is simply another illustration of motion
in opposite directions.
EXERCISES
1. In this exercise the sign indicates the direction of rotation.
Construct the following angles with ruler and protractor, start-
ing with the initial line in the horizontal position : -f 30, -(- 45,
+ 90, +43, +212, -30, -45, -90, -53, -182, -36.
2. Find the final position of a line which, starting at OX
(horizontal), swings successively through the following rota-
tions : + 72, - 38, + 112, - 213, + 336, - 318, - 20,
+ 228.
3. Do you see a short cut in finding the final position of
the line in Ex. 2 ?
GENERAL .MATHEMATICS
210. Business relations. Finally, the idea of positive
and negative numbers may be further illustrated by the
gain or loss in a transaction ; by income and expenditure ;
by a debit and a credit account ; by money deposited and
money checked out; and by the assets and liabilities of
a business corporation. Thus, a bankrupt company is one
which has not been able to prevent the negative side of the
ledger from running up beyond the limit of the confidence
of its supporters.
EXERCISES
1. The assets of a company are $ 26,460, and its liabilities
are $39,290. What is its financial condition ?
2. A newsboy having $25 in the bank deposits $10.25 on
Monday, checks out $16.43 on Tuesday, checks out $7.12 on
Wednesday, deposits $5 on Thursday, deposits $7.25 on Friday,
and checks out $11.29 on Saturday. What is his balance for
the week ?
3. If a man's personal property is worth $1100 and his real
estate $12,460, and if his debts amount to $2765, what is his
financial standing ?
4. A boy buys a bicycle for $10.25 and sells it for $6. Does
he gain or lose and how much ?
211. Addition of three or more monomials. The following
exercises will help us to see how the addition of monomials
may be extended.
EXERCISES
1. Add the following monomials :
(a) 2 + 3 + (- 4) + (5). (c) (- 4) + 2 + 3 + (- 5).
(b) 3 + 2'+ (- 4) + (- 5). (d) (- 5) + (- 4) + 3 + 2.
2. In what form has the commutative law of addition been
stated ? (Art. 36.) Does it seem to hold when some of the
addends are negative?
POSITIVE AND NEGATIVE NUMBERS 161
3. Show by a geometric construction on squared paper that
the commutative law holds when some of the addends are
negative. *
HINT. First take a line segment a units long (Fig. 154) and
add a segment + b units long (Fig. 154) ; then take the segment
that is + b units long and add the segment Q
that is a units long. The results should L^ ~ a I
be the same.
4. Does a bookkeeper balance the ac- r~~ IfT^
count every time an entry is made or does . -..
he keep the debits and credits on separate
pages and balance the two sums at the end of the month '.'
The process of adding several positive and negative num-
bers can be explained in detail by the following problem :
Add - 10, + 50, - 27, + 18, - 22, - 31, + 12.
Arrange the numb^s as follows : 10
+ 50
-27
+ 18
-22
-31
+ 12
There are three positive terms, 50, 18, and 12, whose sum is 80.
There are four negative terms, 10, 27, 22, and 31, whose
sum is - 90. Adding + 80 to - 90 gives - 10.
The process consists of the following three steps :
1. Add all the positive terms.
2. Add all the negative terms.
3. Add these two sums by the following process : (a) De-
termine how much larger the absolute value of one number is
than the absolute value of the other, (b) Write this number
and prefix the sign of the greater addend.
ltli> GENERAL MATHEMATICS
212. Algebraic addition. The results of the preceding
exercises show that positive and negative numbers may
be added according to the f ollo\f1ng laws :
1. To add two algebraic numbers having like signs find
the sum of their absolute 'values and prefix to this sum their
common sign.
2. To add two algebraic numbers having unlike signs find
the difference of their absolute values and prefix to it the sign
of the number having the greater absolute value.
EXERCISES
1. Show that the sum of two numbers with like signs is the
sum of their absolute values with the common sign prefixed.
Illustrate with a concrete experience.
2. Show that the sum of two numbers having unlike signs
but the same absolute value is zero. Illustrate with some fact
from actual experience.
3. Find the following sums, performing all you can orally :
(a) -5 (d)-7 (g) -f -(j)-17f*
1 - +j +261*
H
(b) +5 (e) -Sa (h) -fai (k) + 62z 2
-28f ae 8
<<0 + 7
(f)
-ftjw
(i)
3.16 x
(1)
-2.3a;m 2
1 +5
16m
-5.28 a;
+ 6.5 xm*
4- 3-
Find the following sums
4- -- 6 5. -f 3 6. +51 7. -242
+ 10 +10 +23 +726
8 -7 -18 58
4 - 4 -7 +24
POSITIVE AND NEGATIVE NUMBERS 163
8. + 7.5
+ 12.5
- 9.5
+ 2.5
9. -- Sx
+ 4z
+ 17 a;
10. + 7x
-lOx
-I2x
12.
+ t
16. + 7|
20. 0.5 x 2
+ |
7
0.23 x*
i
2
-i2i
0.12 a; 2
- 2J
0.07 x 2
13.
T|
17. - 12.18 -
21. +27 a; 2
+ 1
- 11.88
15 ar*
_ 3
+ 13.16
+ 17x 2
+ 1
-14.08-
-12 a: 2
14.
+ 3|
18. +10.05
22. +23 am
2 <s
+ 4.85
14 xm
5J
- 3.25
17 xm
2 I
- 12.35
+ 20 asm
15.
+ 4f
19. -3.1s
23. - 18.25 arm
6 1
-5.4s
+ 17.34 aftw
-6f
+ 7.2s
- 19.64 x 2 m
4 8f
-3.1s
+ 21.17 a; 2 m
11. +24 a
6a
- 7a
3a
213. Drill exercises. The following exercises constitute
a drill in determining the common factor of similar mono-
mials and applying the law for the addition of similar
monomials (Art. 40). We need to recall that the sum of
two or more similar monomials is a number whose coefficient
is the sum of the coefficients of the addends and whose literal
factor is the same as the similar addends. The exercises
are the same as a preceding set (Art. 40) except that in
this case the step in which the coefficients of the addends
are added involves the addition of positive and negative
numbers.
164 GENERAL MATHEMATICS
EXERCISES
In each case (1) point out with respect to what factor the
following terms are similar ; (2) express as a monomial by add-
ing like terms :
1. 3y, -6y, 20y, - 35y.
Solution. The common factor is y.
The sum of the coefficients is 3 + (- 6) + 20 + (- 35) = - 18.
Whence the required sum is 18 y.
2. 5x,7x,9x,+12x,3x.
3. 7 b, - 12 b, - 9 b, + 11 b, - 13 b.
4. 9 ab, - 17 ab, - 11 ab, + 13 06.
5. 8 mnx*, + 12 wmx 2 , 15 wwia 2 , 13 mnx*.
6. 5 a 2 *, - 7 o%, + 9 a 2 *, - 5 ^b.
7. 3 ax, 4 ox, 8 az, 7 or.
8. - 5/,r/, lpq\ - &spq*, ~ Bpf.
9. az, - 14 z, -bz, +12 z.
Solution. The common factor is :.
The sum of the coefficients is a + (- 14) + (-&) + 12.
Since a and fe are still undetermined, we can only indicate that
sum thus : a 14 b + 12.
Whence the sum written as a monomial is (a b 2) z.
10. mx, + 5 x, 7x, + ex.
11. m/, - wf, + 5 y 2 , - 12 y 8 , + c/.
12. tyab, + 4|aJ, 5^ ab, - 6| aft.
214. Addition of polynomials. We have had numerous
examples of the addition of polynomials in dealing with per-
imeters. In applying the principles involved to polynomials
having positive and negative terms we need to recall that
in addition the terms may be arranged or grouped in any
order. Thus,
2 + 3 + 4 = 3 + 2 + 4 (Commutative Law)
5 + (- 3) + 4 = - 3 + (5 + 4) (Associative Law)
POSITIVE AND NEGATIVE NUMBERS 165
In adding polynomials it is convenient to group similar
terms in the same column, much as we do in adding
denominate numbers in arithmetic.
EXERCISES
1. Add the following polynomials and reduce the sum to its
simplest form : 3 yd. + 1 ft. + 6 in., 5 yd. + 1 ft. + 2 in., and
12 yd. + 3 in.
Solution. Writing the similar terms in separate columns we have
3 yd. 1 ft. 6 in.
5 yd. 1 ft. 2 in.
12yd. 3 in.
20yd. 2fE 11 in.
Note that the common mathematical factors are not yards, feet,
inches f but the unit common to all of them, or inches. The problem
may be written as follows :
3 x 36 + 1 x 12 + 6
r> x 36 + 1 x 12 + 2
12 x 36 _ + 3
20 x 36 + 2 x 12 + 11
2. Add 9 tj + 3 x + 2 /, o // + 2 x + (> I, and 3 y + 2 x + 8 i.
9 y + 3 x + 2 i
5 y + 2 x + 6 (
3 y + 2 x + 81
17 y + Tar 4- 16 i
3. Add 27 x s 13xy+ 1C if', - 14 a- 8 + 25 xy + 4 ? /, and
Solution. Writing similar terms in separate columns and adding,
we have
27 X s - 13 xy + 16 y 2
- 14 a- 8 + 25 xy + 4 ?/ 2
- 6 xz/ + 8 y 2
166 GENERAL MATHEMATICS
In the following exercises add the polynomials :
+ 2b + "c 7. ~6x + 15y-l6z- Sw
-9/; -2c -Sx + 22y + l6z-12w
+ 3b 5c 3x+ -ly 5z+ 6w
2b 3c 9x- 8+ 7z 6w
5. 2x + 3y + 4 8. 6r-2s + 3t
5x 6y7z 5r 3t
4z + 5y + 8s 2r+6s-5t
-2x-5y-2z 3s + t
6. 2a + 5b+7c \9\ 2x 3y + z
3a Sb 5c x+ y 3z
10. 2x 6b + 13c, llaj + 19ft 30 c, and 5x +4c.
11. 12A;-10Z + 9m, -27c + 2^-4m, and -
12. 14 e 3 y + 1 z, 5 e + 5 y 3 z, and 6e + 4:y + 2z.
13. 24z-
NOTE. Here certain terms are inclosed in grouping symbols ( )
called parenthesis. These indicate that the terms within are to be
treated as one number or one quantity. Other grouping symbols will
be given when needed (see pp. 175, 177).
14. (6*-
15. 3
215. Degree of a number. The degree of a number is
indicated by the exponent of the number. Thus, x 2 is of
the second degree ; a- 3 , of the third degree ; y\ of the fourth
degree ; etc. The monomial 3 xyh* is of the first degree
with respect to x, of the second degree with respect to y,
and of the third degree with respect to r.
POSITIVE AND NEGATIVE NUMBERS 167
216. Degree of a monomial. The degree of a monomial
is determined by the sum of the exponents of the literal
factors. Another way of saying this is : The number of
literal factors in a term is called the degree of the term.
Thus, 3? is of the second degree ; xy 2 , of the third degree ;
and 4:ry 2 2 2 , of the fifth degree.
EXERCISES
Determine the degree of the following monomials :
1. 2ajy. 3. 3 ft 4 . 5. 2mxy. 7. ^- 9. rV.
3.2 z
2. 2ab s . 4. 5x7/V. 6. 8. rs l . 10. m*x*ifz\
m
217. Degree of a polynomial. The degree of a polynomial
is determined by the degree of the term having the highest
degree. Thus, x^y* + x+%y + 5 is of the fourth degree,
and 5o^ a^ + 7isa third-degree expression.
EXERCISES
Indicate the degree of the following polynomials :
1. x* + 2x 3 - x 4 + 2x* + y + 7. 4. x - 2xif+ if.
2. x + 2xij + if. ' 5. a; 4 + 4.
3. y?1xy->ry i . 6. x 5 + x 3 + x* + 1.
218. Arrangement. A polynomial is said to be arranged
according to the descending powers of x when the term of
the highest degree in x is placed first, the term of next
lower degree next, etc., and the term not containing x
last. Thus, 2 + x-\-x 8 -\-^3^ when arranged according to
the descending powers of x takes the form z 3 +3z 2 + #+2.
When arranged in the order 2 + a; + 3 2 + 3 , the polynomial
is said to be arranged according to the ascending powers of x.
168 GENERAL MATHEMATICS
Find the sura of -3a 2 +2a 3 -4a, -a 2 + 7, 5a 3 -4a
and -2 3 -7-2a 2 .
Arranging according to descending powers and adding, we have
2 a 8 - 3 a 2 - 4 a
a 2 +7
5 a 8 - 4 a + 3
- 2 a 8 - 2 a 2 - 7
5a 8 -6a 2 -8a + 3
Check. One way to check is to add carefully in reverse order, as
in arithmetic.
A second method for checking is shown by the following :
Let a = 2. Then we have
2 a 3 3 a 2 4 a = 4
a 2 + 7 = 3
5 a 8 - 4 a + 3 = 35
- 2 a 3 - 2 a 2 - 7 = - 31
5a 3 -6a 2 -8a + 3= 3
The example checks, for we obtained 3 by substituting
2 for a in the sum and also by adding the numbers obtained
by substituting 2 for a separately in the addends.
EXERCISES
In the following list arrange the polynomials in columns
either according to the ascending or the descending order of
some one literal factor. Add and check as in the preceding
problem.
1. x 2 + if + xy, x 2 -xy + y*, 14 xy + 3x*-2y' 2 .
2. 26xi/, -5y* + 12x*, - IQxy - ISxy + 16?f.
3. 5.3 x 2 - 13.6 xy - 2.3 f, - 0.02 f + 5 xy + 3.2 x 2 .
4. x 8 -Sx 2 -5x-12, 3x* + 3x-5x* + 8.
5. 8 a 3 - 2 a 2 + 3 a - 6, - 3 8 + 2 a 2 + a + 7.
6. 3 ?- 2 + 2 r 8 + 3 r - 5, - r* - 2 r + r s + 1, r 2 - 2.
7. f s 2 - i - r 2 , - s * _ rs + ^ s f -5r*.
POSITIVE AND NEGATIVE NUMBERS 169
8. - 21 oty 4- 40 a? + 55 * 2 ?/ - 14 ^,
+ 2 cx 2 z + 58 x 2 - 23 y 8 *,
9. 4 (m + TO) - 6 (m 2 + w 2 ) + 7 (m 3 + n 3 ) - 8(m 4 + n 4 ),
9 (ra 4 + w 4 ) - 3 (m 8 + w 8 ) + 4 O 2 + re 3 ) - 3 (m + ).
10. 3 (a + i) - 5 (a 2 + &) + 7 (a - i 2 ) - 5 (a 2 + a& + 6 2 ),
6 (a + l>) - 3 (a - 6 2 ) + 4 (a 2 + V) + 2 (a 2 + ab + i 2 ).
219. Subtraction. The following exercises will help to
make clear the process of subtraction.
EXERCISES
1. If the thermometer registers + 24 in the morning and
-f- 29 in the evening, how much warmer is it in the evening
than in the morning ? How do you find the result ?
2. A thermometer registers + 10 one hour and + 14 the
next hour. What is the difference between these readings ?
3. If the thermometer registers 2 one hour and + 3
the following hour, how much greater is the second read-
ing than the first ? The question might be stated : What is
the difference between 2 and + 3 ? In what other form
could you state this question?
4. If you were born in 1903, how old were you in 1915?
State the rule you use in finding your age.
5. The year of Christ's birth has been chosen as zero of
the time scale in Christian countries. Thus, we record a
historic event as 60 B.C. or A. D. 14. Instead of using B.C. and
A.D. we may write these numbers with the minus and plus
signs prefixed. How old was Caesar when he died if he was
born in 60 and died in -f- 14 ?
6. A boy was born in 2. How old is he in + 5 ? Apply
your rule for subtraction in this case.
7. Subtract 3 from + 5 ; 4 from + 5 ; 5 from + 5 ;
- 50 from 4- 25.
170 GENERAL MATHEMATICS
8. A newsboy has 410. How much must he earn during
the day so as to have 850 in the evening? State the rule
which you use to solve problems of this kind.
9. A newsboy owes three other newsboys a total of 650.
How much must he earn to pay his debts and have 200 left?
Apply your rule for solving Ex. 8 to this problem.
10. John is $25 in debt, Henry has #40 in cash. How much
better off is Henry than John ? Apply the rule stated for Ex. 8.
11. What is the difference between 12 and 20? 5 and 45 ?
and 20 ? - 1 and 20 ? - 5 and 25 ? - 12 and 18 ? .
12. Interpret each, of the parts of Ex. 11 as a verbal problem.
13. Through how many degrees must the line OI^ turn
(Fig. 155) to reach the position OR 2 ?
14. Archimedes, a great mathema-
tician, was born about the year 287
and was slain by a Roman soldier
in 212 while studying a geometrical
figure that he had drawn in the sand. FIG. 155
How old was he ?
15. Livy, a famous Roman historian, was born in 59 and
lived to be 76 yr. old. In what year did he die ?
16. Herodotus, the Greek historian, sometimes called the
Father of History, was born in 484 and died in 424. At
what age did he die ?
220. Subtraction illustrated by the number scale. In
subtracting 4 from 6 we find what number must be added
-8 -7 -6 -5 -4 -3 -2 -1 +1 4-2 +3 4-4 -H5 4-6 +7 4-8
< i I I I I I I I I I I I I i i i i >
FIG. 156. THE NUMBER SCALE
to the 4 (the subtrahend) to get the 6 (the minuend).
On the number scale (Fig. 156) how many spaces (begin-
ning at 4) must we count until we arrive at 6 ?
POSITIVE AND NEGATIVE NUMBEKS 171
ILLUSTRATIVE EXAMPLES
1. Subtract 2 from 3.
Solution. Beginning at 2 we need to count 5 spaces to the
right (positive) to arrive at 3. Hence, subtracting 2 from 3 equals 5.
Note that we might have obtained the result by adding + 2 to 3.
2. Subtract + 5 from 2.
Solution. Beginning at 5 on the number scale we need to count 7
to the left (negative) to arrive at 2. Hence, subtracting + 5 from
2 equals 7. Note that we could have obtained the same result
by adding 5 to 2.
This exercise may be stated as a temperature problem ; namely,
What is the difference between 5 above zero and 2 below zero ?
3. Subtract 8 from 2. Interpret as a verbal problem.
Solution. Beginning at 8 on the number scale we need to count
6 to the right (positive). Hence, subtracting 8 from 2 equals + 6.
Notice that the same result is obtained if + 8 is added to 2.
These examples show that since subtraction is the reverse
of addition, we can subtract a number by adding its opposite.
Thus, adding $100 to the unnecessary expenses of a firm
is precisely the same as subtracting $100 gain, or, on
the other hand, eliminating (subtracting) $1000 of lost
motion in an industrial enterprise is adding $1000 to the
net gain.
It is convenient for us to make use of this relation, for
by its use there will be no new rules to learn, but merely
an automatic change of sign when we come to a subtraction
problem, and a continuation of the process of addition.
221. Algebraic subtraction. The preceding discussion
shows that subtraction of algebraic numbers may be changed
into algebraic addition by the following law: To subtract
one number from another change the sign of the subtrahend
and add the result to the minuend.
172 GENERAL MATHEMATICS
Thus, the subtraction example
+ 7a
- 3a
+ 10 a
may be changed to the addition example
+ la
+ 3a
+ 10 a
EXERCISES
Subtract the lower number from the upper number in the
following. Illustrate Exs. 1-11 with verbal problems.
1.
29
9. -14
17. +x
25.
+ 0.81 cc 2
-10
- 3
x
f
-2.62x 2
2.
-55
10. +14
18. x
26.
-.660
-15
+ 3
+ #
-.840
3.
-65
11- -T 9 o
19. +x
27.
+ 0.82 r
+ 15
. 2
X
+ 2.41 r
4.
+ 18
12. -5 a- 2
20. +5.74 a; 8
28.
- 3.34 a
+ 24
+ a-*
-6.26 a- 8
+ 5.37 a
5.
A
13. -126 8
21 -3.15 a 2
29.
+ 2.04y
T2
7b 3
.34 a 2
-4.23y
6.
3
"ff
14. - 9r
22. +6(a + &)
30.
+ 8.92 a;
~ T
- 14 /
8(a + J)
+ 9.17 -jr
7.
3
15. + 7 2
23. _5(a-6)
31.
- 7.42 z
14
-14 a 2
3 (a ft)
- 3.71
8.
14
16. -7m 8
24. -4.36*
32.
-2.417/
- 3
+ 3m 8
8.64*
+ 8.62?/
POSITIVE AND NEGATIVE NUMBERS 173
TRANSLATION INTO VERBAL PROBLEMS
33. Translate each of the following subtraction exercises
into a verbal problem, using the suggestion given :
+ 8246
(a) As assets and liabilities :
+ 5
(b) As gain or loss :
I
(c) As debit or credit :
48
(d) As an angle problem :
-27
14
+ 22
(e) As an age problem (time) :
(f ) As line segments on the number scale :
246
(g) As a bank account :
40
(h) As a latitude problem :
-(- Zo
i go
(i) As a longitude problem:
~l~ 75
i 2
(j) As a problem involving forces :
222. Subtraction of polynomials. When the subtrahend
consists of more than one term the subtraction may be
performed by subtracting each term of the subtrahend
from the corresponding term of the minuend.
For example, when we wish to subtract 5 dollars, 3 quarters, and
18 dimes from 12 dollars, 7 quarters, and 31 dimes, we subtract 5
dollars from 12 dollars, leaving 7 dollars ; 3 quarters from 7 quarters,
leaving 4 quarters ; and 18 dimes from 31 dimes, leaving 23 dimes.
174 GENERAL MATHEMATICS
The subtraction of algebraic polynomials is, then, not
different from the subtraction of monomials and may
therefore be reduced to addition, as in the following
two examples, which are exactly equivalent:
SUBTRACTION ADDITION
7 2 - 14 ab - 11 P 7 a 2 - 14 ab - 11 tf
+ 5 a 2 + 3 ab + 3 ft 2 - 5 a 2 - Sab- 3 b*
2 a* - 17 ab - 14 ft 2 2 a 2 - 17 // - 14 ft 2
The student should change the signs of the subtrahend
in the written form until there is no doubt whatever as
to his ability to change them mentally. The example will
appear as follows:
2 + 2 ab + //
- +
+ a 2 - 2 ab + fe 2
4o6
NOTE. The lower signs are the actual signs of the subtrahend.
They are neglected in the adding process.
Numerous verbal problems have been given with the
hope of giving a reasonable basis for the law of subtraction.
The student should now apply the law automatically in the
following exercises.
EXERCISES
Subtract the lower from the upper polynomial :
1. 4a?-3ab + 6b 2 3. x 3 + 3x 2 y + Sxy 2 + y 3
4:a?-5ab-4:b 2 - 7 x z y + 3 xy* + y*
2. x 2 -5zy+ if 4. 2mn 2 + 5m s n+ 6
-3x 2 -4a;//-3y 2 - 7 mri* - 4 m*n + 18
5. From 10 xy 5 xz -j- 6yz subtract 4 xy 3 xz + 3 yz.
6. From 16 x - 5 mx* + 4 m* subtract 7x* 4 mx* + 12 m 3 .
7. From 2a?-2a?b + al> 2 -2b 3 subtract a 3 - 3 a?b + aft 2 - 6 s .
Subtract as indicated, doing as much of the work as possible
mentally.
8. (4r 8 6 r 8 * + 10 s 8 - 6 rs 8 ) - (2 r 8 + 6 r 2 * + 4 s 8 -f- 3 rs 2 ).
9. (_ 8 m *pq - 4 m s p -10mV) -(-6m*p-8m?pq - 15 mV)-
10. (15 x s - 12 afy + Ty 8 ) - (- 11 z 8 + 8 a"y - 5 y 8 ).
11. ( a 8 - 3J aft 8 - 3 a 8 *) - (- f a& 2 + 5^ 2 6 - 3 a 8 ).
12 . (5f rst - 7 J r 8 * - 8f s 8 * - 6 J * 8 /-) - (4 J s 2 * + 3| ** + 7 J A).
13. (2.3 aW - 4.6 a 4 6 8 + 8.7a 6 * 2 ) -(-1.1 aW - 2.1 a 6 * 8 - 3a 3 6 4 ).
14 . (3 x 2 - 4 x + 5) -f (2 x 2 + 5 x - 3) - (- 2 a 2 + 3 x + 6).
15. (5.2 ofy - 41xy + 2 if) - (31 afy + 3.2 xy + 5 y 2 ).
16. (2.42 a 2 6 2 + 5 ab + 6) - (3.12 a 8 ^ 2 ai - 9).
17. (3 a6 3 - 3 afe 8 ) - (-2 a& 8 + 3 a 3 - 4 a5c 8 ) - (- 4 a 8 -
18. (5x 2 +2a-// + 3y 2 ) + (2* 2 -5a;y-/)-(9x 2
19. Compare the signs. of the terms of the subtrahend in the
foregoing exercises before and after the parenthesis are removed.
20. State a rule as to the effect of the minus sign preceding
a polynomial in parenthesis.
21. What is the rule when the plus sign precedes a poly-
nomial in parenthesis ?
223. Symbols of aggregation. It has been found very
convenient to use the parenthesis for grouping numbers.
Such a symbol indicates definitely where a polynomial
begins and ends. Other symbols used with exactly the
same meaning and purpose are [ ] (brackets) ; { } (braces) ;
and " " (vinculum). Thus, to indicate that a + b is
to be subtracted from x + y we may use any one of
the following ways : (# + #) (# + ft), [x + y\ [a + ft].
{x-}- y} {a + ft}, or x+y a + b. The vinculum is like
the familiar line separating numerator and denominator
,. . . .2 a + ft
oi a fraction, as in - or -
o a ft
170' (JEMERAL MATHEMATICS
Sometimes the symbols are inclosed one pair within
another; thus, 19 - (16 - (9 - 2)}.
In an example like the preceding the common agree-
ment is to remove first the innermost parenthesis. First,
2 is to be subtracted from 9, then the result, 7, is to be
subtracted from 16. This result, 9, is in turn to be sub-
tracted from 19 ; whence the final result is 10.
EXERCISES
1. (live the meaning of the following:
(a) 15 -{4 + (6 -8)}.
( C ) _ 5 x _ [_ 7 x _ {2x-
(d) 3 (a; + y) - 5{x - 2x-3y}.
2. Keep definitely in mind the rules governing the effect of
a minus or a plus sign before a grouping symbol. Perform the
following indicated operations and simplify the results :
(a) 12-{5-(-2x-5)}.
(b) 17 - {- 12 x - 3 x - 4}.
(c) 4 a 2 - (a 2 - 3 a 8 + 3 a 2 - a 8 }.
(d) 2e-[6e-36-4e-(2e-46)].
(f ) 15 a? - {- 3 x 2 - (3 x 2 + 5)} - (20 a 2 + 5).
SUMMARY
224. This chapter has taught the meaning of the follow-
ing words and phrases : positive number, negative number,
algebraic numbers, absolute value of a number (or numerical
value), degree -of a number, degree of a monomial, degree
of a polynomial, descending power, ascending power.
POSITIVE AND NEGATIVE NUMBEKS 177
225. The following symbols were used : -f (plus sign)
and (minus sign) for positive and negative number re-
spectively, ( ) (parenthesis), [ ] (brackets), { } (braces),
and ~~ ~ (vinculum).
226. Positive and negative numbers have been illustrated
by game scores, directed line segments, latitude, longitude,
time, the number scale, forces, the thermometer, angles,
profit and loss, debit and credit, assets and liabilities,
deposits and checks.
227. The sum of two algebraic numbers with like signs
is the sum of their absolute values with their common
sign prefixed.
The sum of two algebraic numbers with unlike signs
equals the difference of their absolute values with the sign
of the number having the greater absolute value prefixed.
The sum of three or more monomials is found most
easily by the following method : (1) add all positive terms,
(2) add all negative terms, (3) add the two sums obtained.
228. To add polynomials add the similar terms (write
similar terms in the same column).
229. To find the difference of two numbers change the
sign of the subtrahend and add.
230. A parenthesis is used for grouping. If preceded
by a plus sign, it may be removed without making any other
changes ; if preceded by a minus sign, it may be removed
if the sign of every term within the parenthesis is changed.
CHAPTER IX
+ 4
POSITIVE AND NEGATIVE NUMBERS. MULTIPLICATION
AND DIVISION. FACTORING
231. Multiplication. The laws of multiplication of num-
bers having plus or minus signs are easily applied to a
considerable number of interesting problems. These laws
are illustrated in the following examples:
ILLUSTRATIVE EXAMPLES
1. Find the product of (+ 4) and (+ 2).
Solution. Geometrically we interpret this as follows : Take a
segment + 4 units long and lay it off two times to the right of
zero on the number scale ; that
is, in its ewn direction (Fig. 157).
Thus, (+2) (+4) = + 8.
2. Find the product of
(-4) and (+2).
Solution, Geometrically this
means : Take a segment 4 units
long and lay it off two times to
the left of zero on the number
scale ; that is, in its own direction
(Fig.158). Thus,( + 2)(-4) = -8.
3. Find the product of
(+ 4) and (- 2).
Solution. Geometrically we in-
terpret this as follows: Take a
segment + 4 units long and lay it off two times to the left of zero ;
that is, opposite its own direction (Fig. 159). Thus, ( 2) (+ 4) = 8.
i i i i i i i i i i i i i i i i
+ 8 r ^ X
-4
FIG. 157
4
L -4
O
FIG. 158
-4 J
I I I I i I
O
FIG. 159
POSITIVE AND NEGATIVE NUMBEKS 179
4. Find the product of (- 4) and (- 2).
Solution. If the first factor were a positive 2, then we should
interpret this geometrically by laying off 4 twice, obtaining a
line segment 8 units long (see
OR V in Fig. 160) just as we did in
Ex. 2. But since it is a negative 2,
we lay it off not in the direction of
OR l but in the opposite direction; p ,
namely, OR (see Fig. 161). Thus,
Note that in
this last case, as
in Ex.1, the signs
FIG. 161
of the multipli-
cand and the multiplier are alike, and the product is positive ;
while in Exs. 2 and 3 the signs of the multiplicand and
multiplier are unlike, and the product is negative.
EXERCISES
1. Find geometrically the products of (+ 2) (+ 5); (- 2)(+5);
(+2) (-5); (-2) (-5).
2. State the law of signs for the product of two algebraic
numbers as suggested by the preceding work.
232. The law of signs for multiplication. The law of
signs for multiplication is as follows:
The product of two factors having like signs is positive.
The product of two factors having unlike signs is negative.
EXERCISES
Find the value of the following products, using the law of
signs. Illustrate the first ten geometrically.
1. (+3) (+5). 3. (-3) (+5). 5. (-2) (+3).
2. (- 3)(- 5). 4. (+ 3)(- 5). 6. (- 2)(- 3).
180
GENERAL MATHEMATICS
7. (-2) (+7).
8. (+2) (+7).
9. (9) (-3).
10. (-4) (-a).
11. (2)(-).
12. (-!)(-!).
13. (-2*) (-3).
14. 2a/>-3.
15. (-3) (-5 a).
16. (-f)(
17. (-f)(
18. -2o-
FIG. 162. THE LAW or MULTI-
PLICATION ILLUSTRATED BY THE
BALANCED BEAM
233 ! . Law of multiplication illustrated by the balance.
The law of signs may be illustrated with a balanced bar
(Fig. 162). A light bar is balanced at M. The points r v r 2 ,
etc. represent pegs or small
nails driven at equal distances.
We shall speak of r v r v etc.
as " first right peg," " second
right peg," etc. and of l v 1 2 ,
etc. as " first left peg," " sec-
ond left peg," etc. with the bar
in a position facing the class as
in Fig. 162. The weights, w,
are all equal ; hence we shall
merely speak of them as " two weights," " three weights,"
etc. instead of mentioning the number of ounces or grams
contained. In Experiments 1-3 the string over the pulley
is fastened on the first left peg.
EXPERIMENTS
1. Hang two weights on / r This tends to turn the bar. How
many must be attached to the hook H to keep the bar level ? Hang
three weights on / r What do you notice about the turning tendency
as compared with the first case ? Answer the same question for four
weights on 1 3 .
1 The entire article may be omitted at the teacher's discretion. The
device has, however, proved useful in the hands of many teachers. The
apparatus may be bought at several of the large book companies or,
better still, made in the shop by a member of the class, using a part of
a yardstick for the lever and small nails for pegs.
POSITIVE AND NEGATIVE NUMBERS 181
2. Hang one weight on l r How many must be placed on the hook
to keep the bar level? Hang one weight on / 2 ; remove it and hang
one weight on / 3 ; on / 4 ; and so on. What do you notice about the
turning tendency in each case ? What two things does the turning
tendency seem to depend on ?
3. With the string passing over the pulley fastened to /j ho.w
many weights must be put on the hook to balance two weights
on / 8 ? three weights on 1 2 ? one weight on l t ? two weights on Z 4 ?
three weights on / 4 ?
4. Repeat Experiments 1~3 for the pegs on the right side, with the
pulley string fastened to r r What seems to be the only difference?
Results of experiments. The experiments show that
1. The turning tendency (Jorce) varies as the number of
weights hung on a peg on the bar. Thus, the more weights
hung on any peg, the stronger the force.
2. The turning tendency also varies as the distance of
the peg from the turning point.
3. The turning tendency is equal to the product of the
iveights multiplied by the distance of the peg on which the
weight hangs from the turning point.
4. When a weight is hung on a right peg, the bar turns in
the same direction as the hands of a clock ; when a weight is
hung on a left peg, the bar rotates in a direction opposite to
the hands of a clock.
234. Signs of turning tendency ; weight ; lever arm. It
is conventionally agreed that when the bar turns counter-
clockwise (as you face it), the turning tendency is positive ;
while if the bar rotates clockwise, the turning tendency is
negative.
Weights attached to the pegs are downward-pulling
weights and are designated by the minus sign. Weights
attached at H pull upward on the bar and are designated
by the plus sign.
182 GENERAL MATHEMATICS
The distance from the turning point to the peg where
the weight, or force, acts will be called the lever arm, or
arm of the force. Lever arms measured from the turning
point toward the right will be marked + ; those toward
the left, . For example, if the distance from M to peg i\
is represented by + 1, then the distance from M to r 3 will
be + 3 ; the distance from M to / 2 will be 2 ; and so on.
235. Multiplication of positive and negative numbers. By
means of the apparatus (Fig. 162) the product of positive
and negative numbers is now to be found.
ILLUSTRATIVE EXAMPLES
1. Find the product of (+ 2) (- 4).
Solution. AVe may interpret this exercise as meaning, Hang four
downward-pulling, or negative, weights on the second peg to the
right (positive). The bar turns clockwise. The force is negative;
hence the product of (2) ( 4) is 8.
2. Find the product of (- 2) (- 4).
Solution. Hang two downward-pulling, or negative, weights on the
fourth peg to the left (negative). The bar turns counterclockwise.
The force is positive ; hence the product of ( 2) ( 4) is + s.
3. Show that (+ 3) (+ 4) = + 12.
HINT. Fasten the string over the pulley to the fourth peg to the
right and hang three weights on the hook.
4. Show 'that (-3) (+2) = -6; that (+ 2) (- 3) = - 6. How
does the beam illustrate the law of order in multiplication ?
5. Compare the results of Exs. 1-4 with the law of signs in
multiplication (Art. 232).
It is hoped that the law of signs is made reasonablv clear
by means of these illustrations. The student should now
proceed to ajjply the law automatically.
POSITIVE AND NEGATIVE NUMBERS 183
EXERCISES
State the products of the following, doing mentally as much
of the work as possible :
1. (+4) (-6).
2. (-4) (+6).
3. (+4) (+6).
4. (-4) (-6).
5. (+2) (+5).
6. (+3) (-4).
7. (-5) (-2).
8. (-3) (-7).
9. (-5) (+6).
11. (-3.1) (-5). 21.
12- (-f)(f). 22. (-6X-S).
23. (-8)(-
24. (-c)(-
25. (-
26. (-
17. (+6j-)(+6j). 27. (51) (-^ 2 )
18". (+6j)(-6-i). 28. (-9)(+x 2
16. _6-
19. (_6i)(+6-i-).
10. (-12) (-13). 20.
29. (-1) 3 .
30. -2 3 .
236. Multiplication by zero. The product of 3 x means
+ + = 0.
EXERCISES
1. Show geometrically that ax 0=0.
2. Show by the beam (Fig. 162) that a x 0=0; that x a=0.
3. State a verbal problem in which one of the factors is
zero. In general both a x and x a equal zero. Hence the
value of the product is zero when one of the factors is zero.
4. What is the area of the rec-
tangle in Fig. 163 ? How would the area
change if you were to make the base
smaller and smaller ? What connection
has this with the principle a x = ?
5. How would the area of the rec-
tangle in Fig. 163 change if I> were not changed but a were
made smaller and smaller ? What does this illustrate ?
b
FIG. 163
184 GENERAL MATHEMATICS
237. Product of several factors. The product of several
factors is obtained by multiplying the first factor by the
second, the result by the third, and so on. By the law of
order in multiplication the factors may first be rearranged
if this makes the exercise easier. This is often the case in
a problem which involves fractions.
EXERCISES
1. Find the value of the following products :
(a) (+2) (-3) (-5) (-4).
00 (- -BX-f) (-!!)()
2. Find the value of (-1) 2 ; (- 1) 3 ; (-1) 4 ; (-2) 2 (-2) 3
(-2)*; (-2) 5 ; (- 3) 2 (- 3) 8 (- 3)*; (_4) 2 (-4) 8 .
3 . Find the value of 3 a- 4 5 x s + x* 12 x 5 when x 2.
4. Find the value of x 8 3x*y + Secy 2 + y* when x = 3
and y = 2.
5. Find the value of z 8 + 3 x 2 + 3 x + 1 when x = 10.
6. Compare (- 2) 3 and - 2 s ; - 3 s and (- 3) 8 ; (- 2) 4 and
- 2 4 ; (- a) 8 and - 8 ; (- a) 4 and - a*.
7. What is the sign of the product of five factors of which
three are negative and two are positive? of six factors of
which three are negative and three are positive ?
8. What powers of 1 are positive ? of 2 ? of x ?
State the rule.
238. Multiplication of monomials. Find the product of
The sign of the product is determined as in Art. 232 and is found
to be + .
By the law of order in multiplication the factors may be arranged
as follows : , QN
2(- 3) (- o) zxxxyyy,
which is equal to 3
POSITIVE AND NEGATIVE NUMBERS 185
Hence, to find the product of two or more monomials
1. Determine the sign of the product.
2. Find the product of the absolute values of the arith-
metical factors.
3. Find the product of the literal factors.
4. Indicate the product of the two products just found.
EXERCISES
Simplify the following indicated products, doing mentally as
much of the work as possible :
2. (-2*) (-3*) (4*).
3. (5 xV 2 ) (- 3 *V) ( 2 ay* 3 ) (- 7 ay* 2 ).
4. (- 2 a 2 bcd) (5 aPcd) (- 7 abc 2 d) (- 2 alcd*).
5. ( 3 mnx) ( 5|- m*nx) ( 2 oj).
7. (- 3J 06) (- 5 J a 2 ^ 2 ) (+ 8 oft).
8- (5i?V)(l^X
9. (I Vty (6 jac^^) (
10. (3 ^) (6 m 2 m ^ } (
11. (1.3 xV*)(- 2 aV u * u
12. (1.1 a-y 2 ) (1.1 mxy 2 ) (10 i 2 o; 2 ).
13. ( _ 2 yX-3) 2 (-4)(2) 2 .
14. (- 3) 2 (- 2) 2 (- I) 3 .
15. (5)(-4)(3) 2 (0)(2) 2 .
16. (- a)\- a}\- a 2 ) (- a 3 ).
17. (aj + y)(a5 + y)(a5 + y) 6 .
18. 3 (aj + y) 8 (x - 7/) 2 (x + y)'(x - y}*.
19. Find the value of 1+ (- 5) (- 3) - (- 2) (4) - (- 3) (3;
186 GENERAL MATHEMATICS
20. What is a short method of determining the. sign of the
product containing a large number of factors ?
NOTE. It is agreed that when an arithmetical expression contains
plus or minus signs in connection with multiplication or division
signs, the multiplication and division shall be performed first. This
amounts to the same thing as finding the value of each term and then
(tili/in</ in- s it lit racling as indicated.
239. Multiplication of a polynomial by a monomial. We
shall now see how the process of algebraic multiplication
is extended.
INTRODUCTORY EXERCISES
1. Keview the process of finding the product of a(x + y + z)
in Art. 122.
2. Illustrate by a geometric drawing the meaning of the
product obtained in Ex. 1.
3. How many parts does the whole figure contain ?
4. What is the area of each part ?
The preceding exercises serve to recall the law that a
polynomial may be multiplied by a monomial by multiplying
every term of the polynomial by the monomial and adding the
resulting products.
DRILL EXERCISES
Find the products as indicated and check by substituting
arithmetical values for the literal numbers :
Solution. a 2 - 2 ab + 3 b* = W
_ 3_a= _6
3 a 3 - 6 a 2 6 -f 9 a& 2 = 114
Check. Let a = 2 and b = 3. Then the same result is obtained by
substituting in the product as by substituting in the factors and
then multiplying the numbers. Note that the check is not reliable
AND NEGATIVE NUMBERS 1ST
if we let a literal number in a product containing a power of that
literal number (as a: in a product a; 5 4 x) equal 1, for if x = 1, then
x 5 4 x also equals x 3 4 x, x 2 4 x, x 9 4 x, etc. Explain.
2. 5x(2x* -3x -7).
3.
4.
5. 5.1 4 (i
6.
7. (?V - 3 mV + 4 mV - 9 wV) 3.5 ?V.
8.
9.
10.
11.
240. Product of two polynomials. In Art. 126 we found
the product of two polynomials to be the sum of all the partial
products obtained by multiplying every term of one polynomial
by each term of the other. After reviewing briefly the case
for positive terms we shall proceed to interpret the above
law geometrically even when negative terms are involved.
ILLUSTRATIVE EXAMPLES
1. Find the product of (c + d)(a + It).
Solution. The area of the whole rectangle in Fig. 164 is expressed
by (a + &) (c + d). The dotted line suggests a M d
method for expressing the area as the sum of
two rectangles ; namely, a (c + c?) + b (c + d).
If we use the line MN, the area may be
expressed as the sum of four rectangles ;
namely, ac + ad + be + bd. Each expression c wo
equals the area of one of the rectangles; hence FIG. 164
(a + 6) (c + d) = a(c + d) + b(c + d) = ac + ad + be + bd.
1.88
GENERAL MATHEMATICS
3x y
FIG. 165
G a B
FIG. 166
2. Illustrate, by means of Fig. 165, the law for the
multiplication of two polynomials. % x y
3 . Find the product of (a t>)(c + d).
Solution. In this case one of the factors
involves a negative term.
The product (a b)(c + d) is repre-
sented by a rectangle having the dimen-
sions (a - b) and (c + rf) (Fig. 166). The
rectangle ABEF=ac + ad. Subtracting
from this the rectangles be and bd, we
obtain the rectangle A BCD.
Therefore (a b) (c + d) = ac + ad
be bd, each side of the equation repre-
senting the area of rectangle A BCD.
* 4 . Findthe product of (a b) (a fy .
Solution. Let ABCD (Fig. 167), repre-
sent a square whose side is (a 6) feet.
Show that the area of ABCD equals
EFGC + GHIB - FKDE - KHIA.
Then
(a - ?/) (a - b) = (a - b) 2 Why ?
= a 2 + 6 2 ab ab
= a 2 - 2 ab + IP.
*5. Sketch a rectangle whose area is (m + ri) (r s) ; whose
area is 24 b 2 6 be.
DRILL EXERCISES
Apply the law of multiplication to two polynomials in the
following exercises. Check only the first five.
Solution.
D
(a-b)'
A
B
b 1
f
: G
1
FIG. 167
x z + 2 xy + y-
x + y _
x 3 + 2 x' 2 y + xy 2
+ 2 xy* + y*
x* + 3 x*y + 3 xy* + y*
Check by letting x = 2 and y = 3.
POSITIVE AND NEGATIVE NUMBERS 189
2. (rs + tm) (rs - tm). 7. (2 a + 3 ft) (2 a - 3 4).
3. (a 8 + ax 3 + a s ) (a + a). 8. (i aft - ftc) (f aft + f ftc).
4. (a 2 + 4 a; + 3) (a -2). 9. (a + 6 - c) 2 .
5. (cc 2 -3cc + 5)(2x + 3). 10. (a - b + c - df.
6. (k 2 + 3 Ar + 1) (A - 2). 11. . (- 2 a + 3 1> - 4 c) 2 .
12. Comment on the interesting form of the results in
Exs. 9-11.
13. (0.3 a + 0.4 b - 0.5 c) (10 a - 30 b + 40 c).
14. (2 if - 12 zy + 5 x 2 ) (2 if - 5 .r 2 ).
15. o 2 + *z/ + z/ 2 ) (a - y) 0* + y;.
16. (9x 2
17. (x +
18. Comment on the form of the results in Ex. 17.
19. (r 2 + rs - s 2 ) (r 2 + rs + .<?).
20. (Sr 2 + 5 r + 6) (3 r 2 + 3 .* - 6).
21. (3x + 2.y) 3 -(3x-2y) 8 .
22. (3 x 2 + T/ 2 ) 2 - (3 x z - iff -x*(x-
23. (2a-3^) 2 -(2a + 36) 2 + (2
24. (0.3 a - 0.4 4) 3 - (0.5 a + 0.6 i) 2
- (0.3 a - 0.4 ft) (0.3 a + 0.4 ft).
26. (5 + 3) 2 - (5 - 3) 2 - (5 + 3) (5 - 3).
27. Why may 3527 be written 3 10 s + 5 10 a + 2 . 1 + 7 ?
28. Multiply 352 by 243.
HINT. Write in the form 3 10 3 + 5 10 + L>
2 10 2 + 4 10 + 3
29. Write 56,872 as a polynomial arranged according to the
descending powers of 10.
30. Find the product of 5 and 3427 by the method suggested
in Ex. 28.
190 GENERAL MATHEMATICS ,
241. Product of two binomials. We shall now see how
the algebraic product of two binomials may be obtained
automatically. The following exercises will help the stu-
dent to discover and understand the method.
EXERCISES
Find by actual multiplication the following products :
1. (2x + 3)(4z + 5). 4. (4x + 6) (4 a; + 5).
Solution. 2* + 3 5. (3* -2) (3* -2).
4x+5 6. (x + 2) (* + 9).
8* 2 + 12* 7. (2a;+l)(aJ4-6).
_ +l* + lr ' 8. (b + 3) (6 + 5).
8 x 2 + 22 x + 15
9. (a -7) (a -3).
2. (3 + 5)(2a-8). 10 ' (3 * + 8) (a + 2).
11. (3x + 4)(2a;-3).
Solution. 3 a + 5
2a -8 12 ' (*-3)(*-10).
13. (a- 8 - 9) (*" + 9).
- 24 a - 40 14. (or 2 - 5) (aj 2 + 10).
6a 2_14 a _ 40 15> (3 a; -5) (4 a; -2).
16. (2y 3)(5y 8).
3. (o ?/ + 4) (o ?/ 4).
17. (i
Solution, o M + 4 1 o /
-My -16 20. (3a; + 4y)(3a;-'4y).
-16 21. (4a + 26)(7a-5i).
22. 53 x 57.
Solution. 53 x 57 = (50 + 3) (50 + 7)
= 50 3 + (7 + 3)50 + 21.
23. 61 x 69. 24. 52 x 56. ' 25. 37 x 33.
26. Can you see any way of formulating a rule for finding
the products of two binomials ?
POSITIVE AND NEGATIVE NUMBERS 191
If we agree to use the binomials ax 4- b and ex 4- d to
represent any two binomials where a, b, e, and d are
known numbers like those in the products above, then we
may discover a short cut in multiplying ax 4- b by ex 4- d.
ILLUSTRATIVE EXAMPLE
Find the product of (ax 4- V) and (<-x -\- <Z).
Solution. ax + b
Y
CX + d
bcx
. + adx + bd
acx 2 + (be + ad ) x + bd
The arrows show the cross-multiplications or cross-products whose
sum is equal to the middle term. It is seen that the first term of the
product is the product of the first terms of the binomials, that the last term
is the product of the last terms of the binomials, and that the middle term
is the sum of the tivo cross-products.
EXERCISES
Using the rule stated above, give the products of the fol-
lowing binomials :
1. (2 a + 3) (3 a 4- 5). 6. (3 a - 2fi)(3a - 2 ft).
Solution. The product of the 7. f x _ 7) (4 x + 9).
first terms of the two binomials
is 6 a 2 , the product of the last
terms is 15, and the sum of the 9. (x + *>)(x + 8).
cross-products is 19 a. Therefore , ^ / 7 9 v.
the product is 6 a + 19 a + 15.
2. (4,, + 3) (2. + 1). 11. (4, + 3)(3,-4).
3. (2s -7) (3* + 2). 12 - (*4-9&)(aj + 7ft).
4. (3 x 4- 4) (3x4- 4). 13. (2 aj + 4 y) (3 x - 5 y).
5. (7 x - 2) (7 x + 2). 14. (5 a + 4) (4 a - 2).
192 GENERAL MATHEMATICS
15. (7a + 26)(7o-2i). 18. (3 a - 7 ft) (3 a - 7 6).
16. (5a + 4i)(5 + 46). 19. (6 xy + 2) (3 xy - 5).
17. (3 a: + 2) (12 a: -20). 20. (7 ab + 5c) (60* - 8c).
21. Do you notice anything especially significant about the
product of two binomials that are exactly alike? Explain
by using the product of x + y and a- + y, a b and a b
(compare with Ex. 1, Art. 127).
22. Do you notice anything especially significant about the
product of two binomials that are the same except for the
signs between the two terms ? Explain by using the product
of ra + n and m n.
23. Try to formulate a rule for obtaining automatically the
products referred to in Exs. 21 and 22.
242. Special products. We have seen in Art. 241 how
the multiplication of two binomials may be performed
automatically. Such products are called special products.
The student should observe tha f , Exs. 21 and 22, Art. 241,
furnish examples of such products. For example, the
product of x'+y and x -+- y is equal to x 2 + 2 xy + y*, and
is called the square of the sum of x and y\ while the
product of a b and a b is equal to a 2 2 db + 6 2 , and
is called the square of the difference of a and b. Further,
the product of m + n and m n is equal to m z n*, and
is called the product of the sum and difference of m and n.
EXERCISES
1. Find automatically the following special products and
classify each:
(a) (x + 3)(x + 3). (c) (2x + 4) (2 a; + 4).
(b) (y_2)(y-2). (d) (3* -6) (3* -6).
(e) (2x + 4) 2 . (g) (2 x + 4 y) 2 . (i) (2 a + 4 ft).
(f ) (4 x - 2) 2 . (h) (5 x - 2 y}\ (j) (3 a - 2 ) 2 .
193
2. State a rule for finding automatically the square of the
sum of two numbers.
3. State a rule for finding automatically the square of the
difference of two numbers.
The preceding exercises should establish the following
short cuts for finding the. square of the sum and the square
of the difference of two numbers:
1. The square of the sum of two numbers is equal to the
square of the first number increased by two times the product
of the first number and the second number plus the square of
the second number. Thus, (a -f- ft) 2 = a 2 + 2 ab + b 2 .
2. The square of the difference of two numbers is equal
to the square of the first number decreased by two times the
product of the first number and the second number plus the
square of the second number. Thus, (a 6) 2 = a 2 2 ab -+- 6 2 .
EXERCISES
1. Find automatically the following products :
(a) (x + 5) (x - 5). (d) (c + 7) (c - 7).
(b) (a + 4) (a - 4). (e) (y - 11) (y +.11).
(f) (,. + 8)(r-8).
2. Study the form of the results in Ex. 1. Give the products
in the following list at sight :
(a) (y - 10) (y + 10). (d) (5 -x)(5+ x).
(b) (2 x + 5) (2 x - 5). (e) (y - J) (// + ).
(c) (3d-4)(3<* + 4). (f) (5x-y)(5x + y).
3. Write the sum of x and y ; the difference ; find the prod-
uct of the sum and the difference. Check by multiplication.
4. State the rule for finding the product of the sum and the
difference of two numbers.
194 GENERAL MATHEMATICS
The preceding exercises should establish the following
short cut for finding the product of the sum and the
difference of two numbers:
1. Square each of the numbers.
2. Subtract the second square from the first.
DRILL EXERCISES
Find the following products mentally :
1. (x + 2) (aj + 2). 11. (x - 1) (a; + 1) (* 2 + 1).
2- (U + 3) (y - 3). 12. (w - c) (w + c) (w* - c 2 ).
3. (-4)(s-4). 13.- (10 a; + 9) (10 a; + 9).
4. (2w-5)(2tt; + 6). 14. (y*if - 0.5) (a; 2 / + 0.5).
5. ( - 2 i) (s + 2 J). 15. (11 +/<7/i 2 ) (11 +/^ 2 )-
6. (3s + 2o)(3 + 2a). 16. (a 5 + ^ 5 ) (a 5 - 5 ).
7. (3r-4)(3r + 4). 17. (20 + 2) (20 - 2).
8 . (Ja + j6)(Ja.-ii). 18. (30 + 1) (30 - 1).
9- (iy-*)(t*y + *)- 19 - ( 18 )( 22 >
10. (x - 1) (a + 1). 20. (31) (29).
243. Division. The law for algebraic division is easily
learned because of the relation between division and mul-
tiplication. ^ We recall from arithmetic that division is the
process of finding one of two numbers when their product
and the other number are given and also we remember
that quotient x divisor = dividend.
These facts suggest the law of division. Thus we know
that +12-=- + 2 = -f6 because (+ 2)(+ 6)= +12.
POSITIVE AND NEGATIVE NUMBERS 195
EXERCISES
1. Since (2) (- 6) = - 12, what is - 12 -s- 2 ?
2. Since (- 2) (+ 6) = - 12, what is 12 -s- - 2 ?
3. Since (- 2) (- 6) = + 12, what is + 12 H- - 2 ?
4. Since (+ )(+ &) = -f &, what is (+ aV) -+ a?
5. If the signs of dividend and divisor are alike, what is
the sign of the quotient ?
6. If the signs of dividend and divisor are unlike, what is
the sign of the quotient ?
244. Law of signs in division. The work of the preced-
ing article may be summed up in the following law: If
the dividend and divisor have like signs, the quotient is posi-
tive ; if the dividend and divisor have unlike signs, the
quotient is negative.
245. Dividing a monomial by a monomial. We shall now
have an opportunity to apply the law learned in the pre-
ceding article.
EXERCISES
Find the quotient in the following, doing mentally as much
of the work as possible :
1. (+l5)-(-3)=? 10. (-10ar)'-5-(-2oj)=?
2. (-15) -L(_3)=? 11- (-/>)* (-)=?
3. (- 15) -t- (+ 3) = ? 12- (- *) * (+ V) = 'f
4. (+15)-i-(+3)=? 13- (t *)--(- &) =?-
5. (- 18) -5- (- 3) = ? 14. (- 0.5 x) + (-$x)= ?
6 . (-12) -=-(-12)=? 15. (- 1.21 x 2 ) ^ (- 1.1 *) = ?
7. (+5)-f-(+5)=? 16. (_)-*.(-) = .?.
8. (+*)^(+*)=? 17. (f)-Kf)=?
9. -2a-f.a=? 18 ' (-|)^(-f) =?
196 GENERAL MATHEMATICS
19 . $).,.(_{).? 31. (-*V(*)=?
20- (-1) ^ (- 1) = * 32. (- * 2 ) + (^> -
21. (f)-Kf)='.' 33. (-9^)-(3)=?
22. (?) + (- t)=- 34 ' (-3*)-l-(-ft)=?
23. (_2) + (+-J)=? 35. (6fe)-K-2je)=
36. (+) + (-)=.
25. (+12./-)-(-x) = ? 38. <_
28. (4- ) -*- (- ) = V 39. < t- ww/) -*-( <) = ?
27. (-*)4-(-J*)=? 40. (-//Ar 8 ) -5-(-oA)='.'
28. (- .r*) -i- (- x) = ? 41. ( 7 r) -I- (- 22) = V
29. (^ - (- x) = ? 42. (oi r) -5- (- 3j r) = ?
30. ( _ x 4 ) -s- (- x) = ? 43. 24 a;// -t- X .? = ?
XOTK. The algebraic solution of the more difficult problems of
this type are best interpreted as fractions, since a fraction is an
_>4 ,-
indicated quotient. Thus, 24 -/// -=-3 x may be written The
'^ x 24 j-ii
problem now is one of reducing to lower terms. Thus, in -
both numerator and denominator may be divided by 3 x. The result
8w
is - > or 8 y units.
In algebra, as in arithmetic, the quotient is not altered if dividend
aad divisor are both divided by the same factor. Dividing dividend
and divisor by the highest common factor reduces the quotient
(or fraction) to the simplest form (or to lowest terms).
Solution. The sign of the quotient is negative. Why?
The numerical factors can be divided l>y S : r 5 and x 1 are divisible
by x 2 ; y 3 and / s are divisible by y 3 ; m- and w 3 are divisible by ? 2 .
Hence ***?# = ^ = - f!.-
8 x-im? m m
45.
POSITIVE AND NEGATIVE NUMBERS 197
-S-jfb 343
xz
46.^
9
47.
48.
49.
50.
49 a;
12
246. Dividing a polynomial by a monomial. The division
process will now be extended.
EXERCISES
1. Divide 6 x 2 + 4 xij + 8 xz by 2 a-.
As in dividing monomials, this quotient may be stated as a rec-
tangle problem. Find the length of the base of a rectangle whose
area is 6 x" + 4 xy + 8 xz and whose altitude is 2 x. Indicating this
quotient in the form of a fraction,
we have
6 x z + 4 xy + 8 xz
3X
42
FIG. 168
Dividing numerator and denominator
by 2 x, the result is 3x + 2 y + 4~- Show that the problem may now be in-
terpreted by a rectangle formed by three adjacent rectangles (Fig. 168).
2. Show that the total area of three adjacent flower beds
(Fig. 169) may be expressed in either
of the following forms :
or 5 (3 + 5 + 4).
Which form is the better ? Why ?
3. Find the following quotients, obtaining as many as you
can mentally :
9 a 2 -6 a 5 . 27i + 6 2
5x3
5x5
5x4
354
Fir.. 169
(a)
(g)
00
3a
-4*V
(e) :
(f)
6 r/
198 GENERAL MATHEMATICS
247. Factoring ; prime numbers. To factor a number is
to find two or more numbers which when multiplied
together will produce the number. Thus, one may see by
inspection that 2, 2, and 3 are the factors of 12. 'In like
manner, x + y and a are the factors of ax 4- ay.
A number which has no other factors except itself and
unity is said to be a prime number ; as, 5, x, and a + b.
A monomial is expressed in terms of its prime factors,
thus : 1 5 ax*i? = %.5-a-x.x.y.y.y.
The following is an example of the method of express-
ing a factored polynomial:
In algebra, as in arithmetic, certain forms of number
expression occur very frequently either as multiplications
or as divisions so much so, that it is of considerable
advantage to memorize the characteristics of these num-
bers that we may factor them by inspection and thus be
able to perform the multiplications and divisions auto-
matically. In this text we shall study two general types
of factoring.
248. Factoring Type I. Taking out a common monomial
factor. Type form ax + bx + ex = x (a + b + e).
A number of this type we shall call a number containing
a common monomial factor. The products obtained in the
exercises of Art. 239 are numbers of this type. Although
this type of factoring is not difficult, nevertheless it is im-
portant and should be kept in mind. We shall learn that
many verbal problems lead to equations which can readily
be solved by a method which depends upon factoring.
Factoring also enables us to transform formulas into their
most convenient form.
POSITIVE AND NEGATIVE NUMBERS 199
The method of factoring this type consists of the
following steps:
1. Inspect the terms and discover the factor which is
common to all the terms.
2. Divide by the common monomial factor. The result
obtained is the other factor.
3. In order to find out whether he has factored correctly
the student should multiply the two factors together.
NOTE. In all factoring problems the student should first look to
see if the number contains a common monomial factor.
EXERCISES
Factor the following by inspection and check your work
by multiplication :
1 . bx 5 b be.
Solution. Each term has the factor b. Divide the expression by b.
The quotient is x 5 c.
Check. b (x 5 c) = bx 5 b be.
Therefore the factors of bx 5 6 be are b and x 5 c.
2. 5a-5b. 8. x* - x 3 .
3. 4cc + 4?/. 9. 25 x 2 - 5 x 8 .
4. 5xa-lQxb. 10. 2 x 2 + 4 xy + 2 f.
5. 5 ax 2 - 10 axif. ' 11. d 2 b + ab' 2 + a 8 .
6. 2rx 8 -8?y. 12. 4a; 2 -8^ + 47/ 2 .
7. 3 x 2 - 6 x. 13 . a*a? - 2 aVy 2 + 4 aary .
14. 3 a 2 -15 a + 18.
249. Factoring Type II. The " cut and try " method of
factoring. Type form acx z +(bc+ad')x+bd=:(az+b')(cz+d'~).
The products obtained in the exercises of Art. 241 can all
be factored easily by inspection. The method of factoring
200 GENERAL MATHEMATICS
such products is known as the "cut and try" or "trial and
error" method. The method consists simply of guessing
the correct pair of factors from all of the possible ones
and then verifying the result by multiplying the factors
together. The method is illustrated by the following
example :
Factor 2 y? + 9 ./ + 10.
Solution. There are four possible pairs of factors, as shown below :
2 x + 10 2 x + 10 x + 5 2 x + .5
x+ 1 2r+ 1 -2x + 2 x + 2
It is clear that the last pair is the correct one, since the
sum of the cross-products is 9 x. Of course the correct
pair of factors may be found at any stage of the "cut
and try" method, and while the process may seem slow at
first, practice soon develops such skill that the factors can
easily be found.
It is very important for the student always to be sure
that the factors he has obtained are prime numbers. Such
factors are called prime factors. Incidentally it is impor-
tant to remember that there are some numbers that are
not factorable, because they are already prime numbers.
For example, a^+16 and 2^ + 2 a; +12 are not factorable.
See if you can explain why they are not factorable.
From what has been said the student will see that in
all factoring problems it is important to hold in mind
three things; namely:
1. Try to discover a common monomial factor.
2. Find the prime factors by the "cut and try" method.
3. Check by multiplying the factors together.
POSITIVE AND NEGATIVE NUMBERS 201
EXERCISES
Find the prime factors of the following expressions :
1. 6x*-x-2.
Solution. Since the x and the 2 are both negative, the last terms
of the factors are opposite in sign. The possible combinations of
pairs of factors (regardless of signs) are shown bclov. :
' Qj; 2 Qx 1 :!./ 2 3x 1
The third pair is seen to be correct, provided we write them
8 x - 2 and 2 x + 1. Therefore 6 x 2 - x - 2 - (3 x - 2) (2 x + 1).
2. 2 a 2 + 9 a + 9. 23. 2 a- 2 - a- - 28.
3. x 2 + 2 xy + >/ 2 . 24. 5 x 2 - 33 .r + 18.
4. 2 -16. 25. 16 a- 6 - 25 v/ 4 .
5. cc 2 + 2x-3. 26. 18 x 2 + 21 x - 15.
6. x 2 - 2 x - 3. 27. 6 a 2 - a - 2.
7. 4 a; 2 + 16.x + 16. 28. 16-36a- 6 .
8. 7 ar + 9 x + 2. 29. a 2 - 4 a - 45.
9. 9z 2 - 30 a; + 25. 30. 2 x 2 ?/ + 11 ccy + 12 y.
10. 7/ 4 -s 2 . 31. 2-2x 2 .
11. x 2 -5a; + 4. 32. 24 #d + 138crf - 36d.
12. 5 if" 80. 33. (i~ 6 cib 55 & .
13. 6o; 2 -19x + 15. 34. 2 // 2 - 5 </6 + 6.
14. y 2 - 5 y - 6. 35. 100 *V - 49.
15. x 2 -!. 36. 20x 2 -./--99.
IS. 1 ??i 4 (3 factors). 37. 15 4 j- 4 a- 2 .
17. 3 a- 2 -IT a- +10. 38. 18 - 33 x + 5 a; 2 .
18. 2 - 9 2 . 39. 289 W - 81 d*.
19. 49 -x 2 . 40. 8 a 2 -2.
20. 3ar 4a;+l. 41. 3 a; 2 + 4 x + 2.
21. w* a 4- . - 1.2. 42. 19x + 22rK- 2 - 31.
22. 5 .r 2 -17 x- 12. 43. .r 2 + 20 r + 84.
202 GENERAL MATHEMATICS
250. Factoring perfect trinomial squares. Type form
a? 2a6 + 2 = O ) 2 . Numbers like 4z 2 + 16z + 16 or
x* 2 xy + y 2 , which are obtained by multiplying a bino-
mial by itself, are called perfect trinomial squares. They
are special cases of the second type of factoring discussed
in Art. 249. We have already seen perfect trinomial squares
where all the terms are positive in the problems of Ex. 2,
Art. 127. See if you can formulate a short method of
factoring perfect trinomial squares.
EXERCISES
Factor the following perfect trinomial squares by a short
method :
1. a 2 + 2ab + b 2 . 5. 9 x 2 + 42 xy + 49 if.
2. m 2 -2mn + n 2 . 6. 64 a 2 -32ab + tf.
3. 9z 2 + 12av/ + 4?/ 2 . 7. 4 x 2 tf - 12 xy + 9 z*.
4. 16 a 2 - 40 ab + 25 b 2 . 8. 9 aty 4 + 30 afyV + 25 4 .
251. Factoring the difference of two squares. Type form
ai &=(a + fr)(a 6). Numbers of the form a 2 b 2 are
called the difference of two squares. The products obtained
in the exercises on page 194 are numbers of this type.
This is a special case of the type discussed in Art. 249.
ORAL EXERCISES
1. What is the product of (x + 3) (a; 3)? What then are
the factors of x 2 9 ?
2. State the factors of the following :
(a) x 2 - 4. (c) r 2 - 4 s 2 .
(b) c 2 -25. (d) 25 a*
3. Show by means of Fig. 170 on the following page that
a 2 - b 2 = (a + ft) (a - b).
POSITIVE AND NEGATIVE NUMBERS
208
The equation a 2 b 2 = (a b) (a + &) asserts that a
binomial which is the difference of two squares may be
readily factored as follows : M v
One factor is the sum of
the square roots of the terms
of the binomial, and the other
the difference of the square
roots of the terms of the
binomial.
FIG. 170
Thus, to factor 49 # 2 6 2
first find the square root of
each term ; that is, 7 and ab. Then, according to the rule,
one factor is 7 + <d> and the other 7 ab. Obviously, the
factors may be given hi reverse order. Why ?
EXERCISES
Factor the following binomials. Check by multiplication
when you are not absolutely certain the result is correct.
7. 16 a* -25 6*.
8. 81rt 2 -16s 2 .
9. 25 a; 6 -36s 4 .
10. 49 - 36 a- 6 .
11. 1-
12. \ -a- 2 .
1. a- 2 -16.
2 ,.-1 ( ,i
3. //--I.
4. I-./- 4 .
5 _ ,2 Q j ,2
6. 9 - /.
19. 225 a 6 - wV7t M .
20. x 4 -y 4 .
21. 2o?iV- 81m 4 .
22. a- 8 y 8 .
23. 625 2 i 4 - 256 a\
24. 64 a; 6 -9.
25. Ca + i) a -9.
13. 100 4 a; 2 -36.
14. 289m 2 - 81 ?r.
15.
16.
17. 196 -100 6V.
18. 361 r 2 ^ 2 - 196.
26. 9 (a + xf -16.
27. (x 3 - ?/) 2 - x 6 .
28.
29. 0.25 a -0.64J a .
30. 0.25 r/ 2 -^-
lo
204 GENERAL MATHEMATICS
Knowledge of the special products considered above
enables us to multiply certain arithmetic numbers with
great rapidity. Thus the product of 32 by 28 may be
written (30 + 2) (30 - 2) = (30)2 _ (2 ) = 896.
EXERCISES
1. Give mentally the following products :
(a) 18-22. (e) 32-27. (i) 67-73. (in) 75-85.
(b)17-23. (f)37-43. (j)66-74. (n) 79 - 81.
(c) 26 - 34. (g) 38 - 42. (k) 68 . 72. (o) 42 - 38.
(d)29-31. . (h) 47 -'53. (1)75-65. (p) 95 - 75.
2. Find the value of the following :
(a) 71 2 -19 2 . (c) 146 2 -54 2 . (e) 1215 2 - 15 2 .
(b) 146 2 -46 2 . (d) 312 2 -288 2 . (f) 2146 2 -10 2 .
252. Different ways of carrying out the same calculations.
The preceding problems show that the formula a 2 ft 2
= (a J) (a + J) provides us with a method of making
calculations easier. In fact, the expressions which are
linked by the equality sign in a 2 b 2 = (a ) (a + &)
simply represent two different ways of carrying out the
same calculations, of which the one on the right is by
far the easier.
253. Distinction between identity and equation. An equal-
ity such as a 2 b 2 = (a 5) (a + />) is called an identity.
It represents two ways of making the same calculation.
The statement is true for all values of a and I. The
pupil should not confuse the meaning of an identity with
that of an equation. Thus a 2 4 = (x 2) (x + 2) is true
for all values of x, but y? 4 = 32 is a statement that is
true only when x = 6 or x = 6 ; that is, it is a state-
ment of equality in some special situation ; it may be
POSITIVE AND NEGATIVE NUMBEKS 205
the translation of an area problem, a motion problem, an
alloy problem, etc., but it always represents some concrete
situation, whereas x 2 4 = (x 2) (x + 2) is an abstract
formula for calculation and is true for all values of x.
EXERCISES
1. Tell which of the following are equations and which are
identities :
(a) 4 x* - 16 = 20.
(c) 9z 2 + 12* + 4 = (3z + 2V. _ x 2 -9
(d) 4
2. Solve, by factoring, the following equations :
(a) ax + bx = ac + be. (c) 5 a 2 x 4 b*x = 10 <>'
ca da 20
(b)c + f/ = _ + _. r( d > __ s
/ 6 N 1 2 , ^
1 ; 2x-2 3-3" r 4x-4
6
,o
+ 2
* 254. Calculating areas. The following exercises furnish
applications of the preceding work of this chapter.
EXERCISES
1. Show that the shaded area A in Fig. 171 may be ex-
pressed as follows: A =(S s)(S + s), where 5 is a side
of the large square and s a side of the
small square.
2. A carpet 20 ft. square is placed in a
room 25 ft. square. The uncovered border
strip is to be painted. Find the area of
the strip. Find the cost of painting this
area at 80 cents per square yard. Write a
formula to be used in calculating the cost
of painting similar strips at c cents per
yard, the carpet to be x feet square and the room r feet square.
206
GENERAL MATHEMATICS
3. A metal plate is cut as shown in Fig. 172. If 'a =10 and
b = 2, what is the area of the plate ? In what two ways may
the calculating be done ? What is the ^ a ^
volume of metal if the piece is | in.
thick ? What is the weight if a cubic
inch of the metal weighs 20 grams ?
Write a general formula for a plate
cut in the form of the figure, t inches
thick and weighing g grams per
square inch. Write the result in a
form which is easily calculated.
-
FIG. 173
... . . FIG. 172
4. A design pattern is cut in
the form shown in Fig. 173. Calculate the area. Make a
verbal problem illustrating this formula.
5. We can make an application of our
knowledge of factoring in problems re-
lated to circles, as will be seen by solving
the following :
(a) The area of a circle whose radius
is r is Trr 2 . What is the area of a circle
whose radius is R?
(b) How can you find the area of the ring shaded in
Fig. 174? Indicate the area.
(c ) Simplify the result of (b) by remov-
ing the monomial factor.
(d) What is the area of a running
track in which R = 100 and r = 90 ?
(e) Calculate the area of the shaded
ring in Fig. 174 if R = 5.5 in. and r = 5 ;
if R = 3.75 and r = 0.25.
FIG. 174
6. Allowing 500 Ib. to a cubic foot, find
the weight of a steel pipe 20 ft. long if R = 12 in. and r 11 in.
HINT. Find a rule or formula for the volume of a cylinder.
POSITIVE AND NEGATIVE NUMBERS
207
3C
FlG
7. Find the weight of an iron rod 6 ft. long cast in the
form shown in Fig. 175 if a = 2 in., b \ in y and c = \ in.
HINT. Allow 500 Ib. per cubic foot.
255. Division of polynomials
illustrated by arithmetical num-
bers. The process of dividing
one polynomial by another may
be clearly illustrated by a long-
division problem in arithmetic ;
for example, we shall consider 67,942 -r- 322. Ordinarily we
divide in automatic fashion, adopting many desirable short
cuts which, though they make our work more efficient,
nevertheless obscure the meaning.
In multiplication it was pointed out that because of our decimal
system the 9 in 67,942 does not stand for 9 units, but for 900 units
or 9 10 2 units. Similarly, the 7 means 7000, or 7 10 3 , etc.
If we arrange dividend and divisor in the form of polynomials,
the division may appear in either of the following forms :
60000 + 7000 + 900 + 40 + 2
60000 + 4000 + 400
3000 + 500 + 40
3000 + 200 + 20
300 + 20 + 2
300 + 20 + 2
300 + 20 + 2
200 + 10 + 1
3-10 2
10
6 10* H
^4-10 3 H
-4-10 2
2-10 2 + 10 + 1
3 10 3 -
3-10 3 -
h5-10 2 + 4-10
h2-10 2 + 2-10
3 -10 2 + 2-10 + 2
3 -10 2 + 2 -10 + 2
The student should study the two preceding examples
carefully in order to be better able to understand the simi-
larity of these with the division of algebraic polynomials
which we shall now discuss.
208 GENERAL MATHEMATICS
256. Division of algebraic polynomials. The division of
algebraic polynomials arranged according to either the
ascending or the descending power of some letter is similar
to the preceding division of arithmetical numbers ; thus :
8 .y 4 + 2 y 3 + 4 y~
4 y 3 + 5 y 2 + 3 y
<r l
a s + a 2 /;
a + b
a?-ab
- a*b - ab*
+ aft 2 + b s
+ ab* + b*
257. Process in division. From a study of the preceding
exercises we see that in dividing one polynomial by another
we proceed as follows:
1. Arrange both dividend and divisor according to ascend-
in;! '' descending powers of some common letter.
2. Divide the first term of the dividend by the first term of
the divisor and write the result for the first term of the quotient.
3. Multiply the entire divisor by the first term of the quotient
and subtract the result from the dividend.
4. If there is a remainder, consider it as a new dividend
and proceed as before.
The student should observe that the process in division
furnishes an excellent review of the other fundamental
processes, inasmuch as they are necessary in almost every
division problem. They should therefore be mastered as
soon as possible.
POSITIVE AND NEGATIVE NUMBERS 209
258. Checking a division. We shall now illustrate the
method of checking a division :
Divide x s 3 x 2 y + 3 xy 2 y s by x y.
X s 3 x 2 y + 3 xy 2 y s \x y
x 3 x 2 ;/ | x' 2 2 xy -f y 2
2 x 2 y + 3 xy 1
xy 2 - y 9
xy 2 y 3
First method of checking. Since the division is exact
(that is, there is no remainder), multiply the divisor by the
quotient. If the product equals the dividend, the problem
checks. This may be expressed in cases where there is no
remainder by the formula q x d = D, or q
tZ
How would you check if there were a remainder? If
the answer is not obvious, try to check similar problems in
long division in arithmetic.
Second method of checking. Assume values for x and y.
Let x = 5 and y 2.
Substituting in the example,
and d 3.
D 27
Substituting, the formula q = becomes = 9, or 9 = 9.
d o
Since 9 (or q) = 9 ( or ), the problem checks.
\ /
259. Importance of a thorough drill in division. In the
process of division practically all the principles of the last
two chapters are involved. Hence the following exercises
are important as a means of reviewing the fundamental
laws of addition, subtraction, multiplication, and division.
210 GENERAL MATHEMATICS
EXERCISES
Divide, and check by either method :
1. (x 2 - llx + 30) -=-(x- 5).
2. (// - f - 4y + 4)-<y - 3y + 2).
3. ( ft s + 7 + 18 a + 40) -r- (a 1 + 2 a + 8).
4. (9-9x + 8x 2 -4x 8 )-i-(3-2.r).
5.
6. (27 x 8 - 54x 2 // + 36xy 2 -
7. (27 x 8 + 54 afy + 36 xy* + 8 y 8 ) - '(3 x + 2 y).
8. x 8 --x-7/.
9.
10.
11. (1& + 8 m - 32 m 2 + 32 m s - 15 m 4 ) -=- (3 + 4 m - 5 m 2 ).
12. (x 8 + 2xy + xz + yz + if) + (x + y + ).
13. (14x + 2x 4 + llx 2 + 5x 8 - 24)--(2x 2 +'3x - 4).
14. (r 8 + 65 r - 15 r 2 - 63) -s- (r - 7).
15. (25 a - 20 a 2 + 6 a 8 - 12) -- (- 4 a + 2 a 2 + 3).
16. (8x - 4 + 6x 4 + 8x 8 - Ilx 2 )-r-(4x 2 + 2x 3 - x + 2).
17. (9 x 2 // 2 - 6 x 8 ?/ + x 4 - 4 ?/ 4 ) ^ (x 2 - 3 xy +
18. (25 x 4 - 60 x 2 / + 36 y 4 ) -=- (5 x 2 - 6 y 2 ).
19. (4 x 4 + 12 a-y + 9 ?/) H- (2 x 2 + 3 y 2 ).
20. (a 5 -1) -=-(-!); (a a -l)-s-(a-l).
21. (a 5 -7/ 5 )-(a-7/); (a*-,ft + (a-y).
. 22. (25m 4 -49/i 4 )^-(5m 2 + 7w 2 ).
23. (25 m 4 - 49 w 4 ) -- (5 m 2 - 7 w 2 ).
24. (0.027 aW + c 8 ) -h (0.2 aft + c).
25. (8 a 8 - Z> 8 ) -- (4 a 2 + 2 ab + 6 2 ).
POSITIVE AND NEGATIVE NUMBERS 211
1 2 3 x
260. Division by zero. The quotients - - - -, ,
etc. have no meaning, for a number multiplied by
cannot give 1, 2, 3, x, etc. (see the definition of division in
Art. 243). The quotient - is undetermined, as every num-
ber multiplied by equals 0. Therefore we shall assume
that in all quotients hereafter the divisor is not zero nor
equal to zero.
EXERCISES
1. The following solution is one of several that are some-
times given to show that 1 = 2. Find the fallacy.
Two numbers are given equal, as x y.
Then x - y - 0, Why ?
and 2 (a; - y) = 0. Why ?
Then x - y = 2 (x - y). Why ?
Dividing both sides by x y, 1 = 2.
2. Give a similar argument which seems to show that 2
equals 5.
SUMMARY
261. This chapter has taught the meaning of the follow-
ing words and phrases : turning tendency, force, lever arm,
multiplication, division, factoring, factors, prime number,
number containing a monomial factor.
262. The law of signs in multiplication was illustrated
(1) geometrically with line segments and (2) by means
of the balanced beam.
263. The following agreements were used :
1. A weight pulling downward is negative; one pulling
upward is positive.
2. A force tending to rotate a bar clockwise is negative ;
counterclockwise, positive.
212 GENERAL MATHEMATICS
3. A lever arm to the right of the point where the bar is
balanced is positive ; to the left, negative.
The turning tendency (or force) acting upon a balanced
bar is equal to the product of the weight times the lever arm.
264. Law of signs in multiplication: The product of
two numbers having like signs is positive ; the product of two
numbers having unlike signs is negative.
265. The chapter has taught and geometrically illus-
trated the following processes of multiplication:
1. The multiplication of several monomials.
2. The multiplication of a monomial by a polynomial.
3. The multiplication of polynomials by polynomials.
The order of factors may be changed without changing
the product.
The value of a product is zero if one of the factors
is zero.
266. Law of division : The quotient of two numbers hav-
ing like signs is positive ; the quotient of two numbers having
unlike signs is negative.
Arithmetical numbers may be arranged in the form of
polynomials according to powers of 10.
The process of dividing one polynomial by another is essentially
the same as the process of dividing arithmetical numbers.
In all problems of the text the divisor is not zero.
267. The chapter has taught the following forms of
division :
1. The division of a monomial by a monomial.
2. The division of a polynomial by a monomial.
3. The reduction of a fraction to lowest terms.
4. The division of a polynomial by a polynomial.
POSITIVE AND NEGATIVE NUMBERS 213
268. Division has been illustrated geometrically. Two
methods for checking division were taught.
269. The following types of factoring were taught:
Type I. Taking out a common monomial factor,
ax + bx -f ex = x (a + b + c).
Type II. The " cut and try " method,
acx 2 + (be + ad~) x+bd = (ax + 6) (ex + d).
CHAPTER X
GRAPHICAL REPRESENTATION OF STATISTICS; THE
GRAPH OF A LINEAR EQUATION
270. Facts presented in the form of a table. The follow,
ing table of facts shows in part the recreational interests
of the boys and girls of certain Cleveland (Ohio) high
schools. Thus, of 4528 boys, 4075 play baseball ; of 3727
girls, 2608 play baseball ; 7402 children out of a total of
8255 attend the movies regularly; and so on.
TABLE OF RECREATIONAL INTERESTS
BOYS
GIRLS
TOTAL
Number of students
4528
3727
8255
Number who play baseball
4075
2608
6683
Number who play basketball
Number who play tennis
3018
1811
1390
1864
4408
3675
Number who belong to Camp Fire Girls .
Number who wrestle ....
1358
621
621
1358
Number who attend movies
4010
3392
7402
Number who attend movies daily ....
754
485
1239
EXERCISE
Study the preceding table until you understand the mean-
ing of the columns of figures.
271. Pictograms; graphs. Tables made up of columns of
figures are common in newspapers, magazines, and books,
but a table like the preceding is not the best device for
214
REPRESENTATION OF STATISTICS 215
expressing the meaning of an array of facts. The ordinary
mind cannot see the relations if the list is at all extended ;
hence it often happens that the real meaning of a series of
facts is lost in a complicated table. Newspapers, magazines,
trade journals, and books, realizing this fact, are beginning
to add to tables of statistics pictures which show their
meaning and their relationships more clearly than can be
done by columns of figures.
The significance of the facts of the preceding table is
far more vividly expressed by the pictures of Fig. 176.
Thus the pictures show that of the high-school girls one
out of every two (50 per cent) plays tennis ; two out of
every dozen (16| per cent) are Camp Fire Girls ; of the
high-school boys six out of every twenty (30 per cent)
wrestle ; 85 per cent of all the elementary-school and
high-school boys attend the movies regularly; and so on.
The pictures constitute a more powerful method of
teaching numerical relations, because they teach through
the eye. For this reason they are called graphic pictures,
pictograms, or simply graphs.
EXERCISES
By means of the pictograms in Fig. 176 answer the following
questions :
1. What per cent of the Cleveland boys play tennis ? of
the Cleveland girls ? With which group is tennis the more
popular ?
2. Assuming that every sound-bodied boy should learn to
wrestle, does your class make a better or a worse showing in
per cents than the Cleveland boys ?
3. Are a larger per cent of the girls of your class Camp
Fire Girls than is the case in the Cleveland high schools ?
High-School Girls
High-School Boys
High-School Girls
High-School Boys
High-School Girls
Do not play Tennis
Do not play Tennis
We Never Do
We Never Do.
We Never Do
We play Baseball I Never Do
o^o High-School Girls ^^^^^^U^AA,
Belong to Camp Fire Do not belong to Camp Fire
High-School Boys
Wrestle Do not Wrestle
Elementary- and High-School Boys
Attend Movies
Daily Nonattendance
High-School Boys \j
Help!
High-School Girls ., laU_^W^_
S. 0. S, Board of Education !
Do not attend Movies
FIG. 176. SHOWING HOW PICTOGRAMS ARE CSED TO EXPRESS FACTS
(Adapted from Johnson's "Education through Recreation ")
216
REPRESENTATION OF STATISTICS
217
4. Compare your class's swimming record with the Cleve-
land record.
5. Continue the discussion with your classmates until it is
clear to you just how the figures of the pictograms represent
the facts of the table.
272. The circle pictogram. The circle is frequently used
to show quantitative relations. It shows two things:
(1) the relation of each magnitude to each of the others ;
(2) the relation of each magnitude to the sum of all.
The graph in Fig. 177 shows that in a certain year
Portland (Oregon) spent 30.7$ out of every dollar on
its school system. This is prac-
tically j 3 ^ of all the money spent
by the city and about 1^ times
as much as the next highest
item. Compare the part spent
for education with each of
the other parts shown in the
figure.
The scale used in making
circle pictograms is based on
the degree. The angular space
around the center of the circle
(360) is divided into parts so
as to express the numerical re-
lations ; for example, since almost ^ of the money is
spent for education, an angle of ^ of 360, or 108, is
constructed with the protractor at the center of the circle
and the sides of the angle extended until they intersect
the circle. The sector (the part of a circle bounded by
two radii and an arc) formed shows the part of the city's
money that goes toward education.
FIG. 177. THE CIRCLE
PICTOGRAM
(From Cubberly's " Portland
Survey ")
218 (iKXERAL MATHEMATICS
EXERCISES
1. In Fig. 177 how does the amount paid as interest com-
pare with the amount paid for police and fire protection ?
2. Do you think it would have been more profitable in the
long run for Portland to pay cash for all public improvements ?
HINT. A definite answer to this problem may be obtained if
several members of the class will solve Ex. 3 and report to the class.
*3. What will it cost a city to build a $100,000 high-school
building if $20,000 of its cost is paid in cash and the remainder
paid by issuing 4 per cent bonds of which $4000 worth are to be
retired (paid) annually ? (All interest due to be paid annually.)
NOTE. The problem must not be interpreted as an argument
showing that bonding (borrowing) is never justifiable.
*4. The discussion of the method of paying the expenses
of the United States for the first year of our participation in
the European War was sharply divided between two groups.
One group favored a large amount of borrowing by the issu-
ance of bonds, while the other advocated a pay-as-you-go
policy, that is, raising the money by taxation. Debate the
merits of the two plans.
5. Show the following facts by means of a circle divided
into sectors :
TABLE SHOWING DISPOSITION OF THE GROSS REVENUE OF
THE BELL TELEPHONE SYSTEM FOR THE YEAR 1917
ITEMS PER CENT
Salaries, wages, and incidentals 50
Taxes . 7
Surplus 4
Materials, rent, and traveling expenses .... 20
Interest 7
Dividends 12
Though widely used, the circle divided into sectors is
not a quite satisfactory method of showing the ratio of
REPKESENTATION OF STATISTICS
219
numbers. In fact, the objections are so serious that the
method of construction was given to protect the student
against false conclusions. The method is not inaccurate
when the parts which constitute a unit are shown by the
use of one circle. It frequently
happens, however, that the com-
parison is made by circles differ-
ing in size. In such a case, since
the eye tends to make the com-
parison on an area basis, the ratio FlG - 178 - CIRCLES DRAWN ON
P.I v 1111 AN AREA BASIS SHOWING THE
of the two numbers should be ex- BB op BANK DEPOSITORS
pressed by the ratio of the areas
of the two circles, and statistical authorities so recommend.
In Fig. 1 78 the circles are drawn on an area basis, but the
right-hand circle appears less prominent than the figures
justify, thus causing the reader
to underestimate the ratio. In
Fig. 179 the circles are drawn
on a diameter basis. The right-
hand circle appears more prom-
inent than the figures justify,
thus causing the reader to over-
estimate the ratio. This feature
is frequently utilized by those who make dishonest use
of circle diagrams. The conclusion is that a comparison
between circles differing in size should be avoided alto-
gether. Better graphic methods will be taught. Space is
given here to circle pictograms because of their extensive
use in many fields.
EXERCISE
Test the accuracy of circle pictograms which you may find
in magazine articles and advertisements. Discuss their value
with your classmates.
FIG. 179. CIRCLES DRAWN ON
A DIAMETER BASIS
220
GENERAL MATHEMATICS
1911
1899
273. Area pictograms. The picture of the
two traveling men given here is intended to
show the increase in the passenger traffic of
the railroads. The two men are compared
on the basis of height.
The 1911 man, on
account of his far
greater area, looks
more than 2^ times
as large as the 1899
man. The men should
be compared on the
basis of area. 1 This
type too should be
avoided because it tends to deceive the ordinary reader.
14,591,000 One Mile 32,837,000 One Mile
FIG. 180. A POPULAR TYPE OF PICTOGRAM,
TO BE AVOIDED
1911
32.837000
ONE MILE
mmmmM
1899
14.59I.OOO
ONE MILE
FIG. 181. A MORE ACCURATE METHOD OF PORTRAYING FACTS
EXERCISES
1. Why would it be difficult to make a drawing on the
basis of area?
2. Do you know any method which could be used to check
a drawing made on the basis of area ? (See Art. 109.)
274. Volume or block pictograms. Cubes, parallelepipeds,
and spheres are frequently used in comparing relative
volumes ; for example, pictures of bales of hay or cotton
1 Brinton, in his excellent text, ' ' Graphic Methods for Presenting Facts,' '
presents a chart (Fig. 181) drawn from the same facts as that in Fig. 180.
Note that the facts are portrayed much more clearly and accurately.
REPRESENTATION OF STATISTICS 221
are used to show the output of the states producing these
articles. The comparison should be made on a basis of
volume, but often there is no way for the reader to tell
on what basis the drawing was constructed, whether by
height, area, or volume. Certainly it would be difficult to
check the statement made in such a case.
275. Limitations of area and volume pictograms. The
student will need to remember that in a correctly con-
structed area graph the quantities represented should vary
directly as the number of square units within the out-
lines of the figures. Thus, in the comparison of passenger
service relative size should not be determined by the
relative heights of the men but by the number of square
units within the outlines. Hence a rough method of
checking is to transfer the pictures of the traveling men to
squaredpaper by means of tracing paper and compare the
number of square millimeters in the area of each with the
corresponding facts of the table. Similarly, in accurate
volume or block graphs the quantities should vary as the
number of cubic units.
Many who use this form of statistical interpretation
carelessly fail to observe these principles, and the diffi-
culty of a check makes this form of graph a convenient
device for those who would dishonestly misrepresent the
facts. The general public is not always able to interpret
the graphs correctly even if they have been properly drawn.
Because of these limitations it is somewhat unfortunate
that this type of graph is so extensively used in bulletins
and current magazines.
EXERCISE
Try to obtain and present to the class an advertisement
illustrating the misuse of an area or volume pictogram.
222
GENERAL MATHEMATICS
276. Practice in interpreting the bar diagram. Fig. 182
shows one of the suggestions of the Joint Committee on
Standards for Graphic Presentation. The diagram, Fig. 182,
(a), based on linear measurement, is called a bar diagram.
We shall study this topic further in the next article. Review
Year
1900
1914
FIG
Tons
270.588
555,031
(a) (b)
182. BAR DIAGRAMS SHOW FACTS BETTER THAN AREA AND
VOLUME PICTOGRAMS
the objections to the other two diagrams (the squares and
blocks shown in Fig. 182, (b)). Where it is possible the
student should represent quantities by linear magnitudes,
as representation by areas or volumes is more likely to
be misinterpreted.
EXERCISES
1. Study Fig. 183 and determine to what extent the two
horizontal bars are helpful in expressing the ratio of the two
numbers given.
2. Would the UXfl 1 Cotton, $820,320,000
bars in Fig. 183
be sufficient with-
out the illustra- 1HI" Wheat, $561,051,000
tions at the left
of the numbers ? ^ IG ' ^' ^ F AIR DIAGRAM. (AFTER BRINTON)
3. With the aid of compasses check the accuracy of Fig. 184.
Note that the figures are written to the left of the bars. In many
woe $4.409,136 ^1 bar diagrams the figures
^^^^^^^^^^ are written to the right
1912 28 soo 139 ^^^^m ^ ^ e bars. Can you
FIG. 184.1 DIAGRAM SHOWJNG .N Ex- think of a serious ob .l ec -
PORTS OF AUTOMOBILES. (AFTER BRINTON) tion to that method ?
1 See paragraph 7 under Art. 277
REPRESENTATION OF STATISTICS
223
4. Why is there a space left between the bars for 1906 and
1911 in Fig. 184 ? Do you see any other way to improve the
diagram ? (See Art. 277.)
5. Draw on the blackboard a figure similar to Fig. 184,
adding a bar for the
year 1917. (The sum
for this year is about
900,000,000.)
6. Explain Fig. 185.
7. Show that the
bars of Fig. 186 reveal
more clearly than the following table the rank of the United
States in respect to wealth. These are the 1914 estimates.
United States $150,000,000,000
FIG. 185. DIAGRAM SHOWING DEATH
KATE FROM TYPHOID IN 1912 PER HUN-
DRED THOUSAND POPULATION
Great Britain and Ireland
Germany
France
Russia
Austria-Hungary . . .
Italy
85,000,000,000
80,000,000,000
50,000,000,000
40,000,000,000
25,000,000,000
20,000,000,000
8. Show that it would have been as accurate and more con-
venient to draw the preceding diagrams on squared paper.
75 100 125 150
UNITED STATES 150
GREAT BRITAIN or
AND IRELAND- 5
GERMANY 80
FRANCE - - 60
! <; ' f j
i ; - 1 ' \ '
';>>' \ ;
: . U
RUSSIA -.40
' -- i
AUSTRIA-HUNGARY. 25
ITALY 20
. . .;>; ..
mm
FIG. 186. COMPARATIVE WEALTH OF NATIONS IN 1914
9. The table for the wealth of nations contains estimates
prepared at the beginning of the European War (1914-1918).
-2'24 - GENERAL MATHEMATICS
These estimates are now far from facts. The student should
attempt to get the latest estimates from the " World Almanac "
and construct a bar diagram which will present the situation to
date and will enable him to make an interesting comparison.
10. Discuss bar diagrams similar to those given on pages
222-223 which you may find in Popular Mechanics Magazine,
Motor, Popular Science Monthly, and Industrial Management.
For the time being limit yourself to the simpler diagrams.
277. How to construct a bar diagram. An understanding
of how to construct bar diagrams and how to interpret
those he may find in newspapers and magazines should
be a part of the education of every general reader, just as
it is of every engineer, physician, statistician, and biologist. 1
As civilization advances there is being brought to the atten-
tion of the reading public a constantly increasing amount
of comparative figures of a scientific, technical, and statis-
tical nature. A picture or a diagram which presents such
data in a way to save time and also to gain clearness is
a graph. The bar diagram is a widely used method of con-
veying statistical information graphically. The solution of
the introductory exercises along with the discussion of such
supplementary graphs as may have seemed profitable for the
class to discuss will help the pupil to understand the follow-
ing outline of the method of constructing a bar diagram :
1. The bars should be constructed to scale. To obtain
a convenient unit first inspect the size of the smallest
and the largest number and then choose a line segment to
1 Neither pupils nor teachers should be misled by the apparent sim-
plicity of this work. The details are of the greatest importance. It will
be helpful to obtain the reports of the Joint Committee on Standards for
Graphic Presentation. This is a competent committee of seventeen, which
has expended considerable effort on these elementary phases. The pre-
liminary report may be had from the American Society of Mechanical
Engineers, 29 W. 39th St., New York ; price, 10 cents.
KEPRESENTATION OF STATISTICS 225
represent a number such that it will be possible to draw an
accurate bar for the smallest number and a bar not too long
for the largest number. The lines in Fig. 186 on page 223
are so constructed that the relation between the lengths
of any two is the same as the relative size of the quan-
tities represented. A line segment 1 mm. long represents
$2,500,000,000 of wealth. In the table on page 223 the
United States is- estimated as possessing three times as
much wealth as France, and so the line segments repre-
senting the wealth of the United States and France are
respectively 60 mm. and 20 mm. long.
2. The scale and sufficient data should appear on the
diagram.
3. Each bar should be designated.
4. The bars should be uniform in width.
5. The diagram, should have a title or legend.
6. Accuracy is the important characteristic.
7. The space between the bars should be the same as the
width of the bars, except in a case like Fig. 184, where a
larger space indicates that the three bars do -not represent
consecutive years.
8. In general the zero of the scale should be shown. How-
ever, there are exceptions ; for example, in graphing the tem-
perature of a patient we are particularly concerned with how
much above or below the normal the patient's temperature
is. Hence, in a case like this we should emphasize the
normal temperature line.
EXERCISES
1. Present the facts of the table given on page 226 by means
of a bar diagram, using the scale 1 mm. = f 2,000,000. The table
is arranged to show the twenty heaviest buyers of American
goods, as indicated by the value of exports from the United
States during the fiscal year 1914.
GE;NEKAL MATHEMATICS
AMERICA'S TWENTY BEST CUSTOMERS
(From the report of the Bureau of Foreign and Domestic Commerce)
VALUE OF
PURCHASES
VALUE OF.
PURCHASES
1. England . .
$548,641,399
11. Argentina . .
$45,179,089
2. Germany .
344,794,276
12. Mexico
38,748,793
3. Canada . . .
344,716,981
13. Scotland . . .
33,950,947
4. France . . .
159,818,924
14. Spain . . . .
30,387,569
5. Netherlands .
112,215,673
15. Russia ....
30,088,643
6. Oceania . .
83,568,417
16. Brazil ....
29,963,914
7. Italy . . .
74,235,012
17. China ....
24,698,734
8. Cuba . . .
68,884,428
18. Austria-Hungary
22,718,258
9. Belgium . .
61,219,894
19. Panama . . .
22,678,234
10. Japan . . .
51,205,520
20. Chile ....
17,482,392
*2. If possible, ascertain the facts to date (see "World Alma-
nac "), graph results as in Ex. 1, and compare the two diagrams.
Account for unusual changes. Are new customers appearing
among the " twenty best " ? Have old ones dropped out ?
3. Present the statistics of the following table by means of a
bar diagram showing the comparative length of rivers. (Use the
scale 1 cm. = 400 mi. ; the lengths given in the table are in miles.)
LENGTH
LENGTH
Missouri-Mississippi . .
4200
Volga
2400
Amazon
3800
Mackenzie
2300
Nile
3766
Plata . . ...
2300
Yangtze
3400
St Lawrence
2150
Yenisei
3300
Danube
1725
Kongo
3000
Euphrates
1700
Lena
3000
Indus
1700
Niger
3000
Brahmaputra
1680
Ob*
2700
Ganges
1500
Hoangho
2600
Mekon^
1500
Amur
2500
Rio Theodoro
950
REPRESENTATION OF STATISTICS
227
4. Represent the statistics of the following table by bar
diagrams. The estimates of the leading crops in the United
States for the year 1917 are here compared with the revised
figures for the crops of the preceding nine years. The pupil
should note that each column is a separate problem.
REPORT OF THE UNITED STATES DEPARTMENT OF
AGRICULTURE FOR 1917
YEAR
CORN
WHEAT
OATS
BARLEY
RYE
COTTON
Bushels
Bushels
Bushels
Bushels
Bushels
Bales . '
1917 ....
3,159,494,000
650,828,000
1,587,286,000
208,975,000
60,145,000
10,949,000
1916 ....
2,583,241,000
639,886,000
1,251,992,000
180,927,000
47,383,000
12,900,000
1915 ....
2,994,793,000
1,025,801,000
1,549,030,000
228,851,000
54,000,050
12,862,000
1914 ....
2,672,804,000
891,017,000
1,141,060,000
194,953,000
42,778,000
15,136,000
1913 ....
2,446,988,000
763,380,000
1,121,768,000
178,189,000
41,381,000
14,552,000
1912 ....
3,124,746,000
730,267,000
1,418,377,000
223,824,000
35,664,000
14,104,000
1911 ....
2,531,488,000
621,338,000
922,298,000
160,240,000
33,119,000
16,101,000
1910 ....
2,886,260,000
635,121,000
1,186,341,000
173,832,000
34,897,000
12,075,000
1909 ....
2,552,190,000
683,350,000
1,007,129,000
173,321,000
29,520,000
10,513,000
1908 ....
2,668,651,000
664,002,000
807,156,000
166,756,000
31,851,000
13,817,000
278. Bar diagrams used to show several factors. We
shall now see how bar diagrams may be used to show
several factors.
INTRODUCTORY EXERCISES
1. Fig. 187, on page 228, differs from those in Art. 277
in that it presents two factors. What is the scale of the dia-
gram ? Note that the bars representing new buildings extend
from the top to the bottom of the black bar. Try to account for
the heavy losses by fire in 1904 and 1906. Why is the bar so
short for new buildings in 1908 ? (The values are given in
millions of dollars.) Criticize this diagram according to the
principles of Art. 277.
2. Give the facts of Fig. 187 for the twelfth year; the
eighteenth year.
228
GENERAL MATHEMATICS
JNew Building
Fire Losses
1901 1902 1903 1904
1905
1906 1907 1908 1909 1910 1911
FIG. 187. DIAGRAM OF YEARLY VALUES OF NEW BUILDINGS, AND OF ALL
BUILDINGS LOST BY FIRE IN THE UNITED STATES, 1901-1911, INCLUSIVE
(Courtesy of W. C. Brinton)
3. Fig. 188 shows the business relations involved when a city
bonds itself to buy some present need or luxury. The parts
of a single bar (say the tenth) show the following: (a) the
interest paid to date (the black portion) ; (b) the amount
of the $75,000 still outstanding (the plain portion) ; (c) the
part of the debt that has been paid (the crosshatched portion).
Show that a public bond issue is not only a debt but that
it " conies dangerously near to a perpetual tax,"
REPRESENTATION OF STATISTICS
229
19 20
100.000
90.000
80,000
70,000
60.000
50.000
40,000
30.000
20,000
10.000
"-
FIG. 188. BAR DIAGRAM USED TO SHOW MONEY TRANSACTIONS
INVOLVED IN PAVING FOR A $75,000 SCHOOL BUILDING
(Adapted from Ayres's " Springfield Survey ")
The preceding exercises show how a bar diagram may
be used to compare several factors of some problem
which are more or less related. If the pupil is especially
interested in this side of the subject, he may do the
following exercises. The topic is not particularly impor-
tant, however, because another method which we shall
presently stftdy is much more efficient.
EXERCISES
*1. Go to the township, county, or city-hall authorities and
find out how one or more of your public improvements is
being paid for ; that is, find out (a) if bonds were issued ;
(b) how many dollars' worth are retired (paid for) each year ;
(c) how much interest must be paid each year. Construct a
230 GEXEKAL MATHEMATICS
bar diagram similar to the one reprinted from the Ayres report,
showing what it will ultimately cost your community to pay
for the project.
*2. A certain county in Indiana built 20 mi. of macadam-
ized road by issuing $40,000 worth of 4 per cent nontaxable
bonds. Twenty $100 bonds were to be retired each year. By
means of a bar diagram show how much it ultimately cost this
township to build its turnpike.
*3. Ten years after the turnpike referred to in Ex. 2 was
built it was practically worn out. Did this township lend to
posterity or borrow from it ? Give reasons for your answer.
279. Cartograms. Statistical maps which show quanti-
ties that vary with different geographic regions are some-
times called cartograms. The student will doubtless find
examples in his geography. He should also examine the
" Statistical Atlas of the United States," published by
the Census Bureau. Various colors and shades are used
to help express the meaning.
When the cost of color printing is prohibitive the same
ends may be attained by Crosshatch work. The student
should examine rainfall maps containing cartograms and
which are often printed in newspapers.
A special form of cartogram is the dotted map. If \ve
wish to show the density of population of a city, we may
take a map of-the city and place a dot within a square
for every fifty people living in the square. The scale
should be so chosen that the dots will be fairly close
together in the sections whose population is of greatest
density. Space is not given here to illustrating this type,
but the pupil will have no difficulty with the exercises
that follow.
REPKESENTATION OF STATISTICS
231
EXERCISES
1. Obtain at least five different forms of pictograms and
cartograms from newspapers, magazines, trade journals, or
government bulletins. Explain very briefly what each intends
to show.
2. Discuss the merits or defects of the graphs of Ex. 1.
280. Interpreting (or reading) graphic curves. The intro-
ductory exercises given below will furnish the student with
some practice in the interpretation of graphic curves.
1915-
-1916-
\7_
-1917-
FIG. 189. SHOWING RAILWAY-STOCK FLUCTUATIONS, BY MONTHS, IN
THE AVERAGE PRICE OF TWENTY-FIVE OF THE LEADING STOCKS ON
THE NEW YORK STOCK EXCHANGE
(Adapted from the New York Times)
INTRODUCTORY EXERCISES
1. Explain the curve in Fig. 189, noting the highest price,
the lowest price, the cause of the upward trend in 1915, the
cause of the downward movement in 1917, and the cause of
the sharp break upward in the closing days of 1917.
232
GENERAL MATHEMATICS
2. Explain the curve in
Fig. 190. Check the graph
for the early years. Give a
reason for such results as you
may find.
3. Fig. 191 is a temperature
chart of a case of typhoid fever,
(a) Explain the rise and fall
of the curve, (b) What is the
meaning of the dots ? Do these
points mark the tops of bars ?
(c) What assumption does the
physician make when he con-
nects these paints by a curve ?
(d) Note that this diagram does
not have a zero scale ; why was
it omitted? The chart would
be improved if it had an empha-
1910 1911 1912 1913 1914 1915 1916 1917 1918
90
(40
4.3
FIG. 190. NUMBER OF CARS OF
DIFFERENT TYPES
(Adapted from Motor)
sized line representing normal temperature (98.4). Why ? Con-
struct a line in color in your text for the normal-temperature line.
/7
/jajt&t
V&
II IS
It/7
107*
106'
105
104*
103
102*
101"
100'
?
"Itf
YU590 bl
&S> 1
Si/eitr-va
FIG. 191. A "TEMPERATrRE CHART OF A CASE OF TYPHOID FEVER
KEPKESENTATIOX OF STATISTICS
233
4. Explain the heavy line (temperature) of Fig. 192.
5. Explain the light line (wind) of Fig. 192.
6. If your study of science has familiarized you with the
term " relative humidity," explain the dotted line of Fig. 192.
SPCCSPCSPCSPCCRSSCS PCPC RPCRPCRR CRRRRR PCPC
2 4 6 8 10 12 14 16 18 20 22 24 26 28 80
FIG. 192. WEATHER RECORD IN ST. Louis FOR OCTOBER, 1917
The heavy lines indicate temperature in degrees F. The light lines indicate
wind in miles per hour. The broken lines indicate relative humidity in percen-
tage from readings taken at 8 A.M. and 8 P.M. The arrows fly with the pre-
vailing direction of the wind. S, clear ; PC, partly cloudy ; C, cloudy ; R, rain.
(From the Healing and Ventilating Magazine, December, 1917)
The curves of the preceding exercises are called graphic
curves. The graphic curve is particularly useful in com-
paring the relation that exists between ttvo quantities ; for
example, the relation between the price of wheat and the
annual yield ; the relation between the price of beef and
the amount produced or exported; the relation between
office expenses and the size of the corporation ; the tariff
necessary in order that an article may be manufactured in
this country ; and so on.
281. How the graphic curve is drawn. The graphic curve
in Fig. 193 represents the growth of the population in the
United States from 1790 to 1910, as shown by the table
on page 234.
(1) The horizontal line OM represents the time line.
1 mm. represents 2 yr. ; that is, 10 yr. is represented by
234
GENERAL MATHEMATICS
YEAK
POPULATION
YEAR
POPULATION
1790 . . .
3,929,214
I860
31 443 321
1800
5,308,483
1870
38 558 371
1810
7 239 881
1880
50 155 783
1820
9,638 453
1890
62 947 714
1830
12,860,702
1900
75 994 575
1840
1 7 063 363
1910
91 972 266
I860
23,191,876
two small spaces. (2) The vertical scale represents the
population in millions. Two small spaces represent ten
million. Therefore a bar about 1.6 mm. long is placed on the
1790 line, and a second bar a little over 2 mm. long is placed
on the 1800 line.
Similarly, bars
were placed on
the other lines.
(3)The upper end
points of the ver-
tical segments
(bars) are joined
by a curve. In
so far as the
bars are con-
cerned the fig-
ure does not
differ from an
ordinary bar diagram. We may assume, however, that in-
crease in population between any two periods was gradual;
for example, we may estimate that in 1795 the popula-
tion was reasonably near some number halfway between
3,900,000 and 5,300,000 ; that is, about 4,600,000. Simi-
larly, we may estimate the population in the year 1793.
upr
so
y
J n
1
/
I
/
f
/
CO
/
rn
J/
x
x*
X
-
^
x
^
X
X
X
*
^
r^
- '
r
1
0-
-J
O-
|
s
J
_c
S-
u
-0
5
I-
J
R-
j
n-
-0
0-
-j
:\
.
-35-.
FIG. 193. THE GRAPHIC CURVE
REPRESENTATION OF STATISTICS
235
This assumption leads us to draw the smooth curve which
enables us to estimate the change in population without
knowing the exact length of the bars. By means of the
curve predict what the population will be in 1920. In what
way will the accuracy of your prediction be affected by the
European War?
EXERCISES
1. The following table shows the total monthly sales of a
bookstore through a period of two years.
January ....
February ....
1917
$2125
2237
1918
$2329
2416
July
August
1917
$2271
2231
1918
$2380
2350
March
2460
2479
September ....
2542
2620
April
2521
2590
October ...
2725
2831
May .
2486
2580
November ....
2345
2540
June
2393
2482
December .
2825
3129
Draw a graphic curve for each year on the same sheet of
graphic paper. Draw the two curves with different-colored
ink or else use a dotted line for one and an unbroken line for
the other. Explain the curves.
2. On a winter day the thermometer was read at 9A.M. and
every hour afterward until 9 P. ai. The hourly readings were
- 5, 0, - 2, - 8, - 10, - 10, - 5, 0, - 5, - 4, - 2,
3, 7. Draw the temperature graph.
3. Using a Convenient scale and calling the vertical lines
age -lines, graph these average heights of boys and girls :
AGE
BOYS
GIRLS
AGE
BOYS
GIRLS
2yr.
1.6ft.
1.6ft.
12 yr.
4.8ft.
4.5ft.
4
2.6
2.6
14
5.2
4.8
6
3.0
3.0
16
5.5
5.2
8
3.5
3.5
18
5.6
5.3
10
4.0
3.9
20
5.7
5.4
GENERAL MATHEMATICS
At what age do boys grow most rapidly ? At what age do
girls grow most rapidly ? Is it reasonable to assume that the
average height of a boy nineteen years old is 5.65 ft.?
4. The standings of the champion batters from 1900-1907,
inclusive, are here given for the National and American
leagues.
The National League :
0.384 0.382 0.367 0.355 0.349 0.377 0.339 0.350
The American League :
0.387 0.422 0.376 0.355 0.381
0.329 0.358 0.350
Graph the data for each league to a convenient scale, both
on the same sheet. Tell what the lines show.
5. Draw a temperature chart of a patient, the data for
which are given below (see Fig. 191).
Hour . . .
IJl'.-M.
7
8
9
10
11
12
1 A.M.
2
3
4
3
Temperature
100.5'
101
1015
103.2"
102.5
101.4
101.3'
101.3
101.2
101.2
101
100.7
6. If possible, get a copy of a temperature curve such as is
commonly kept in hospitals and explain the graph to the class.
The class will profit more by your discussion if the curve
presents the data for a long period.
7. Be on the lookout for graphic curves which convey infor-
mation of general interest to your class. In nearly every news-
paper you will find tables of statistics which may be plotted to
advantage. Glance occasionally through the "Statistical Ab-
stract of the United States " or the " Statistical Atlas of the
United States" (published by the Bureau of Foreign and
Domestic Commerce), Popular Mechanics Magazine, Popular
Science Monthly, Scientific American, and so on, with the pur-
pose of finding interesting graphs. If a lack of time prevents
class discussion, post these graphs on bulletin boards.
REPRESENTATION OF STATISTICS
237
282. Continuous and discrete series. We may represent
a continuous change in wealth, in population, in the growth
of boys and girls, etc. by a smooth curve. Thus, if we read
four reports of deposits made in a country bank as $20,000
on January 1, $25,000 on April 1, $18,000 on June 1,
and $19,000 on September 1, we assume that there was
a gradual increase of deposits from January 1 to April 1,
a rather rigorous withdrawal from April 1 to June 1, and a
slow rally from June 1 to September 1. This is precisely
the way a physician treats the temperature of a patient,
even though he may take the temperature but twice per day.
However, the data of every table cannot be considered as
continuous between the limits. This fact is clearly illus-
trated by the following table of Fourth of July accidents :
YEAR
KILLED
INJURED
TOTAL
1909 .... ...
215
5092
5307
1910
131
2792
2923
1911
57
1546
1603
1912
41
945
986
1913
32
1163
1195
1914
40
1506
1546
If we were to draw a continuous curve, it would not
state the facts. Though a few Fourth of July accidents may
occur on the third or on the fifth of July, we are certain
that a continuous curve would not represent the facts for
the rest of the year. Such a collection of items is said to
be a discrete, or broken, series. A record of wages paid in
a factory is likely to be a discrete series, for the wages are
usually (except in piecework) a certain number of dollars
per week, the fractional parts* being seldom less than 10 <.
We should find very few men getting odd sums, say, $18.02
per week. Hence there would be many gaps in the series.
238 GENERAL MATHEMATICS
283. Statistics as a science defined. We have now pro-
gressed far enough for, the student to understand that the
term " statistics " refers to a large mass of facts, or data,
which bear upon some human problem. One of the chief
uses of statistics as a science is to render the meaning of
masses of figures clear and comprehensible at a glance.
Statistics gives us a bird's-eye view of a situation involv-
ing a complex array of numerous instances in such a way
that we get a picture which centers our attention on a
few significant relations. Such a view shows how one fac-
tor in a complicated social or economic problem influences
another ; in short, it enables us to understand the relation
between variable (changing) quantities.
284. The uses of statistics. Statistical studies do not
exist merely to satisfy idle curiosity. They are necessary
in the solution of the most weighty social, governmental, and
economic problems. Do certain social conditions make for
increase in crime and poverty ? The sociologist determines
statistically the relations bearing on the question. Are cer-
tain criminal acts due to heredity ? The biologist presents
statistical data. Is tuberculosis increasing or decreasing?
Under what conditions does it increase ? Reliable statistics
presented by the medical world guide our public hygienic
policies. Further possibilities of statistical studies in the
medical world are suggested by the recent work of Dr. Alexis
Carrel. The work of Dr. Carrel has been widely discussed.
Though authorities disagree concerning some of the details,
all will probably agree that the mathematical attack on the
problem of war surgery is a distinct scientific advance.
What insurance rates ought we to pay? Statistical
investigations have determined laws for the expectation of
life under given conditions which for practical purposes
are as accurate as the formula for the area of a square.
REPRESENTATION OF STATISTICS 239
The business world at times trembles under the threat
of gigantic strikes that would paralyze all business. Are
the demands of the men unreasonable ? Are the corpora-
tions earning undue dividends ? The public does not
know and will not know until a scientific group of citizens
present reliable statistics of earnings and expenditures.
There is now in existence in Washington a nonpar-
tisan tariff commission which consists of five members ap-
pointed by the president, which collects statistics and
makes recommendations to Congress from time to time.
It is now thought that this commission may tend to do
away with the. old haphazard methods of handling tariff
questions.
How rapidly and with what degree of accuracy should
a fourth-grade pupil be able to add a certain column of
figures ? The educator is able to present an answer based
on tests of more than 100,000 fourth-grade children for
that particular problem.
Because of the numerous trained enumerators which
they employed to cover the world's output, the large spec-
ulators on the Chicago Board of Trade knew with absolute
certainty for days in advance of the record-breaking jump
in the price of wheat in August, 1916.
We might continue indefinitely to present evidence
showing that the intelligent reader in any field will profit
by a knowledge of the elementary principles of statistical
methods.
285. Frequency table ; class limits ; class interval. In
the investigation of a problem it is necessary that the
data be tabulated in some systematic fashion that will
enable us to grasp the problem. Suppose we measured
the length of 220 ears of corn. Let us say the smallest
ear measures between 5 in. and 6 in., the longest between
240
GENERAL MATHEMATICS
12 in. and 13 in. We could then group the ears by inches,
throwing them into eight groups, and tabulate the results
somewhat as follows:
FREQUENCY TABLE SHOWING LENGTH OF EARS OF CORN
LENGTH IN INCHES
NUMBER
OF EARS
LENGTH IN INCHES
NUMBER
OF EARS
6-5.99
2
9- 9.99
74
6-6.99
4
10-10.99 .....
47
7-7 99
20
11-11.99 ...
21
8-8.99
48
12-12.99
4
70
CO
Such an arrangement of data is called a frequency table.
The ears of corn have been di- niiiimiiiiiiiiiiiimmiiimiimni
vided into classes. The table
should be read as follows :
rt There are two ears measuring
somewhere between 5 in. and
6 in., four between 6 in. and
Tin.," etc. The boundary lines
are known as class limits, and
the distance between the two
limits of any class is designated 40
as a class interval. The class
interval in the preceding case
is 1 in. Class intervals should
always be equal.
The facts of the table are
shown by the graph in Fig. 194. 10
This graph is the same as the
bar diagram (Fig. 186) which
we have drawn, with the excep-
tion that in this case the bars
cover the scale intervals.
so
20
7 8 9 10 11 12 13
FIG. 194. DISTRIBUTION OF FRE-
QUENCY POLYGON OF 220 EARS
OF CORN
REPRESENTATION OF STATISTICS
241
EXERCISES
1. What is your guess as to the shape of the graph if the
corn were classified into half-inch groups ? into fourth-inch
groups ?
2. Are there many very long ears of corn ? many very short
ears ? How do you tell ?
3. Into what class interval does the largest group fall ?
4. The following table of frequency shows the lengths
of 113 leaves picked from a tree purely at random. (See
King's Elements of Statistical Method," p. 102.)
LENGTH OF LEAF IN
CENTIMETERS
NUMBER OF
LEAVES IN
EACH GROUP
LENGTH OF LEAF ix
CENTIMETERS
NUMBER OF
LEAVES IN
EACH GROUP
3-3.99 ....
4
8-8.99 ....
6
4-4.99 ....
5
9-9.99 ....
4
5-5.99 ....
13
10-10.99 . . .
3
6-6.99 ....
56
11-11.99 . . .
2
7-7.99 ....
19
12-12.99 . . .
1
The table should be read as follows : " There are four leaves
between 3 cm. and 4 cm. long, five between 4 cm. and 5 cm.," etc.
Graph as in Fig. 194 the data of this table.
5. The frequency table on page 242 shows the weights of
1000 twelve-year-old boys, 1000 thirteen-year-old boys, and
1000 fourteen-year-old boys. (The weights are taken from
Roberts's "Manual of Anthropometry" and include 9 Ib. of
clothing in each case. Figures are reduced to the thousand
basis.)
6. Study each of the groups for Exs. 4 and 5 as to
(a) Where the smallest classes are found.
(b) Where the largest class is located.
(c) The gradual rise and fall of the figures.
( i KNERAL M ATHEMATICS
FREQUENCY TABLE OF WEIGHTS SHOWING 1000 TWELVE-
YEAR-OLD BOYS, 1000 THIRTEEN-YEAR-OLD BOYS, AND 1000
FOURTEEN-YEAR-OLD BOYS
WEIGHTS IN
POUNDS
NUMBER OF
12-YEAR-OLD BOYS
NUMBER OF
13-YEAR-OLD BOYS
NUMBER OF
14-YEAR-OLD BOYS
49-56
4
56-63
24
6
63-70
118
38
'3
70-77
233
100
38
77-84
273
225
41
84-91
221
256
130
91-98
79
187
228
98-105
36
112
^47
105-112
12
43
118
112-119
17
82
119-126
16
29
126-133
12
133-140
18
NOTE. The fact that the third column in the table lacks 54 boys of
totaling 1000 is because 54 boys in this group weighed over 140 pounds.
The 49-56 means from 49 up to but not including 56'. Wherever the
last number of a class is the same as the first number of the next class,
the first class includes up to that point, but does not include that point.
Construct the graph (similar to Fig. 194) for each of the
groups of the table above.
7. The following test on the ability to use the four funda-
mental laws in solving simple equations was given to 115 first-
year high-school students, the time given for the test being
fifteen minutes.
DIRECTIONS TO PUPIL
Find the value of the unknown numbers in each of the following
equations. Do not check your results. Work the problems in order
if possible. If you find one too difficult, do not waste too much time
on it, but pass on to the next. Be sure that it is too difficult, however,
before you pass on. Do not omit any problem which you can solve.
243
1. x + 3 = 7.
2. 2 #=4.
3. 2 fc + 7 = 17.
4. z - 2 = 3.
5. 2 x - 4 = 6.
. = ,
7 5 y _ 15
THE TEST
13. 16 y + 2 y - 18 y + 2 ij = 22.
14. 7x + 2 = 3z + 10.
.
o
10. 3.c + 4| =9.
17. 5.3 y + 0.34 = 2.99.
18. 0.5x-3 = 1.5.
19. 3 x -9^ = 17.5
20. 7 y + 20 - 3 y = 60 + 4 y + 40 - 8 //.
21.
66 18 3
11. + 1 = 6.
4
12. |-4=10.
24.
25 a;
The results of the test are given by the following table
of frequency. The student should study it carefully.
NUMBER OF
EXAMPLES
NUMBER OF
STUDENTS
ATTEMPTING
NUMBER OF
STUDENTS
SUCCESSFUL
NUMBER OF
EXAMPLES
NUMBER OF
STUDENTS
ATTEMPTING
NUMBER OF
STUDENTS
SUCCESSFUL
13
1
11
1
1
14
4
15
2
1
15-
4
8
3
16
10
10
4
17
13
3
5
18
13
7
6
3
19
8
1
7
4
20
22
2
8
1
5
21
13
9
9
22
18
10
14
23
6
1
11
13
24
1
12
1
7
244 GENERAL MATHEMATICS
Explanation. The table consists of two parts, of which the first
part is the first, second, fourth, and fifth columns, which should be
read, " Of the one hundred and fifteen students taking the test one
tried but 8 examples, one attempted 12, one attempted 13, four
attempted 14," etc. The second part consists of the first, third,
fourth, and sixth columns and should be read, " Of the one hundred
and fifteen students taking the test one solved only 1 problem cor-
rectly, one solved only 2 correctly, three solved 6 correctly, four
solved 7 correctly," etc.
8. Construct a graph showing the facts of the table given
on page 243 for the number of students attempting.
9. Under the directions of your instructor take the test
in Ex. 7.
10. Ask your instructor to give you a frequency table show-
ing the number of attempts and successes in the test taken by
your class and determine how the work done by your class
compares in speed and accuracy with that done by the one
hundred and fifteen students in the test described in Ex. 7.
286. Measure of central tendencies ; the arithmetic
average. A frequency table and a frequency graph help
us to understand a mass of facts because they show us the
distribution of the items, so that we see where the largest
groups and the smallest groups fall. The graph shows us
the general trend of the facts. The large groups assume
importance. We need terms to describe the central tend-
ency. Often the word " average " is used to meet this
need. Such a term is helpful in making a mass of facts
clear. Thus, a group of farmers could not possibly learn
much about a field of corn if we read a list to them show-
ing the length of every ear in a field. But they would
get some idea of what yield to expect if told that the
average length of an ear is 91 in. They could certainly
give a fair estimate of the yield if in addition we told
REPRESENTATION OF STATISTICS 245
them that on tha average a row in a square 40-acre field
grew 620 stalks. We shall presently learn that the word
" average," as commonly used, is not correct. The phrase
" arithmetic average " means the quotient obtained by divid-
ing the sum of all the items by the number of items. Thus,
to find the average mark obtained by your class on a test
we need to add the marks of all the students and divide
by the number of students in the class. If two or more
students obtain the same mark (say 70), we can shorten
the first step of the process by multiplying the mark by the
number of times it occurs instead of adding 70 five times.
This means that in a frequency table a student must re-
member to multiply each item by its frequency before adding.
When the size of the item is only approximately known,
the mid-point of the class interval is taken to represent
the size of each item therein. To illustrate, suppose that
we should try to find the average number of problems
attempted in the simple-equation test. We shall suppose
that three students report that they attempted 6 problems.
This does not really mean that all three exactly completed
6 problems when time was called. In all probability one
had made a slight start on number 7, the second was about
in the middle of number 7, and the third had almost
completed 7. Of course the number of students is too
small to make this certain, but if we should take a larger
number of students (say thirteen), in air probability there
would be as many who were more than half through with
the seventh problem as there would be students less than
half through. Hence, to find the average we say that the
thirteen students attempted 6i and not 6, as they reported.
To illustrate:
Find the average number of equations attempted by a class
on the simple-equation test if two students report 5 problems
24G GENERAL MATHEMATK S
attempted, four report 6, five report 7, three report 8, and two
report 9.
Solution. 2x5 = 11
4 x 6 = 26
5 x 7 1 = 37.:.
:5 x 8i = 2').-,
2x9* = 19
16 119
110 -=-16 =7.4.
Therefore the average number of problems attempted by the
class is 7.4.
The point is that the series of facts in the table is not
a discrete series, as one would at first be inclined to think,
but a continuous series.
EXERCISES
*1. Calculate the average number of equations attempted by
your class in the simple-equation test (Art. 285).
2. Using the table of Ex. 5, Art. 285, find the average
weight of the twelve-year-old boys ; of the thirteen-year-old
boys ; of the fourteen-year-old boys.
3. Find the average length of the 113 leaves in Ex. 4,
Art. 285.
4. Find the average length of the 220 ears of corn of the
first table in Art. 285.
5. Find the average number of Fourth of July accidents
for the six years of the table of Art. 282.
6. Compare the averages of the champion batters for the
last eight years (Ex. 4, Art. 281).
7. At Minneapolis the 7 A.M. temperature readings for
the ten days beginning February 1 were as follows : 5, 3,
5, 3, 7, 9, 8, 2,0, 6. Find the average 7 A.M.
temperature reading for the period.
REPRESENTATION OF STATISTICS 247
8. Find the average of the following temperatures: 8 A.M.,
-6; 9 A.M., -5; 10A.M., -2; 11 A.M., -1; 12 M.,
-2; IP.M., -4; 2 P.M., -6; 3 P.M., -7; 4 P.M., -7;
5 P. M., - 5 ; 6 P. M., - 2 ; 7 P. M., - 1.
9. Find the average church contributions according to the
following frequency table.
TABLE OF CHURCH CONTRIBUTIONS
INDIVIDUAL
CONTRIBUTIONS
NTMBER OF
CONTRIBUTORS
INDIVIDUAL
CONTRIBUTIONS
NUMBER OF
CONTRIBUTORS
No contribution .
1 cent
2
23
10 cents . . .
25 cents
13
9
5 cents
42
$50
1
287. Disadvantages of the arithmetic average. Some of
the preceding exercises suggest that there are certain
objections to the arithmetic average. For example, it
means little to say that the average church contribution
in Ex. 9, Art. 286, is 62 cents. People ordinarily use
the word " average " thinking it means the most usual
occurrence ; that is, the common thing. As a matter of fact
nobody gave 62 cents, and only one person gave as much
as that. The objection to the arithmetic average is that it
gives too much emphasis to the extreme items. To illustrate
more fully: A boy who has just finished an elementary
surveying course learns that the average weekly wage of a
railway-surveying group is $23. This is very encduraging
until an analysis shows him that the chief engineer gets $55
a week ; his assistant, |30 ; and all others but $15. To say
that the average weekly earning of ten men working in an
insurance office is $30 a week may be misleading, for one
man may be a $5000-a-year man, in which case the usual
salary is much lower than $30 per week. Other objections
248 GENERAL MATHEMATICS
to the arithmetic average are the following: (a) it cannot
be located either in a frequency table or in a frequency
graph ; (b) it cannot be accurately determined when the
extreme items are missing; (c) it is likely to fall where
no item exists (for example, a sociologist may discover
that the average-size family in a given community has 4.39
members, though such a number is evidently impossible).
For these reasons it is desirable to have some other
measure of tht central tendency of a group.
288. Central tendency ; the mode. One of the most use-
ful measures is the mode. It may be denned as the scale
interval that has the most frequent item, or we may say it
is the place where the longest bar of a bar diagram is
drawn. The term describes the most usual occurrence, or
the common thing. The popular use of the term " average "
approximates the meaning of the word. When we hear of
the average high-school boy he is supposed to represent
a type one who receives exactly the most common mark
of his classes, is of the most common athletic ability,
spends the most common amount of time in study, shows
the most common amount of school spirit, wastes the most
common amount of time, is of the most common age, etc.
It is obvious that no such high-school boy can be found.
Though a boy may possess some of these attributes he is
certain to differ from the common type in others.
The word " average " is thus incorrectly used for " mode,"
which means the common type. Thus the mode in the
church-contribution table (Ex. 9, Art. 286) is five cents.
More people in this church gave a nickel than any other
coin. The mode in the frequency 'table for the simple-
equation test for attempts (Ex. 7, Art. 285) is twenty
examples. In the test more students were at this point
when time was called than at any other point.
REPRESENTATION OF STATISTICS 249
EXERCISES
1. Find the mode for your class in the frequency table for
successes in the simple-equation test of Ex. 7, Art. 285.
2. From the table of Ex. 5, Art. 285, find the mode for the
weight of the twelve-year-old boys ; of the thirteen-year-old
boys ; of the fourteen-year-old boys.
NOTE. The student is expected merely to glance at the tables
to see at what scale interval the greatest number of items are found.
3. From the table of Ex. 4, Art. 285, find the mode for the
113 leaves.
4. From the first table of Art. 285 find the mode for the
220 ears of corn.
5. From the table in Art. 282 find the mode for the Fourth
of July accidents.
289. Advantages and disadvantages of the mode. The
advantages of the mode may be summarized as follows:
(a) the mode eliminates the extremes. In the results
of an examination the mode is not affected by the occa-
sional hundred or zero marks. The salary of the super-
intendent of a division of a shop does not affect the mode ;
(b) to the ordinary mind the mode means more than
does an average. It means more to say that the modal size
of classes in a high school is 15 than to say that the average
size is 17.24, first, because there is not a single class that
actually has the latter number ; and, second, because a few
large freshman classes in city high schools may tend to
increase considerably the average number. In making laws
we shall do the greatest good to the greatest number if we
keep the mode in mind and not the average. Unfortunately
street cars are built for the average number carried,
250 GENERAL MATHEMATICS
not for the modal number ; hence the " strap hanger."
The manufacturer of ready-made clothing fits the modal
man, not the average man. The spirit of a community's
charity fund is far more evident in the mode than in the
average.
A disadvantage of the mode is that there are a large
number of frequency tables to which it cannot be easily
applied. In such cases we have an irregular group with no
particular type standing out, and the mode is difficult to
find, as will be illustrated presently.
290. Central tendency ; the median. If a number of ob-
jects are measured with reference to some trait, or attribute,
and then ranked accordingly, they are said to be arrayed.
Suppose that your instructor gives an examination which
really tests mathematical ability, and that after the results
are announced the students stand in line, taking the
position corresponding to their marks on the examination ;
that is, the student with least mathematical ability at the
foot of the class, the one next in ability next to the foot,
etc. The class is then arrayed. If any group of objects is
arrayed, the middle one is known as the median item. If
your class had twenty-three pupils standing in the order
of their ability, the twelfth pupil from the foot or the head
of the class is the student whose mark is the median mark.
There are just as many below as above him in ability. The
median is another measure of the central tendency of a
group. If there is an even number of items, the median
is said to exist halfway between the two middle items.
Thus, if your class had twenty-two pupils, the mark half-
way between that of the eleventh and that of the twelfth
student from either end would be called, the median mark.
The meaning is further illustrated by the exercises given
on pages 251 and 252.
KEPKESENTATION OF STATISTICS
251
1. Find the median wage in the following table of the
weekly wage of the workers in a retail millinery shop. 1
WEEKLY WAGE
NCMBER OF
WORKERS
.
WEEKLY WAGE
NUMBER OF
WORKERS
$4-5 .
g
$11-12
o-O
18
12-13
5
(5-7
16
13-14 . . ' .
7-8
12
14-lo
6
8-9
7
15-16
18
9-10
8
16-17 ....
5
1(H11
5
The table above shows the wages of one hundred girls.
We are asked to find a weekly wage so that we shall be
able to say that one half the girls in this shop receive less
than this sum and one half receive more than this sum;
that is, we want a measure of the group.
In the first place, the student should notice that the
arithmetic average $10.05 seems too high to be represent-
ative, for there are too many girls working for smaller
sums. In the second place, the mode is unsatisfactory.
The wage $15 to $16 seems to be a mode, but there are
more girls working for about $5, $6, or $7 ; hence neither
the arithmetic average nor the mode has very much mean-
ing, so we proceed to locate the median.
There are one hundred girls in the shop ; hence we must
find a wage halfway between that of the fiftieth and that
of the fifty-first girl from the lowest wage. Adding the
number of the first four groups of girls (3 + 15 + 16+12)
gives us forty-six girls and takes us to the 8-dollar wage.
We need to count four more of the next seven, who are
1 For actual facts see "Dressmaking and Millinery," in "The Cleve-
land Survey," page 63. The table was adapted to meet the purposes of
the exercise.
252 GENERAL MATHEMATICS
getting between #8 and $9. The table assumes that the
series is continuous (piecework ?) ; hence the next seven
are distributed at equal distances between $8 and $9. We
may think of the seven girls as being distributed graphi-
cally, as shown in Fig. 195.
The graph makes clear the assumption that the first girl
(the forty -seventh) earns a sum which is between 8 and
$8| ; we assume that the wage is at the mid-point of this in-
terval, or $8-Jj. Similarly, the second girl (the forty-eighth)
earns a sum between $8| and
$8f and we assume the wage
to be at the mid-point of this
interval, or |8^. In like man-
ner the wage of the forty- F 195
ninth girl is $8 T 5 , the fiftieth
girl |8 T 7 j, and the fifty-first girl $8 T 9 ? . Midway between
the mid-points of the fiftieth and the fifty-first wage is
halfway between $8 T 7 and $8-^, or $8|. Hence the
median is $8 plus $|, or $84, for this wage is halfway
between the wage of the fiftieth girl and that of the fifty-
first. The student should study this graph until this
point is clear. He should note that the average is found
by calculating, the mode by inspection, and the median by
counting. Merely count' along the imagined scale until a
point is found that divides the item into two equal groups.
Since a wage problem usually involves a discrete series
(why?), a more practical illustration of the principle is
given below.
2. Find the median for the attempts of the one hundred and
fifteen students in the simple-equation test in Ex. 7, Art. 285.
Solution. We must find the number of equations attempted by
the fifty-eighth student from either end, for he will be the middle
student of the one hundred and fifteen. Counting from the top of
REPRESENTATION OF STATISTICS 253
the table (p. 243), we get fifty-five pupils who hive finished 19
equations. We need to count 3 more to get the fifty-eighth pupil.
There are twenty-two more who were somewhere in the twentieth
equation when time was called. If we assume, as we did in finding
modes, that the twenty-two students are at equal spaces through-
out the twentieth equation, then the median is 19 equations plus
? \ of an equation, or just over 19.1 equations.
EXERCISES
1. State the rule for finding the median, as developed in
the two preceding exercises.
2. In Ex. 5, Art. 285, find the median weight for 1000
twelve-year-old boys; 1000 thirteen-year-eld boys; 1000
fourteen-year-old boys.
3. In Ex. 4, Art. 285, find the median leaf and its measure
in the array of 113 leaves.
4. Find the median for the 220 ears of corn (Art. 285).
291. Limitations of statistics. There is a common saying
among nonscientific people that anything can be proved by
means of statistics. Experience lends conviction to the
homely saying " Figures' do not lie, but liars will figure."
This belief is due to the fact that figures have deceived the
public either by being dishonestly manipulated or by being
handled unscientifically. A table dishonestly manipulated
or based on unreliable data appears at first glance as con-
vincing as the work of a trained scientist. The public does
not find it possible to submit every piece of evidence to a
critical study and resents such deceptions as those referred
to above.
As a beginning the student should determine (1) the
reliability, and training of those 'who gathered the facts;
(2) how and when they were gathered ; (3) to what extent
254 GENERAL MATHEMATICS
the statistical studies have been exposed to the critical
judgment of trained experts; (4) to what extent similar
studies show similar results.
292. Law of statistical regularity. In calculating the
value of the farm lands in Indiana it is by no means
necessary to evaluate and tabulate every acre in the state.
To find out the average size of a twenty-five-year-old man
in New York it is not necessary .to measure and tabulate
every man in the city. The " Statistical Abstract of the
United States" (published by the Bureau of Foreign and
Domestic Commerce) states the value in dollars of hogs,
sheep, and cattle- produced in 1918, but this does not mean
that this total is obtained by tabulating every individual
animal. To find out how fast on an average a twelve-
year-old Chicago boy can run 100 yards we would not need
to hold a stop watch on every boy. In fact, a few chosen
in each class in each school building would probably give
us an average that would be identical with an average
obtained from the whole group. This is due to a mathe-
matical law of nature which states that if a reasonable num-
ber of individual cases are chosen " at random " from among
a very large group, they are almost sure, on the average, to pos-
sess the same characteristics as the larger groups. The phrases
" at random" and "reasonable number" make the law appear
somewhat vague. King, in "Elements of Statistical Method,"
illustrates the law as follows : "If two persons, blindfolded,
were to pick, here and there, three hundred walnuts from
a bin containing a million nute, the average weight of the
nuts picked out by each person would be almost iden-
tical, even though the nuts varied considerably in size."
Gamblers use the principle just illustrated when they have
determined how many times a given event happens out of
a given number of possibilities. They are thus able to ply
REPRESENTATION OF STATISTICS '255
their craft continuously and profitably on a small margin
in their favor. This principle is the basis of all insur-
ance ; thus it is possible to predict with a great degree
of accuracy how many men of a given age out of a given
one thousand will, under ordinary conditions, die during
the next year. The law of statistical regularity is very
extensively employed in the Census Bureau. The totals
are usually estimates based on careful study of sufficient
representative cases.
However, the student should be critical of the phrase
'" at random." It is not asserted that any small group
will give the same results as a measurement of the whole
group. Thus, if we measured the height of the first
four hundred men that passed us as we stood at the
corner of Randolph and State Streets, Chicago, we could
not be sure of getting an average that would accurately
represent the city. Any number of events might vitiate the
results ; for example, the Minnesota football team might
be passing by, or a group of unusually small men might
be returning from some political or social meeting limited
to one nationality. The sampling should be represent-
ative ; that is, sufficiently large and at random (here and
there). The larger the number of items, the greater the
chances of getting a fair sample of the larger group of
objects studied.
*293. The law of inertia of large numbers. This law
follows from the law of statistical regularity. It asserts
that when a part of a large group differs so as to show a
tendency in one direction, the probability is that an equal
part of the same group has a tendency to vary in the opposite
direction; hence the total change is slight.
294. Compensating and cumulative errors. The preced-
ing laws are also involved in a discussion of errors. If
256 GENERAL MATHEMATICS
the pupils in your school were to measure carefully the
length of your instructor's desk, the chances would be that
as many would give results too large as too short.
The estimates of a thousand observers of crop conditions
which are summarized or graphed in a volume such as
the " Statistical Atlas " (published by the Department of
Commerce) tend to approximate actual conditions. These
are illustrations of compensating errors. " In the long run
they tend to make the result lower as much as higher."
This type of error need not concern us greatly, provided
we have a sufficient number of cases.
However, we need to be on our guard against a con-
stant or cumulative error. If we use a meter stick that is
too short, we cannot eliminate the error by measuring a
very long line. A watch too fast could not eventually be
a correct guide. A wholesaler who lost a little on each
article sold could not possibly square accounts by selling
large quantities.
The value of a mass of facts involving a constant error
is seriously vitiated. Hence the student should be con-
stantly critical in his effort to detect this type.
EXERCISE
*Draw as accurately as possible on the blackboard a line
segment a certain number of inches in length. Ask as many
as from forty to fifty schoolmates, if possible, to stand on a cer-
tain spot and estimate the length of the line. Find how many
estimated the line too long ; how many estimated it too short.
HINT. The work must be done carefully. Have each student
estimate four times ; that is, estimate, look away, estimate, etc. Reject
estimates of all students who do not comply seriously with your
request. How many estimated the line too long ? How many esti-
mated it too short ? Report the results to your class.
KEPRESE^sTATION OF STATISTICS
257
*295. Normal distribution. The student has observed
the regular rise and fall of the numbers of the frequency
tables and the regular rise and fall of the graphs which
express these relations. It appears that the large majority
of the items are usually grouped, and as the distance from
this point of grouping becomes greater, the items become
rapidly fewer in number. If we measure ability to solve a
set of 24 simple equations, we shall discover that the large
majority of any representative class can attempt a little
more than 19 in 15 minutes. Only a few, possibly none,
will try all 24 problems; and only a few, probably none,
will have attempted
less than 6.
measure any
If we
single
FIG. 196. FREQUENCY CHART OF SCALLOP
SHELLS PILED ACCORDING TO THE NUMBER
OF RIBS. (AFTER BRINTON)
human trait, we shall
discover the same
tendency in the graph
of the results toward
a bell-shaped curve.
Whether we measure ability to spell, ability to add a column
of figures, ability to throw a baseball, the distance boys can
broad-jump, always there are a few very good at it and
a few very poor at it, with the tendency of the great
majority to possess but mediocre ability in a particular trait.
Nature shows the same tendency. The length of leaves,
the height of cornstalks, the length of ears of corn, etc.
give curves similar to the preceding. Note the piles of
scallop shells in Fig. 196. The shells are sorted into
separate piles according to the number of ribs. The piles
(from left to right) have respectively 15, 16, 1.7, 18, 19,
and 20 ribs. How do you think these piles would have
looked if a greater number of shells (say several hundred)
had been used?
258
GENERAL MATHEMATICS
The same tendency is observed in economic affairs.
Thus, if we measured the income of the ordinary agri-
cultural community, we should find out of a thousand
persons only a few whose income is less than $300 per year,
only a few, if any, with an income over $2500, and the
rest grouped and
tapering between
these limits.
When the rise
and fall is regu-
lar (that is, the
curve falls regu-
larly on both sides
from the mode),
the distribution is
likely to approxi-
mate what we call
a normal distribu-
tion, and the curve
is called a normal
distribution curve.
A normal dis-
tribution is illus-
trated by the
table and diagram
(Fig. 197) given here, which represents the heights from
actual measurement of four hundred and thirty Eng-
lish public-school boys from eleven to twelve years
of age. 1
It will be seen that the numbers conform to a very
uniform rule: the most numerous groups are in the middle,
at 53 in. and 54 in., while the groups at 51 in. and 56 in.
1 From Roberts's "Manual of Anthropometry," p. 18.
FIG. 197.
PHYSICAL PHENOMENA ILLUSTRATE
NORMAL DISTRIBUTION
REPRESENTATION OF STATISTICS
259
are less in number, those at 50 in. and 57 in. are still fewer,
and so on until the extremely small numbers of the very
short and very tall boys of 47 in. and 60 in. are reached.
It is shown that the modal, or typical, boy of the class
and age given is 53.5 in., and since he represents the most
numerous group, he forms the standard.
The curve would probably be smoother if more boys
were measured or grouped into half -inch groups. As it is,
it approximates very nearly a normal distribution.
Of course it is not asserted that every distribution is
of this type. There is merely a tendency in chance and
in nature to produce it. There are many causes which
make distribution irregular, as we shall presently see.
*296. Symmetry of a curve. The graph in Fig. 198
shows the height of 25,878 American adult men in inches.
This curve, like the one
of Art. 295, is more reg-
ular than most curves
which we have studied.
It would probably be
much smoother if the
class interval were one-
fourth inch. If we draw
a perpendicular AK from
the highest point of the curve, we may think of this as an
axis around which the rectangles are built. The curve
to the right of this axis looks very much like the part to
the left. In this respect we say the curve is almost
symmetrical.
Symmetry of figures may be illustrated by the human
head, which is symmetrical with respect to a plane midway
between the eyes and perpendicular to the face ; thus the
left eye and the left ear have corresponding parts to the
FIG. 198. HEIGHT OF MEN. (AFTER
THORNDIKE, " MENTAL AND SOCIAL
MEASUREMENTS," p. 98)
260
GENERAL MATHEMATICS
right of this ti.ris of symmetry. Note that the parts are
arrayed in reverse order.
Other familiar illustrations of symmetry are (1) the
hand and the image obtained by holding the hand in front
of a plane mirror ; (2) words written in ink and the im-
print of those words on the blotting paper with which they
are blotted : (3) our clothes, which are largely built on the
principle of .symmetry ; (4) the normal distribution curve.
In architecture, in art, and in higher mathematics the
principle of symmetry is very important.
*297. Skewness of a curve. The term "skewness" de-
notes the opposite of symmetry and means that the items
are not symmetrically distributed. The curve is not of the
bell-shaped form. It is higher either above or below the
mode than a sense of symmetry would have us expect.
To illustrate: Snppose that the incomes of all the people
living in a certain community were tabulated as follows :
INCO.MK ix DOLLARS
XfMBER OF
PERSONS
INCOME ix DOLLARS
N I'M HER OF
PERSONA
0-500 . . .
20
3500-4000 . . .
4
500-1000 . . .
36
4000-4500 . . .
3
1000-1500 . . .
20
4500-5000 . . .
o
1500-2000 . . .
12
5000-5500 . . .
1
2000-2500 . . .
6
5500-6000 . . .
1
2.300-3000 ...
5
6000-6500 ...
1
3000-3500 . . .
4
The graph (Fig. 199) of this table is not symmetri-
cal, but is skewed toward the lower side. The meaning of
skewness is clearly shown by the graph. The graph no
longer presents the normal, symmetrical, bell-shaped form ;
the base is drawn out to a greater extent on the one side
than on the other.
REPRESENTATION OF STATISTICS
261
Distribution is often affected by laws which are dis-
turbing factors in the situation. Thus, in investigating
the wages of carpenters we should expect a few to get
high wages, say 90^ per hour, and a few very low, say 40 <
per hour, and the rest to be grouped, according to ability,
between these limits. However, by agreement between
unions and contractors, carpenters' wages are fixed in most
FIG. 190. GRAPH SHOWING SKE\VNESS OF A CURVE
cities at a price somewhere between 60^ and 85<. Hence
we should have but one interval in a distribution table,
for a particular city, say Minneapolis, showing that all
carpenters get 75 < per hour.
R
298. The graph of constant cost relations. 1 Graphs may be
constructed and used as " ready reckoners " for determining
1 Teachers may find it desirable to take up the graphing of formulas
from science at this point; for example, the graph of the centigrade-
Fahrenheit formula. However, the authors prefer to use a simpler
introductory exercise here for the sake of method and to take up
more purposeful formulae in the next chapter.
262
GENERAL MATHEMATICS
costs of different quantities of goods without computation.
This is shown by the following example:
If oranges sell at 30 $ a dozen, the relation between the num-
ber of dozens and the cost may be expressed by the equation
c = 30 c/, where d is the number of dozens and c the cost
per dozen. If values are given to d, corresponding values
may be found for c, as given in the following table :
d
1
2
3
4
5
10
11
c
30
(iO
90
120
150
300
330
On squared paper draw two axes, OX and OF, at right
angles. On OY let a small unit represent 1 doz., and on
OX let a small unit represent 10 $. Then, on the 30 < line
10
50 100 150 200 250 300
FIG. 200. THE GRAPH OF A COST FORMULA
X
mark a point representing 1 doz. On the 20$ line mark
a point representing 3 doz. Draw a line through the points
thus marked. It is seen that this line, or graph (Fig. 200),
is a straight line.
By looking at this price curve we can get the cost of
any number of dozens, even of a fractional number. For
example, to find the cost of 6 doz. observe the point where
263
the horizontal line six small units up meets the price
curve ; observe the point directly beneath this on the
axis OX', this is eighteen small units from and hence
represents $1.80. Similarly, the cost of 8 doz. is seen
to be $2.55.
EXERCISES
1. By means of the graph in Fig. 200 determine the cost
of the following: 9 doz.; 11 doz.; 2^ doz.; 3^ doz.; 10|doz.;
5| doz. ; 3^ doz.
2. If eggs sell at 45$ a dozen, draw the price graph.
3. On the price graph drawn for Ex. 2 find the cost of 4 doz. ;
3 doz.; 10 doz. ; 3^doz.; 5^ doz. ; 4^ doz.
4. Draw a price graph for sugar costing 10| $ a pound.
5. On the graph drawn for Ex. 4 find the cost of 11 lb.;
31 lb.; 6|lb.; 10 lb.
6. Construct a graph which may be used in calculating the
price of potatoes at $2.10 per bushel.
7. Use the graph of Ex. 6 to find the cost of 3 bu.; 4|- bu.;
2 bu. 3 pk. ; 5 bu. ; 5 bu. 3 pk.
8. Since the graphs in Exs. 1~7 are straight lines, how
many of the points would have to be located in each case in
order to draw the line ? Should these be taken close together
or far apart, in order to get the position of the price graph
more nearly accurate ? Why ?
299. Graphs of linear equations ; locus ; coordinates. As
shown in the preceding sections the relation between two
quantities may be expressed in three ways: (1) by an
ordinary English sentence, (2) by an equation, or (3) by
a graph. The graph is said to be the graph of the equation.
A graph may be constructed for each equation that we
264
GENERAL MATHEMATICS
have studied to date. The process of drawing the graph
of an equation will be given in this article.
Let the equation be y = 2 x + 3, which we shall suppose
is the translation of some sentence which states some
definite practical rule; for example, the cost of sending a
package by parcel post into a certain zone equals two cents
per ounce plus three cents. We want to draw a graph
for the equation y = 2 x + 3.
EXERCISES
1. What is the value of y in the equation y = 2x + 3 when
x equals ? when x equals 1 ? when x equals 2 ? when x equals
3 ? when x equals 2 ? when x equals 3 ?
2. Fill in the following table of values of x and y for the
equation y = 2 x + 3.
X
1
2
3
4
5
6
7
- 1
- 2
- 3
- 4
y
3
5
7
We are now ready to transfer the data of Ex. 2 to
squared paper. The process does not differ very much
from our work in frequency tables except that usually in
graphing equations we need to consider both positive and
negative numbers. For the sake of method we shall extend
the discussion to cover this point. Two axes, AA'' and
)')"' (Fig. 201), are drawn at right angles^and meet
at 0. Corresponding to each set of values of x and y a
point is located, the values of x being measured along or
parallel to XX', and the values of y along or parallel to
YY'. -Positive values of x are measured to the right of
YY' and negative values to the left ; positive values of y
are measured above XX' and negative values below A'A"'.
REPRESENTATION OF STATISTICS
265
m
I
For example, the point A corresponding to x ^= 1 and y = 5
is obtained by measuring one space to the right and five
spaces upward. The point B corresponding to x = 1 and
y = 1 is obtained by measuring one space to the left and
one space upward. The
point C corresponding
to x = 3 and y = 3
is obtained by measur-
ing three spaces to the
left and three spaces
downward. The x and
y values of each point
are called the coordi-
nates of that point.
Continue finding
points which represent
corresponding parts of
numbers in the table.
tt soon becomes ap-
parent that all the points seem to lie on a straight line.
Hence we need to plot only two points to be able to
draw the line. However, we are more certain to discover
possible errors if we plot three points. Why?
, EXERCISES
1. Find the values of x and y at the points D and E
(Fig. 201) by inspection. Determine whether they satisfy
the equation of the graph.
2. Select any point in the line and determine whether the
values of x and y at this point satisfy the equation.
3. Select any point not on the graph, find the values of x
and ylat this point, and determine whether they satisfy the
equation of the graph.
FIG. 201. GRAPH OF A PARCEL POST
FORMULA
260 GENERAL MATHEMATICS
4. Select any point of the graph and determine whether the
values of x and y satisfy the equation.
5. How many points could one find on the line ?
The preceding exercises illustrate the following facts :
(a) The coordinates of every point on the line satisfy the
equation.
(b) The coordinates of every point not on the line do not
satisfy the equation.
These two facts can be proved rigidly in advanced
mathematics, and they enable us to say that the straight
line found is the lo<nis (the place) of all points whose coordi-
nates satisfy the given equation. It is important to observe
that the idea of a locus involves two things, specified under
(a) and (b) above.
Since it appears that the graph of an equation of the
first degree having two unknowns is a straight line, equa-
tions of the first degree are called linear equations.
A line may be extended indefinitely in either direction,
and there are an indefinitely large (infinite) number of points
upon a straight line. Since the coordinates of each point on
the line satisfy the equation of the line, there are an infinite
number of solutions of a linear equation with two unknowns.
This fact is evident, also, because for every value of one of
the unknowns we can find a corresponding value for the
other unknown.
ORAL EXERCISES
1. What is the location (locus) of all points in a plane
which are at a distance of 5 ft. from a given point P in the
plane? at a distance of 7% ft. fromP? at a distance of x feet,?
2. What is the locus of all points in space at a distance
of 10 ft. from a given point ? 1 cm. from a given point ?
x yards from a given point ?
REPRESENTATION OF STATISTICS
267
3. What is the locus of all points in a plane 3 in. distant from a
given straight line in the plane ? 5| in. distant ? y inches distant ?
4. What is the locus of all points in space at a distance of
4 in. from a given straight line ? 6 cm. ? y feet ?
5. What is the locus of all points in a plane equally dis-
tant from two given parallel lines in the plane ?
*6. What is the locus of all points in space equally distant
from two given parallel lines ?
7. What is the locus of all points in a plane 6 in. distant from
each of two given points in the plane which are 10 in. apart ?
8. What is the locus of all points 3 ft. from the ceiling of
your classroom ?
*9. What is the locus of all points in a plane 5 in. distant
from a line segment 7 in. long in the plane ?
*10. What is the locus of all points in space 5 in. distant
from a line segment 10 ft. long ?
300. Terms used in graphing a linear equation. Certain
terms used in mathematics ,in connection with graphical
representation will now
be given and illustrated
by Fig. 202. The lines
XX' and YY', drawn at
right angles, are called
axes (XX 1 the horizontal
axis and YY' the verti-
cal axis). The point is
called the origin. From P,
any point on the squared
-x
2
paper, perpendiculars are
drawn to the axes: the FlG - 202 ' I""MRATISG K TERMS
USED IN PLOTTING A POINT
distance PM is called the
ordinate of P, and the distance PN is called the abscissa of P ;
2b'8 GENERAL MATHEMATICS
together they are called the coordinates of P. The axes are
called coordinate axes. The scale used is indicated on the
axes. The distances on OX and Y are positive ; those on
OX' and on OY' negative. The abscissa of P is 2 and the
ordinate is 2*-; the point P is called the point (%, 2).
Notice that the abscissa is written first and the ordinate
second. Finding a point on a graphic sheet which corre-
sponds to a given pair of coordinates is called plotting
the point.
EXERCISES
1. What is the abscissa of point A ? B ? C ? D ? E ? (Fig. 202).
2. What is the ordinate of the point A? B? C? D? E :'
(Fig. 202.)
3. Give the coordinates of points A, B, C, D, E (Fig. 202).
4. On a sheet of graphic paper draw a set of coordinate axes
intersecting near the center of the paper, and plot the following
points : (2, 4), (5, 2), (4, - 2), (- 3, 4), (- 3, - 2),(- 2J, 3J).
5 . Compare the process of plotting points with the numbering
of houses in a city.
6. On a sheet of graphic paper locate the points A (2, 2),
B(5, 3), C(2, 7), and D(5,.8). What kind of figure do you
think is formed when the points A, B, C, and 7)are connected?
Draw the diagonals of the figure, and find the coordinates of
the point where the diagonals intersect.
301. Summary of the method for the process of graphing
a linear equation. With Art. 300 in mind we shall now
illustrate and summarize the process of graphing a linear
equation.
Draw the graph of 4 x 3 y = 6.
(a) Solve the equation for either unknown in terms of the other: thus,
REPRESENTATION OF STATISTICS
269
This throws the equation into a form from which the corre-
sponding pairs of values are more easily obtained.
(b) Let
Then
And let
Then
x 3, etc.
That is, build a table of corresponding values as follows : (Try to
get at least two pairs of integral numbers. Why?)
6 + 3 77
(c) Plot the points
corresponding to the
f t f.j, FIG. 203. GRAPH OP THE LINEAR EQUATION
pairs oj numbers oj the . _ ,
table (see Fig. 203).
(d) To check, choose a point on the line drawn and determine
whether its coordinates satisfy the given equation or plot a third
pair of numbers in the table. This third point also should fall on
the- line drawn.
(e) The two points plotted should not be too near each other. Why f
270 GENERAL MATHEMATICS
>
EXERCISES
Draw the graphs of the following equations, each on a sep-
arate sheet of squared paper :
1. x + y = 7. 5. 5x- 4y = 20. 9. 5x-2y = -3.
2.2x y = &. 6. 3x + 5y = l5. 10. x + 5 y 12.
3. 3x- 2y = l2. 7. 5x-2y = W. 11. 2x = 3-4/.
4. 3x + 2y=6. 8. 6x-4y = 3. 12. 3y = 4 8ar.
HISTORICAL NOTE. Statistics has attained the dignity of a sci-
ence during the last fifty years. Its growth goes hand in hand with
national organization. Even in a crude tribal organization the ruler
must needs know something of its wealth to determine the taxes or
tribute which may be levied. Our earliest statistical compilations
(some time before 3000 B.C.) presented the population and wealth
of Egypt in order to arrange for the construction of the pyramids.
Many centuries later (about 1400 B.C.) Rameges II took a census
of all the lands of Egypt to reapportion his subjects.
In the Bible we read how Moses numbered the tribes of Israel
and of the census of the Roman emperor, Augustus Caesar, in the
year which marked the birth of Christ.
The Greeks and Romans and the feudal barons of the Middle
Ages made many enumerations for the purposes of apportioning
land, levying taxes, classifying the inhabitants, and determining the
military strength. In all cases except that of the Romans some
special reason existed for collecting the data. The Romans col-
lected such data at regular intervals.
During the Mercantile Age of western Europe the feeling grew
that it was the function of a government to encourage the measures
aimed to secure a balance of trade. In order to decide correctly
concerning the needs of commercial legislation, more detailed infor-
mation was necessary than had hitherto been gathered. The growth
in a centralized monarchy further stimulated statistical study. That
monarch was most successful who could in advance most accurately
compare his resources with his rivals'.
In 1575 Philip II of Spain made extensive inquiries from the prel-
ates concerning their districts. In 1696 Louis XIV required reports
on the conditions of the country from each of the general intendants.
REPRESENTATION OF STATISTICS 271
Prussia began in modern times the policy of making periodic
collections of statistical data. In 1719 Frederick William I began
requiring semiannual reports as to population, occupations, real-
estate holdings, taxes, city finance, etc. Later these data were col-
lected every three years. Frederick the Great also was a vigorous
exponent of the value of statistics. He enlarged the scope of statis-
tics in general by including nationality, age, deaths and their causes,
conditions of agriculture, trade, manufactures, shipping, in fact,
anything that might possibly contribute to national efficiency.
A provision in our constitution of 1790 initiated the decennial
census. One country after another has adopted some form of regular
enumeration, until, in 1911, China took her first official census.
In recent times the censuses have grown extremely elaborate.
In 1900 the United States established a permanent Census Bureau
whose function it is to study special problems in the light of the
data collected and to publish the results of this study. Most leading
nations also have special bureaus which attempt to keep the sta-
tistics of a nation u\> to date by means of scientific estimates. An
example of such a bureau is our National Bureau of Statistics.
Many states have established bureaus to meet the needs of the
state. Recently a movement has gained momentum to establish
municipal bureaus to collect and study the data of the community
and to instruct the public as to the significant results obtained by
means of elaborate reports. An example of this idea is illustrated
by the Survey Committee of the Cleveland Foundation.
SUMMARY
302. Chapter X has taught the meaning of the following
words and phrases: pictogram, cartogram, bar diagrams,
graphic curve, frequency table, class interval, central tend-
ency, arithmetic average, mode, median, normal distri-
bution, random sampling, compensating errors, constant
or accumulating errors, symmetry, symmetry of a curve,
skewness of a curve, price graph, linear equation, locus,
axes, horizontal axis, vertical axis, ordinate, abscissa, coor-
dinates, coordinate axes, plotting a point.
L'7_' GKNKKAL MATHEMATICS
303. The graphic curve may be used to show the rela-
tion between two quantities. Specific directions were given
showing how a graphic curve is drawn.
304. Continuous and discrete series were illustrated and
explained.
305. Statistical studies are necessary to solve our social,
governmental, and economic problems. The intelligent
reader will profit by a knowledge of the elements of
statistical methods.
306. Tabulating the facts bearing on a problem in the
form of a frequency table enables one to get a grasp on
the problem.
307. The word " average " as generally used may mean
arithmetic average, mode, or median. All are measures of
the central tendency of a mass of statistical data. The
arithmetic average is found by figuring, the mode is found
by inspection, and the median by counting.
308. The law of statistical regularity was illustrated.
309. The law of inertia of large, numbers was stated.
310. The graph of goods purchased at a constant cost
may be used as a " ready reckoner."
311. The chapter has taught how to plot points on
squared paper.
312. The graph of a linear equation is a straight line,
The coordinates of every point on the line satisfy the equa-
tion, and the coordinates of every point not on the line do
not satisfy the equation. This illustrates the locus idea.
313. The chapter has taught the method of graphing a
linear equation.
CHAPTER XI
GAINING CONTROL OF THE FORMULA; GRAPHICAL
INTERPRETATION OF FORMULAS
314. The formula. The formula has been defined as an
equation which is an abbreviated translation of some prac-
tical rule of procedure. Thus, I = Prt is a formula because
it is an equation which is an abbreviated form of the fol-
lowing practical rule for finding interest : To find the interest
(expressed in dollars) multiply the principal (expressed in
dollars) by the product of the rate (expressed hi hundredths)
and the time (expressed in years).
The formula is applied extensively in shop work, engi-
neering, science, and, in fact, in every field of business and
industry where the literature is at all technical. The sym-
bolic form of the rule of procedure is not only more easily
understood than the sentence form but is more easily applied
to a particular problem.
315. Applying the interest formula. A formula is applied
to a problem when the known facts of the problem are sub-
stituted in place of the letters of the formula. A formula
may be used when all the letters except one appear as
known facts in the problem. The pupil should study the
following illustration :
What is the interest on $200 at 5% for two years ?
Solution. Substituting the known facts in the formula,
7 = 200.^-2.
Simplifying the right member,
/ = $20.
273
274 GENERAL MATHEMATICS
EXERCISES
1. Find the interest on $425 at 4% for 2 yr.
2. Find the interest on $640 at 4% for Syr.
4.5 9
HINT. Substitute ^ , or , for r. Why ?
3. Find the interest on $820 at 4% for 2yr. 3 mo. 5 da.
HINT. Reduce 2 yr. 3 mo. 5 da. to days, divide this result by 360,
and substitute for /. Why?
316. Other types of interest problems conveniently solved
by special forms of I=Prt. The method of solving other
types of interest problems is illustrated by the following
problem :
How much money must be invested at 5% for 2 yr. so as to
yield $180 interest ?
NOTE. This problem differs from Ex. 3, Art. 315, in that rate,
time, and interest are given and the problem is to find P (the prin-
cipal). It may be solved by substituting the three numbers given
for the corresponding three letters of the formula. Why ? However,
it will be found on trial to be far more convenient if we first solve
for P in / = Prt.
Solution. Dividing both members of the equation by rt, P.
rt
This may be translated into the following rule of arithmetic :
To find the principal divide the interest by the product of the prin-
cipal and the rate. P = is only a special form of I = Prt, but
constitutes complete directions for finding the principal when the
other three factors are given.
In the proposed problem we obtain, by substituting,
Thus the principal is $1800.
CONTROL OF THE FORMULA 275
EXERCISES
1. What principal must be invested at 4|-% for 2 yr to
yield $81?
2. What is the principal if the rate is 6%, the time 4yr.
3 da., and the interest $120 ?
3. What is the rate if the principal is $500, the time 3 yr.,
and the interest $90 ?
Here P, t, and / are given ; r is the unknown. Hence we solve
/ = Prt for r.
Dividing both members by P and then by t or by (Pf),
J_
Pt~
I
Substituting the known facts in r = ,
Pt <<\
1)0 6
= - = 6%.
500 3 100
4. Translate r = into a rule of procedure for finding
the rate.
5. What is the rate if the interest is $85.50, the time
l|-yr., and the principal $950?
6. What is a fourth type of interest problem ? Find a
formula most convenient for the solution of such type
problems.
7. Show how to obtain this formula from the form 7 = Prt.
8. Translate into a rule of arithmetic. ,
9. Into what two parts can 1500 be divided so that the
income of one at 6% shall equal the income of the other
at 4% ?
10. How can a man divide $1800 so that the income of part
at 4% shall be the same as that of the rest at 5,% ?
11. A certain sum invested at 4|% gave the same interest in
2 yr. as $4000 gave in 1^ yr. at 4^ . How large was the sum ?
276
GENERAL MATHEMATICS
317. Solving a formula. The process of deriving t =
-LV
from / = Prt is called solving the formula for t. Similarly,
deriving the form P is called solving the formula
/ L
for P. The special form obtained is not only the most
convenient form for the particular problem, but it may be
used to solve the whole class of problems to which it
belongs. The solving of the formulas of this chapter are
of the practical kind and will involve little more than
the applications of the axioms of Chapter I.
318. Graphical illustration of interest problems. The
relation between any two of the factors that appear in
an interest formula may be represented graphically.
EXERCISE
How does the yearly interest vary on principals invested
at 5% ?
Substituting T 5 for r, and 1 for t,
then /= T(j P.
Note that this is a linear equa-
tion involving 7 and P which may
be plotted by the method of Art. 301.
The table below was used to make the
graph in Fig. 204.
/ P
$2.50
$50.00
$5.00
$100.00
$10.00
$200.00
10 15 20 25
Interest
Let one small unit on the horizontal
lines represent $1 of interest, and one
. .
large unit on the vertical lines repre- 1N CALCULATING INTEREST ON
sent $50 of principal invested. . PRINCIPALS INVESTED AT 5%
. n/< n
G. 204. GRAPH TO BE USED
' CONTROL OF THE FORMULA 277
Use OX as the line for plotting interests and OY for plotting prin-
cipals. Then the point corresponding to ($2.50, $50) means 2^ small
spaces to the right and 1 large space up. Since we know that the
graph will be a straight line, the line OR may be safely drawn as
soon as tw'o points are plotted.
EXERCISES
\. Look at the graph in Fig. 204 and tell offhand what
interest you would expect to collect at 5% for 1 yr. on $300;
on $350 ; on $400 ; on $60 ; on
2. Determine by looking at the graph in Fig. 204 how much
money you would need to invest at 5% to collect $18 interest
in 1 yr.; $20 interest; $27.50 interest; $14 interest.
3. How would you go about finding the interest on $12.50
by means of a graph ? on $2000 ?
4. Check some of the answers given by calculating the
interest by the usual method.
5. Graph the equation / = T ^p and use the graph to cal-
culate interest on sums lent at 6%.
6. Let P = $100 and r = T f 7 m the formula 1 = Prt, thus
obtaining I = 6t. Graph 7 = 6 1 and use the graph to deter-
mine the interest on $100 at 6% for 2 yr. ; for 2| yr. ; for 3 yr. ;
for 4 yr. ; for 5 yr. ; for 2 mo.
7. If possible, report in detail the methods used by your
family banker to calculate interest. On what principles do the
various " short cuts " rest ?
*319. Formulas involving the amount. In the exercises
that follow we shall study some formulas a little more diffi-
cult to solve, but they can be understood if the funda-
mental laws in solving equations are carefully applied.
278 GENERAL MATHEMATICS
EXERCISES
1. If $400 is invested at 4%, what is the amount at the
end of 1 yr. ? of 2 yr. ? of 3 yr. ?
2. If $1200 is invested at 3%, what is the amount at the
end of 2 yr. ?
3. What is the rule for finding the amount when principal,
rate, and time are given ?
4. Using A for the amount and Prt for the interest, translate
the preceding rule into a formula.
5. The formula for the amount may also be written in the
form A = P (1 + rt~). Prove.
6. Solve A = P (1 + rt) for P.
HINT. Dividing both members of the equation by the coefficient
of P, namely, (1 + rf), we obtain P = -
7. Translate into a rule of arithmetic the formula obtained
in Ex. 6.
8. Find the principal if the rate is 6%, the time 3 yr., and
the amount $472.
9. Find the principal if the rate is 5%, the time 3yr., and
the amount $1150.
10. Solve the equation .4 = 7* + Prt for t. Translate the
resulting formula into words.
11. Find the time if the principal is $2500, the amount
$2725, and the rate 3%.
12. Solve the equation A = P + Prt for r and translate the
resulting formula into words.
13. Find the rate if the principal is $1500, the amount
$1740, and the time 4 yr.
14. Summarize the advantages of solving interest problems
by formulas.
CONTROL OF THE FORMULA 279
320. Evaluating a formula. The process of finding the
arithmetical value of the literal number called for in a
formula is called evaluating the formula. The foregoing
exercises show that the process consists of
1. Substituting the known numbers in the formula.
2. Reducing the arithmetical number obtained to the simplest
form.
NOTE. A drill list involving these processes is given in Art. 329.
321 . Summary of the discussion of a formula. Cultivating
and gaining control of a formula means
1. Analyzing an arithmetical situation so as 'to see the
rule of procedure.
2. Translating the rule into a formula.
3. Solving the formula for any letter in terms of all the
others.
4. Evaluating the formula.
These steps will now be illustrated in the solution of
motion problems. We shall then proceed to solve short
lists of exercises which should develop power in these steps.
322. The formula applied to motion problems. In solving
the following problems try to observe the steps summarized
in Art. 321.
ORAL EXERCISES
1. If a 220-yard-dash man runs the last 50 yd. in 5 sec., at
what rate is he finishing ?
2. If an automobile makes 75 mi. in 2|-hr., how fast is it
being driven ?
3. Express the distance covered by a train in 8hr. at an
average rate of 20 mi. per hour; of 12^ mi. per hour; of x miles
per hour ; of x -f 3 mi. per hour.
280 GENERAL MATHEMATICS
4. Express the distance covered by a train in t hours at the
rate of / miles per hour.
5. Express the time it takes an automobile to go 150 mi. at the
rate of 10 mi. per hour ; of 15 mi. per hour ; of 20 mi. per hour;
of m miles per hour ; of 2 m miles per day ; of 2 m + 4 mi. per day.
6. How long does it take to make a trip of d miles at the
rate of r miles per hour ?
7. The rate of a train is 30 mi. an hour. If it leaves the
station at 1 A.M., how far away is it at 2A.M.; at 3 A.M.; at
1 A.M.: at 5A.M.; etc.? How far away is it at 3.15A.M.;
at 4.30A.M.; at 6.45A.M.?
8. Denoting the distance traveled by d, find d when the rate
is 45 mi. an hour and the number of hours is six.
323. Distance, rate, time. The preceding exercises show
that a problem involving motion is concerned with distance,
rate, and time. The number of linear units passed over by
a moving body is called the distance, and the number
of units of distance traversed may be represented by d.
The rate of uniform motion, that is, the number of units
traversed in each unit of time, is called the rate (or speed)
and is represented by r. The time, t, is expressed in
minutes, hours, days, etc.
ORAL EXERCISES
1. Illustrate by familiar experiences that distance equals the
rate multiplied by the time ; that is, that d = rt.
2. Show how to obtain t = from d = rt.
r
3. Translate t = into a rule for finding the time.
4. Show how to obtain r = - from d = rt.
5. Translate r = - into a rule for finding the rate.
CONTROL OF THE FORMULA 281
6. A motor cycle goes 110 mi. in 5 hr. and 30 min. Assum-
ing the rate to be uniform, what is the rate ? Which of the
formulas did you use ?
7. Sound travels about 1080 ft. per second. If the sound of
a stroke of lightning is heard 2.5 sec. after the flash, how far
away is the stroke ? Which form of the formula is used ?
8. How many seconds would it take the sound to reach the
ear if a tree 2376 ft. distant were struck by lightning ? Which
form of the formula is used ?
WRITTEN EXERCISES
1. A motor boat starts 10 mi. behind a sailboat and runs
14 mi. per hour, while the sailboat makes 6 mi. per hour. How
long will it require the motor boat to overtake the sailboat ?
Let x be the number of hours it takes the motor boat to overtake
the sailboat.
Then, according to the data of the problem :
for the motor boat, for the sailboat,
t = x. t = x.
r = 14. r = 6.
Hence d = 14 x. Hence d 6 x.
Since the motor boat must go 10 mi. more than the sailboat, the
following equation expresses the conditions of the problem :
14 x = 6 x + 10.
Solve the equation to find the value of x, which turns out to
be llhr.
2. A and B live 22^ mi. from each other. In order to meet
A, B leaves home an hour earlier than A. If A travels at the
rate of 4 mi. an hour and B at the rate of 3| mi. an hour, when
and where will they meet ?
3. A northbound and a southbound train leave Chicago at
the same time, the former running 4 mi. an hour faster than
the latter. If at the end of 1^ hr. the trains are 126 mi. apart,
find the rate of each.
282 GENERAL MATHEMATICS
4. In running 280 mi. a freight train whose rate is | that
of an express train takes 3-|- hr. longer than the express train.
Find the rate of each.
5. An automobile runs 10 mi. an hour faster than a
motor cycle, and it takes the automobile 2 hr. longer to run
150 mi. than it takes the motor cycle to run 60 mi. Find the
rate of each.
6. A man rows downstream at the rate of 8 mi. an hour
and returns at the rate of 5 mi. an hour. How far downstream
can he go and return if he has 5| hr. at his disposal ? At what
rate does the stream flow ?
7. Chicago and Cincinnati are about 250 mi. apart. Suppose
that a train starts from each city toward the other, one at the
rate of 30 mi. per hour and the other at the rate of 35 mi. per
hour. How soon will they meet ?
8. A train is traveling at the rate of 30 mi. an hour. In
how many hours will a second train overtake the first if the
second starts 3 hr. later than the first and travels at the rate
of 35 mi. an hour ?
*9. A and B run a mile race. A runs 20 ft. per second, and
B 19^ ft. per second. B has a start of 32 yd. In how many
seconds will A overtake B ? Which will win the race ?
* 10. A bullet going 1500 ft. per second is heard to strike
the target 3 sec. after it is fired. How far away is the target ?
(Sound travels at the rate of about 1080 ft. per second.)
11. A motor cyclist rode 85 mi. in 5 hr. Part of the distance
was on a country road at a speed of 20 mi. an hour and the
rest within the city limits at 10 mi. an hour. Find how many
hours of his ride were in the country.
12. Two boats 149 mi. apart approach each other, leaving at
the same time. One goes 10 mi. per hour faster than the other,
and they meet in 2 hr. What is the rate of each ?
CONTROL OF THE FORMULA
283
30
45
60
324. Graphical illustration of a motion problem. Many
motion problems can be conveniently illustrated graphi-
cally, as the student will discover if he solves the following
exercises.
EXERCISES
1. In the Indianapolis races De Palma drove his car at a
rate varying but little from 90 mi. per hour. Draw a graph
showing the relation between the distance and d t
time of De Palma's performance.
Substituting 90 in d = tr,
d = 9Qt.
Note that d 90 1 is a linear equation which
may be graphed (see table and Fig. 205). Ten small
units on the vertical axis represent 30 mi.; ten small units on the
horizontal axis represent ^ hr.
2. Determine from the graph in Fig. 205 how many miles
De Palma drove in 2 hr. ; in 1^ hr. ; in 1 hr. 24 min. ; in 40 min. ;
in 4 min.; in 2 hr. 12 min.
3. Determine by the graph
in Fig. 205 how long it took
De Palma to go 50 mi. ; 40 mi. ;
GO mi.; 75 mi.; 140 mi.; IGOmi.;
10 mi.
Obviously the preceding re- %
suits could be calculated either
by arithmetic or by the formula.
However, the graph has the advan-
tage of revealing all the results
in vivid fashion.
i l 1J
Time in Hours
FIG. 205. THE GRAPH OF A MOTION-
PROBLEM FORMULA
4. Draw a graph showing
the distances traversed by a passenger train running uniformly
at the rate of 40 mi. per hour for the first ten hours of its trip.
*5. Find out, if possible, what use railroad officials make of
graphs in arranging schedules.
284 GENERAL MATHEMATICS
325. Circular motion. Circular motion is of frequent
occurrence in mechanics. A familiar illustration is found
in the movement of the hands of a clock.
EXERCISES
1. At what time between 3 and 4 o'clock are the hands of
a clock together ? h
Solution. Let x (Fig. 206) equal the
number of minutes after 3 o'clock when
the hands are together ; that is, x equals
the number of minute spaces over which
the minute hand passes from 3 o'clock
until it overtakes the hour hand.
x
Then equals the number of minute
FIG. 206. CLOCK PROBLEMS
spaces passed over by the hour hand. ILLUSTRATE A TYPE OF ClR .
Why? CULAK MOTION
Since the number of minute spaces
from 12 to 3 is 15, and since the whole is equal to the sum of its
parts, it follows that
Whence x = 16 ^ 4 T min.
Therefore the hands are together at 16^ min. after 3 o'clock.
2. At what time between 4 and 5 o'clock are the hands of
a clock together ?
HINT. Draw a figure similar to the one for Ex. 1.
Notice that the formula for a clock problem is x = + m,
\.t
where m equals the number of minute spaces the minute hand must
gain in order to reach the desired position.
3. At what time between 2 and 3 o'clock are the hands of
a clock 15 min. apart ?
HINT. Draw a figure, think the problem through, and then try
to see how the formula in Ex. 2 applies.
CONTROL OF THE FORMULA 285
4. At what time between 2 and 3 o'clock are the hands of
a clock 30 minute spaces apart ?
5. What angle is formed by the hands of a clock at 2.30?
*6. At what time between 5 ,and 6 o'clock are the hands of
a clock 20 min. apart ? How many answers ? How may these
results be obtained from the formula of Ex. 2 ?
326. Work problems. The work problem is another
type of problem easily solved by formula.
EXERCISES
1. One pipe will fill a tank in 3 hr. and a second pipe can
fill it in 4 hr. How long will it take to fill the tank if both
pipes are left running ?
Let n = the number of hours it will take both pipes to fill the tank.
Then = the part of the tank filled in 1 hr.,
n
i = the part of the tank filled by the first pipe in 1 hr.,
and ^ = the part of the tank filled by the second pipe.
Hence - +'- = - Why?
3 4 n
Multiplying by 12 n, 4 n + 3 n = 12,
or 7 n = 12.
Whence n = If hr.
2. A can lay a drain in 5 da. and B can do it in 7 da. How
long will it take both working together ?
3. One boy can drive his car over a trip in 8 hr. and a
second boy can make the trip in 5 hr. How long would it take
them to meet if each started at an end ?
NOTE. It is clear from the foregoing problems that any numbers
would be used just as 3 and 4 are used in Ex. 1. Hence a formula
may be obtained, as is shown by Ex. 4.
286 GENERAL MATHEMATICS
4. A can do a piece of work in a days and B can do it in
b days. How long will it take them to do it together ?
Let n = the number of days it will take
them together.
Then - = the amount of work they can do
n , ,
in 1 da.,
- = the amount A can do in 1 da.,
a
and - = the amount B can do in 1 da.
b
1,1 1
Hence _ + _ _.
a b n
Multiplying by aim, l>n + an = ah,
(J> + a)n ah,
ab
a + I >
NOTE. Any problem of the type of Ex. 1 on page 285 may be
solved by using the equation n = as a formula. Thus, to solve
a + b
Ex. 1 let a = 3, b 4. Then n = = = 1^ hr.
5. One boy can make a paper route in 2 hr. and his friend
can make the route in 1^ hr. How long will it take the two
together ? (Solve by formula.)
*6. Suppose that in Ex. 1 on page 285 the second pipe is an
emptying pipe, how long will it take to fill the tank if both
pipes are running ? What form does the formula take ?
*7. A can sweep a walk in 7 min., B in 8 min., and C in
10 min. How long will it take them working together ? What
form does the formula for a work problem take ?
*8. A could lay a sidewalk in 3 da., B in 4 da., and C in
4.5 da. How long does it take them when working together ?
Solve by substituting in the formula for Ex. 7.
CONTROL OF THE FORMULA 287
327. Translating rules of procedure into formulas. Write
each of the following in the form of a formula :
1. The area of a triangle equals the product of half the
base times the altitude.
2. The area of a rectangle equals the product of its base
and altitude.
3. The area of a parallelogram equals the product of its
base and altitude.
4. The area of a trapezoid equals one half the sum of the
parallel bases multiplied by the altitude.
5. The volume of a pyramid equals one third the base
times the altitude.
6. The length of a circle is approximately equal to twenty-
two sevenths of the diameter.
7. The circumference of a circle is equal to TT times the
diameter.
8. The area of a circle is TT times the square of the radius.
9. The product equals the multiplicand times the multiplier.
10. The product obtained by multiplying a fraction by a
whole number is the product of the whole number and the
numerator divided by the denominator.
11. The quotient of two fractions equals the dividend multi-
plied by the inverted divisor.
12. The square root of a fraction equals the square root of
the numerator divided by the square root of the denominator.
13. The square of a fraction is the square of the numerator
divided by the square of the denominator.
14. The rule for calculating the cost of one article when
you know that a certain number of them cost so much ; write
the cost of m articles.
15. The rule for expressing years, mouths, and days as years.
288 GENERAL MATHEMATICS
16. The rule for calculating the area of three adjacent rooms
of different lengths but the same width.
17. The rule for calculating the area of the floor of a
square room.
18. The rule for finding the cost of a telegram.
19 . The rule for finding the area of a figure cut from cardboard,
given its weight and the weight of a square unit of cardboard.
20. The rule for finding the amount of available air for each
person in a classroom, given the dimensions of the room and
the number in the class.
21. The rule for finding the weight of a single lead shot,
given the weight of a beaker with a given number of shot in
it and the weight of the empty beaker.
22. The rule for predicting the population of a town after
a certain number of months, given the present population, the
average number of births, and the average number of deaths.
23. The rule for finding the distance apart, after a given
time, of two cars which start from the same point and travel
in opposite directions at different speeds.
24. The same as in Ex. 23 except that the cars are m miles
apart at starting.
25. The same as in Ex. 23 except that the cars go in the
same direction with different speeds.
26. The reading on a Fahrenheit thermometer is always 32
greater than ^ of the reading on a centigrade thermometer.
27. The reading of a centigrade thermometer may be cal-
culated by noting the reading on the Fahrenheit, subtracting
32, and taking | of this result.
328. Graphic representation of the relation between the
readings on centigrade and Fahrenheit thermometers. The
last two exercises deal with two types of thermometers
that are used to measure temperature. Fig. 207 shows that
CONTROL OF THE FORMULA
289
F
Boiling 100'
90 f
80'
70'
60'
50'
40'
30'
20'
10'
Freezing 0'
-10'
-17.78'
the only fundamental difference is the different graduations
of the scale. In the Fahrenheit thermometer the place
where the mercury stands if immersed in freezing water is
32 ; on the centigrade it is zero. This defines the freezing
point on each. The boiling point is
marked 212 on the Fahrenheit and
only 100 011 the centigrade. Hence
the Fahrenheit thermometer has 180
divisions of the scale in the interval
from freezing to boiling, while the
centigrade has but 100. This means
that a unit on the centigrade is
longer, or for any space on the centi-
grade there are i|$ times, or | times,
as many Fahrenheit units. Hence
the number of units in a Fahrenheit
reading equals -| times as many cen-
tigrade units plus 32 (which are
below the freezing point). Stated
as a formula, F=fC + 32.
We could, of course, always trans- BETWEEN THE CENTI-
late the reading on one thermometer GRAI)E AND FAHRENHEIT
... ,. THERMOMETERS
to the corresponding reading on the
other by means of this formula. It is far easier, how-
ever, to graph the linear equation F = |C + 32, which
will reveal possible relations, as is shown in Fig. 208.
NOTE. The pupil should construct the graph independently of
the text, using a table similar to the following :
C. F.
Let C.= 0, then F. = 32.
Let C. = 25, then F. = 77.
Let C. = 15, then F. = 59.
212
194
-176
-158
140
-122
-104
86
68
50
32
14
FIG. 207. THE RELATION
290
CKNKRAL MATHEMATICS
EXERCISES
1. Determine by the graph the corresponding Fahrenheit
readings for the following centigrade readings : 5, 10, 20,
30, - 5, - 10, - 15, - 25.
2. Determine by the graph the centigrade readings cor-
responding to the following Fahrenheit readings : 80, 70,
60, 30, 20, 10, - 5, - 10.
3. In the formula F = f C
-f- 32 substitute in each case
the two numbers you think are
corresponding readings. The
error should be very small.
4. Normal room tempera-
ture is 68 F. What is it
centigrade ?
5. The normal temperature
of the human body is 98.4 F.
What is it centigrade ?
6. What temperature centi-
grade corresponds to F.?
7. Could you go skating at FlG 20g A GRAPH T0 BE rSK1) JN
15 C. ? CHANGING CENTIGRADE READINGS
TO FAHRENHEIT AND VICE VERSA
8. In your general-science
course you are told that mercury freezes at 40 F. What
is this centigrade ?
9. Would your classroom be comfortable at 25 C. ?
329. Evaluating a formula. Find the value of the letter
called for in each of the exercises given on page 291.
When no explanation is given, it is assumed that the
student recognizes the formula.
CONTROL OF THE FORMULA
291
EXERCISES
1. Given C = | (F - 32). Find C. if F. = 0; 32;
212; 100.
2. Given F = |C.+ 32. Find F. if C. = 0; 100;
-20; 60.
3. Given d = rt. Find d if / = 87.5 mi. per hour and
t = 12 hr. ; if r = 10^ ft. per second and t 10 sec.
4. Given r = - Find r if d = 1 mi. and t = 4 min. 16 sec. ;
{/
if d = ^ mi. and t = 2.07 sec.
5. Given A=P+ Prt. Find A if P = $240, r = 4-}%, and
*=lyr. 2 mo. 3 da.; if P = $128, r=6%, and = 2yr. 3 da.;
if P = $511, r = 6%, and t = 20 yr.
6. Given V=lwh. Find F (see Fig. 209)
if I =12.2 ft., w = 8.3 ft., and h = 6.4 ft.;
if = 9.3 in., iv = 5.6 in., and h = 1 in.
7.
Find Tiff =63 ft.,
FIG. 209. RECTAX-
w = 2.4 ft., and h = 1.6 ft.; if I 3 cm., GULAR PARALLELE-
iv = 2.1 cm., and h = 1.4 cm. PIPED
8. Given c = ^- d. Find c if d = 1 f t. ; 1 in. ; 4 in. ; 10 in. ;
5-1 in.
9. Given A = Trr 2 (TT = - 2 T 2 -). Find .4 if r = lin.; 5ft.;
10 yd. ; 7 m. ; 8.5 cm.
10. The volume F of any prism (Fig. 210)
is equal to the product of its base B and
its altitude h. Find V if B = 246.12 sq. in.
and h = 12 in.; if B = 212.44 sq. in. and
& = 2| ft.
11. The lateral surface L of a right
prism equals the perimeter of the base FlG 2 io. PRISM
P times the altitude h. Find L if P =
126 in. and h = 11 in.; if P = 21.6 in. and h = 0.35 in.
292
GENERAL MATHEMATICS
CYLINDER
12. The volume of a right cylinder (Fig. 211) is equal to
the product of its base and height. The formula is V = TrrVi,
where r is the radius of the circular base. Find Vii r=10.2 cm.
and h = 9 cm. ; if r = 6 in. and h 12 in.
13. The lateral surface of a right cylinder
equals the product of the altitude and the
circumference of the base. The formula
usually given is S = Ch. Find S if C = *f ^ in.
and h = 10 in.
14. The entire surface T of a right cylinder
equals the circumference of the circular base
times the sum of the altitude and the radius of Fj( . 2 n
the base ; that is, T = 2 irr(r + h). Find T if
A =10 in. and r = 5 in.; if A = 2ft. and r=l ft.
15. The volume V of any pyramid (Fig. 212)
is equal to one third the product of its base
B and its altitude h ; that is, V = Find
o
V if = 200sq. in> and A =12 in.; if B =
24.6 sq. in. and h = 2 ft.
16. The lateral area S of a regular pyramid
is equal to one half the product of the perimeter P of its base
Pi
and its slant height ; that is, S = Find A if P = 10. 6 in.
t
and I = 8.2 in. ; if P = 4.3 cm. and I = 15 cm.
17. The lateral area S of a right circular
cone (Fig. 213) is equal to one half the prod-
uct of its slant height I and the circumfer-
ence C of its base. Write the formula for S,
and find S if 1= 14.6 in. and C = 10in. ; if
I = 3.6 ft. and C = 31.416 ft. FIG. 213. RIGHT
18. The lateral area S of a right circular cone ClRCULAR CONE
is Trrh, where r is the radius of the base and h is the slant height.
Find S if r = 10 in. and h = 10 in. ; if r = 10 in. and h = 26.2 in.
FIG. 212. PYRAMID
CONTROL OF THE FORMULA 293
19. The entire surface T of a right circular cone equals the
lateral area I plus the area of the base ; that is, T = trrl + Trr 2 ,
or 7rr(l + r). Find T if I = 10 in. and r = 5 in. ; if I = 12.6 in.
and r = 6 in.
20. An object falling from rest falls in a given time a
distance equal to the product of 16 and the square of the
number of seconds it has fallen; that is, d = 16t 2 . Find
d if t = 1 sec. ; 2 sec. ; 3 sec. ; 4 sec.
21. An object thrown downward travels in a given time a dis-
tance equal to the product of 16 and the square of the number
of seconds it has fallen, plus the product of the velocity with
which it is thrown and the number of seconds it is falling. The
formula is S = 16 1* + Vt. Find S if t = 3 sec. and V= 13 ft. per
second; if t = 5 sec. and F=100ft. per second.
22. The volume of a sphere (Fig. 214) equals
the cube of the radius multiplied by ^ TT. Find V
if y = 1 in.; 10 in.; 5 in.; 10ft.; 12ft.
23. The surface S of a sphere is equal to FIG. 214. THF,
4 7H- 2 . Find S if r = 10 in. ; 12 ft. ; 6j ft.
*24. The force of pressure P of the wind, in pounds per
square foot, is given by the equation P = 0.005 F 2 , where V is
the velocity pf the wind in miles per hour. What would be the
total pressure of this wind against the side of a wall 25 ft.
high and 80 ft. long of a wind blowing 30 mi. per hour ?
*25. Show that the formula for the length I of a belt passing
around two pulleys of the same size whose radii are each r feet,
and the distance between whose centers is d feet, is 1= 2 7rr+ 2 d.
Find I when r = l and d = 4|.
26. In a price list the cost of sewer pipe per foot of length
is given by the formula C = 0.4 d 2 + 14, where d is the diam-
eter of the pipe in inches and C the cost in cents. What will
be the cost of 20 ft. of pipe 2 in. in diameter ?
294 GENERAL MATHEMATICS
330. Practice in solving for any letter. It is often desir-
able to solve a formula for some particular letter in that
formula. Too often the student will recognize a formula
provided it stands in the form in which it is commonly
written, but will not appreciate its meaning if it is written
in a different way. For example, how many students
V
would recognize the formula c = as the well-known
formula V=abc? It is the same formula except that it
is in a different form. If the student realizes this, it helps
him to gain control of the formula. The following exer-
cises will furnish practice in solving for particular letters.
EXERCISES
Solve each of the following formulas for the letter or letters
indicated :
ab Bh
1. .-I = for a ; for />. 11. T = for h.
2. r = aJcforc; fora; forb. Pi.
12. .1 = for /.
3. <? = rtforr; for t.
D 13. A = P + Prt for t.
4. ^ = Y for,/;for/>. 14 . r = |(F- 32) for F.
5. WA*WJJ*to**W t 15 Sss9 * fmi ,. lory.
6- V- for w; for A. i 6 . s = 2 irrh + 2 ir>* for h.
7. C = 2 . 5 (7 for r/. 17. .1 = l 9 ' 2 for // ;
8. r = 41 ,. for r. for ^ ; for 6 a .
9. V = Bh for B. 18 . c = _A_ for E . for ; , .
10. V=*i*h for A. for r.
CONTROL OF THE FORMULA 295
SUMMARY
331. This chapter has taught the meaning of the following
words and phrases : formula, solving a formula, evaluating
a formula, applying a formula, centigrade, Fahrenheit.
332. A formula is a conveniently abbreviated -form of
some practical rule of procedure.
333. A clear understanding of a formula implies:
1. An analysis of some arithmetical situation so as to
arrive at some rule of procedure.
2. Translating the rule into a formula.
3. The ability to solve for any letter in terms of the
other letters in the formula.
4. The ability to apply the formula to a particular
problem and to evaluate the formula.
334. The preceding steps were illustrated in detail by
applications to interest problems, to problems involving
motion, to work problems, to thermometer problems, and
to geometric problems.
335. The graphical interpretations suggested economical
methods of manipulating a formula. For example :
1. Simple-interest problems were solved by the formulas
1 = Prt and A = P + Prt.
2. A problem involving uniform motion in a straight
line was solved by the formula d = rt.
3. The relation between centigrade and Fahrenheit
readings was expressed by the formula (7 =!(/* 32).
336. While the important thing in this chapter is the
power of manipulating and evaluating a formula, the stu-
dent was given the meaning of most of the formulas in
order to have him realize from the very outset that both the
formulas and their manipulation refer to actual situations.
296 GENERAL MATHEMATICS
HISTORICAL NOTE. The development of the formula belongs to
a very late stage in the development of mathematics. It requires a
much higher form of thinking to see that the area of any triangle
can be expressed by A = than to find the area of a particular
A
lot whose base is two hundred feet and whose altitude is fifty feet.
Hence, it was very late in the race's development that letters were
used in expressing rules.
The early mathematicians represented the unknown by some 'word
like res (meaning " the thing "). Later, calculators used a single letter
for the unknown, but the problems still dealt with particular cases.
Diophantus, representing Greek mathematics, stated some problems
in general terms, but usually solved the problems by taking special
cases. Vieta used capital letters (consonants and vowels) to represent
known and unknown numbers respectively. Newton is said to be
the first to let a letter stand for negative as well as positive numbers,
which greatly decreases the number of formulas necessary.
While the race has had a difficult time discovering and under-
standing formulas, it takes comparatively little intelligence to use a
formula. Many men in the industrial world do their work efficiently
by the means of a formula whose derivation and meaning they do
not understand. It is said that even among college-trained engineers
only a few out of every hundred do more than follow formulas or
other directions blindly. Thus, it appears that for the great majority
only the immediately practical is valuable. However, we can be
reasonably sure that no one can rise to be a leader in any field by
his own ability without understanding the theoretical as well as
the practical.
The formula is very important in the present complex industrial
age. A considerable portion of the necessary calculation is done by
following the directions of some formula. Therefore to meet this
need the study of the formula should be emphasized. In discussing
the kind of mathematics that should be required Professor A. R.
Crathorne (School and Society, July 7, 1917, p. 14) says: "Great
emphasis would be placed on the formula, and all sorts of formulas
could be brought in. The popular science magazines, the trade
journals and catalogues, are mines of information about which the
modern boy or girl understands. The pupil should think of the
formula as an algebraic declarative sentence that can be translated
ARCHIMEDES (287-212 B.C.)
(Bust in Naples Museum)
298 GENERAL MATHEMATICS
into English. The evaluation leads up to the tabular presentation
of the formula. Mechanical ability in the manipulation of symbols
should be encouraged through inversion of the formula, or what the
Englishman calls 'changing the subject of the formula.' We have
here also the beginning of the equation when our declarative sentence
is changed to the interrogative."
Archimedes (287-212 H.C.), a great mathematician who studied
in the university at Alexandria and lived iu Sicily, loved science
so much that he held it undesirable to apply his information to
practical use. But so great was his mechanical ability that when
a difficulty had to be overcome the government often called on
him. He introduced many inventions into the everyday lives of
the people.
His life is exceedingly interesting 1 . Read the stones of his detec-
tion of the dishonest goldsmith ; of the use of burning-glasses to
destroy the ships of the attacking Roman squadron ; of his clever
use of a lever device for helping out Hiero, who had built a ship
so large that he could not launch it off the slips ; of his screw for
pumping water out of ships and for irrigating the Nile valley. He
devised the catapults which held the Roman attack for three years.
These were so constructed that the range was either long or short
and so that they could be discharged through a small loophole
without exposing the men to the fire of the enemy.
When the Romans finally captured the city Archimedes was killed,
though contrary to the orders of Marcellus, the general in charge of
the siege. It is said that soldiers entered Archimedes' study while he
was studying a geometrical figure which he had drawn in sand on
the floor. Archimedes told a soldier to get off the diagram and not
to spoil it. The soldier, being insulted at having orders given to
him and not knowing the old man, killed him.
The Romans erected a splendid tomb with the figure of a sphere
engraved on it. Archimedes had requested this to commemorate his
discovery of the two formulas : the volume of a sphere equals two-
thirds that of the circumscribing right cylinder, and the surface of
a sphere equals four times the area of a great circle. You may also
read an interesting account by Cicero of his successful efforts to
find Archimedes' tomb. It will be profitable if the student will
read Ball's " A Short History of Mathematics," pp. 65-77.
CHAPTER XII
FUNCTION ; LINEAR FUNCTIONS ; THE RELATED IDEAS OF
FUNCTION, EQUATION, AND FORMULA INTERPRETED
GRAPHICALLY; VARIATION
337. Function the dependence of one quantity upon
another. One of the most common notions in our lives is
the notion of the dependence of one thing upon another.
We shall here study the mathematics of such dependence
by considering several concrete examples.
EXERCISES
1. Upon what does the cost of 10yd. of cloth depend?
2. If Resta drives his car at an average rate of 98.3 mi. per
hour, upon what does the length (distance) of the race depend ?
3. A boy rides a motor cycle for two hours. Upon what
does the length of his trip depend ?
4. How much interest would you expect to collect in a
year on $200?
5. Upon what does the length of a circular running track
depend ?
6. A man wishes to buy wire fencing to inclose a square lot.
How much fencing must he buy ?
7. State upon what quantities each of the following depends :
(a) The amount of sirloin steak that can be bought for a dollar.
(b) The number of theater tickets that can be bought for
a dollar.
(c) The height of a maple tree that averages a growth of
4 ft. per year.
299
300 GENERAL MATHEMATICS
(d) The time it takes you to get your mathematics lesson if
you solve one problem every three minutes.
(e) The value of a submarine as a merchant vessel.
(f) The rate of interest charged by your local bank.
(g) The perimeter 4 x 4 of the rectangle in Fig. 215.
The preceding exercises illustrate the dependence of one
quantity upon another. We have had numerous other
examples of dependence in the chapters on statistics and
the formula. In fact, every practical formula implies that
the value of some quantity depends upon one or more
others. Thus the circumference of a circular running track
depends upon the diameter. When a quantity depends
upon another quantity for its value, it is said to be a func-
tion of the latter. Thus the area of a circle is a function
of the diameter because it depends upon the diameter for
its value ; the amount of sirloin steak that can be bought
for a dollar is a function of the price per pound ; and the
expression 4 x 4 is a function of x because its value
changes with every change in the value of x.
See if you can illustrate the idea of function by ten
familiar examples not given above.
338. Variable. A number that may change, assuming a
series of values throughout a discussion, is called a variable.
It is not obliged to vary it is " able to vary." Thus the price
of wheat and the number s in the equation A = s 2 are variables.
339. Dependent and independent variables. In the formula
C = ird the number d is said to be the independent variable.
In a discussion or in the construction of circles we may
take its value equal to any number we please. On the
other hand, the value of C is automatically fixed once the
value of d is determined. Because of this fact C is called
the dependent variable.
FUNCTION INTERPRETED GRAPHICALLY 301
EXERCISES
1. What is the value of C in the equation C ird if d 2?
if d = 5? if d = W?
2. Illustrate the ideas of dependent and independent vari-
ables with examples chosen from the text or from your own
experience.
340. Constant. The number TT in the formula C = Trd
differs from C and d inasmuch as it never changes at any
time in the discussion of circles. This number is approxi-
mately -2y 2 -, or 3.1416, whether we are dealing with small or
large circles. We therefore call a number like this, which
has a fixed value, a constant. Obviously any arithmetical
number appearing in a formula is a constant ; thus the 2 in
A = and the - and the 32 in .7^= ^ 7+ 32 are constants.
2
EXERCISE
Turn to Chapter XI, on the formula, and find five formulas
that illustrate the idea of a constant.
341. Graph of a function. A graph may be constructed
showing how a function changes as the value of the
independent variable changes. x
The rectangle in Fig. 215 is a
picture (either enlarged or re- x-z x-z
duced) of every rectangle whose
length exceeds its width by two
units. We shall now proceed to FlG 2 i5
show graphically that the perim-
eter varies with every change in the value of x. The table
on the following page gives the corresponding values for
the length x and for 4 x 4, the perimeter.
302
GENERAL MATHEMATICS
X
3
4
5
6
7
8
9
10
4x-4
8
12
16
20
24
4x-4
--20
If we plot the points corresponding to (3, 8), (4, 12),
(5, 16), etc., using the horizontal axis to plot the values of x
and the vertical axis to plot the values of 4 x 4, we obtain
the points as shown on the straight line AB in Fig. 216. The
line shows that as x increases, the value of 4 x 4 increases
accordingly.
EXERCISES
1. Tell in your own words how the graph in Fig. 216
shows that the function 4 x 4 increases as x increases.
2. Determine from the graph
the perimeters of . rectangles
whose lengths are as follows :
8 in.;, 9 in.; 10.5 in. ; 11 in.
3. Determine from the graph
the length of the rectangles
whose perimeters are as follows :
30 in.; 25 in.; 20 in. ; 18 in.;
10 in. ; 3 in. ; in.
4. Suppose you chose to make
a particular rectangle 10 in. long.
How long would the perimeter be?
How does the graph show this ?
5. How long would you make
a rectangle of the same shape
as the one in Fig. 215 so as to
have its perimeter 16 in. ? How
does the graph show this ?
6. Relying on your past experience, tell how many rectangles
you could construct in the shop or construct in your notebook
"whose length shall exceed their width by two units ".
FIG. 216. GRAPH SHOWING THAT
THE PERIMETER OF THE RECTAN-
GLE IN FIG. 215 is A FUNCTION
OF THE LENGTH
FUNCTION INTEEPBETED (GRAPHICALLY 303
342. Linear function. Since the graph of the expression
4 x 4 is a straight line, the function is called a linear func-
tion. If we let y represent the value of the linear function,
we get the corresponding linear equation y 4 jc 4.
EXERCISE
Give five examples of linear functions.
343. Solving a family of equations by means of the
graph. The' graph of the function 4 x 4 may be used
to solve all equations one of whose members is 4 r 4
and the other some arithmetical number or constant. For
example, if in the equation y = 4x 4 we let ;z/ = 16,
then the equation 4 # 4 = 16 may be interpreted as rais-
ing the question, What is the value 'of x that will make
4. r 4=16? In order to answer this question we find
16 on the ?/-axis (the vertical axis), pass horizontally to
the graph of 4 x 4, and read the corresponding value
of x. The corresponding value of x is seen to be 5. Hence
4 x 4 = 16 when x = 5.
As a verbal problem the equation 4 a* 4=16 may be
translated into the following interrogative sentence: What
shall be the length of the rectangle in order that it may
have a perimeter of 16 ? A glance at the graph is sufficient
to determine the answer ; namely, 5.
EXERCISES
Solve by graph, and check the following equations :
1. 4z-4 = 20. 5. 4z-8 = 2.
HINT. Add 4 to both mem-
2. 4ic 4 = Z4. .
bers so as to obtain the equation
3. 4.r - 4 = 12. 4z-4 = 6.
4. 4- 4 = 6. 6. 4x-5 = 13.
304 <;KM-:KAL MATHEMATICS
7. 4.r-9 = 10. 9. 4ir+ 2 = 12.
8. 4.r + 6 = 26. 10. 4a: + 5 = 19.
HIM. Subtract 10 from both - - J. _i_ 1 <" _ <><i
members so as to obtain the
result 4 x - 4 = 16. 12. 4 ./ - 4 = 0.
344. The graphical solution of the function set equal to
zero. Problem 12, Art. 343, is an interesting special case
for two reasons: (1) It gives us an easy method of find-
ing the value of x in the equation 4^ 4 = 0. We need
only refer to the graph and observe where the line crosses
the o>axis. The line is seen to cross where # = 1. This value
of x checks because 4-14 = 0. Hence x = 1 is a solu-
tion of the equation 4 x 4 = 0. (2) It furnishes us a
graphic method for solving all linear equations in one
unknown because every equation in one unknown can be
thrown into a form similar to 4 x 4 = 0. Show how this
may be done with the equation 3:r + 7 = :r + 12.
EXERCISES
1. Solve graphically the equation 3x + 7 = x + 12.
HINT. The equation may be written in the form 2 x 5 = 0. Why ?
Graph the function 2x 5 just as we graphed 4# 4 (Fig. 216).
See where the graph of 2 x 5 crosses the x-axis. This is the
correct value of x.
Check by substituting this value of x in the original equation
3 x + 7 = x + 12.
2. Solve the following linear equations by the graph, and
check the results :
(a) ox + 2 = 2x + S. (e) 2.ox + 9 = 3x + 7.
(b) 6x - 5 = 4x + 2. (f) |_ 7 = x - 5.
(c)5a; + 8 = 8x-4. x x
(d) 11 x - 9 = 14* + 7. ') 3 4 = 2~ 7 '
FUNCTION INTERPRETED GRAPHICALLY 305
HISTORICAL NOTE. Perhaps the most important idea which a
student can learn from an elementary study of mathematics is the idea
of a function. This is given far greater prominence in European than
in American texts. Indeed, it is remarkable that though the function
idea is generally admitted to be a fundamental notion in mathematics,
the teaching of the notion is often neglected.
From the simple illustrations in this text it will appear that many
familiar facts and principles of natural phenomena can be expressed
as functions. A study of these facts in the form of a mathematical
function throws much additional light on them. Thus, a clear
understanding of the principles of light, sound, and electricity could
not be obtained without a study of the mathematical functions by
which these principles are expressed. The dependence of one quantity
upon another is one of the fundamental notions of human experience.
It is valuable to learn to treat this notion mathematically.
Teachers of mathematics will also remember Professor Felix Klein
as one who has improved mathematics teaching by insisting that it
be made more meaningful to pupils of high-school age by introducing
and emphasizing the notion of a function.
345. Direct variation. If a man walks at the rate of
3 mi. an hour, the distance d which he walks in a given
time t is found from the equation d = 3 t. The following
table gives the distance passed over in 1 hr., 2 hr., 3 hr., etc.
and shows
1. That the greater the time, the greater the distance
passed over.
2. That the ratio of any time to the corresponding
distance is I.
Number of hours . .
1
2
3
4
5
6
7
8
Distance passed over
3
6
9
12
15
18
21
24
When two numbers vary so that one depends upon the
other for its value, keeping constant the ratio of any value
of one to the corresponding value of the other, then one
306 GENERAL MATHEMATICS
is said to vary directly as the other or to be rtiwtly pro-
portional to the other. Thus the number r is said to vary
x
directly as y if the ratio - remains constant, as x and y both
x
change or vary. The equation - = k expresses algebraically,
i7
and is equivalent to, the statement that JT varies directly
X
as y. The equation - = k is often written jc = ky. Show
why this is correct. ^
EXERCISES
Translate the following statements into equations of the
form = /. :
y
1. The cost of 10 yd. of dress goods is directly propor-
tional to the price per yard.
Solution. I 'sing c for the total cost and j> for the price per yard,
- = 10, or c = 10,,.
P
This illustrates direct variation, for the greater^the price per yard,
the greater the total cost.
2. The railroad fare within a certain state is 3 cents per
mile. Write the equation, showing that the distance is directly
proportional to the mileage.
3. The weight of a mass of iron varies directly as the
volume.
4. If a body moves at a uniform rate, the distance varies
directly as the time.
5. The length (circumference) of a circle varies directly as
the diameter.
6. The distance d through which a body falls from rest
varies directly as the square of the time t in which it falls.
A body is observed to fall 400 ft. in 5 sec. What is the
constant ratio of d to t 2 ? How far does a body fall in 2 sec. ?
FUNCTION INTERPRETED GRAPHICALLY 307
Solution. The equation for d and t is
j 2 = k. Why.'
In this problem *gg- = k ;
hence k = 16.
Substituting k = 16 and t' 2 = 2 2 in - = k,
Solving, f/ = 64.
This solution shows that & may be determined as soon as one
value of t and the corresponding value of d is known. Once deter-
mined, this value of k may be used in all problems of this type.
Thus, in all falling-body problems k = 16 (approx.).
7. How far does a body fall from rest in -3 sec. ? in 5 sec. ?
8. If a: varies directly as y, and x = 40 when y = 8, find
the value of x when y = 15.
9. If ic varies directly as a-, and -w = 24 when x = 2, find
w when x = 11.
10. A stone fell from a building 576 ft. high. In how 1 many
seconds did it reach the ground ? (Use the method of Ex. 6.)
11. The speed of a falling body varies directly as the time.
Write the equation for the speed V and the, time t. A body
falling from rest moves at the rate of 180 ft. per second five
seconds after it begins to fall. What will be the speed attained
in nine seconds ?.
12. The time t (in seconds) of oscillation of a pendulum
varies directly as the square root of the length I ; that is,
- = /,-. A pendulum 39.2 in. long makes one oscillation in
v/
one second. Find the length of a pendulum which makes an
oscillation in two seconds.
13. The simple interest on an investment varies directly as
the time. If the interest for 5 yr. on a sum of money is $150,
what will be the interest for 6 yr. 4 mo. ?
308
GENERAL MATHEMATICS
14. The weight of a sphere of a given material varies directly
as the cube of its radius. Two spheres of the same material
have radii of 3 in. and 2 in. respectively. If the first sphere
weighs 6 lb., what is the weight of the second ?
346. Graphing direct variation. Direct variation between
two quantities may be represented graphically by means
of a straight line. Turn back to Chapter XI, on the formu-
las, and find three graphs illustrating direct variation.
-20
FIG. 217. GRAPH OF C = ird SHOWING DIRECT VARIATION
An interesting example is furnished by graphing the
C
equation = TT (where TT = 3.14). This equation says that
the circumference of a circle varies directly as its diameter.
Complete the following table, and graph the results so
as to obtain the graph in Fig. 217. Interpret the graph.
d
1
2
3
4
5
6
7
8
C
3.1
6.3
9.4
FUNCTION INTERPRETED GRAPHICALLY 309
EXERCISES
Graph the following examples of direct variation :
1 . v = 16 1. (Velocity equals 16 times the number of seconds.)
2. t = 5 /. (Turning tendency equals the weight of 5 times
the lever arm.)
3. A = 3b. (The area of a rectangle whose altitude is 3
varies directly as the base.)
347. Inverse variation. We shall now consider a new
and interesting kind of variation. Suppose a gardener
wishes to seed 64 sq. ft. of his garden in lettuce. If he
makes it 16 ft. long, the width must be 4 ft. (Why?) If he
makes it 32 ft. long, the width need be only 2 ft. (Why ?)
How many possible shapes do you think the gardener
might choose for his lettuce bed ? The following table
will help you to answer this question if you remember
that it has been decided that the area shall be 64 sq. ft.
Length . . .
40
36
32
25
20
1(5
8
4
>
\
Width . . .
1.6
1.8
2
2.56
The table shows that the length must vary so as to leave
the area constant, and that because of this fact the
greater the length, the smaller the width. The length is
thus said to vary inversely as the width or to be inversely
proportional to the width. Algebraically speaking, a number
x varies inversely as y if the product xy remains constant
as both x and y vary ; that is, if xy = k. The student may
Jc k
also find this equation written x = - or y = -
y x
310 GENERAL MATHEMATICS
EXERCISES
1. Express each of tin- following statements by means of
equations :
(a) The time needed to go a certain distance varies inversely
as the rate of travel.
(b) The heat of a stove varies inversely as the square of
the distance from it.
(c) Tke rate at which a boy goes to the corner drug store
varies inversely as the time it takes him.
18
2. If .u = , show that w varies inversely with z.
ir
3. If y varies inversely as x, and x = 12 when y = 4, find
the value of y when x = 2.
Solution. By definition of inverse variation,
xy = k.
In the first case, x = 12,
and y = 4.
Therefore 1'J 4 = k,
or k = 48.
In the second case, x = 2.
Therefore 2 y = 4S, since /. is constant.
Then y = 24.
4. If x varies inversely as ?/, and x = 12 when y = 13, find
the value of x when y = 2.
5. When gas in a cylinder is put under pressure, the vol-
ume is reduced as the pressure is increased. The physicist
shows us by experiment that the volume varies inversely as
the pressure. If the volume of a gas is 14 cc. when the pres-
sure is 9 lb., what is the volume under a pressure of 16 Ib. ?
6. The number of men doing a piece of work varies inversely
as the time. If 10 men can do a piece of work in 33 da., in
how many days can 12 men do the same piece of work ?
FUNCTION INTERPRETED GRAPHICALLY 311
40
30
348. Graphing inverse variation. We shall now proceed
to show how inverse variation may be represented graph-
ically. Two cities are 48 mi. apart. Trains running at
various rates carry the traffic between the two cities.
Suppose we attempt
to find out how
long it will take a
train which moves
uniformly at the
rate of 24 mi. per
hour to make the
trip, then 6 mi. per
hour, 8 mi. per hour,
etc. The following
table contains some
of the values by
means of which the
points in Fig. 218
were plotted. The
equation represent-
ing the situation is
48 = rt. When the
points of the .table
are plotted, it is clear that they do not lie on a straight
line, as was the case in direct variation ; but if they are
connected, the result is the curved line of Fig. 218. This
line is one of two branches of a curve called a hyperbola.
20
Rate
FIG. 218. GRAPH SHOWING INVERSE
VARIATION
r
48
32
30
24
20
16
12
8
6
4
_
3
2
1
t
1
1.6
1.6
2
312 GENERAL MATHEMATICS
EXERCISES
1. Determine from the graph in Fig. 218 the time it takes
a train whose rate is 31 mi. per hour ; 20 mi. per hour ; 25 mi.
per hour.
2. Determine from the graph in Fig. 218 how fast a train
runs which makes the trip in 1|- hr.; 2| hr.; 5^ hr.; 8^ hr.
3. Graph the equation ^?y = 144 (see Ex. 5, Art. 347) and
interpret the graph.
4. The hyperbola is an interesting mathematical curve. See
if you can help your class learn more about it by consulting
other books.
349. Joint variation. In the interest formula /= Prt,
I depends for its value on P, r, and L A change in any
one of these letters causes a corresponding change in the
value of /. We express this by saying that the interest
varies jointly as the principal, rate, and time. The alge-
braic equation which defines joint variation is x = kyz.
EXERCISES
1. Turn to Chapter XI, on the formula, and find five illus-
trations of joint variation.
2. If z varies jointly as x and y, and if = 60 when x = 3
and y = 2, find z when x = ^ and y = ^.
*3. Write the following law as a formula: The safe load L
of a horizontal beam supported at both ends varies jointly as
the width w and the square of the depth d and inversely as the
length I between the supports.
*4. A beam 12 ft. long, 4 in. wide, and 8 in. deep when
supported at both ends can bear safely a maximum load of
1920 Ib. What would be the safe load for a beam of the same
material 10 ft. long, 3 in. wide, and 6 in. thick ?
FUNCTION INTERPRETED GRAPHICALLY 313
SUMMARY
350. This chapter has taught the meaning of the fol-
lowing words and phrases : function, linear function, vari-
able, dependent variable, independent variable, constant,
direct variation, inverse variation, joint variation.
351. A clear understanding of the dependence of one
quantity upon another is very important in the everyday
affairs of life.
352. A linear function -may be treated algebraically, or
geometrically by means of a graph.
3/
353. The equation expressing direct variation is - = k.
tj
It is often written x = ky. Its graph is a straight line.
354. The equation expressing direct variation is xy = &,
k k
x = -> or y = - Its graph is a curved line which is
y x
one branch of a hyperbola.
355. The equation expressing joint variation is x = kzy.
CHAPTER XIII
SIMILARITY; CONSTRUCTION OF SIMILAR TRIANGLES;
PROPORTION
356. Construction of similar, triangles ; first method ;
introductory exercises. The following exercises will help to
form a basis for the work of this chapter. The student
should study them carefully.
EXERCISES
1. On squared paper lay off a line segment AB of any con-
venient length. At .1 construct, with the protractor, an angle
of 32. At B construct an angle of 63 and produce the sides
of the two angles so as to form a triangle. Call the vertex
angle C.
2. Compare the shape of the triangle ABC that you have
drawn with the shape of those drawn by your classmates.
3. What was done in Ex. 1 to insure that all members of
the class should get triangles of the same shape ?
4. With the protractor measure angle (.' in your figure. How
else might you have determined its size ?
5. Compare the size of angle C in your figure with angle C
in the figures drawn by your classmates.
6. Show that any angle C drawn by any member of the
class ought to be equal to any other angle C drawn.
7. Are the triangles drawn by the class of the same size ?
Are any two necessarily of the same size '.'
314
CONSTRUCTION OF SIMILAR TRIANGLES 315
357. Similar triangles. Triangles having the same shape
are called similar triangles. Similar triangles are not neces-
sarily of the same size. They may be constructed by making
two angles of one equal to two angles of the other, as was
done in Ex. 1, Art. 356. If two angles 6f one are equal
to two angles of the other, it follows that the third angles
are equal. (Why ?) The symbol for similarity is ~->. Thus
AABC^AA'B'C" is read "triangle ABC is similar to
triangle A'B'C'" The results of Art. 356 may be summed
up in the following geometric theorem : If two angles of
one triangle are equal respectively to two angles of another
triangle, the triangles are similar.
358. Second relation of parts in similar triangles. The
student should be able to discover a second method of con-
structing similar triangles if he studies and understands
the following exercises.
INTRODUCTORY EXERCISES
1. In the triangle ABC drawn for Ex. 1, Art. 356, letter the
side opposite angle C with a small letter c, the side opposite
angle B with a small letter I, and the side opposite angle A with
a small letter a.
2. Measure the lengths of the sides a, l>, and c in Ex. 1
to two decimal places. Find the ratio of a to I, of 1> to c, of
a to c (in each, case to two decimal places).
3. Compare your results in Ex. 2 with the results obtained
by the other members of your class. What conclusion do you
make with reference to the ratios of the sides '.'
359. Construction of similar triangles ; second method.
The results of Exs. 1-3, Art. 358, may be summarized as
follows : In similar triangles the ratios of corresponding sides
are equal. The work of Art. 358 suggests a second method
for constructing similar triangles.
3 It) GENERAL MATHEMATICS
EXERCISES
1. Draw a triangle. Draw a second triangle whose sides
are respectively twice as long as the sides of the first
triangle.
2. Compare the triangles constructed in Ex. 1 as to shape.
Are they similar ? Find the ratio* of the corresponding sides.
3. Draw a triangle with sides three times as long as the
corresponding sides of another triangle. Are they similar?
Give reasons for your answer. How do the ratios of the
corresponding sides compare ?
4. Draw a triangle ABC. Bisect the lines AB, AC, and EC.
Call the halves x', y\ and z'. Construct a second triangle, using
the segments x', y', and z' as sides. Compare the two triangles
as to shape. Are they similar ? What are the ratios of the
corresponding sides ?
The preceding exercises suggest the following theorem :
Two triangles are similar if the ratios of the correspond-
ing sides are equal. This gives us a second method of
constructing similar triangles ; namely, by making the
ratios of their corresponding sides equal.
360. Construction of similar triangles ; third method.
We shall study a third method of constructing similar
triangles which is suggested by the following exercises :
EXERCISES
1. Construct a triangle with two sides 4.6cm. and 6.2cm.,
and with the protractor make the included angle 70. Con-
struct a second triangle with two sides 9.2 cm. and 12.4 cm.
and the included angle 70. Compare the triangles as to shape.
What is the ratio of the corresponding sides ? Measure the
corresponding angles.
CONSTRUCTION OF SIMILAR TRIANGLES 317
2. If convenient the class sholild divide itself into sections,
,ne first section constructing a triangle with two sides and the
included angle as follows : a = 12, b = 18, and C = 40; the
second section taking a = 8, b = 12, and C = 40; and the third
section taking a = 4, b = 6, and C 40. Compare the triangles
drawn by the three sections as to shape. What is the ratio
of the corresponding sides ?
The preceding exercises support the geometric theorem:
Two triangles are, similar if the ratio of two sides of one equals
the ratio of two corresponding sides of the other, and the angles
included between these sides are equal. This theorem suggests
the third method of constructing similar triangles.
361. Summary of constructions for similar triangles. Two
triangles are similar
1. If two angles of one triangle are constructed equal respec-
tively to two angles of the second triangle.
2. If the sides of the triangles are constructed so that the
ratios of their corresponding sides are equal.
3. If the triangles are constructed so that the ratio of two
sides of one is equal to the ratio of two sides of the other and
the angles included between
these sides are equal.
362. Similar right tri-
angles. We shall now
prove the following the-
orem: The perpendicular
to the hypotenuse from the
vertex of a right triangle divides the triangle into two triangles
that are similar to each other (see Fig. 219).
Proof. x = /.x'. Why?
Z. V = Z/. Why?
.-. \ADC ^ AEDC. Why?
D
FIG. 219
318 GENERAL MATHEMATICS
EXERCISES
(Exs. 1-4 refer to Fig. 219)
1. Show that A.I IK' A ABC.
2. Show also that A BCD ^ A.-lLv '.
3. Translate the results of Exs. 1 and 2 into a geometric
theorem.
4. State a theorem expressing the results of this article.
363. Similar polygons. In later work in mathematics
we learn that similar polygons also have corresponding
angles equal and that the ratios of the corresponding sides
are equal. This rests on the fact that two similar polygons
may be divided into sets of similar triangles by drawing
corresponding diagonals us
in Fig. 220.
Similar figures are of
frequent occurrence. The
plans of construction work, FJG 22Q SIMILAK POLYGOXS
drawings in shop, a sur-
veyor's copy of a field triangle, blue prints, a photograph,
enlarged and reduced pictures, are all examples. The rela-
tion of the different parts in all the foregoing is shown by
magnifying or reducing all parts to a definite scale. Thus,
you may be able to determine by looking at a photograph
of a man that he has large ears, although in the picture
the actual measurement of either of his ears may be less
than a centimeter. One can tell by looking at the plan
of a house whether the windows are large or small, be-
cause the relation is brought out by the fact that all parts
are reduced to the same scale ; that is, the ratios of the
corresponding parts are equal. See if you can find examples
that will illustrate the last statement
CONSTRUCTION OF SIMILAR TKI ANGLES 319
Similar triangles may be regarded as copies of the same
triangle magnified or minified to a scale, or both may be
regarded as scale drawings of the same triangle to dif-
ferent scales. We shall study the geometric relations more
in detail in the^next chapter.
364. Algebraic problems on similar figures. The fact
that the ratios of the corresponding sides of similar poly-
gons are equal furnishes us with an algebraic method of
finding distances.
EXERCISES
1. In the similar triangles of Fig. 221, if a = 3 in., a' = 9 in.,
and b = 3 in., how long is // ?
2. In the similar triangles of Fig. 222, if n = 6 -mm.,
a' = 8 mm., and I = 8 mm., how long
is b 1 ?
3. In the similar triangles of Fig. 223,
if .a' =10.5 mm., b =12 mm., and // = a/\h
15 mm., how long is a ?
Flo 221
4. The sides of a triangle are 16, 20,
and 26. The shortest side of a similar
triangle is 22. Find the other sides.
5. The sides of a triangle are 2.3 cm.,
2.7 cm., and 3 cm. The corresponding
sides of a similar triangle are a-, y, and
12cm. Find the values* of x and y.
6. Two rectangular boards are desired.
One is to be 4 in. wide and 6 in. long, .
the other is to be IS^in. long. How wide / \
should the second board be? FK , 223
7. At a certain time of day a foot rule
casts a shadow 10 in. long. How long is the shadow of a yard
stick'at the same time? Draw a diagram and prove your work.
820
( ; K.\ KK A I. MATHEMATICS
8. In Fig. 224 the pole, the length of its shadow, and
the sun's rays passing over the top of the pole form a triangle.
The shadow of the pole is measured, and is found to be 60 ft.
long. At the same time the shadow of a vertical stick 2^ ft.
high is measured, and is found to be 7|- ft. long. How may we
determine the height of the pole
without actually measuring it ?
c
\
\
\
h
\
\
\
60'
FIG. 22 1
7.5'
FIG. 225
Solution. The stick? the shadow, and the sun's rays form a
triangle similar to the first triangle (see Fig. 225). Why?
If we let h denote the height of the pole, we get
h 6 Why?
Then
2.5 7.5
h = 20.
z
9. The shadow of a chimney is 85.2 ft. long. At the same
time the shadow of a man 6 ft. 2 in. tall is 9 ft. 2 in. How high
is the chimney ? @
10. Draw a triangle ABC on squared
paper as in Fig. 226. Through a point D on
AC draw line DE II AB. Measure the seg- A'
ments CD, DA , CE, and EB to two decimal
CD Cf
places. Find the ratios and Howdo
DA EB
these ratios compare ? What does this show ?
11. Draw a scalene triangle on squared
paper, making the base coincide with one
of the horizontal lines. Letter the triangles
as in Fig. 227. Choose any line parallel to the base and letter
it DE as in Fig. 227. Find the ratios and -7-- How do
DA A B
these ratios compare? State your conclusion as a theorem.
FIG. 227
CONSTR0C-TION OF SIMILAR TUI ANGLES
n
FIG. 221
12. Suppose that in Fig. 227 Z>C =2.5 mi., DA = 7.5 mi.,
and CE = 9.ini., how long is EB ?
13. In triangle ABC, Fig. 227, the line DE has been drawn
parallel to the base AB. Prove that
the small triangle CDE is similar to
the large triangle ABC.
14. IiiFig.228,if^U;=3,DE=5,
and AB = 25.5, how long is BC ?
15. Divide ; a line segment into
two equal parts and show that your
construction is correct.
Construction. Let AH, Fig. 229, be
the line segment. Draw A C through A,
making any convenient angle with AB.
On AC lay off AD and DE, each
equal to 1 unit. Join E to B. Draw
DF II EB. Show that ,1 F = FB.
HINT, Use Ex. 10, p. 320.
FIG. 229. How TO BISECT
A LINE
16-. Divide a line segment into
two parts whose ratio is |.
Construction. Let AB, Fig. 230, be
the given segment. Draw A C, making
any convenient angle with AB, as
shown. On AC lay off AD = 2 units
and DE = 3 units. Join E and B.
Through D draw DF II EB. Show
AF 2
I)'
Fir,. 230
that
FB 3
17. Divide a line segment into
3 1 2 a
parts having the ratios-; ^; ^; y
FIG. 231
18. The distance across a swamp
(Fig. 231) is to be found. A point C is located in the same
line with A and B. At C and B lines CD and BE are drawn
322 GENEKAL MATHEMATICS
perpendicular to CB, and the line AD is drawn. The lengths
of CB, DE, and EA are found to be 80 ft., 90 ft., and 250 ft
respectively. Find the distance AB.
19. Show that the distance A B across
the swamp could also be found by meas-
uring the lines shown in Fig. 232.
365. Proportion. The preceding
exercises dealing with similar triangles
were solved by means of a special
type of equation expressing the fact that two ratios in the
geometric figure were equal. Thus in Fig. 233 the line AB
is divided into two parts whose ratio is | (see the method
of Ex. 16, Art. 364). In this construction it turns out that
4F 2
3- Wh .v'-' A. f. , B
~2*" '' ''
?. Why? Vs/
DE 3
AF AD
and =- Wh y ? Fio. 233
Such an equation, which expresses equality of two ratios,
is called a proportion. The line segments AF, FB, AD, and
DE are said to be proportional, or in proportion. This
means that AF divided by FB will always equal AD
divided by DE.
A proportion may thus be defined as an equation which
expresses the equality of two fractions ; as, T 8 :r = f. Another
ct c
example of a proportion is - = - This may be read
" a divided by b equals c divided by <?," or " a is to b as
c is to rf," or " a over b equals c over d." Sometimes it is
written a : b = c : d, but this form is not desirable.
CONSTRUCTION OF SIMILAR TRIANGLES 323
EXERCISE
Is the statement f = f a proportion ? Give reasons for
your answer. Is f = -^ a proportion ? Explain your answer.
366. Means and extremes. The first and last terms in
a proportion are called the extremes and the second and
third terms the means. Thus, in the proportion - = a and
d are the extremes and b and c the means.
EXERCISES
1. Compare the product of the means with the product of
the extremes in the following proportions :
What statement can you make concerning the products ?
2. Make up several proportions and compare the product
of the means with the product of the extremes.
367. Theorem on the relation between the means and the
extremes of a proportion. Exs. 1-2, Art. 366, illustrate a
well-known law or theorem ; namely, that in a proportion
the product of the means equals the product of the extremes.
a. c
If the given proportion is - = - then the law is algebraically
stated thus : ad = be.
The theorem may be proved as follows :
a c
Let - = - represent
b d
members by bd we get
ft f*
Let - = -. represent any proportion. Multiplying both
b d
abd _ cbd
Reducing each fraction to lowest terms,
ad be.
324 GENERAL MATHEMATICS
Since a proportion is a special kind of equation, there
are special laws which often make a proportion easier
to solve than other equations which are not proportions.
The law given on page 323, Art. 307, is one of the many
principles of proportion convenient to use. Thus, instead
4 16
of finding the L.C.D. in the equation - = and solving
o . X
in that way, we simply use the preceding law, and say
4 x = 48.
x = 16.
The law is also a convenient test of proportionality,
since it is usually simpler to find the products than to
reduce the ratios to lowest terms.
EXERCISES
1. Test the following statements to see if they are
proportions :
3 _ 15 ' . l5 J 12/ 42 w _^ 21 m
' 7 ~~ 35 ' ' 2^5 ~~ 1.4 ' 5 n ~ 2.5 n '
5 _ 8 2 a _ 5a 11.5 _ 7.7
\~J ^ 77' v*) o ^~K " (*) o " o~o *
<- 11 ox t .o x o.o Z.Z
2. Find the values of the unknowns in the following pro-
portions, and check by substituting in the original equations :
(b) - .= ^ Solution. >/- + y - 20 = y z 9.
66_l ' y -2 = -.
} 10"5'
f f -\\ y~ 12 3 11 + 5 11 -3
< d >niT5' Check - irT3 = irri
, 3+1 = 3^ + 2 i-> _ 8
5 14 14" 7'
CONSTRUCTION, ! (3F SIMILAR TRIANGLES 326
] a - 13 = a - 14 : ^ ~T~ ~3
I IL ' . /i\ x i O _ J-
' /T _I- .^ ~~ _1_ 1 U/ . K '
3. If 5 and 3 are each added to a certain number, and
1. and 2 are each subtracted from it, the four numbers thus
obtained are in proportion. Find the number.
4. Show how to divide a. board 54 in. long into two parts
whose ratio is y^.
5. What are the two parts of a line segment 10 cm. long
if it is divided into two parts whose ratio is | ?
6. The acute angles of a right triangle are as 2 is to 5 ;
that is, their ratio is -|. Find the angles.
7. If 10 be subtracted from one of two complementary
angles and added to the other, the ratio of the two angles thus
formed is ^. Find the angles.
8. If 1| in. on a railroad map represents 80 mi., what
distance is represented by 2| in.?
9. Two books have the same shape. One is 4^ in. wide;
and 7-g^ in. long. The other is 18 in. long ; how wide as it ?
10. The records of two leading teams in the American
League were Boston won 68, lost 32 ; Chicago won 64, lost 36.
If the teams were scheduled to play each other ten more gauit-s.
how many must Chicago have won to have been, tied with
Boston ?
11. If 1 cu. ft. of lime and 2 cm. ft. of sand are used in
making 2.4 cu. ft. of mortar, how much of each is needed to
make 96 cu. ft. of mortar ?
368. Proportion involved in variation. Many problems
in physics, chemistry, general science, domestic science,
astronomy, and mathematics may be solved by either
variation or proportion. In fact, the whole theory of
826 GENERAL MATHEMATICS
proportion is involved in our discussion of variation, but
this fact is not always so obvious to a beginner. The fact
that problems may be stated both in terms of variation and
in terms of proportion makes it necessary for the student
to recognize clearly the relation between variation and pro-
portion. This relation will be illustrated in the following
list of exercises.
EXERCISES
Solve by either variation or proportion :
1. If 11 men can build a cement walk in 82 da., how long
will it take 15 men to build it ?
(a) Solution as a variation problem :
mt = k. (The time it takes to build
the walk varies inversely as
the number of men.)
Then 11 82 = k.
Hence k = 902.
Using this value of k in the second case,
mt = 902 ;
but m = 15.
Whence 15 t = 902,
and t = * T V = 60 T 2 S da.
(b) Solution as a proportion problem. The number of men is not
in the same ratio as the time necessary to build the walk, but in
inverse ratio ; that is,
This proportion means " the first group of men is to the second
group of men as the time it takes the second group is to the time it
takes the first group."
Substituting the three known facts,
II -!i
15 7 82'
Whence 15 d, = 902,
and c/ = T* = 60 da.
CONSTRUCTION OF SIMILAR TRIANGLES 327
2. If 200 ft. of copper wire weighs 60 lb., what is the weight
of 125 ft. of the same kind of wire ?
3. The cost of wire fencing of a certain kind varies as the
number of yards bought. If 12 rods cost $12.80, how much
can be bought for $44.80?
4. Two men are paid in proportion to the work they do.
A can do in 24 da. the same work that it takes B 16 da. to do.
Compare their wages.
5. A farmer has a team of which one horse weighs 1200 lb.
and the other 1500 lb. If" they pull in proportion to their
weight, where must the farmer place the clevis on a four-foot
doubletree so as to distribute the load according to the size
of the horses ?
369. Different arrangements of a proportion. The stu-
dent will be interested in seeing in how many different
forms a proportion may be arranged. This he may learn
by solving the exercises that follow.
EXERCISES
1. Arrange the numbers 3, 6, 7, and 14 in as many propor-
tions as possible. Do the same for the numbers 2, 5, 8, and 20.
2. Can you write two ratios that will not be equal, using
the numbers 2, 5, 8, and 20 as terms of these ratios ?
3. How do you decide which arrangement constitutes a
proportion ?
The preceding exercises suggest that a proportion such
as - = - may take four forms, as follows :
o a
(a) The given proportion
328 GENERAL MATHEMATICS
(b) The form obtained by alternating the means in (a):
a b
~c = d'
(c) The form obtained by alternating the extremes in (a):
(d) The form obtained by alternating both the means
and extremes in (a) :
d b
The last form can be obtained simply by inverting the
ratios- in (a).
We know that the proportions given above are true, for by
applying the test of proportionality we see that the product
of the means equals the product of the- extremes in each
case. Furthermore, any one of them could have been
obtained by dividing the members of the equation ad = be
by the proper number. Thus, to get = - we must divide
both members of the equation ad = be by ab. Why ?
. ad l<-
Then = :
ab ab
from which T = - , or form (c).
b a
370. Theorem. The preceding discussion illustrates the
use of the theorem which says that if the product of two
numbers is equal to the product of two other munbers, either
pair may be made the means and the other pair the extremes
of a proportion.
CONSTRUCTION OF SIMILAR TKIANGLES 329
EXERCISES
1. Start with the equation ad = be and obtain the forms
a c a b d b
7 = 3' - = 3 ' und - = - This completes the proof of the
(> a c a c a
theorem just stated. Why ?
2. Write the four possible forms that can be obtained from
the following products :
(a) 5 21 = 7 15. (o) 3 a . 4 b = a 12 b.
(b) 3 28 = 12 7. (d) 15 - 7 t = 105 /. . (
= . she, that =.
FIG. 234
4. If two equilateral polygons have the same number of
sides, the corresponding sides are proportional (see Fig. 235).
i n
Proof. =1, Why?
JJ o
and 1 -
TIV . A , r , .>
Therefore ; = \v hy ?
J5( /> C
By alternating the means,
,4 BC
A'E' B'C'
and so on for the other corresponding sides.
371. Mean proportional. In the proportion -= -, b is
called the mean proportional between a and c (note that
b appears twice in the proportion).
330 GENERAL MATHEMATICS
EXERCISES
1. What is a mean proportional between 4 and 9 ?
HINT. Let x = the number.
TU * X
Ineii - = -.
x y
From which x- = 36.
Then r = 6.
2. Show that the value of b in the proportion - = - is given
C
by the equation b = Vac (read " & equals -f- or the square
root of ac"~).
3. "What is a mean proportional between 2 and 18 ? between
10 and 40 ? between 2 and 800 ?
4. Find a mean proportional between a? and i 2 ; between
3? and y 3 .
372. How to pick out corresponding sides of similar
triangles. The similar triangles of Fig. 236 are -placed so
that in certain cases a line is a side in each of two similar
triangles. Thus, AC is a
side of A ADC and also of
the similar triangle ABC.
This suggests that the
line may occur twice in
the proportion of the cor-
responding sides. In this FlG 236
way it is seen that the
line becomes a mean proportional between the other two.
This analysis can be checked only by actually writing
the proportion of pairs of corresponding sides of similar
triangles. In order to do this correctly the student must
remember that he (nrrexpondinri sides of similar triangles
CONSTRUCTION OF SIMILAR TRIANGLES 381
are the sides which lie opposite equal angles. Hence, from
the fact that A ADC > A ABC we may write the following
proportion :
AD (opposite Z z in A ADC) _ ^4 C (opposite Z 2: in A ADC)
A C (opposite Z z' in A AB (7) ^(opposite Z.C in AACB)
, AD AC
1 hat is, =
AC AB
AC is thus seen to be the mean proportional between AD
and AB.
Show in a similar way that BC is a mean proportional
between the hypotenuse AB and the adjacent segment BD ;
BD BC .
that is, show that - = -
BC AB
We may summarize the preceding exercises and dis-
cussion by the theorem: In a right triangle either side
including the right angle is a mean proportional between the
hypotenuse and the adjacent segment of the hypotenuse made
by a perpendicular from the vertex of the right angle to the
hypotenuse.
373. Theorem. If in a right triangle a perpendicular is
drawn from the vertex of the right angle on the hypotenuse,
the perpendicular is a mean proportional
between the segments of the hypotenuse.
The truth of the preceding theorem will
be seen from the following: A D
In AABC (Fig. 237) Z C is a right Fio. 237
angle, and CD AB. |f = ff because A ADC - A CDS,
and the corresponding sides are therefore in proportion.
332 ; :...-. ..GENERAL MATHEMATICS-, ...;_
EXERCISES
1. Write out the complete proof for the preceding theorem.
2. Find the altitude drawn to the hypotenuse of a right
triangle if it divides the hypotenuse into two segments whose
lengths are 4 in. and 16 in. respectively. Find also each leg
of the right triangle.
*3. In a right triangle ABC (right-angled at C) a perpen-
dicular is drawn from C to AB. If CD = 8, then A I) = 4.
Find the length of AB.
374. Construction of a mean proportional. The theorem
of Art. ^73 on page 331 furnishes us with a method of con-
structing a mean proportional between any | b [
two line segments, as will now be shown.
In Fig. 238 we are given two line seg- ' - '
ments a and b. The problem is to con-
struct a mean proportional (say x units long) between a and b.
CL CC
We know that the equation - = - will represent the
x b
situation,
Construction. On a working line, as A K in Fig. 239, we lay off a
from A to B and b from B to C: With AC as a diameter we
construct a semicircle. At
B we erect a perpendicular ^-- ~~~
^
B OK
intersecting the circle at D. /'
Then BD is the required
mean proportional. ' /'
I /'
Proof. Connect A with D [^_ _ r JL
and C with D. Then BD is l
the required mean propor- FIG. 239. MEAN PROPORTIONAL
tional between a and b pro- CONSTRUCTION
vided we can show that Z-D
is a right angle. (Why?) We shall proceed to show that ZD is a
right angle by proving that if any point on a circle is connected with
the ends of a diameter, the angle formed at that point is a right angle.
CONSTRUCTION OF SIMILAR TRIANGLES 333
In Fig. 240 we have a given circle constructed on the diameter
AC and a point D connected with the ends of AC. We must
show that ZZ) is a right angle.
Connect D and 0.
Then Z z = Z .s + Z .s', (1)
because an exterior angle of a tri-
angle is equal to the sum of the
two nonadjacent interior angles;
and
Z y = Z t + Z I'
(2)
for the same reason.
By adding (1) and (2),
FIG. 240
Since
But
and
Therefore
Then
Z x + Z y = Z ,s- + Z / + Z t + Z ('.
Z x + Z y - 180,
Z < = Z <'.
2 Z.v + 2 Z< = 180.
Zs + Z< = 90.
ZZ)--90.
Why?
Why ?
Why?
..Why?
Why?
Why?
Then, if in Fig. 239 Z D = 90, the proportion - is true,
and BD is a mean proportional between a and
BD b
Give reasons.
EXERCISES
1. Explain how a mean proportional between two given
line segments may be constructed.
2. Construct a mean proportional be-
tween 9 and 16, 4 and 16, 4 and 9, 16
and 25, 25 and 36. A
3. In triangle ABC, Fig. 241, Z.C
is a right angle, CDA.AB, AD = 2, and DB = 6. Find the
lengths of AC and CB.
334 GENERAL MATHEMATICS
4. Find the mean proportional between the line segments
m and n in Fig. 242.
5. Measure in and n in Fig. 242 and the mean proportional
constructed in Ex. 4. Square the value of m
the mean proportional and see how the
value compares with the product of m , n |
and n - Fir.. 242
*6. Construct a square equal in area to a
given rectangle ; to a given parallelogram ; to a given triangle.
a c
375. Fourth proportional. In the proportion T = -^' d is
\Jv
called the fourth proportional to , 6, and c. There are
two methods of finding the fourth proportional to three
given numbers a, 5, and c.
Algebraic method. Let x represent the value of the fourth
proportional.
TI, a c
Then - = -
ft x
(by definition of a fourth proportional).
Solving for x, ax = be.
be
x _ A c F G V
a
FIG. 243. How TO CONSTRUCT
Geometric method. Take three given A FOURTH PROPORTIONAL
lines, as a, b, and c in Fig. 243, and
draw any convenient angle, as Z.BAC. On AB lay off AD = a,
DE-l. On the other line A Clay oSAF=c. Draw DF. Then draw
EG II DF as shown. Then FG is the required fourth proportional.
See if you can show why the construction is correct.
EXERCISES
1. Check the construction above by measuring the four
a c
segments to see if = -
o x
2. Construct a fourth proportional to three given line seg-
ments 2 cm., 3 cm., and 5 cm. long respectively.
CONSTRUCTION OF SIMILAR TKIA^ULES 33o
3. Show algebraically that the segment obtained in Ex. '1
should be 1\ cm. long.
4. Construct a fourth proportional to three segments 5 cm.,
6 cm., and 9 cm. long respectively.
5. Check your work in Ex. 4 by an algebraic method.
* 376. To find the quotient of two arithmetical numbers by
a special method. To find || in per cent we need to solve the
22 x
equation ^ = w
(Why?) This pro-
portion suggests simi-
lar triangles. If we
take a horizontal line
OM (Fig. 244) as
a dividend line on
squared paper, and
ON perpendicular- to
OM as a divisor line,
then lay off OA on
OM equal to 22 units,
and at A erect a per-
pendicular and mark
off AB equal to 70
units, we can solve
our problem provided
we draw another line DR 100 units above OM and parallel
to it. Call DR the quotient line.
Stretch a string fastened at so that it passes through
J5, meeting the quotient line at C.
Di
de
nd -Li
-M-
FIG. 244
Then
- or
NOTE. The proof is left to the student.
Therefore 22 is approximately 31% of 70.
330 ( J KN K1J A L M ATH.EMATK 'S
EXERCISES
*1. Point out the similar triangles in the device for express-
ing quotients "used in Fig. 244. Read the sides which are
proportional.
*2. A gardener planted 12 A. of potatoes, 8 A. of beans,
13 A. of onions, 3 A. of celery, and 5 A. of cabbage. By means
of the device used in Fig. 244 show the distribution of his
garden in per cents.
377. Verbal problems solved by proportion. We have
said that many problems of science, the shop, and engineer-
ing can be solved by proportion. We shall proceed to
study how to solve some of these problems by using our
knowledge of proportion.
In the study of turning tendency, Art. 233, we recognized
the following familiar principle of the balanced beam : The
left weight times the left lever arm equals the right weight times
the right lever arm. As a formula
I~- 7-77 A F B
tins may be written w l l l = u' 2 l z . i . .
100
Ib.
60
Ib.
This principle is already familiar
to all who have played with a
seesaw. They discovered long ago Fi G 245
that a teeter board will balance
when" equal products are obtained by multiplying the
weight of each person by his distance from the point of
support (fulcrum).
If, in Fig. 245, B weighs 60 Ib. and is 5 ft. from the ful-
crum F, then A, who weighs 100 Ib., must be 3 ft. from
the fulcrum. Thus, 60 5 = 100 3 is a special case of
general law w l l 1 = w%l z .
If we divide both members of the equation iv^^ = W 2 1 2 by
7 nn
w^ly we get - = - , which is in the form of a proportion.
'2 w i
CONSTRUCTION OF SIMILAR TRIANGLES 337
The student will learn that in shop work many problems
dealing with the lever or balanced beam may be solved by
some form of the preceding proportion.
BEAM PROBLEMS
1. A lever (Fig. 246) 10 ft. long carries weights of 40 Ib.
and 50 Ib. at its ends. Where should the fulcrum be placed
so as to make the lever balance ?
Solution. Let x = the number of feet from F to A ,
and 10 x = the number of feet from /' to B.
Then 50 x = 40(10 -x). Why? A x F 10-x B
50x = 400-40x. A A
90 x = 400.
50
Ib.
\40\
lib.
FIG. 246
2. A, who weighs 80 Ib., sits 6 ft. from the fulcrum. If B
weighs 100 Ib., at what distance from the fulcrum should A
sit in order to balance B ?
3. A and B together weigh 220 Ib. They balance when A is
5 ft. and B 6 ft. from the fujcrum. Find the weight of each.
4. A and B are 4 ft. and 6 ft. respectively from the fulcrum.
If B weighs 60 Ib.,
how much does A
weigh ?
5. How could
you weigh yourself
without a scale ?
6. AB in Fig. 247
is a crowbar 8| ft.
long supported at F, \ ft. from A . A stone presses down at
A with a force of 2400 Ib. How many pounds of force must
be exerted by a man pressing down at />' to raise the stone ?
(Disregard the weight of the crowbar.)
FIG. 247
GESEKAL MATHEMATK >
7. In attempting to raise an automobile (Fig. 248) a man lifts
with a force of 150 Ib. at the end of a lever 10 ft. long. The distance
from the axle to F is 2 ft.
What force is exerted up-
ward on the axle as a result
of the man's lifting ?
8. Find ^ if 1 9 = 18 ft.,
>/- 2 = 62 Ib., and w 1 = 51 Ib.
9. Find J a if ^ = 40 in.,
/- 2 =26 Ib., and ^=38 Ib.
FIG. 248
MIXTURE AND ALLOY PROBLEMS
1. How much water must be added to 10 gal. of milk,
testing \fo butter fat, to make it test 4% butter fat?
Solution. Let x = the number of gallons of water added.
Then x + 10 = the number of gallons of diluted milk,
51
and 10 = the amount of butter fat in the undiluted milk.
I $ff ( x + 10) = the amount of butter fat in the diluted milk.
Since the amount of butter fat remains constant,
-$1 10= * (x
100
110 a: + 10
200
Why 'i
Whv '{
25
4 x + 40 = fjo.
x = 3 1, the number of gallons of water to be added.
2. A physician has a 25% mixture of listerine in water. How
much water must he add to it to make it a 15 mixture ?
Solution. Consider an arbitrary quantity of the mixture, say
100 oz.
Let x = the number of ounces of water added to every
100 oz. of the mixture.
Then 100 + x = the number of ounces in the new mixture.
CONSTRUCTION OF SIMILAR TRIANGLES 339
Since 25% of the original mixture is listerine,
25
= the per cent of listerine in the
new mixture.
And since 15% of the new mixture is to be listerine,
25 = JL5_
100 + x 100 '
1500 +15 a: = 2500.
15 x = 1000.
Hence 66$ oz. of water must be added to every 100 oz. of the
original mixture.
3. How much water should be added to a bottle contain-
ing 4oz. of the original mixture in Ex. 2 to make it a 15%
mixture ?
4. If a patent medicine contains 30% alcohol, how much
of other ingredients must be added to 12 qt. of it so that
the mixture shall contain only 20% alcohol ?
5. How many quarts of water must be mixed with 30 qt.
of alcohol 82 % pure to make a mixture 70 % pure ?
6. "\Yhat per cent of evaporation must take place from
a 5% solution of salt and water (of which 5% by weight
is salt) to make the remaining portion of the mixture a 7%
solution ?
7. Two grades of coffee costing a dealer 25$ and 30$ per
pound are to be mixed so that 50 Ib. of the mixture will be
worth 28$ per pound. How many pounds of each kind of
coffee must be used in the mixture ?
8. In a mass of alloy for watch cases which contains 80 oz.
there are 30 oz. of gold. How much copper must be added
in order that in a case weighing 2 oz. there shall be -}>/.
of gold ?
340 GENERAL MATHEMATICS
Solution. Let x = the number of ounces of copper to be added.
Then 80 4- x = the number of ounces in the new alloy.
- = the ratio between the whole mass of alloy
30
and the gold.
y = the ratio of a sample of the new alloy to the
gold in the sample.
ri-,1 80 + X \
Then _ = _.. Why?
80 + z = 120. Why?
Hence 40 oz. of copper should be added.
9. In an alloy of gold and silver weighing 80 oz. there are
10 oz. of gold. How much silver should be added in order that
10 oz. of the new alloy shall contain only ^ oz. of gold ?
10. Gun metal is composed of tin and copper. An alloy of
2050 Ib. of gun metal of a certain grade contains 1722 Ib. of
copper. How much tin must be added so that 1050 Ib. of the
gun metal may contain 861 Ib. of copper ?
*378. Specific-gravity problems. A cubic foot of glass
weighs 2.89 times as much as a cubic foot of water (a
cubic foot of water weighs 62.4 Ib.). The number 2.89 is
called the specific gravity of glass. In general, the specific
gravity of a substance is defined as the ratio of the weight
of a given volume of the substance to the weight of an equal
volume of water at 4 centigrade. What would it mean,
therefore, to say that the specific gravity of 14-karat gold
is 14.88 ? A cubic centimeter of distilled water at 4 cen-
tigrade weighs just 1 gm. Since the specific gravity of
14-karat gold is 14.88, one cubic centimeter of gold weighs
14.88 gm., 2 cc. weighs 29.76 gm., etc. In short, the weight
of an object in grams equals the product of its volume in
centimeters times its specific gravity.
CONSTRUCTION OF SIMILAR TRIANGLES 341
EXERCISES
1. How many cubic centimeters of distilled water (specific
gravity equal to 1) must be mixed with 400 cc. of alcohol
(specific gravity equal to 0.79) so that the specific gravity of
the mixture shall be 0.9?
HINT. Find the weight of the two parts and set the sum equal
to the weight of the mixture.
2. Would you accept half a cubic foot of gold on the con-
dition that you carry it to the bank ? Explain your answer.
3. Brass is made of copper and zinc. Its specific gravity is 8.5.
How many cubic centimeters of copper (specific gravity 8.9) must
be used with 100 cc. of zinc (specific gravity 7.15) to make brass ?
4. What is the specific gravity of a steel sphere of radius
1 cm. and weight 32.7 gm. ?
379. Proportionality of areas. The geometric exercises
to be given in this article are important. The student
should study them carefully, prove them, and try to
illustrate each.
EXERCISES
1. Prove that the areas of two rectangles are to each other
as the products of their correspond-
ing dimensions.
Proof. Denote the areas of the rec-
tangles by #j and 7? 2 , as in Fig. 249,
and their dimensions as shown. '
R l = a l b l . Why?
R z = a 2 b 2 . Why?
Therefore 5l = 2LI. Why?
R 2 a A FIG. 249
It is important to note that the last proportion is obtained by
dividing the members of the first equation by those of the second.
842 GENERAL MATHEMATICS
2. If two rectangles (Fig. 250) have equal bases, they are
to each other as their altitudes. (Follow the method of Ex. 1.)
3. If two rectangles have equal alti-
tudes, they are to each other as their bases.
4. The area of a rectangle is 48 sq. ft.
and the base is 11 yd. What is the area of
a rectangle having the same altitude and
a base equal to 27.5 yd. ?
5. .Prove that the areas of two parallelo-
grams are to each other as the products of
their bases and altitudes.
6. The areas of two triangles are to
each other as the products of their bases and altitudes.
7. The areas of two parallelograms having equal bases are
to each other as their altitudes ; the areas of two parallelograms
having equal altitudes are to each other as their bases.
8. The areas of two triangles having equal bases are to
each other as their altitudes ; the areas of two triangles having
equal altitudes are to each other as their bases.
9. Prove that triangles having equal bases and equal alti-
tudes are equal.
*10. Construct the following by means of Ex. 9: a right tri-
angle, an isosceles triangle, an obtuse-angled triangle, each equal
to a given triangle.
SUMMARY
380. This chapter has taught the meaning of the fol-
lowing words and phrases : similar triangles, similar
polygons, proportion, means, extremes, fulcrum, mean
proportional, fourth proportional, alloy, specific gravity.
381. Polygons that have the same shape are similar.
382. In similar triangles the corresponding angles are
equal and the corresponding sides are in proportion.
CONSTRUCTION OF SIMILAR TRIANGLES 343
383. Two similar triangles may be constructed by
1. Making two angles of one equal to two. angles of
the other.
2. Making the ratios of corresponding sides equal.
3. Making the ratio of two sides of one equal to the
ratio of two sides of the other, and the angles included
between these sides equal.
384. A proportion expresses the equality of two ratios.
385. A convenient test of proportionality is the theorem
that says the product of the means equals the product of
the extremes.
386. If ad = be, we may write the following four
proportions :
- a c a b d c ... d b
387. The fact that the ratios of corresponding sides of
similar polygons are equal furnishes us with an algebraic
method of finding distances.
388. Inaccessible distances out of doors may often be
determined by means of a proportion.
389. Beam problems and mixture, alloy, and specific-
gravity problems may be solved by equations which take
the form of proportions.
390. If a line is drawn parallel to the base of a triangle,
the triangle cut off is similar to the given triangle, and the
corresponding sides are in proportion.
391. The following important theorems about the area
of two parallelograms have been proved :
1. The areas of two parallelograms are to each other
as the product of their bases and altitudes.
344 GENERAL MATHEMATICS
2. The areas of two parallelograms having equal bases
are to each other as their altitudes, and the areas of two
parallelograms having equal altitudes are to each other as
their bases.
392. Three theorems similar to those in Art. 391 were
proved for the areas of rectangles and triangles.
393. If in a right triangle a line is drawn from the
vertex of the right angle perpendicular to the hypotenuse,
1. The triangle is divided into two similar triangles
which are each also similar to the given triangle.
2. The perpendicular is a mean proportional between
the two segments of the hypotenuse.
3. Either side about the right angle is a mean propor-
tional between the whole hypotenuse and the adjacent
segment.
394. The following constructions have been taught:
1. How to construct a mean proportional.
2. How to construct a square equal to a given rec-
tangle, parallelogram, or triangle.
3. How to construct a right triangle or an isosceles
triangle equal to a given scalene triangle.
4. How to construct a fourth proportional.
5. How to divide a line segment into two parts which
have a given ratio.
CHAPTER XIV
INDIRECT MEASUREMENT; SCALE DRAWINGS;
TRIGONOMETRY
395. Scale drawings. Up to this point we have made
several uses of scale drawings. In Chapter X the relation of
FIG. 251. SCALK DRAWING OF A LIBRARY TABLE
(Courtesy of Industrial Arts Magazine)
quantities was shown by line segments whose proportional
lengths represented the relative size of the quantities.
345
34l> GEXE11AL MATHEMATICS
In another form of graphic work, scale drawings have
helped us to understand the meaning of functions, equa-
tions, and formulas. In addition to the foregoing, scale
drawings are probably familiar to the student in the form
of shop drawings, geography maps, blue prints, maps in
railroad guides, and architects' plans.
The shop drawing in Fig. 251 illustrates a use of a scale
drawing, which we shall now study in some detail.
The figure shows that a scale drawing gives us an accu-
rate picture of the real object by presenting all the parts
in the same order of arrangement and showing the relative
sizes graphically by means of proportional line segments.
Obviously this fact rests on the principle of similarity, and
the ratio between any two line segments in the plan equals
the ratio between the lengths of the two corresponding
parts of the library table (Fig. 252).
By means of the scale drawing we are able to deter-
mine the dimensions of parts of the table even though
they are not given on the plan. In fact, in the case of
architects' and surveyors' scale drawings we are able to
measure lines which in the real object are inaccessible.
This last procedure illustrates precisely the use which
we want to make of scale drawings in this chapter. In
many cases we shall want to measure distances that can-
not be measured directly with steel tape or other surveying
devices; for example, (1) the heights of towers, buildings,
or trees ; (2) the width of ponds, lakes, or rivers ; (3) the
length of boundary lines passing through houses, barns, or
other obstructions.
We can usually determine such distances by following
the method set forth in the following outline:
1. Measure enough actual lines and angles in the real
object so that a scale drawing of the object can be made.
TRIGONOMETRY
347
2. Draw the figure to scale, preferably on squared paper.
3. Measure carefully with the compasses and squared
paper the lines in the figure which represent the inaccess-
ible lines of the actual object that is being considered.
FIG. 252. THE FINISHED LIBRARY TABLE
(Courtesy of Industrial Arts Magazine)
4. Translate the measurements obtained in (3) into the
units used in measuring the lines of the actual figure.
EXERCISES
1. A man walks from his home around a swamp (Fig. 253).
He starts from his home at A, walks 0.95 mi. north, then 1.2 mi.
east, then 0.35 mi. south. How far from B i 2ml
home is he ?
Solution. Let 2 cm. represent 1 mi. Make a
drawing on squared paper of the distances as
shown in Fig. 253. Then on the squared paper
a side of every small square represents 0.1 mi.
(Why ?) The required distance is the number
of miles represented by the segment A D, which
is 13.9 small units long. Hence AD represents 1.39 mi. Why?
348 GENERAL MATHEMATICS
2. Show how the four steps given in the outline of Art. 395
are followed in the solution of the preceding problem.
3. A man starting at a point S walks 48 yd. north and then
56 yd. east. Find the direct distance from the stopping-point
to the starting-point. (Let 1 cm. = 10 yd.)
4. A man walks 92 yd. south, then 154 yd. east, and then
132 yd. north. How far is he from the starting-point ? (Use
1 cm. for every 12 yd.)
5. Two men start from the same point. One walks 15 mi.
west, then 9 mi. north ; the other walks 12 mi. south, then
16 mi. east. How far apart are they ?
6. Draw to scale a plan of your desk top and find the
distance diagonally across. (Use the scale 1 cm. = 1 ft.)
7. A baseball diamond is a square whose side is 90 ft.
By means of a scale drawing, find the length of a throw from
" home plate " to " second base."
8. The broken line ABC
(Fig. 254) represents a coun-
try road. Find out how much
nearer it would be to walk diag-
onally across country from A to A 3 ' 3 mi> B
C than it is to follow the road. FlG - 254
9. A roadbed is said to have a 6% grade when the level
of the road rises 6 ft. in 100 ft. measured horizontally. ' Draw
to scale a roadbed 520 yd. long which has a 6% grade.
10. The sides of a triangular chicken lot are 20 ft., 16 ft.,
and 18 ft. respectively. Make a scale drawing of this lot on
squared paper and estimate the area by counting the small
squares and approximating the remaining area.
11. In a map drawn to the scale of 1 to 200,000 what lengths
will represent the boundaries of a rectangular-shaped county
40 mi. long and 20 mi. wide ? Give the answer to the nearest
hundredth of an inch.
TRIGONOMETRY
349
12. A railroad surveyor wishes to measure across the swamp
at A B represented in Fig. 255. He measures the distance from
a tree A to a stone C and finds it to be 110 yd.
Find the distance across the swamp if the
angle at C is 70 and if BC = 100 yd.-
NOTE. The lines BC and AC are meas-
ured by means of a steel tape (Fig. 256) or a
surveyor's chain (Fig. 257). The angle at C is
measured by means of a transit (Fig. 43). Chaining pins (Fig. 258)
are used by surveyors to mark the end-points of the chain or tape.
FIG. 256. STEEL
TAPE
FIG. 257. SURVEYOR'S
CHAIN
FIG. 258. CHAIN-
ING PINS
* 13. If available examine a steel tape, chain, and the pins used
by surveyors and report to class on the length, graduations, etc.
14. In Fig. 259, S represents a water-pumping station in
Lake Michigan. A and B represent two Chicago buildings on
the lake shore. Reproduce the measurements to scale and
determine the distance of S from each of the two buildings.
15. In Fig. 260" a swimming course AB across a small lake
is represented. Find A B by means of a scale drawing.
350 GENERAL MATHEMATICS
16. A triangular lot has these dimensions: .4.6 = 20 yd.;
BC 40 yd.; A C = 30 yd. Make a scale drawing of the lot on
squared paper and determine its area. (Since the formula for
the area of a triangle A = calls for an altitude, the student
L
will draw one from A to BC and then apply the formula.)
*17. In order to measure the distance between two pump-
ing stations A and B in Lake Michigan a base line C/>=18.8
chains long (1 chain = 66 ft.) was measured along the shore.
The following angles were then measured: Z. A CD = 132 ;
Z.BCD= 50; Z.CDA = 46; ZCZ> = 125. Draw the figure
to scale and find the distance from A to B in feet.
*18. Two streets intersect at an angle of 80. The corner
lot has frontages of 200 ft. and 230 ft. on the two streets, and
the remaining two boundary lines of the lot are perpendicular
to the two streets. What is the length of these two boundary
lines ? What is the area of the lot ?
HINT. Construct the two perpendiculars with compasses. Then
draw a diagonal so as to form two triangles and construct their
two altitudes as in Ex- 16, above.
*19. A class in surveying wishes
to determine the height of a smoke-
stack as shown in Fig. 261. The
transit is placed at B, and the angle
ij is found to be 62: then at A,
and the angle x is found to l>e 32.
Line AB is 48 ft. long and is measured along level ground.
The transit rests on a tripod 3^ ft. high. Find the height
of the chimney.
396. Angle of elevation. The angles x and y which
were measured in Ex. 19 are called angles of elevation.
The angle KAH in Fig. 262 shows what is meant by an
angle of .elevation. To. find .the angle of elevation, the
^x* .
A
...-' /
/
B -. y
I
?B0n
TRIGONOMETRY
transit is placed at A as in Fig. 262. The telescope
of the transit is first pointed horizontally toward the
smokestack. The farther end is then ^
raised until the top of the chimney
K is in the line of sight. The angle
KAH, through which the telescope A < JL tal Une \ {
turns, is the angle of elevation of K
. . , FIG. 262. ANGLE OK
irom A, the pouit or observation. ELEVATION
EXERCISES
By means of scale drawings, compasses, and protractor
solve the following exercises :
1. When the angle of elevation of the sun is 20 a building
casts a shadow 82 ft. long on level ground. Find the height of
the building.
2. Find the angle of elevation of the sun when a church
spire 80 ft. high casts a shadow 120 ft. long.
3. A roof slopes 1 in. per horizontal foot. What angle does
the roof make with the horizontal ?
4. A light on a certain steamer is known to be 30 ft. above
the water. An observer on the shore whose instrument is
4 ft. above the water finds the angle of elevation of this
light to be 6. What is the distance from the observer to
the steamer ?
5. What angle does a mountain slope make with a horizontal
plane if it rises 150 ft. in a horizontal distance of one tenth
of a mile ?
6. The cable of a captive balloon is 620 ft. long. Assuming
the cable to be straight, how high is the balloon when^all
the cable is out if, owing to the wind, the cable makes an angle
of 20 with the level ground (that is, the angle of elevation
is 20) ?
352 GENERAL MATHEMATICS
7. On the top of a building is a flagpole. At a point A on
level grour/4 70 ft. from the building the angle of elevation of
the top of the^lagpole is 42. At the same point, A, the angle
of elevation of the top of the building is 32. Find the height
of the flagpole. How high is the building ?
397. Angle of depression. A telescope at M in the top
of a lighthouse (Fig. 263) is pointed horizontally (zero
reading), and then the farther end is lowered (depressed)
until the telescope points to a boat at B. The angle HMB,
through which the telescope turns, is the angle of depression
of the boat from the point M. In Fig. 26 3, Z HMB = ^MBC.
Why is this true ?
EXERCISES
1. If the height of the lighthouse (Fig. 263) is 220 ft.
above water, and the angle of depression of the boat, as seen
from M. is 40, what is the distance
. . . H horizontal line M
of the boat from R if R C is known
to be 40ft.?
2. A boat passes a tower on which
is a searchlight 220 ft. above sea level.
Find the angle through which the B
beam of light must be depressed from FlG - 263 - ANGLE OF
,i , , i DEPRESSION
the horizontal so that it may shine
directly on the boat when it is 300 ft. from the base of
the tower.
3. How far is the boat from the base of the tower if the
angle of depression" is 51 ? 30 ? Xote that the height of the
lighthouse is known, and that the distance of a boat out at sea
depends on the size of the angle ; that is, the distance is a
function of the angle. In other words, the lighthouse keeper
needs only to know the angle of depression to determine the
distance of a boat at sea.
TRIGONOMETRY
353
4. From the top of a mountain 2500 ft. above the floor of
the valley the angles of depression of two barns in the level
valley beneath, both of which were due east of the observer,
were found to be 27 and 56. What is the horizontal distance
between the two barns ?
5. From the top of a hill the angles of depression of two
consecutive milestones on a straight level road, running due
south from the observer, were found to be 23 and 47. How
high is the hill ?
398. Reading angles in the horizontal plane ; bearing of
a line. The special terms used in stating problems in sur-
veying and navigation make the problems seem unfamiliar,
although the mathematics is frequently not more difficult
N
N
N
N
w-
-EW-
-E
S s s s
FIG. 264. ILLUSTRATING THE BEARING OF A LINK
than iii the exercises given in Art. 397. Thus the acute
angle which a line makes with the north-south line is the
bearing of the line.
The bearings of the lines indicated by the arrows in
Fig. 264 are read as follows : 75 east of north, 46 west
of north, 20 east of south, and 30 west of south. These
are written more briefly as follows: N. 75 E., N.46W.,
S.20E., and S.30W.
354
GENEKAL MATHEMATICS
EXERCISES
1. Read the bearings of the arrow lines in Fig. 265.
2. W T ith a ruler and protractor draw lines having the fol-
lowing bearings :
(a) 26 east of south. (d) 37^ west of south.
(b) 39 east of north. (e) 33 west of north.
(c) 40 west of north. (f) 3 east of south.
3. Write in abbreviated form the bearings of the lines in Ex. 2.
N N N N ,V
*f
W- V -E W- E W--
-E W^- E W- -h^--
s s s
FIG. 265
399. Bearing of a point. The bearing of a point B
(Fig. 266) from a point is the bearing of the line OB
with reference to the north-south line through 0.
EXERCISES
1. In Fig. 266 read the bear-
ing of
(a) A from O. (d) from B.
(b) from A. (e) C from O.
(c) B from O. (f) from C.
2. Point A is 6.4 mi. east
and 9.8 mi. north of B. Find
the distance from A to B. What
angle does AB make with the
north-south line through B?
What is the bearing of B
from A? of A from B?
3. Sketch the figure for Ex. 2
and show why the angles appearing as results for Ex. 2 are equal.
TRIGONOMETRY 355
4. The bearing of a fort B from A, both on the seacoast, is
N. 55 W. An enemy's vessel at anchor off the coast is observed
from A to bear northwest; from B, northeast. The forts are
known to be 8 mi. apart. Find the distance from each fort to
the vessel.
400. The limitations of scale drawings. By this time the
student probably appreciates the fact that a scale drawing
has its limitations. He would probably not agree to buy
a triangular down-town lot whose altitude and area had
been determined by a scale drawing. If a millimeter on
the squared paper represents 0.1 of a mile, a slight slip
of the pencil or compasses means disaster to accuracy.
Scale drawing is used extensively by the surveyor and
engineer in the following ways: (1) as a method of esti-
mating probable results; (2) as a help to clear thinking
about the relations of lines and angles involved in a geomet-
ric drawing ; (3) as a valuable check on results obtained by
more powerful methods. But as a matter of fact we need a
more refined method to determine lines and angles where
a high degree of accuracy is desirable. We shall now pro-
ceed to consider a far more efficient method of determining
such lines and angles. Most students will find the method
fascinating, because the solution is simple and the results
obtained are as accurate as the lines and angles which are
directly measured.
TRIGONOMETRY
401. Similar right triangles. A few exercises on similar
right triangles will help the student to understand the
new and more accurate method of determining lines and
angles. This method may be used independent of scale
drawings, is shorter in most cases, and lays the foundation
for future mathematical work.
GENERAL MATHEMATICS
a
b O
<;. 207
EXERCISES
1. With the protractor construct a right-angled triangle
having an angle of 37. Letter the figure as suggested in
Fig. 267. Measure the lines a, b, and c. Let
2 cm. represent 1 unit. Find the value of the
ratios - - > and to two decimal places.
c c l>
2. Compare your result with other mem-
bers of the class. Did all members of the class
use the same length for the bases ? Are any
two of the triangles drawn necessarily of the same size ? Show
why the result obtained for
c
should be the same number as
the results of your classmates.
3. Prove that if two right
triangles have an acute angle
of one equal to an acute angle of the other, the ratios of their corre-
sponding sides are equal. Write two proportions. (Use Fig. 268.)
4. Could you draw a right triangle with angle A =37 in
which - does not equal approximately 0.60, or ~ ? Prove.
HINT. The fact that - = - means that in every right triangle the
side opposite a 37 angle is approximately as long as the hypotenuse.
5. A balloon B (Fig. 269) is fastened by a cable 200 ft. long.
Owing to the wind the cable is held practically straight and
makes a 37 angle with the horizontal. How high is the balloon ?
Solution. This triangle is similar to every tri- B
angle drawn by the class in Ex. 1. Prove.
Therefore
= O.GO.
200
a = 120 ft.
\37
90;
Solving,
Note that the solution is exceedingly simple
(only two equations) and that the accuracy of
the result does not now depend upon the accuracy of Fig. 269.
b
FIG. 269
TRIGONOMETRY 357
402. Sine of an angle. The ratio - (Fig. 270) is called the
sine of the angle A. The abbreviation for "sine" is"sin." This
definition may be written sin A - Thus,
c
sin 37 = = O.GO (approx.). Do you think '
we would have obtained the same value for -
c A b C
if in Ex. 1 we had made the angle 47 ? FHJ. 270
EXERCISES
1. Find the sine of 20, using the definition given in Art. 402.
HINT. As in Ex. 1, Art. 401, construct the triangle, measure a
and c, and find the value of - to two decimal places.
c
2. Find the sine of each of the following angles : 10, 15,
2u, 32, 47, 68, 87. Compare each result with the results
of your classmates.
403. Table of sines. The preceding exercises show that
the sine of the angle changes with the angle ; that is,
sin 68 is not equal to sin 37. By taking a large sheet of
graphic paper and a very large unit we could get a fairly
good table, but it would be too much trouble to do 'this
for every problem. Such a table has been very carefully
calculated for you in the first column of the table in
Art. 410.
EXERCISE
Turn to the table in Art. 410 and .see how efficient yo\i
have been by comparing your results for Ex. 2, Art. 402, with
the table.
404. Cosine of an angle. The exercises given in this
article will introduce another trigonometric ratio.
358 GEJSEKAh MATHEMATICS
INTRODUCTORY EXERCISES
1. Construct a right-angled triangle with angle .1 (see
Fig. 270) equal to 43. Measure b and c. Find the quotient
-.to two decimal places. Compare the results with those of
the other members of the class.
2. Show that all results ought to agree to two decimal
places. The ratio - (Fig. 270) is called the cosine of the angle A .
C
The abbreviation for " cosine " is " cos." Thus, cos 43 = 0.73
(approx.). This means that in an}- right-angled triangle the
side adjacent to the angle 43 is about -j 7 ^- as long as the
hypotenuse.
3. Find the cosine of 5, 18, 25. 35, 47, 65, 87.
4. Compare the results for Ex. 3 with the table of cosines
in Art. 410.
405. Tangent of an angle. We shall now introduce a
third important ratio connected with similar right triangles.
Historically the tangent ratio came first. We shall have
occasion to learn more about it.
INTRODUCTORY EXERCISES
1. In Fig. 270, what is the value of - ? Compare your result
with the results obtained by other members of the class.
2. Show that all the results obtained for - in Ex. 1 should
b
agree.
The ratio - is called the tangent of angle. A. In speak-
ing of the tangent of 43 we mean that the side opposite
angle A is y 9 ^ (approx.) of the length of the side adjacent.
The abbreviation for " tangent" is "tan." Thus, tan 45 = 1.
TRIGONOMETRY 359
EXERCISES
1. Find the tangent of 11, 36, 45, 57, 82.
2. Compare the results of Ex. 1 with the table of tangents
in Art. 410.
406. Trigonometric ratios. Solving a triangle. The ratios
a b , a
-> -> and - are called trigonometric ratios. We shall now
proceed to show that the use of these ratios greatly
simplifies the solution of many problems in-
volving indirect measurements. By their use
any part of a right triangle can be found if
any two parts (not both angles) besides the
right angle are given. This process is called
solving the triangle.
FIG. 271
407. Summary of definitions. The follow-
ing outline will be found convenient to the student in
helping him to remember the definitions (see Fig. 271):
a side opposite
1. sm A=- - C-
c lii/potennse
b side adjacent
'2. cos A = - =
i' hypotenuse
a side opposite
3. tan A = - = -
o side adjacent
408. Trigonometric ratios clear examples of the function
idea. Either by your own crude efforts at building a table
or by a study of the table of ratios given, it is easy to see
that the value of the ratio changes as the angle changes:
that is, a trigonometric ratio depends for its value upon
360 GENERAL MATHEMATICS
the size of the angle. Hence the ratios furnish us with
one more clear example of the function idea. We may
therefore refer to them as trigonometric functions.
HISTORICAL NOTE. Trigonometric ratios are suggested even
in the Ahmes Papyrus (c. 1700 B.C. ?), which, as has been stated,
may itself be a copy of some other collection written before the
time of Moses. In dealing with pyramids Ahmes makes use of
one ratio that may possibly correspond roughly to our cosine and
tangent.
The first to make any noteworthy progress in the development of
trigonometry was Hipparchus, a Greek, who lived about 150 n.r.
He studied at Alexandria, and later retired to the island of Rhodes,
where he did his principal work. He was able to calculate the length
of a year to within six minutes.
The Hindus contributed to the early development of the science,
from about A.D. 500, and the Arabs added materially to their work
from about A.D. 800 to A.D. 1000
Regiomontanus (or Johann Miiller, 1436-1476), a German, freed
the subject from its direct astronomical connection and made it an
independent science.
In the sixteenth century the subject developed slowly, but in the
seventeenth century it made a very decided advance, due to the
invention of logarithms, mentioned later, and to the great improve-
ment of algebraic symbolism which made it possible to write trigo-
nometric formulas in a simple manner. Trigonometry in the form
that we know it may be said to have been fully developed, except
for slight changes in symbols, in the seventeenth century.
409. Table of trigonometric ratios of angles from 1 to 89.
The student should now become familiar with the table
on the following page. The ratios are in most cases only
approximate, but are accurate enough for all ordinary
work.
410. The use of a trigonometry table. The problems
beginning on page 362 are intended to furnish the student
practice in the use of the table.
TRIGONOMETRY
361
Angle
Sine
(oPP.\
\hvp.)
Cosine
tad\
\hupj
Tangent
(opp.\
\adj.)
Angle
Sine
(oM>.\
\hvP.J
Cosine
/adJA
\hyp.)
Tangent
(opp.\
\adj.)
1
2
3
4
.000
.017
.035
.052
.070
1.000
1.000
.999
.999
.998
.000
.017
.035
.052
.070
45
46
47
48
49
.707
.719
.731
.743
.755
.707
.695
.682
.669
.656
1.000
1.036
1.072
1.111
1.150
5
6
7
8
9
.087
.105
.122
.139
.156
.996
.995
.993
.990
.988
.087
.105
.123
.141
.158
50
51
52
53
54
.766
.777
.788
.799
.809
.643
.629
.616
.602
.588
1.192
1.235
1.280
1.327
1.376
10
11
12
13
14
.174
.191
.208
.225
.242
.985
.982
.978
.974
.970
.176
.194
.213
.231
.249
55
56
57
58
59
.819
.829
.839
.848
.857
.574
.559
.545
.530
.515
1.428
1.483
1.540
1.600
1.664
15
16
17
18
19
.259
.276
.292
.309
.326
.966
.961
.956
.951-
.946
.268
.287
.306
.325
.344
60
61
62
63
64
.866
.875
.883
.891
.899
.500
.485
.469
.454
.438
1.732
1.804
1.881
1.963
2.050
20
21
22
23
24
.342
.358
.375
.391
.407
.940
.934
.927
.921
.914
.364
.384
.404
.424
.445
65
66
67
68
69
.906
.914
.921
.927
.934
.423
.407
.391
.375
.358
2.145
2.246
2.356
2.475
2.605
25
26
27
28
29
.423
.438
.454
.469
.485
.906
.899
.891
.883
.875
.466
.488
.510
.532
.554
70
71
72
73
74
.940
.946
.951
.956
.961
.342
.326
.309
.292
.276
2.747
2.904
3.078
3.271
3.487
30
31
32
33
34
.500
.515
.530
.545
.559
.866
.857
.848
.839
.829
.577
.601
.625
.649
.675
75
76
77
78
79
.966
.970
.974
.978
.982
.259
.242
.225
.208
.191
3.732
4.011
4.331
4.705
5.145 .
35
36
37
38
39
.574
.588
.602
.616
.629
.819
.809
.799
.788
.777
.700
.727
.754
.781
.810
80
8.1
82
83
84
.985
.988
.990
.993
.995
.174
.156
.139
.122
.105
5.671
6.314
7.115
8.144
9.514
40
41
42
43
44
.643
.656
.669
.682
.695
.766
.755
.743
.731
.719
.839
.869
.900
.933
.966
85
86
87
88
89
.996
.998
.999
.999
1.000
.087
.070
.052
.035
.017
11.430
14.301
19.081
28.636
57.290
45
.707
.707
1.000
90
1.000
.000
NOTE. The abbreviation hyp. means "hypotenuse"; adj. means "the
side adjacent to the angle"; opp. means "the side opposite the angle."
362 GENEliAL MATHEMATICS
EXERCISES
1. A balloon B (Fig. 272) is anchored to the ground at a
point .1 by a rope, making an angle of 57 with the ground.
The point C on the ground directly under the balloon is 146 ft.
from A. Assuming the rope to be straight, find
the height of the balloon.
Solution. Let a = height of balloon.
Then - = tangent of 57.
146
But by the table, Art. 410,
tan 57 = 1.54.
Hence -- = 1.51. A 146/
Fir 272
Solving for a. a = 224.84 ft.
NOTK. The figure does not need to be drawn accurately, for our
results are obtained independently of it. The solution is brief and
depends for its accuracy upon the accuracy of the angle 57, the accu-
racy of the length of the line A C, and the accuracy of the tangent table.
2. The angle of elevation of an aeroplane at a point A on
level ground is 53. The point C on the ground directly
under the aeroplane is 315 yd. from A. Find the height of
the aeroplane.
3. The length of a kite string is 210yd. and the angle of
elevation of the kite is 48. Find the height of the kite,
supposing the line of the kite string to be straight.
4. A pole 20 ft. in length stands vertically in a horizontal
area, and the length of its shadow is 16.78 ft. Find the angle
of elevation of the sun.
HINT. Find the value of the tangent - Then look in the table
o
to see what angle has a tangent corresponding to the value of
- It may be necessary for you to approximate, since the table is
b
not calculated for minutes. Ask your instructor to show you a more
complete table of trigonometric ratios.
TRIGONOMETRY
5. A tree is broken by the wind so that its two parts form
with the ground a right-angled triangle. The upper part makes
an angle of 55 with the ground, and the distance on the
ground from the trunk to the to.p of the tree is 57 ft. Find
the length of the tree.
6. A circular pool has a pole standing vertically at its
center, and its top is 50 ft. above the surface. At a point in
the edge of the pool the angle subtended by the pole is 25.
Find the radius and the area of the pool.
7. A ladder 35 ft. long leans against a house and reaches
to a point 19.6 ft. from the ground. Find the angle between
the ladder and the house and the distance the foot of the
ladder is from the house.
8. Measure two adjacent edges of your desk or of a rec-
tangular table, say your study table. Find the angles that the
diagonal makes with the edges (1) by drawing an accurate
figure and measuring the angle with a pro-
tractor ; (2) by use of the trigonometric ratios.
9. The tread of a step on a certain stairway
is 11 in. wide ; the step rises 8 in. above the next
lower step. Find the angle at which the stairway
rises (1) by means of a protractor and an accurate
figure ; (2) by means of a trigonometric ratio.
10. To find the distance across a lake (Fig. 273)
between two points A and C, a surveyor measured off 71 ft. on
a line EC perpendicular to A C. He ^,Z>
then found ZC^=53. Find A C.
11. The Washington Monument
is 555 ft. high. How far apart are g x A C
two observers who from points due Fjo 274
west of the monument observe its
angles of elevation to be 20 and 38 respectively ? (See Fig. 274.)
HINT. Find. If.
FIG. 273
Then
x + value of A C
= tan 20
364
GENERAL MATHEMATICS
*12. A man standing on the bank of a river observes that the
angle of elevation of the top of a tree on the opposite bank is
56 ; when he retires 55 ft. from the edge of the river the
angle of elevation is 32. Find the height of the tree and the
width of the river.
*13. From the
summit of a hill
(Fig. 275) there are
observed two con-
secutive milestones on a straight horizontal road running from
the base of the hill. The angles of depression are found to be
13 and 8 respectively. Find the height of the hill.
FIG. 275
HINT. Construct TC _L CM Z .
Let CM l = x.
Then
- = tan 77, (Why?)
h
and
x + 1
h
= tan 82. (Why?)
(1)
(2)
Subtracting (1) from (2), - = tan 82 - tan 77.
Consult the table on page 36, substitute, and solve for /;.
*14. A railroad having a hundred-foot right of way cuts
through a farmer's field as shown in Fig. 276. If the field is
rectangular and the measurements
are made as shown, find the num-
ber of square rods occupied by the
right of way and the assessed damage
if the land is appraised at $200
an acre.
FIG. 276
.15. A ship has sailed due south-
west a distance of 2.05 mi. How far is the ship south of
the starting point ? How far is it west of the starting point ?
16. From the top of a mountain 4260 ft. above sea level the
angle of depression of a distant boat is 41. How far is the
boat from the summit of the mountain ?
*17. Sketch the figure and solve the right-angled triangle
ABC when
(e) I = 92.5, c = 100 ft.
(f ) a = 15.2, c = 50 ft.
(g) .1 = 40, c = 80 ft.
(h) B = 82, c = 100 ft.
(a) A = 30, a = 30 ft.
. (b) B = 42, b = 60 ft.
(c) A = 64, f, = 22 ft.
(d) a = 35, I = 85 ft.
411. A trigonometric formula for the area of a triangle.
It can be shown that the area of a triangle equals half the
product of any two sides multiplied by
the nine of the included angle ; that is,
ab sin A
~2~
Solution. In Fig. 277 construct the
altitude CD.
Then T (the area) = y. (Why?) (1)
But - = sin A (see the definition of "sine"). (2)
Whence h=bsinA. (Why?)
Substituting the value of 7t in (1),
_ be sin A
i ^ *
(3)
EXERCISES
1. A boy discovers that his father's drug store completely
covers their triangular lot and that it extends 60 ft. and 80 ft.
on two sides from a corner. With a field protractor he measures
the angle between the streets and finds it to be 58. He then
tries to find the area of the lot. What result should he get ?
* 2. Prove that the area of a parallelogram equals the product
of two sides and the sine of the angle included between these
two sides.
366 GENERAL MATHEMATICS
SUMMARY
412. This chapter has taught the meaning of the fol-
lowing words and phrases : scale drawing, surveyor's chain,
steel tape, angle of elevation, angle of depression, bearing
of a line, bearing of a point, sine of an angle, cosine of an
angle, tangent of an angle, trigonometric ratios or trigo-
nometric functions, solving a triangle.
413. Scale drawings were used as a means of indirect
measurement.
414. A scale drawing is useful in making estimates
of angles, lines, and areas, in getting a clear picture in
mind of the relation of the parts that make up the figure,
and in checking the accuracy of an algebraic solution.
However, it is not as brief and accurate as the algebraic-
sol ution.
415. If two right triangles have an acute angle of one
equal to an acute angle of the other, the ratios of their
corresponding sides are equal.
416. The chapter contains a table of trigonometric ratios
of angles from 1 to 89 and .correct to three decimal places.
417. Trigonometric ratios furnish us with a powerful
method of solving triangles.
418. The area of a triangle may be expressed by the
,. , be sin A
formula T=
419. The area of a parallelogram equals the product of
two sides and the sine of the angle included between these
two sides.
CHAPTER XV
THEORY AND APPLICATION OF SIMULTANEOUS LINEAR
EQUATIONS ; CLASSIFIED LISTS OF VERBAL PROBLEMS
420. Two unknowns ; solution by the graphic method.
In solving verbal problems it is sometimes desirable to
use two unknowns. This chapter aims to teach three
methods which the pupil may apply to such problems. The
graphic method is
shown in the dis-
cussion of the fol-
lowing problem:
In a baseball game
between the Chicago
Cubs and the New
York , Giants, the
Cubs made four
more hits than the
Giants. How many
hits did each team
make?
-1-5-
-10
x-
Fir.. 278
If we let x repre-
sent the number of
hits made by the Cubs
and y the number made by the Giants, then the equation x = y + 4
expresses the condition as set forth in the problem.
Obviously there are any number of possible combinations such
that the number of hits made by one team may be four more than
the number made by the other team. This is clearly shown in the
graph of the eqxiation x = ?/ + 4 in Fig. 278.
367*
3b'S
GENERAL MATHEMATICS
EXERCISES
1. From the graph in Fig. 278 find the number of hits
made by the Giants, assuming that the Cubs made 6; 8;
10; 15; 20.
2. Show that every point (with integral coordinates) on the
line will give a possible combination of hits such that x y + 4.
NOTE. By this time the student is no doubt convinced that a
definite solution of the problem as stated is impossible, because it
involves two unknowns and we have given but one fact. Another fact
which should have been
included in the problem
is that the total number of
hits made by both teams
was 18.
If we write the equa-
tion x + y =18, expressing
'this second fact, and study
it by means of the graph,
Fig. 279, we see that there
is more than one possi-
ble combination such that
the total number of hits
made is 18.
-20
-10
FIG. 279
3. Find from the graph
in Fig. 279 the number
of hits made by the Giants if the Cubs made 4 ; 6 ; 9 ; 12 ; 15.
4. Show .that every point (with integral coordinates) on the
line will give a possible solution for the equation x + y = 18.
XOTE. We have not been able to obtain a definite solution,
because we have been considering the two facts about the ball game
separately. The two equations express different relations between
the two unknowns in the ball game. This means that we must find
one pair of numbers which will satisfy both equations, and to do
this we must graph both equations on the same sheet to the same
scale, as shown in Fig. 280.
SIMULTANEOUS LINEAR EQUATIONS 369
5. Find a point in the graph of Fig. 280 that lies on both
lines. What does this mean ?
6. What are the a:- values and y- values of this point ?
7. Show that the pair of values obtained in Ex. 6 will satisfy
both equations.
(1)
y
4
9
5
20
16
-20
x = y + 4, (1)
x + y = 18. (2)
FIG. 280. THE GRAPH OF A PAIR OF SIMUL-
TANEOUS LINEAR EQUATIONS
The preceding exercises show that the point of inter-
section, namely x = 11, y = 7, is on both lines, and hence
the solution of the problem is x = 11, y = 7. Thus the
number of hits made by the Cubs was 11, and the number
made by the Giants was 7.
421. Simultaneous linear equations. A pair of linear equa-
tions which are satisfied at the same time (simultaneously)
by the same pair of values are called simultaneous linear
equations. The graph is a pair of intersecting straight lines.
422. System of equations. A pair of equations like
those in Art. 420 is often called a system of two linear
equations. A system of linear equations having a common
'61 it GENERAL MATHEMATICS
solution is said to be solved when the correct values of the
unknowns are determined. In the graphic method the
coordinates of the point of intersection furnish the solu-
tion. The following is a summary of the graphic method
of solving a pair of simultaneous linear equations:
1. Graph loth equations to the same scale.
'2. Find the point of intersection of the two Hnrs obtained
in 1.
3. Estimate as accurately as possible the x-raluc find the
y-value of this point.
4. Check by substituting in both equation*.
EXERCISES
1. Solve the following systems by the graphic method and
check each :
M* + y=s7 ' , v2x + 3y = 23, 5y-3* = 19,
W - ' ) }
3* + 2^=27, 2y + 3x=13,
^5z-4*/ = l. ( ; 5y-6x=-S. '
2. What difficulties did you have in finding the correct
values for the coordinates of the points of intersection in the
problems of Ex. 1 ?
423. Indeterminate equations. A single equation in two
unknowns is satisfied by an infinite (unlimited) number
of values, but there is no one pair of values which satisfy
it to the exclusion of all the others ; for example, the equa-
tion x + y = 4 is satisfied by as many pairs of values as
are represented by each distinct point on the graph of
the equation x + y = 4. Such an equation is called an
in determinate equation.
SIMULTANEOUS LINEAR EQUATIONS 371
EXERCISES
1. Find three solutions for each of the following indeter-
minate equations :
(a) x + y = 7. (c) y z = 6. (e) 5 x z = 2.
(b)m + 3n = 5. (d) 2 x - 4 ?/ = 3. (f) 3z - 4 >/ -1= 0.
424. Contradictory equations. It sometimes happens that
even though we have two equations in tyvo unknowns, it
is still impossible to obtain a distinct or a unique solution,
as is shown by the following example:
Find two numbers such that their difference is 12 arid such
that twice the first diminished by twice the second is equal to 14.
-20
-1-0
20-
&Z
f
FIG. 281. THE GRAPH OF A PAIR OF CONTRADICTORY EQUATIONS
If we let x denote one number and y the other, then from the
first condition,
x - y = 12.
From the second condition, '2 .c '2 y 14.
(1)
(2)
In order to study the problem fxirther we will construct the
graphs of (1) and (2) with reference to the same coordinate
axes (Fig. 281).
372 GENERAL MATHEMATICS
EXERCISES
1. What relation seems to exist between the two lines of
the graph in Fig. 281 ?
2. Are there, then, any two numbers which will satisfy the
conditions of the problem given on the preceding page ?
A system of equations which expresses a contradictory
relation between the unknowns is called a system of
contradictory, or inconsistent, equations. The graph consists
of two (at least) parallel lines. The definition suggests
that in a verbal problem one of the given conditions is
not true.
425. Identical equations. A type of problem which has no
unique solution but admits of many solutions is illustrated
by the following problem :
Divide a pole 10 ft. long into two parts so that 3 times the
first part increased by 3 times the second part is equal to 30.
If we let x and ?/ denote the length of the two parts, the conditions
of the problem are represented by the equations
x + y = 10, (l)
and 3 x + 3 y = 30. (2)
EXERCISES
1. Graph the equations x + y = 10 and 3 x + 3 y = 30 to
the same scale. Interpret the graph.
2. Divide the equation 3sc + 3?/ = 30 by 3 and compare
the result with the equation x -\- y = 10.
Equations like (1) and (2), above, which express the
same relation between the unknowns, are called identical,
dependent, or equivalent equations. Like an indeterminate
equation, they have an infinite number of solutions but
SIMULTANEOUS LINEAR EQUATIONS 373
no distinct solution. Their graphs coincide. If a verbal
problem leads to two identical equations, one condition
has been expressed in two different ways.
426. Outline of systems of equations and their number of
solutions. We have seen that a linear system of equations
in two unknowns may be
1. Determinant and have a distinct solution. (Tlie lines
intersect.)
2. Contradictory and have no distinct solution. {The lines
are parallel.}
3. Identical and have an infinite number of solutions. ( The
lines are coincident.}
EXERCISES
1. Classify the following systems according to the preceding
outline by drawing graphs of each system :
= , - = ,
; 6 x + 8 y = 10. ; 2 x + 3 y = 3.
2. Could you have classified the four systems in Ex. 1
without graphing them ? Explain.
427. Algebraic methods of solving systems of equations in
two unknowns. It is often difficult (sometimes impossible)
to judge the exact values in a graphic solution. The graphic
method helps us to see what is meant by a solution, but
i,t is not, in general, as exact and concise a method as the
algebraic methods which we shall now illustrate.
428. Elimination. To solve a system without the use
of graphs it will be necessary to reduce the two equations
in two unknowns to one equation in one unknown. This
process is called elimination.
374 GENEKAL MATHEMATICS
429. Elimination by addition or subtraction. The two
problems which follow illustrate the method of elimination
by addition or subtraction.
Solve x + y = 6, (1)
Solution. Multiplying (1) by 3 so as to make the coefficients of
// numerically the same in both equations,
3 x + 3 y = 18 ("j)
2ar-3y = 2 (4)
Adding, or =20
Substituting 4 for .c in (1), 4 + ^ = 6.
Solving for y, y = 2.
4 + 2 = 6,
Check. 8-6 = 2.
This method is called elimination by addition. Why ?
Solve 3;r-4y = l, (1)
2 a- -f 5 y = 7. (2)
Solution. Multiplying (1) by 2 and (2) by 3 so as to make the
coefficients of x numerically equal,
6 x - 8 y = 2 (3)
t;., -f 15 y = -21 (4)
Subtracting, - 23 y = 23
Substituting 1 for // in (1), 3 x + 4 =1.
3 a: =-3.
a
-c = 1.
Hefeee y = 1,
and x = 1.
This method is called elimination by subtraction. Why ?
SIMULTANEOUS LINEAR EQUATIONS 375
430. Outline of elimination by addition or subtraction.
To solve two simultaneous linear equations involving two
unknowns by the method of addition or subtraction, proceed
as follows :
1. Multiply, if necessary, the members of the first and
second equations by such numbers as irill make the coefficients
of one of the unknowns numerically the same in both equations.
2. If the coefficients have the same signs, subtract one
equation from the other; if they have opposite signs, add
the equations. This eliminates one unknown.
3. Solve the equation resulting from step 2 for the unknown
which it contains.
4. Substitute the value of the unknown found in step 3 in
either equation containing both unknowns and solve for the
second unknown.
5. Check the solution by substituting in both of the given
equations the values found.
EXERCISES
Solve and check the following systems by the method of
addition or subtraction :
' 3 x + 2 y 7, ox 3 '// = 14, 3 x o // = 23,
- 2z + 3y=8. " 2a; + 4//=10. '' 1 x + <j = - 35.
4z-3?/ = -l, 7a- + 9y=-lo, 2ar + 3y = 0.
! ' 5* + 2y=16: " 5ar-9y = -21. '' 3x + 2i/ = o.
llx-7y=-6, . :--2y = 54
'
11. x = y y
' x + y 8.
x ?/ = , m n _
+ =
in . _
' $(,.+11,-- 28. 3 10
376
GENERAL MATHEMATICS
60
FIG. 282
GEOMETRIC EXERCISES FOR ALGEBRAIC SOLUTION
13. The combined height of a tower and flagpole is 110 ft. ;
the height of the tower is 70 ft. more than the length of the
flagpole. Find the height of the tower and the length of
the flagpole.
14. A rectangular field is 25 rd. longer than it is wide. The
perimeter of the field is 130 rd. Find the dimensions of the field.
15. The perimeter of a football field is 320 yd. and the
length is 10 yd. more than twice
the width. What are its dimensions ?
16. The circumference of a circle
exceeds the diameter by 75 ft. Find
the circumference and the diameter.
(Use the formula C = - 2 y 2 d.~)
17. Two parallel lines are cut by
a transversal forming eight angles,
as shown in Fig. 282. Find x and y, and all of the angles.
18. The interior angles on the same side of a transversal
cutting two parallel lines are (5 x 3 y) and (x + y). Their
difference is 70. Find x and y.
19. Two adjacent angles of a parallelogram are represented
/o \ o
by 3 (2m - ri) and i-g- + 4 n } ,
and their difference is 30. Find m, n,
and the angles of the parallelogram.
20. The difference between the
acute angles of a right triangle is
43. Find the acute angles. 2g3
21. A picture frame 1 in. wide
(Fig. 283) has an area of 44 sq. in. The picture inside the
frame is 4 in. longer than it is wide. Find the dimensions
of the picture.
SIMULTANEOUS LINEAR EQUATIONS 377
431. Elimination by the method of substitution. The
following problem illustrates the method of elimination
by substitution :
Solve m -f n = 5, (1)
2 m -5n=-ll. (2)
Solution. Solving for i in terms of n in the first equation,
w = 5 - n. (3)
Substituting this value of m in (2),
2(o-i)-5ti=-ll. (4)
10-2-5n=-ll.
- 7 n = - 21.
n-=.3. .
Substituting 3 for n in (1), m + 3 = 5,
m = 2.
Hence m = 22, and n = 3.
Check. 2 + = 5,
4 - 15 = - 11.
The preceding method is called the method of substitution.
432. Outline of steps in the solution of a system by the
substitution method of elimination. To solve a system of
linear equations containing two unknowns by the method
of substitution, proceed as follows:
1. Sol v<' one equation for one unknown in terms of the other.
2. Substitute for the unknown in the other equation the
lvalue found for it in step 1.
3. Solve the e<juation resulting in step 2 for the unknown
which it contains.
4. Substitute the value of the unknown obtained in step 3
in any equation containing both unknowns and solve for the
second unknown.
5. Check the solution by substituting 'in the given equations.
378 GENERAL MATHEMATICS
EXERCISES
Solve the following problems by the method of substitution:
3ff + 46 = ll, -2ar-5?/? = 17, 3m + 2n = 130,
' oa-b = 3. ' 3x + 2><; = 2. ' 5 m - 6 n = 30.
7s + 3y = l, 2* + 3y = 8, ^ y _
"llz-5y = 21. > '3z-5y = 42. 5 + 2 '
9* + 7-=-12, 7ar-3y=8, *' x _ //
15*-2r=21 8 + 2y = l34. 3 2
NUMBER-RELATION* PROBLEMS
Solve the following problems by the method of substitution,
and check by one of the other methods :
9. Find two numbers whose sum is 150 and whose difference
is 10.
10. Find two numbers whose difference is 15 such that when
one is added to twice the other the sum is 295.
11. Find two numbers such that 8 times the first plus 4
times the second equals 100, and 3 times the first plus 7 times
the second equals 87.
12. The quotient of two numbers is 2 and their sum is 54.
Find the numbers.
13. The value of a certain fraction is 4-. If 2 is added to the
numerator and 7 to the denominator, the value of the resulting
fraction is ^. Find the fraction.
14. The sum of the two digits in a two-place number is 8.
If 18 be subtracted from the number, the resulting number
will be expressed by the original digits in reverse order. Find
the number.
Solution. Let u represent the digit in units' place and t the digit
in tens' place.
Then 10 t + u represents the original number.
From the first condition, t + u = 8.
SIMULTANEOUS LINEAR EQUATIONS 379
From the second condition,
10 1 + u - 18 = 10 u + t. (2)
Simplifying (2), t - u = 2. (3)
Solving (1) and (3), 2 t = 10.
t = 5.
Substituting 5 for / in (1), = 3.
Therefore the number is 53.
15. The tens' digit of a two-place number is three times
the units' digit. If 54 be subtracted from the number, the
difference is a number expressed by the digits in the reverse
order. Find the number.
16. If a two-digit number be decreased by 13, and this dif-
ference divided by the sum of the digits, the quotient is 5.
If the number be divided by one fourth of the units' digit,
the quotient is 49. Find the number.
433. Summary of methods of elimination. This chapter
has taught the following three methods of solving a system
of simultaneous linear equations :
1. The graphic method.
2. Elimination by addition or subtraction.
3. Elimination by substitution.
EXERCISES
Some of your classmates may be interested to learn a fourth
method called elimination by comparison. Turn to a standard
algebra and report to your classmates on this method.
MIXTURE PROBLEMS
Solve the following problems by any method :
1. A grocer has two kinds of coffee, one worth 300 per
pound and another worth 200 per pound. How many pounds
of the 30-cent coffee must be mixed with 12 Ib. of the 20-cent
coffee to make a mixture worth 240 per pound?
380 GENERAL -MATHEMATICS
2. A grocer makes a mixture of 20-cent nuts and 32-cent
nuts to sell at 28 a pound. What quantities of each grade of
nuts must he take to make 60 Ib. of the mixture ?
3. How much milk testing 5% butter fat and cream testing
25% butter fat must be mixed to make 30 gal. that test 22%
butter fat ?
4. How much milk testing 3.7% butter fat and cream testing
25.5% butter fat must be mixed to make 15 gal. that test 20%
butter fat ?
5. "What numl)er of ounces of gold 75% pure and 85% pure
must be mixed to give 10 oz. of gold 80% pure ?
G. An alloy of copper and silver weighing 50 oz. contains
5 oz. of copper. How much silver must be added so that 10 oz.
of the new alloy may contain ^ oz. of copper ?
7. The standard daily diet for an adult requires about 75 g.
of protein and 100 g. of fat. Mutton (leg) contains 19.8%
protein and 12.4% fat. Bread (average) contains 9.2% protein
and 1.3% fat. Find how many grains each of bread and mutton
are required to furnish the required amount of protein and fat
in a standard ration for one day.
HINT. Let x and y represent the number of grains required of
mutton and bread respectively.
Then 0.198 x + 0.092 y = 75, (1)
and 0.121 x + 0.013 y = 100. (2)
Solve the equations (1) and (2) simultaneously.
If x or y turns out negative, we know that it is not possible to make
up a standard diet out of the two foods mentioned.
8. The table on the following page gives the amounts
of protein and fats in the various foods often used in the
daily diet.
SIMULTANEOUS LINEAR EQUATIONS 381
FOOD
PER CENT
OF PROTEIN
PER CENT
OF FAT
Mutton (le (r )
19.8
12.4
Beef (roast)
22.3
28.6
Pork (chops)
Esfsrs .
16.6
13.4
30.1
10.5
Bread (average)
9.2
1.3
Beans (dried)
Cabbage
22.5
1.6
1.8
0.3
Rice
8.0
0.3
Find three pairs of food combinations that will make a
standard diet and determine the number of grams required of
each in the following list :
(a) Mutton and rice. (f ) Bread and cabbage.
(b) Eggs and rice. (g) Beans and cabbage.
(c) Bread and eggs. (h) Bread and beans.
(d) Pork and bread. ( i ) Beef and bread.
(e) Pork and beans. (j) Beef and rice.
434. Systems of equations containing fractions. The fol-
lowing list of problems offers no new difficulties. The
student merely needs to remember to remove the fractions
in each equation by multiplying through by the L.C.M. of
the denominators in each, thus reducing each equation to
the standard form ax + by = c, where a represents the co-
efficient of x, b the coefficient of y, and c the constant term.
5
: 6'
3
x y
or
4 3 12
The first equation may be written
x + y=i:
Similarly, (2) reduces to x y = \^
(1)
(2)
(3)
Why?
382 GENERAL MATHEMATICS
EXERCISES
Reduce to the standard form, and solve :
1.
2.
3.
X + If
a; 3 y
m + 1 1
3
2x +
3
'/ 3 ?/ 4- x
>
5
4. + l 2'
m-1 1
3
7 5
4
4
1
71-1 4
-f 3 .s + 4 5
5
jr. y
3
, ^ + y
!'
2
5. 5 10 2 '
f> * ~
2
3
./ i ./
3
Zt .) .s
7 3
0.3 x + 0.7 y = 5.7,
2 x 4 y = 4.
3
x 4- 5
1 !~
435. Linear systems of the type -+- = c\ work
. y
problems. In the problems of Art. 434 we have seen the
advisability of reducing each of the equations in a system
to the standard form by eliminating the fractions and
collecting similar terms. There are some problems, how-
ever, in which it is advisable to solve without eliminating
the fractions. An example will illustrate what is meant.
Two pipes can fill ^ of a cistern if the first runs 2 hr.
and the second runs 3 hr., but if the first runs 3 hr. and the
second runs 2 hr., they can fill ^ of the cistern.
Solution. Let x = the number of hours it will take the first
pipe alone to fill the cistern,
.
and y = the number of; hours it will take the
second pipe alone to fill the cistern.
Then - = the part of the cistern the first pipe can
fill in 1 hr.
and - = the part of the cistern the second pipe
y can fill in llir.
SIMULTANEOUS LINEAR EQUATIONS 383
From the first condition.
; + r^r (1)
From the second condition,
- + - = 7! (2)
x y lo
Multiplying (1) by 3 and (2) by 2,
6 9 _ 27
and - + - = ~ (4)
x y 15
5 27 28
Subtracting, - = - (5)
(Note that this is a linear equation in one unknown.)
Solving, y - G,
x = 5.
Problems like the preceding are called work problems.
WORK PROBLEMS
1. A and B can build a fence in 4 da. If A works 6 da. and
quits, B can finish the work in 3 da. In how many days can
each do the work alone ?
2. A tank can be filled by two pipes one of which runs 3 hr.
and the other 7 hr., or by the same two pipes if one runs 5 hr.
and the other 6 hr. How long will it take each pipe alone to
fill the tank ?
3. A mechanic and an apprentice receive $4.40 for a job of
work. The mechanic works 5 hr. and the apprentice 8 hr.
Working at another time, and at the same rate per hour, the
mechanic works 10 hr. and the apprentice 11 hr., and they
receive &7.30. What are the wages per hour for each?
384 GENERAL MATHEMATICS
4. Solve the following problems without getting rid of the
fractions, and check :
1 1 _5 47_15
H 7; > . i
* x y 6 3 x y 4
'J-i-i 'M = i.
x i/ (y x y
2 3 = 19 11 _ 9 = 26
2 //i n 15 , s f 5
' A . = i ' 13_3 = 8
m n 5 s t 5
436. Review list of verbal problems. The following
problems review types studied in earlier chapters. In actual
practice many problems may be solved by using either one
or two unknowns. In general it is advisable to use one un-
known, but sometimes it is easier to translate the problem
into algebraic language if two unknowns are used. It will
be helpful if some member of the class will show the two
methods in contrast.
MOTION PROBLEMS
1. A crew can row 8 mi. downstream in 40 min. and 12 mi.
upstream in 1 hr. and 30 min. Find the rate in miles per hour
of the current and the rate of the crew in still water.
Solution. Let x = the rate of the crew in still water,
and y = the rate of the current.
Then, if we express the rates in miles per hour,
x-y = = S. (2)
(TO
Adding, 2 x = 20.
Hence x = 10, the rate of the crew in still water,
and y = 2, the rate of the current.
SIMULTANEOUS LINEAR EQUATIONS 880
2. A boatman rowed 10 mi. upstream and 4 mi. back in
3' lir. If the velocity of the current was 2 mi. per hour, what
was his rate of rowing ?
3. The report of a gun traveled 367yd. per sec. with tin-
wind and 353 yd. per sec. against the wind. Find the velocity
of sound and the rate at which the wind was blowing.
4. An ae'roplane flying with a wind blowing at the rate of
2 mi. an hour consumes 2 hr. 30 min. in going a certain dis-
tance and 2 hr. 42 min. 30 sec. in returning. Find the distance
and the rate of the aeroplane in still air.
BEAM PROBLEMS
NOTK. Beam, or lever, problems which involve two unknowns
are readily solved by means of the laws of leverages and forces
discussed in Art. 233.
5. Two boys carry
a bag of coal weigh-
ing 25 Ib. by hanging M
it on a pole 8 ft. long k I n ,__ J~
at a point 2 ft. from I
the middle of the pole. +x JJ5 + y
How much of the -load F 284
does each boy carry ?
Solution. Let x and y represent the number of pounds carried
by each boy.
Then, using M, Fig. 284 (the middle of the pole), as the turning
point,
4 y - 50 - 4 x = 0. (Why ?)
From which // x = 12. .1. (1)
x + y = 25. (Why ?) (2)
Adding (1) and (2), 2 ?/ = 37.f>.
r~i8f.
Substituting for y in (2), 18;} ./ = (i\.
386 GENERAL MATHEMATICS
6. Two weights balance when one is 14 in. and the other
10 in. from the fulcrum. If the first weight is increased by
2 Ib. and placed 10 in. from the fulcrum, the balance is main-
tained. Find the number of pounds in each weight.
" --3Two weights balance when one is 12 in. and the other
is 10 inT^em the fulcrum. The balance is maintained if the
first weight isriKived 3 in. nearer the fulcrum and if 3 Ib. is
subtracted from the s&soml. Find the weights.
8. An iron bar 6 ft. loug^^^ighing 20 Ib. is used by two
boys, one at each end, to carry a Io9ji-oi^oO Ib. How many
pounds must each boy carry if the load hangs~2~ttr-feaio_the
right end ? (Consider the weight of the entire bar as hanging
at the middle of the bar.)
9. A wagon bed 12 ft. long is loaded with 20 cakes of ice
weighing 250 Ib. The bed extends 2 ft. over the front axle of
the running gears and 3 ft. behind the rear axle. Find the load
supported by each axle.
10. The material in a 30-foot bridge weighs 3600 Ib. The
bridge supports two loads : 700 Ib. at 3 ft. from one end, and
1500 Ib. at 5 ft. from the other end. Find the loads borne by
the two supports.
11. Three men have to carry an oak beam 15 ft. long weigh-
ing 250 Ib. Two of the men lift at the ends of an iron bar
placed crosswise beneath the beam, and the third man lifts at
the rear end of the beam. Where must the iron bar be placed
in order that each man will carry one third of the load ?
* RECREATION PROBLEMS
12. A man upon being asked the age of his two sons,
answered, " If to the sum of their ages 18 be added, the result
will be double the age of the elder ; but if 6 be taken from the
difference of their ages, the remainder will be equal to the age
of the younger." Find the age of each.
SIMULTANEOUS LINEAR EQUATIONS 387
13. In a guessing game the leader says, " If you will add
10 years to your age, divide the sum by your age, add 6 to the
quotient, and tell me the result, I will tell you your age." How
did he find it ?
14. A baseball team has played 40 games, of which it has
won 28. How many games must it win in succession in order
to bring its average of games won up to 0.750 ?
15. A girl has worked 12 problems. If she should work 13
more problems and get 8 of them right, her average would be
72%. How many problems has she worked correctly thus far ?
16. Two bicycle riders ride together around a circular track,
one along the outside edge, where the radius of the circle is
R, and the other along the inside edge, where the radius is r.
One revolution of the pedals carries the former's bicycle 20 ft.
and the latter's 25 ft. Write a formula expressing the differ-
ence between the number of pedal revolutions made by the two
cyclists in going around the track once ; five times ; n times.
17. If 10 marbles of one size are dropped into a bucket of
water, and the water rises a, inches, and 15 equal marbles of
another size are dropped into the same bucket, and the water
rises b inches, write a formula showing how many times larger
one of the first size is than one of the second size.
18. An automobile .tire when fully inflated has a radius of
18 in. Owing to a leakage of air, this is reduced to 17 in. Indi-
cate how many more revolutions per mile are necessary because
of the leakage. If the original radius is R and the reduced
radius r, what is the formula which could be used to calculate
the difference of revolutions per mile ?
19. Divide $183 into two parts, so that ^ of the first part
shall be equal to y 3 ^ of the second part.
20. Each of two brothers wanted to buy a lot valued at
$240. The elder brother said to the younger, " You lend me
| of your money, and I can purchase the lt-" "But," said the
088 GENERAL MATHEMATICS
younger brother, ''you lend me | of your money, and I ran
purchase the lot." How much money did each have '.'
21. A group of boys bought a touring car. After paying for
it they discovered that if there had been one boy more, They
would have paid $30 apiece less, but if there had been our
boy less, they would have paid $60 apiece more. How many
boys were there, and what did they pay for the car?
22. The Champion American League team one year (1914)
won 46 more games than it lost. The team standing second
played 153 games, winning 8 less than the first and losing
9 more than the first. Find the number of games won and
lost by each team.
23. It is said that the following problem was assigned by
Euclid to his pupils about three centuries B.C. : " A mule and
a donkey were going to market laden with wheat. The mule
said to the donkey, ' If you were to give me one measure,
I would carry twice as much as you ; if I were to give you one
measure, our burdens would: be equal.' What was the burden
of each ? "
* MISCELLANEOUS PROBLEMS
24. A bar 30 in. long is balanced by a 40-pound weight at one
end and a 32-pound weight at the other end. Find the position
nl' the support.
25. A man has a f 2000 exemption frOm income tax, but pays
at the rate of 6% on the rest of his income. He finds that after
paying income tax his actual income is $3410. On what amount
does he pay the 6% tax ?
26. A chemist has the same acid in two strengths. If 16 1.
of the second are mixed with 241. of the first, "the mixture is
42% pure, and if 61. of the first are mixed with 41. of the
second, the mixture is 43% pure. Find the per cent of purity
of each acid. Problems like this are often given as practical
problems. Find out from some chemist why it is not practical.
SIMULTANEOUS LINEAR EQUATIONS 389
27. After a strike a corporation decided to raise the wages
of each laborer from .? to y by the formula y = tux + i>, where
ra and b are to be determined by the facts that a man who
made $2 is to receive $2.30,. and one who made $2.20 is to
receive $2.41. Find ra and b, also the new wage of a man
who formerly received $3, $4, $4.20.
28. At what market price must one-year 5% bonds be offered
for sale in order that the buyer, by holding them until maturity,
nitiy make 6% on his investment?
HINT. The profit made must come from two sources : the interest
on the par value, which is $5, and the excess of the maturity value
over the price paid for the bonds. If x is the price paid, then
100 - x + 5 = 0.06 x.
SUMMARY -
437. This chapter has taught the meaning of the fol-
lowing words and phrases : simultaneous equations, linear
systems of equations, indeterminate equations, contradic-
tory equations, identical equations, elimination.
438. This chapter has taught the following methods of
solving a system of equations in two unknowns:
1. Solution by graph.
2. Solution by addition or subtraction.
3. Solution by substitution.
439. The student has been taught how to solve systems
involving fractions, and systems of the type - -f - = c.
x y
440. The following types of verbal problems have been
introduced: geometric problems, number-relation problems,
mixture problems, work problems, motion problems, beam
problems, and recreation problems.
CHAPTER XVI
GEOMETRIC AND ALGEBRAIC INTERPRETATION OF
ROOTS AND POWERS
441. Introductory work; square root. The following
exercises are introductory to the work of the chapter.
EXERCISES
1. What number multiplied by itself equals 9? 16? 121?
169? x 2 ? if?
2. How many answers are there to each part of Ex. 1?
(Why?)
3. One of the two equal factors of a number is called the
square root of the number. What is the square root of 49 ?
4 4y 2
of 64? of 625? ofar 2 ? of 4^? of J? o^-J
9 9y*
4. The positive square root of a number is indicated by a
sign ( V ) called the radical sign, and either the radical sign
alone (V ) 01> the radical sign preceded by the plus sign
(+ V ) means the positive square root of the number under-
neath the sign. The number underneath the radical sign is
called the radicand. The negative square root is indicated by
the radical sign preceded by the minus sign ( V ) With
the preceding definitions in mind give the value of the fol-
lowing: V25; Vl6; VlOO; -Vl21; V(X25 ; -Vl44; Vj;
9 ' \25iy 2
5. Express the following statement by means of a formula:
A number y equals the square of another number x.
390
INTERPRETATION OF ROOTS AND POWERS 391
6. If x = 1 in the formula y = ar 2 , what is the value of y?
If x = 2, what is the value of y ?
7. Calculate the corresponding values of y in the formula
y = x' for each of the following values ofic: 1 ; 2 ; -(-3;
Q. i 1 . i 2 . 2
~~ t> y ~7~ 7)~" j T "3* 5 "*""" Q
8. Fill in the proper values in the following table of
squares and square roots for use in the next article.
X
i
- 1
2
- 2
3
4
5 6
b
I
i
i
y
i
i
i
4
.
i
X
4
T
t
s
i i
: j
- i
1
V
t
\
i
-
y
i.
-I ^
^
>
y
r
~J
i
J
\
/
\
/
j
\ L
/
L
1
V
^
\
^
A
/
m
J
^
/
1 ^
^ k
^i
\^
. '
"s^ L
*
^ ^
^^
tSi
^ ^
- JF
"%i. -
5 J
'T
" ~ 5 "
n ~
-'
r-i- p
i;
-,
5
-:t~
t
_t :
i^-^
f"
FIG. 285. DEVICE FOR FINDING SQUARES AND SQUARE ROOTS
442. Graph of y = x*\ a device for finding squares and
square roots. The values of the preceding table have
been plotted in Fig. 285. Values for x were laid off
392 GENERAL MATHEMATICS
horizontally on the 2>axis, and the corresponding values
for y vertically on the ^-axis. The points were then con-
nected by a smooth curve, as shown. This curve serves
as, a device for determining squares and square roots, as
we shall now see.
EXERCISES
1. Determine by the graph in Fig. 285 the square root of 1 ' :
9; 2,0; 22; 3; 2. How many answers' do you obtain for the
square root of 9 ? 4 ? 25 '.' for the square root of each number
shown?
2. By means of the graph in Fig. 285 find the square of 2;
1.4; 2.2; 2.4; 3.3; 5.6; 3.9; 1.7.
3. Check your results for Ex. 2 by squaring the numbers
given. The squares should be approximately those you found
by means of the graph.
4. How would you make a graph that would give you
squares and square roots more accurately than the graph in
Fig. 285 ?
443. A positive number has two square roots. The graph
shows that the square root of 4 equals either -j- *2 or 2 ;
that is, there are two answers for the square root of a
positive number. Thus, (3) (3) =9, as also does
( -H 3) ('+ 3 ). Note the symmetry of the curve in Fig. 285
and see if you can explain it.
444. Quadratic surd. The indicated square root of a
number which is not a perfect square is called a quadratic
* i\l : for example, v3, v20, V.r.
445. Quadratic trinomial. Trinomials like a 2 -f 2 ab + b 2
and .?-' 2 2 .ry + y 2 are of the second degree, and are called
ijuadratic trinomials. The word " quadratic " comes from
INTERPRETATION OF ROOTS AND POWERS 393
the old word " quadrature," which means a geometric
square ; hence quadratic means that a 2 4- 2 ab -f 5 2 and
./ 2 '2xy + y 2 are of the second degree. They are the
a
ab
b b 2
2
ab
ab
ab
a + b
FIG. 286. GEOMETRIC REPRESENTATION OF (a + b) 2 .
squares of a + b and ^ y respectively, as we have already
seen earlier in this book. In Fig. 286 the geometric square
of a + b is shown.
EXERCISES
1. See if you can point out where we have worked with
problems dealing with the square of binomials like a + b
and x y.
2. Where, earlier in the book, have we learned how to take
the square root of trinomial squares by factoring ?
3. Find the value of the following :
Vra 2 + 2 mn + n* ; V4 "a? + 12 xy + 9 if ; V9 a 2 - 12 6 +T7X
*446. Square roots of algebraic polynomials and arith-
metical numbers. We have already seen in Chapter IX
how the square root of a trinomial is found by reversing the
process of squaring a binomial : that is, .by finding one of
the two equal factors of the trinomial. The same process
may be explained geometrically if the exercises given on
page 394 are carefully solved.
394
GENERAL MATHEMATICS
ab
ab
FIG. 287
ILLUSTRATIVE EXERCISES
1. Find the square root of 16 x~ + 40a-y + 25 y 2
Solution. If this trinomial is a perfect square of some binomial, it
may be illustrated by Fig. 287, in which the side of the largest square
obtained by inspection and corresponding to a 2 is 4 x. Therefore
the side of each rectangle corresponding to ,
each ab is 4 x, and the area correspond-
ing to 2 ab + b 2 must be 40 xy + 25 y' 2 . The
problem therefore consists in determining
the width of the strip which we are adding
on two sides and which corresponds to the
b of the formula. In this case b is 5 y. Now
5 y may be obtained by dividing 40 xy by
the sum of 4 x and 4 x, or 8 x. Hence
doubling the term already found (4 x) the
result 8 x serves as a divisor for determin-
ing the next term. Fig. 287 shows that we double 4 x because
8# is approximately the combined length of the strip to which
we are adding. This is illustrated more clearly in the next problem.
2. Two boys were asked to stake out a square plot of
ground with an area .of 4225 sq. ft. 5
What is the length of a side ?
Solution. The boys' thinking about the
problem might take some such form as
follows :
(a) It is obvious that we can make it
at least 60 by 60. We shall suppose that 60
this is constructed. See the square with
unbroken lines (Fig. 288). This uses up p IGi 288
:5600 sq. ft., leaving 625 sq. ft. We can add
to the square already constructed by adding to two sides and still
keep it a square.
(b) The combined length of the edges to which we are adding
is 120 ft. Hence the approximate length of the strip added is
120 ft. Why approximate?
(c) 120 is contained five times in 625 (with a remainder).
INTERPRETATION OF ROOTS AND POWERS 395
(d) If we make the strip 5 ft. wide, the total length will be 125
(for one strip will be 60 ft. and the other 65 ft.).
(e) 125 is contained exactly five times in 625.
(f) Hence the square must be constructed so as to be 65' by 65';
that is, the square root of 4225 is 65.
3. Find the square root of the polynomial a 2 + 2 ab + b 2
+ 2ac + 2bc + c 2 (see Fig. 289).
Solution. The side of the largest square is a^ therefore the trial
divisor is 2 a. The width of the first strip is b, therefore the divisor
is 2 a + b. Multiplying by b and subtracting the
remainder gives 2 OK + 2 be + c 2 . The length of
the square now constructed is a + b. The edge
to which we are adding is 2 a + 2 b units long
(trial divisor). 2 a + 2 b is contained c times
in 2 ac + 2 be. If we make the strip c units
wide, the total length of the strip to which we
add is 2 a + 2 b + c (complete divisor). (Why ?)
Multiplying and subtracting, the remainder is
zero. The side of the total square is a '+ b + c, or
FIG. 289
Va 2 + 2 ab + b z + 2 ac + 2 be + c 2 = a + b + c.
The work may be arranged as follows :
Largest square, a 2 a 2 + 2 ah + b 2 + 2 ac + 2 be 4- c 2 [ a + b + c
2 ab + b*
2 ah + b 2
First trial divisor, 2 a
First complete divisor, 2 <i + l>
Second trial divisor, 2 a + 2 b
Second complete divisor, 2 a + 2 I + c
2 ac + 2 be + c 2
2 ac + 2 be + c 2
4. Find the first digit in the square root of 177,2^1.
Solution. To determine this first digit we must remember (a) that
the square of a number of one digit consists of one or two digits, the
square of a number of two digits consists of three or four digits, the
square of a number of three digits consists of five or six digits, and so
on ; (b) that the number of digits in the integral part of the square
of a number is twice as large or one less than twice as large as the
number of digits in the integral part of the given number. This
suggests the following device for determining the number of digits
396 GENEKAL MATHEMATICS
in the integral part of the square root of a number. Beginning at
the decimal jioint, mark off toward the left groups of two digits
each. Then the number of digits in the square root will be the
same as the number of groups. Thus, since 177,241 is made up of
three groups of two digits (17'72'41'), the square root of 177,241
contains three digits in its integral part. We are thus able to esti-
mate the largest square as 400 (that is, the first digit is 4) and then
proceed as in Ex. 3. The work may be arranged as follows :
17 72'41 1 400 + 20 + 1
16 00 00
First trial divisor, 800 I 1 72 41
First complete divisor, 820 1 1 64 00
Second trial divisor, 840
Second complete divisor, 841
Therefore Vl77241 = 421.
*447. Steps involved in finding square roots. The follow-
ing steps were used in Exs. 1~4, above ; the student should
study them carefully.
1. Estimate the largest square in the number.
2. Double the root already found for a trial divisor.
3. Divide the first term of the remainder by the trial
ilirt'xor, placing the quotient as the next term of the root-.
4. Annex the term just found to the trial divisor to form a
complete divisor and continue the process until the other fcn/i*
f>f the root are found.
EXERCISES
Find the square roots of the following polynomials :
1. <>' 2 + 2ab + b 2 . -"3. 4 4 +-4 8 + 9a 2 + 4 0+4.
2. 16z* + 2xy + 9//'. 6. x 4 - 2x 3 + 3x 2 - 2 x + 1.
3. 49 if - 14 yz + z\ 7. 1 - 4 a + 6 . 2 - 4 8 + \
4. x*+2x 8 + 3x 2 + 2r + l. 8.
9. x 6 + 4 ax* - 2 aV + 4 aV - 4 <>\r + a 6 .
10. 9 + 12 // + 6 f + 4 ?/ + 4 if + if.
INTERPRETATION OF ROOTS AND POWERS 397
11. ** + 8.
, '.9,
12. 7 + 6a
4
13. 576.
14. 9025.
15. 51,529.
_ 16 4 16. 61,504.
~ x 42' 17. 57,121.
6 a* 2 _8a 3 18. 2.
2 # 2 . NOTE. Write 2.00WOO and
proceed as in Ex. 4, Art. 446.
19. 3.
20. 3.1416.
448. Table of roots and powers. In practical situations
it is convenient to use a table of roots and powers. There
are a number of very useful tables in textbook or leaflet
form, and the student is now in a position where he c,an
easily learn how to use them. A very simple table of
roots and powers is submitted on page 398. It will fre--
quently prove a great convenience to the student in his
work on the following pages.
449. The theorem of Pythagoras. If we study the follow-
ing exercise carefully we shall discover a well-known
geometric theorem which will be useful in later work.
EXERCISE
Construct a right triangle, making the 'sides including the
right angle 3 and 4 units long respectively (see AAJ'>( ', Fig. 290).
Using the same unit, find the length of J />'.
On each side draw a square and divide
each square into unit squares. Counting
these squares, find how the square on the
hypotenuse compares with the sum of the
squares on the other two sides.
The preceding exercise illustrates
,. , Fro. 290
the familiar theorem of Pythagoras:
/// it right triangle the sum of the squares on the sides inclml-
l n < i the- -right angle is-^ual to the .squaw on. tlie '
398
GENERAL MATHEMATICS
TABLE OF ROOTS AND POWKKS
No.
Squares
Cubes
Square
Roots
Cube
Roots
No.
Squares
Cubes
Square
Roots
Cube
Roots
1
1
1
1.000
1.000
51
2,601
132,651
7.141
3.708
2
4
8
1.414
1.259
52
2,704
140,608
7.211
3.732
3
9
27
1.732
1.442
53
2,809
148,877
7.280
3.756
4
16
64
2.000
1.587
54
2,916.
157,464
7.348
3.779
5
25
125
2.236
1.709
55
3,025
166,375
7.416
3.802
6
36
216
2.449
1.817
56
3,136
175,616
7.483
3.825
7
49
343
2.645
1.912
57
3,249
185,193
7.549
3.848
8
64
512
2.828
2.000
58
3,364
195,112
7.615
3.870
9
81
729
3.000
2.080
59
3,481
205,379
7.681
3.892
10
100
1,000
3.162
2.154
60
3,600
216,000
7.745
3.914
11
121
1,331
3.316
2.223
61
3,721
226,981
7.810
3.936
12
144
1,728
3.464
2.289
62
3,844
238,328
7.874
3.957
13
169
2,197
3.605
2.351
63
3,969
250,047
7.937
3.979
14
1%
2,744
3.741
2.410
64
4,096
262,144
8.000
4.000
15
225
3,375
3.872
2.466
65
4,225
274,625
8.062
4.020
16
256
4,096
4.000
2.519
66
4,356
287,496
8.124
4.041
17
289
4,913
4.123
2.571
67
4,489
300,763
8.185
4.061
18
324
5,832
4.242
2.620
68
4,624
314,432
8.246
4.081
19
361
6,859
4.358
2.668
69
4,761
328,509
8.306
4.101
20
400
8,000
4.472
2.714
70
4,900
343,000
8.366
4.121
21
441
9,261
4.582
2.758
71
5,041
357,911
8.426
4.140
22
484
10,648
4.690
2.802
72
5,184
373,248
8.485
4.160
23
529
12,167
4.795
2.843
73
5,329
389,017
8.544
4.179
24
576
13,824
4.898
2.884
74
5,476
405,224
8.602
4.198
25
625
15,625
5.000
2.924
75
5,625
421,875
8.660
4.217
26
676
17,576
5.099
2.962
76
5,776
438,976
8.717
4.235
27
729
19,683
5.196
3.000
77
5,929
456,533
8.774
4.254
28
784
21,952
5.291
3.036
78
6,084
474,552
8.831
4.272
29
841
24,389
5.385
3.072
79
6,241
493,039
8.888
4.290
30
900
27,000
5.477
3.107
80
6,400
512,000
8.944
4.308
31
961
29,791
5.567
3.141
81
6,561
531,441
9.000
4.326
32
1,024
32,768
5.656
3.174
82
6,724
551,368
9.055
4.344
33
1,089
35,937
5.744
3.207
83
6,889
571,787
9.110
4.362
34
1,156
39,304
5.830
3.239
84
7,056
592,704
9.165
4.379
35
1,225
42,875
5.916
3.271
85
7,225
614,125
9.219
4.396
36
1,296
46,656
6.000
3.301
86
7,396
636,056
9.273
4.414
37
1,369
50,653
6.082
3.332
87
7,569
658,503
9.327
4.431
38
1,444
54,872
6.164
3.361
88
7,744
681,472
9.380
4.447
39
1,521
59,319
6.244
3.391
89
7,921
704,969
9.433
4.464
40
1,600
64,000
6.324
3.419
90
8,100
729,000
9.486
4.481
41
1,681
68,921
6.403
3.448
91
8,281
753,571
9.539
4.497
42
1,764
74,088
6.480
3.476
92
8,464
778,688
9.591
4.514
43
1,849
79,507
6.557
3.503
93
8,649
804,357
9.643
4.530
44
1,936
85,184
6.633
3.530
94
8,836
830,584
9.695
4.546
45
2,025
91.125
6.708
3.556
95
9,025
857,375
9.746
4.562
46
2,116
97.336
6.782
3.583
96
9,216
884,736
9.797
4.578
47
2,209
103,823
6.855
3.608
97
9,409
912,673
9.848
4.594
48
2,304
110,592
6.928
3.634
98
9,604'
941,192
9.899
4.610
49
2,401
117,649
7.000
3.659
99
9,801
970,299
9.949
4.626
50
2,500
125,000
7.071
3.684
100
10,000
1,000,000
10.000
4.641
INTERPRETATION OF ROOTS AND POWERS 399
This theorem is one of the most famous theorems of
geometry. Centuries before Christ the Egyptians used a
rope divided by knots so that its three lengths were in
the ratio 3:4:5. This rope was used in land surveying
and also in the orientation of their temples. In fact,
we read of professional " rope fasteners " (surveyors ?).
Furthermore, the proof of the theorem itself has always
appealed to the interest of mathematicians. When we shall
have advanced in our study of mathematics it will be
possible for the student to find many proofs of this theorem
that he can understand. The earliest general proof is
credited to Pythagoras, who lived about 500 B.C.
The student has probably found this theorem to be the
basis for one of the most useful rules of arithmetic. The
proof given in arithmetic classes is usually that given in
the exercise above. However, a general proof demands
that we prove the theorem independent of the accuracy
of the figure (that is, independent of the measurements
and constructions involved). We shall presently give
such a proof. The exercises which follow are intended
to review the material necessary to establish this proof.
EXERCISES
1. In Fig. 291 A A EC is a right triangle, right-angled at C, with
CD AB. Re view the proof which shows that AADC^ A ABC.
c b
2. Prove that in Fig. 291 - = -
and that b 2 = cm.
b;
3. Review the proof which shows
that ABDC-^ A ABC. A\
4. Prove that in Fig. 291 - = -
O n 41
5. Show by using Exs. 2 and 4 that a 2 + tf = <?.
400
( i ION EKAL MATHEMATK s
450. Theorem of Pythagoras proved. No doubt the
student now sees that the theorem of Pythagoras is
proved by Exs. 1~5, aboye. We
shall, however, set up the prooi'
for reference. l>,
(liven the right triangle A /'><'.
right-angled at C, to prove that the
square on the hypotenuse is equal
to the sum of the squares on the
sides including the right angle. In terms of Fig. 292 this
means to prove that c 2 = a 2 -\- l>~.
Proof
STATEMENTS
REASONS
In Fig. 292 draw CI)AJi
and letter the figure as shown.
rr,
Then
and
(1)
(2)
In (1) and (2) I- = me and
= nc. (3)
By adding the two equations
in (3),
a- + b- = me + ne. (4)
a 2 + ft 2 =e(jn + n). (5)
But m + c. ((5)
.-. n- + //- = ,--.
(7)
Because if in a right triangle
a line is drawn from the vertex
of a right angle perpendicular to
the hypotenuse, either side about
the right angle is a mean pro-
portional between the whole hy-
potenuse and the segment of the
hypotenuse adjacent to it.
Because when four quantities
are in proportion the product of
the me'ans equals the product of
the extremes.
Addition axiom.
By factoring out c.
The whole is equal to the sum
of all its parts.
Bv substitution.
PYTHAGORAS
402 GENERAL MATHEMATICS
HISTORICAL NOTE. Pythagoras (c. 569 u.c.-c. 500 B.C.), the second
of the great philosophers of Greece, is said to hate " changed the study
of geometry into the form of a liberal education." After some wander-
ings, he founded the famous Pythagorean School at Croton, a Dorian
colony in the south of Italy. Here enthusiastic audiences composed
of citizens of all ranks, especially the upper classes, crowded to hear
him. It is said that the women went to hear him in direct violation
of a law against their public appearance.
' Pythagoras divided his audiences into two classes : the Proba-
tioners (or listeners) and the Pythagoreans (or mathematicians).
After three years in the first class a listener could be initiated
into the second class, to whom were confided the main discoveries
of the school.
The Pythagoreans formed a brotherhood in which each member
was bound by oath not to reveal the teachings or secrets of the
school. Their food was simple, their discipline severe, and their
mode of life arranged to encourage self-command, temperance, purity,
and obedience.
The triple interwoven triangle, or pentagram (star-shaped regular
pentagon), was used as a sign of recognition, and was to them a symbol
of health. It is related that a Pythagorean while traveling fell ill and,
although carefully nursed by a kind-hearted innkeeper, was unable
to survive. Before dying, however, he inscribed the pentagram star
on a board and begged his host to hang it up outside. This the host
did ; and after a considerable length of time another Pythagorean,
passing by, noticed the sign and, after hearing the innkeeper's story,
rewarded him handsomely. One motto of the brotherhood was :
"A figure and a step forwards; not a figure to gain three oboli."
The views of society advocated by the brotherhood were opposite
to those of the democratic party of Pythagoras's time, and hence
most of the brotherhood were aristocrats. For a short time the
Pythagoreans succeeded in dominating affairs, but a popular revolt
in 501 B.C led to the murder of many prominent members of the
school, and Pythagoras himself was killed shortly afterwards.
Though the brotherhood no longer existed as a political party, the
Pythagoreans continued to exist a long time as a philosophical and
mathematical society, but to the end remained a secret organization,
publishing nothing, and thus leaving us little information as'to the
details of their history. See Ball's "A History of Mathematics," p. 19.
EXERCISES
1. The base and altitude of a right triangle (Fig. 293) are
6 and 8 respectively. What is the length of the hypotenuse ?
(Use the theorem of Pythagoras.)
2. How long must a rope be run from the
top of a 16-foot tent pole to a point 20 ft. from
the foot of the pole ?
3. A baseball diamond is a square a side of
which is 90 ft. What is the length of a throw
_b IG. t'f>
from " home " to " second " ?
4. Find the formula for the diagonal of a square whose
side is s. Use this formula to determine the diagonal when
s = 10 ; when s = 15.
5. Prove from the Pythagorean theorem that a = Vc 2 6 2
and translate the equation into words.
6. Prove also that b = Vc 2 a 2 and translate the equation
into words.
7. A ladder 20ft. long just reaches a window 15ft. above
the ground. How far is the foot of the ladder from the foot
of the wall if the ground is level ?
8. The hypotenuse of a right triangle is 35 ft. and the
altitude is 21 ft. Find the base.
9. Using the formula of Ex. 6, find the value of a when
c = 22 and b = 20.
10. A tree standing on level ground was broken 24 ft. from
the ground, and the top struck the ground 18 ft. from the stump,
the broken end remaining on the stump. How tall was the tree
before breaking ?
11. Construct on squared paper a right triangle, using
the following pairs of numbers for the base and altitude
404
GENERAL MATHEMATICS
respectively : 1 and 1 ; 1 and V2 ; 2 and 2 ; 2 and 3 ; 4 and 4 ;
4 and 5 ; 1 and 5 ; 2 and 5 ; 3 and 5 ; 12 and 1.
HINT. Use the line segment you obtained for the first part.
12. Calculate for each part of Ex. 11 the length of the
hypotenuse.
451. The theorem of Pythagoras furnishes a method of
constructing with ruler and compasses the square root of a
number. Exercises 11 and 12, Art. 450, suggest a method
of finding the square root of a number by
means of ruler and compasses. The method
is illustrated by the following exercise :
Construct the square root of 42.
The following study (analysis) of the exer-
cise will help us to understand the problem.
Suppose that we have the figure constructed ;
that is, let us imagine that Fig. 294 is the required
figure and that AB is the required length V42.
Now a and b can be of various lengths provided
a- + tf = 42. (Why?) Let us suppose that CB - 6 ;
then how long is l>'l We know that o(> + tr 42
would be the equation from which the value of b can be found.
It is clear that b would have to equal A 7/ 6.
Then the problem merely becomes one of learning how to con-
struct V6. Some members of the class may already know how
to do this, but we shall proceed with our analysis.
"Imagine another triangle, A'B'C' (Fig. 295), so con-
structed that the hypotenuse turns out to be V6 and so
that B'C' is 2 units long; then A 'C' must equal V2. Why?
Our problem finally reduces, then, to a problem of
constructing V2. If we can find this geometrically we
can solve the original exercise, as our analysis has shown.
We already know how to construct v2 by constructing a right
triangle with the two legs about the right angle equal to 1. Then
the hypotenuse equals V2. Why?
INTERPRETATION OF ROOTS AND POWERS 405
We then reverse our analysis as follows :
(a) Construct V2 as indicated above.
(b) Construct a second right triangle with a base of v'2 units long
and an altitude 2 units long. Its hypotenuse will equal Vti. Why?
(c) Construct a third right triangle whose base is \/6 units long
and whose altitude is 6 units long. Its hypotenuse will be \/42 units.
Why?
EXERCISE
Construct with compasses a line segment equivalent to each of
the following: V6 ; VTT ; A/27 ; Vl43; V214 ; 3 VJJ ; 2 V2.
452. Mean proportional construction a method for find-
ing square roots. We shall now see that our mean pro-
portional construction (Art. 374) furnishes us with an easy
method of constructing square roots.
EXERCISES
1. Review the construction (Art. 374) for finding a mean
proportional between two line segments a and b (Fig. 296).
2. Construct the mean proportional a
between 4 and 9 ; 4 and 16. ft
3. Review the proof for the statement
.. , T , FJG. 296
that a mean proportional between two line
segments a and b equals the square root of the product of a and //.
The preceding exercises suggest that the mean propor-
tional construction furnishes a method for finding the square
root of a number. For example,
Find the square root of 12.
On squared paper find the mean propor- .
tional of two factors of 12, for example, 2 /
and 6. The mean proportional x (Fig. 297) {_
is the square root of 12, for K 2-
1 -. Why? FIG. 297. MEAN PROPOR-
y 6 TIONAL METHOD OF FIND-
x 2 = 12^_ Why? ING THE SQUARE ROOT OF
Whence x = Vl2. Whv? A NUMBER
406 GENERAL MATHEMATICS
EXERCISE
Construct the square root of 21 ; 6 ; 5 ; 18 ; 42 ; 84 ; 66 ;
76. Compare the results with the table of Art. 449. Your
results ought to approximate the second decimal place.
453. Large numbers under the radical signs. When the
number under the radical sign is large, the various geo-
metric constructions for finding square roots are neither
convenient nor, in general, sufficiently accurate. In this
case it is of advantage to reduce the given quadratic surd
to an equivalent expression which has a smaller number
under the radical sign. Suppose we wish to find the value
of V5056. The square root is at once evident if we resolve
the number into two equal groups of factors ; thus :
V5056 = V(2 2 3 . 7) (2 2 3 . 7) = V84 84 = 84.
Even when the number is not a perfect square, factor-
ing will often enable the student to- find its square root
much more easily, as will be shown later.
EXERCISES
Find the following indicated square roots :
1. V576. 3. V484. 5. V3600.
2. V1296. 4. V1089. 6. Vl936.
454. The square root of a product. The preceding exer-
cises show that the square root of the product of several
factors, each of which is a square, may be found by taking
the square root of each factor separately, as in the follow-
ing examples:
1. V9- 25=V9V25 = 3- 5=15.
INTERPRETATION OF ROOTS AND POWERS 407
This is true because 9 25 can be written as the product
of two groups of equal factors (3 5) (3 5). Hence, by the
definition of a square root, (3 5) is the square root of 9 25.
2. Vl6 xY = Vl6 V^ Vp = 4 a*f.
This is true because 16 afy 6 may be written as the product
of two groups of equal factors (4 a^?/ 3 ) (4 z 2 ?/ 3 ). Hence
4 a; 2 ?/ 3 is the square root of IGa^y 6 .
The preceding exercises show that the square root of a
product is obtained by finding the square root of each factor
separately and then taking the product of these roots. That
is, in general, _ _
v a b = v a Vft.
This principle may be used to simplify radical surds in
the following manner. Suppose we wish to find the value
of V11858. Then _
V11858 = V2 11- 11 7-7
= 77 V2.
By the table of roots, V% = 1.414.
Then 77(1.414) = 108.878.
Hence V11858 = 108.878.
It will be helpful to observe the following:
(1) The principle enables us to simplify the radicand to
a point where we can easily find the root by the table or
by several geometric constructions.
(2) A quadratic surd is in its simplest form when the
number under the radical sign does not contain a perfect
square factor.
In general, if the expression under the radical sign
contains a factor which is a square, this factor may be
removed by writing its square root before the radical sign.
-108 GENERAL MATHEMATICS
EXERCISES
Change the following so as to leave no factor which is a
square under the radical sign :
1. Vl2.
_6. V2S.
11.
2. V40.
7. V75.
12.
3. VlS.
8. V125.
v 13.
4. V50.
^9. VlOS.
- 14.
' 5. V72.
10. Vo*.
* 15.
455. Value of memorizing square roots of certain numbers.
Exercises 1~9, Art. 454, suggest the manner in which the
square roots of a few small numbers, like 2, 3, 5, are
made to do service in finding the roots of many large
numbers. In fact, many students in other fields find that
memorizing these numbers, which occur again and again
in their problems, increases their efficiency.
EXERCISES
From the table of roots we know that V2=1.414, A/3=1.732,
and V 5 = 2.236. Using these facts, compute each of the fol-
lowing correct to two decimal places :
1. V75. 4. V72. 7. V50 + V75-V6.
2. V80. 5. V98. 8. 2 V32 + V72 - Vl8.
3. V48. 6. V363. 9. V45-\/|.
D
456. The square root of a fraction. A fraction is squared
by squaring its numerator and its denominator separately
9
and indicating the product thus : x - = . Hence, to
bob
extract the square root of a fraction, we find the square
root of its numerator and denominator separately.
For example, VT\ = J, since % % = f V-
INTERPRETATION OF ROOTS AND POWERS 409
EXERCISES
Find the square roots of the following accurate to two places
(use the table for square roots):
1. f. 2. f. 3. TV 4. f. 5. f. 6. V-. 7. _*_. 8 . _2 y .
457. Rationalizing the denominator of a fraction. It
probably has been noticed that the calculation of those
problems of the preceding lists in which the denominator
is a square is far easier than when this is not the case.
This may be illustrated as follows:
[2 _ V2
1.414
1.732
Dividing 1.414 by 1.732 is not an easy division. In fact,
most people would find it impossible to do mentally. On
the other hand, J? = _ = ?dii? = Q.816 involves an easy
M9 V9 8
mental division.
It is of interest to note that the denominator of a
fraction can always be made a square merely by multi-
plying numerator and denominator by the proper factor ;
, Jf = J? = ^ =
i o \ 9 o
thus, = = = = 0.816. Note that the
o
division - is an easy mental division compared with
o
the division ' , which we had above.
A\ r e may summarize this method of taking the square
root of a fraction as follows :
1. To find the square root of a fraction, change the fraction
into an equivalent fraction whose denominator is a perfect
square.
410 GENERAL MATHEMATICS
2. The square root of the netv fraction equals the square
root of its numerator divided by the square root of its
denominator.
3. If desired, express the result in simplest decimal form.
The process of changing a radical expression so as to
leave no denominator under a radical sign is called ration-
alizing the denominator. Thus, -yj - = - V3 ; \J- = -^a-
1 3 o \ a a
EXERCISES
1. Find the value of the following square roots (find values
approximately accurate to the second decimal place) :
(a) J. (b) f . (c) 1 (d) 1 (e) f . (f) f (g) T V
2. Rationalize denominators of the following, and simplify :
3. What is the value of the expressions in Ex. 2 from (e)
to (j), inclusive, if a = 3 and b = 2 ?
458. Addition and subtraction of surds. Sometimes the
arithmetic in a problem may be simplified if the surds are
combined into one term. Thus,
V20 + V45 = 2 V5 + 3 V5
= 5V5, or5(2.236)
= 11.180.
By adding 2 \/5 + 3 V5 just as we add 2_- 4 plus 3 4, we
need only to look up the \/5, whereas V20 + V45 calls for
two square roots.
INTERPRETATION OF ROOTS AND POWERS 411
In the following list simplify each expression as far as
possible without using approximate roots ; that is, leave
your result in indicated form. Practically this is often
better than finding an approximation, for in this manner
you submit results that are absolutely accurate. It leaves
the approximation to the next person, who may find as
many decimal places as the needs of his particular problem
demand.
EXERCISES
1. Vl08 + V75 + Vl2. 8.
O
2. 2V98-VT8. 9. Vf +
3. 6 V288 - 4 Vl8 + Vl28. 10. Vf-
4. 5 V432 - 4 V3 + Vl47. n. 1+^
3
5. 2 V27 + 3 V48 -*3 V75. 12. - f +
6. 3 V20 + 2 Vl25 - Vl80. 13. ^/T _ 2 V2 +
4- V- 14. 4 V28 + 3 Vo5 - Vll2.
459. Multiplication and division of quadratic surds. These
two processes will be treated briefly. In elementary mathe-
matics there are few verbal problems which involve this
process. Further, the principles involved do not offer any-
thing new for us.
Thus, to divide V2 by V5, we may write this in the
form of a fraction and proceed as we ordinarily do
when finding the value of a fraction which involves
quadratic surds.
The rule in multiplication is equally familiar. The
equation Vab=\ / aVb may be read just as well from
right to left, and we have VaV5=VaJ. Thus, V2V5 is
precisely the same as VlO.
n-2 GENERAL MATHEMATICS
EXERCISES
1. Find the product of the following:
(a) V3 V27. (d) V(5Wl2a; 8 ~.
(b) vW* 8 . (e) V|VH-_
(c) V3V5. (f) V|Vf Vf Vl
( g ) ( V2 + Va - Vs)( V2 - Vs + Vs).
Solution. We multiply each term of one polynomial by each term
of the other and simplify the result as follows :
V2 + V - V5
VQ - Va + Vs
2 + Ve - Vio
- Ve - a + Vis
+ VTo + Vis - 5
2 Vlo - 6
(h) (2 Vr + Vs - Vs)( Vs).
(i) (4V2-2V3 + V5)(2 V3).
(j) (Vs - 3 V2 + VTo)(Vs).
(k) (3 V2 + 4 Vs)(2 V2 - S Vs).
(i) ( Vs - VI + Vs)(V3 + VI - Vs).
2. Divide, and express the result in simplest form :
(a) 1 by VS. (c) V6 by Vs.
(b) 24 by Vs. (d) 2 Vl2 - 5 VlS by Vs.
460. Fractional exponents another -means of indicating
roots and powers. Thus far we have not used a fraction
as an exponent, and evidently it could not be so used
without extending the meaning of the word "exponent."
Thus, a? means x x x, but X s evidently cannot mean
that x is to be used as a factor one half of a time.
INTERPRETATION OF ROOTS AND POWERS 413
It is very important that all exponents should be gov-
erned by the same laws; therefore, instead of giving a
formal definition of fractional exponents, we shall lay down
the one condition that the laws for integral exponents shall
be generally true, and we shall permit the fractional ex-
ponent to assume a meaning necessary in order that the
exponent laws shall hold.
Since we agree that ^ x^ x, we see that a?* is one of
the two equal factors of x ; that is, x^ is the square root
of #, or we may write y? = V#.
Similarly, since X s X s x 5 = z, X s is the cube root of x ;
\ 3/
that is, X s = v x.
Again, x* x* x' 5 = X s = xt. This means that X s is the
2 3/
cube root of 2 ; that is, x* = v x 2 .
This discussion is sufficient to show that the fractional
exponent under the laws which govern integral exponents
takes on a meaning which makes a fractional exponent
just another way of indicating roots and powers ; that is,
the denominator indicates the root, and the numerator indi-
cates the power.
Thus, 8^ means to take the cube root of 8 and square
the result, or square 8 and take the cube root of the result.
In either case the final, result is 4. Again, 10* (or 10- 6666 )
means to take the cube root of 100, which, by the table of
Art. 449, is 4.641. Note that a common-fraction exponent
may appear as a decimal fraction.
From this it will be obvious that the student needs only
to become familiar with the new method of writing roots
and powers. The following brief list of problems presents
precisely the same ideas as those in the preceding list
dealing with surds ; only the form is different.
414 GENERAL MATHEMATICS
EXERCISES
1. Write in simplest form :
(a) 4* + 9* + 16* + 25* + 36*.
(b) 1* + 8* + 64* + 0*.
(c) 64* + 9* + 16* + (- 32)* - (- 27)1
(d) 24* + 54' - 6*.
(e) 18* + 32* - Vl28 + V2.
(f) (^*+8*-(f) 4 +(50)*.
(g) (81)* - 2 (24)* + V28 + 2 (63)*.
2. Multiply:
(a) z*z*. (c) 10 2 - 5 10-. 125 10 1 - 25 .
(b) 10* 10* - 10*. (d) 10 3 - 6250 10- 3750 . 10- 0625 .
3. Translate (c) and (d), Ex. 2, into expressions with common
fractions as exponents and estimate how large the numbers
would be if we had a way of finding the result.
NOTE. 10 - 8 " 5 means 10^, or the eighth root of 1000. Another
way to look at it is that 10 is to be raised to the 375th power and
the thousandth root taken. These ideas are of very great importance
in getting a clear understanding of the chapters on logarithms and
the slide rule. In these chapters we shall learn how to find a root
(say the 15th) just as easily as the square root.
461. Zero exponents. Under the laws which govern
integral exponents a = 1, as is shown by the following :
5 -=- a 5 = a. (By the Division Law.)
o 5 -5-o 6 = l. (The quotient of any number divided
by itself is 1.)
Hence a = l. (By the equality axiom.)
Thus the zero power of any number (except zero) is 1.
Thus, 15 =1; (560)= 1; (-6a:) =l; 10 = 1.
462. Negative exponents. If x~ s obeys the same law as
integral exponents, then
_ Q Q Q -|
-3- X = , or ;
1 X 6 X 6 X 6
that is, x~ s
Similarly, if we multiply the numerator and denominator
x~ a 1
of - - by x a , we obtain x~ a =
Tin-, ~~> looAo-^jL, v-'-mf
J^ = 10,000.
See if you can state in simple language the meaning of
a negative exponent.
EXERCISES
Simplify :
1. 10- 2 x 10 3 - 125 x 10 - 0625 x 10- 1 - 03675 .
2. 10^ x lo" x (56) x 10*.
3. (39) (169)"*. 6. 16- 2- 4 . 9. x m x n .
4. 1000 (100)"* 7. 1000- IO- 3 .
i in y b . y c
5. 2 X x 2-* x 2. 8. 144~* 24.
11. X m H-X n .
12. (+625)*; (-125)-*.
13. (x 2 ) 3 ; (x 3 ) 4 ; (x ) 2 .
HINT, (z 2 ) 3 = x 2 x z x 2 .
14. (IO 2 ) 2 ; (IO 2 ) 8 ; (IO 2 ) 4 .
15. (10- 125 ) 2 ; (10 - 0625 ) 4 ; (10- 125 ) 8 .
16. (x m ) n ; translate the formula (x m } n = x mn into an alge-
braic rule.
17. -v/x^; ^x 8 ; ^x 5 .
NOTE. The 3 in vx means "find the cube root."
416
GENERAL MATHEMATICS
18.
19. Since (x m ) n ^ x mn , what is
m
Translate the formula Vic = x" into an algebraic rule.
20. Find the value of :
(a) VlO^. (b) A/10"-' 75 , (c) -v/H) 3 -" 250 . (d) -x/lO 5 - 8750 .
463. Cube of a binomial. We shall now see how certain
other short cuts in finding powers may be illustrated and
explained.
INTRODUCTORY EXERCISES
1. Find the cube of x -f- y by first finding the square of
x -f y and multiplying the result by x + y.
2. By multiplication find (a + &) 3 .
3. Find the cube of (x y); of (a &). Compare these
cubes with the results of Exs. 1
and 2.
4. It will be helpful to your class-
mates if you will make a set of blocks,
as shown in Fig. 298, in order to
show that
(x -f- 7/) 3 = x 3 + 3 x 2 y + 3 xif + ?/ 3 -
How many blocks are needed ?
5. Find the following cubes, doing
as many as you can mentally :
(a) (c + dy. (c) (c-df. (e) (a-4-27/) 8 . (g) (2x +
(b) (m + nf. (d) (m-nY- (f) (2 x - y) s . (h) (2x-3y) s .
Exercises 15 show that the cube of a binomial is equal
to the cube of the first term, plus three times the square of
the first term multiplied by the second, plus three times the
first term multiplied by the square of the second, plus the
cube of the second term.
SIR ISAAC NEWTON
418 GENERAL MATHEMATICS
HISTORICAL NOTE. The first use of positive and negative frac-
tions as exponents is found in a book written by the great English
mathematician and physicist, Sir Isaac Newton (1642-1727). He
discovered the binomial theorem and numerous laws of physics ;
for example, the law of gravitation, the law about lenses and prisms,
and the explanation of the rainbow. Among his numerous books are
" Universal Arithmetic " (really an algebra) and " Principia " (one of
the greatest books of all times).
The biography of Xewton (see Ball's " A Short History of Mathe-
matics," pp. 328-362, and Cajori's "A History of Elementary Mathe-
matics," pp. 238-240) is very interesting and inspiring. As a boy
he was expected to be learning how to take care of his father's
farm, but he spent much of his time studying and trying mechan-
ical experiments. Thus, we read of his constructing a- clock run
by water which kept very fair time. His mother noticing this
sensibly resolved to send him to Cambridge. Here followed a brilliant
career of thirty-five years as student and teacher. As a professor it
was his practice to lecture publicly once a week, and then only from
half an hour to an hour at a time. In the week following he gave
four hours of consultation to students who wished to discuss the
results of the previous lecture. It is said that he never repeated a
course and that one course began at the point where the preceding
course had ended. The result of his second study during this period
has dazed master minds who have attempted to understand all that
Newton accomplished. Perhaps you will later agree with some of
the following tributes to him :
Nature and Nature's laws lay bid in night :
God said, " Let Newton be ! " and all was light. POPE
There, Priest of Nature ! dost thou shine,
Newton ! a king among the kings divine. SOCTHEY
Tafcing mathematics from the beginning of the world to the time when
Newton lived, what he had done was much the better half. LEIBNITZ
I don't know what I may seem to the world, but as to myself, I
seem to have been only as a boy playing on the sea-shore, and diverting
myself in now and then finding a smoother pebble or a prettier shell
than ordinary, whilst the great ocean of truth lay all undiscovered
before me. NEWTON
INTERPRETATION OF ROOTS AND POWERS 419
464. Cube roots. The equation y = $ asserts that y
is the cube of .r, or, inversely, x is the cube root of y.
If the equa-
tion is graphed
we shall ob-
tain a curve
analogous to
the curve for
squares and
square roots
which may be
used to find
cube roots and
cubes. We pro-
ceed to find
corresponding
values for y
and x in order
that we may
plot sufficient
points for the
-x-
-y-
curve.
FIG. 299. GRAPH OF y - X s
EXERCISES
1. What is the value of y in the equation y = y? when x
equals ? when x = + 1 ? when x 1 ? when x = + 2 ?
when x = 2 ?
2. Calculate values as in Ex. 1 and fill the blank spaces of
the following table. If the curve is not obvious, expand the
table until you have enough points to draw the curve.
X
y
+ 1
+ 1
+ 2
+ 8
-2
-8
3
4
6
6
7
i
i
f
1
&.
1.
27
420 GENERAL MATHEMATICS
From this table we may obtain the curve in Fig. 299.
One small square vertically represents 1 unit, and 5 small
squares horizontally represent 1 unit.
From this curve we can read off, approximately, the
cube or the cube root of any number; thus the cube of
2.2 is seen to be about 10.5 (by the table actually 10.64);
the cube root of 13 is seen to be a little over 2.4 (see
the table for values accurate to two decimal places). More
accurate results can be obtained by drawing the curve to
a large scale.
465. Cube roots of arithmetical numbers. An arithmetical
method for finding cube roots based on an algebraic for-
mula for (a + 6) 3 could be devised. But this method is
seldom used, because cube roots as well as higher roots
are more quickly found by means of logarithms. This
method will be taught in the next chapter. In the mean-
time the student may for all practical purposes use the
table in Art. 449.
Furthermore, we could devise analogous rules and curves
for fourth roots, fifth roots, fourth powers, and so on, but
these too are more readily found by logarithms.
466. Indicating higher roots. By means of an index figure
the radical sign is made to indicate other roots than square
roots.
Thus the cube root of 8 or one of its three equal factors
is written V8 = 2. The fourth root of 16 is written Vl6 = 2.
The 3 in v 8 is the index of the root.
Any expression which contains an indicated root is called
a radical expression.
The principles of reducing surds to simpler forms, dis-
cussed in detail for quadratic surds, may be applied to
higher indicated roots.
EXERCISES
1. Simplify the following (remove any factor which is a
perfect power of the degree indicated by the index):
^32; -\/64; ^64 x 6 /; "\/48 4 ; A/16 a
2. Add and subtract as indicated:
(a) A/16 + -^54 - -^250 + -v
(b) -v/54 + -v/128 + -v/1024 +
SUMMARY
467. This chapter lias taught the meaning of the fol-
lowing words and phrases : square root of a number,
quadratic surd, radical sign, radicand, quadratic trinomial
square, index.
468. The graph of the equation y = x 2 was used as a
device for finding squares and square roots.
469. A positive number has two square roots ; thus,
V4 = + 2 or - 2.
470. The square of the sum or difference of two numbers
may be found by the formula
(a A) 2 = a 2 2 db + ft 2 .
This formula was illustrated geometrically.
471. To find the square root of a perfect square tri-
nomial: Extract the square roots of the two perfect square
terms and connect them by the sign of the remaining term.
472. The square of a trinomial consists of the sum of the
squares of its terms plus twice the product of each term by
each succeeding term. By remembering this rule the square
roots of some polynomials may be found by inspection.
422 GENERAL MATHEMATICS
473. The chapter taught a method of finding the square
root of algebraic polynomials and arithmetical numbers.
474. The chapter includes a simple table of square
roots and cube roots.
475. Quadratic surds may often be simplified by apply-
ing the principle vW>=VaVi.
476. We find the square root of a fraction by dividing
the square root of the numerator by the square root of
the denominator; that is, A 7 = -
: b V 6
477. 'Rationalizing the denominator simplifies the cal-
rn /~
culation ; that is, -yj is more difficult than -=-
478. When the same number occurs as the radical in
a series of terms, the terms may be combined by the
rule for adding similar terms. This usually simplifies
the calculation.
479. The theorem of Pythagoras was proved.
480. The theorem of Pythagoras furnishes a method of
constructing the square root of a number.
481. The mean proportional construction furnishes
another method of finding the square root.
482. A fractional exponent is another method of indicat-
ing roots and powers; thus, x% means -$2?. The numerator
indicates the power, and the denominator the root.
483. a is defined as 1.
484. A number with a negative exponent is defined so
as to be equal to the reciprocal of the same number with
a positive exponent; that is, a~ 5 =
INTERPRETATION OF ROOTS AND POWERS 423
485. The cube of a binomial may be found by the
following formula:
(a + 6) 3 = a 3 + 3 a 2 6 + 3 a
486. Cube roots may be found by the table, graph, or
more easily by logarithms and the slide rule (the last two
methods will be shown in the next two chapters).
CHAPTER XVII
* LOGARITHMS APPLIED TO MULTIPLICATION, DIVISION,
ROOTS AND POWERS, AND VERBAL PROBLEMS INVOLV-
ING EXPONENTIAL EQUATIONS
LOGARITHMS
487. Labor-saving devices. In Chapter IV we showed
how extensive calculations even with only four or five
place numbers are apt to become laborious and, in some
cases, inaccurate and involving unnecessary steps. We
showed how to minimize the inaccuracy and how some of
the unnecessary steps may be eliminated, especially with
regard to the processes of multiplication and division by
the so-called " abbreviated method." But in many cases
the work remains long and tedious, even with the use of
these abbreviated methods.
In Art. 449 will be found a table of powers and roots
which are given for the purpose of saving time and labor.
Scientific books include similar tables which help to save time
and conserve our energy. Other labor-saving devices com-
monly used are adding machines, cash registers, graphs, etc.
One of the greatest labor-saving devices by which diffi-
cult problems may be readily solved is the method of
calculation by logarithms. This chapter will be devoted
to a simple explanation of this method. If the student
will study the chapter carefully and solve the problems
correctly, he will get a foundation in logarithmic work
that will be very helpful in subsequent work.
424
LOGARITHMS 425
488. Two methods of multiplying. We are already
familiar with the two methods of multiplying illustrated
by the following examples:
100 x 1000 = 100,000.
10 2 x 10 8 = 10 5
= 100,000.
The student will observe that the product is the same
in both examples, but that the method used in the second
has reduced the operation of actually multiplying the two
numbers to a simple problem in the addition of exponents.
Although the numbers multiplied here are in each case
powers of 10, the method will hold for other bases as well.
However, we shall consider only the base 10 in our sub-
sequent discussion since it is the one commonly used.
Find the product of the following pairs of numbers by
the two methods given above :
10 and 100. 10,000 and 100,000.
1000 and 1000. 1000 and 1,000,000.
489. Powers of 10. From the preceding exercises it is
clear that we can multiply together numbers which are
integral powers of 10 merely by adding the exponents of
these powers. Since every positive number may be expressed
exactly or approximately as a power of 10, we may obtain
the product of any two numbers in a similar manner by
adding the exponents of the powers of 10 which equal the
respective numbers. For example, we may multiply 17.782
and 3.162 by adding the exponents of the powers of 10,
which equal 17.782 and 3.162 respectively.
This raises two important questions: (1) What powers
of 10 equal 17.782 and 3.162? (2) What is the value
of 10 when raised to the sum of these two powers ? It is
426 GENERAL MATHEMATICS
possible for us to work out a table which will give us the
powers of 10 which equal 17.782 and 3.162. The table
below shows how the different values are obtained. The
table is not complete, as will be shown later, but it contains
the approximate values of several integral and fractional
powers of 10 which we need at this point.
We know that 10 = 1, 10 1 = 10, 10 2 = 100, 10 3 = 1000, and so on.
\Vr can find the value of 10- 5 as follows :
10- 5 = 10* = VlO = 3.162 (approx.).
From these values the other values in the table are easily found,
as the student can verify.
TABLE OF POWERS OF 10
10' = 1.0000
I = V8.162 = 1.7782
10- 5 = 10z = VlO = 3.1623
100.75 _ (ioi.5)2 = v'31.62 = 5.6234
10 1 = 10.0000
lQi-25 = lOi . 100.25 = i7.7 8 o
lQi-5 _ i i . 100.5 _ 31.623
lOi- =10* 10<>-'5 = 56.234
10* = 100.000
102.25 _ i i . iQi.25 _ 177.82
10 2 - 5 = lOi . 101.5 _ 316.23
jO-2.75 _ 1 1 . 101-75 _ 562.34
10 =1000.00
We may now resume the solution of the problem proposed
above, namely, multiplying 17.782 by 3.162. We can see
by referring to the table that 17.782 = 10 1 - 25 and that
3.162 =10- 5 . Hence, 17.782 x 3.162 = 10 1 - 25 xlO- 5 = 10 1 - 78 ,
which, by referring to the table, we see is equal to 56.234
(this product is accurate to the second decimal place).
LOGARITHMS 427
EXERCISE
Check the preceding result by actually multiplying 17.782
by 3.162, and account for the difference in results. Is there a
significant difference ? Is the result obtained by actual multi-
plication accurate to more than two decimal places ?
490. Logarithms ; notation for logarithms. In the equa-
tion 17.782 =10 1 - 25 the exponent 1.25 (which indicates
the power to which 10 must be raised to give 17.782) is
called the logarithm of 17.782 to the base 10.
Thus the logarithm of a number to the base 10 is tlie
exponent of the power to which 10 must be raised to equal
that number. From now on we shall assume that the base
is 10 when we speak of the logarithm of a number. The
symbol for logarithm is log. Thus, log 1000 = 3 is read " the
logarithm of 1000 equals 3," the base 10 being understood.
EXERCISE
By means of the table of powers of 10 in Art. 489 find log 1 ;
log 10 ; log 100 ; log 1.78; log 316.23.
491. A logarithm is an exponent. The student will do
well to remember the two ways of thinking of an expo-
nent ; for example, in the equation 10 2 = 100 the 2 can be
thought of (a) as the exponent of the power to which 10
must be raised to equal 100 ; that is, 10 2 = 100 ; (b) as the
logarithm of 100 to the base 10; that is, 2 = Iog 10 100.
EXERCISES
Read the following in two ways :
1. 10 1 = 10. 2. 10 2 = 100. 3. 10 8 = 1000. 4. 10 4 = 10,000.
428 GENERAL MATHEMATICS
492. Characteristic ; mantissa. A glance at the table of
Art. 489 will show that each exponent of 10 (each loga-
rithm of the corresponding numbers to the right) may con-
tain an integral part and a fractional part. For example, in
the equation 10 1 - 25 = 17.782 the 1.25 (that is, log 17. 782)
has 1 for its integral part and 0.25 for its decimal (frac-
tional) part. In 10 2 = 100 the entire logarithm is integral.
(Why?) The integral part of a logarithm is called the
characteristic of the logarithm, and the decimal part is
called the mantissa of the logarithm.
The characteristic of a logarithm of any number can
always be determined at sight. For example:
log 10 = 1, because 10 1 = 10 ;
log 100 = 2, because 10 2 = 100 ;
log 1000 = 3, because 10 3 = 1000 ;
and so on. But these numbers are all integral powers of
10. However, the characteristic of the logarithm of any
other number may be obtained as well by observing what
powers of 10 inclose it. For example, the characteristic
of log 525 is 2 because 525 lies between the second and
the third powers of 10 ; that is, between 10 2 = 100 and
10 3 = 1000 (see the table, Art. 489).
It is not so easy to determine the mantissas (the deci-
mal part) of the logarithms of numbers. We have worked
out a few of these in the table of Art. 489, but to compute
the mantissas for all other numbers in this way would be
a tedious'task. Moreover, the methods necessary to com-
pute them would be beyond us in difficulty. However,
these mantissas have been computed for all the various
powers of 10 (by more advanced methods), and they appear
in the table of mantissas which follows. So that now when
LOGARITHMS 429
we want to know what the logarithm of any number is,
we decide (by inspection) what the characteristic is and
then look in the table for the mantissa.
EXERCISES
1. Look in the table (pp. 430-431) for the decimal part of
the logarithms of 10 ; 15 ; 20 ; 38 ; 86 ; "99.
2. What is the decimal part of the logarithm of 100 ?
3. What is the power to which 10 must be raised to produce
10,000 ?
4. What, then, is the logarithm of 10,000 to the base 10 ?
5. Examine the table carefully and tell what numbers have
integers for logarithms ; that is, those that have zero mantissas.
6. When will a logarithm have a decimal mantissa ?
7. Find the product of 48 and 55.
Solution. By means of the table we see that
48 = 10 1 - 6812 ,
and that 55 = 10 1 - 7404 .
Therefore 48 x 55 = 10 1 - 6812 x 10 1 - 7404
The 3 in the exponent 3.4216 tells us that the product of 48 x 55
is a number between the 3d and 4th powers of 10 ; that is, the 3
tells us where to put the decimal point. We must find the mantissa
0.4216 in the table of logarithms and see what number corre-
sponds to it.
If we look in the table of mantissas we find that 0.4216 is the
mantissa of the logarithm of the number 264. Now since the charac-
teristic of the logarithm is 3, the number must be between the 3d
and 4th powers of 10 ; that is, between 1000 and 10,000. This means
that the decimal point comes after the fourth place, so that we must
add a cipher to 264. Hence the number is 2640.
430
GENERAL MATHEMATICS
TABLE OF MANTISSAS
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0000
0414
0792
1139
1461
0043
0453
0828
1173
1492
0086
0492
0864
1206
1523
0128
0531
0899
1239
1553
0170
0569
0934
1271
1584
0212
0607
0969
1303
1614
0253
0645
1004
1335
1644
0294
0682
1038
1367
1673
0334
0719
1072
1399
1703
0374
0755
1106
1430
1732
15
16
17
18
19
1761
2041
2304
2553
2788
1790
2068
2330
2577
2810
1818
2095'
2355
2601
2833
1847
2122
2380
2625
2856
1875
2148
2405
2648
2878
1903
2175
2430
2672
2900
1931
2201
2455
2695
2923
1959
2227
2480
2718
2945
1987
2253
2504
2742
2967
2014
2279
2529
2765
2989
20
21
22
23
24
3010
3222
,3424
3617
3802
3032
3243
3444
3636
3820
3054
3263
3464
3655
3838
3075
3284
3483
3674
3856
3(196
3304
3502
3692
3874
3118
3324
3522
3-711
3892
3139
3345
3541
3729
3909
3160
3365
3560
3747
3927
3181
3385
3579
3766
3945
3201
3404
3598
3784
3962
25
26
27
28
29
3979
4150
4314
4472
4624
3997
4166
4330
4487
4639
4014
4183
4346
4502
4654
4031
4200
4362
4518
4669
4048
4216
4378
4533
4683
4065
4232
4393
4548
4698
4082
4249
4409
4564
4713
4099
4265
4425
4579
4728
4116
4281
4440
4594
4742
4133
4298
4456
4609
4757
30
31
32
33
34
4771
4914
5051
5185
5315
4786
4928
5065
5198
5328
4800
4942
5079
5211
5340
4814
4955
5092
5224
5353
4829
4969
5105
5237
5366
4843
4983
5119
5250
5378
4857
4997
5132
5263
5391
4871
5011
5145
5276
5403
4886
5024
5159
5289
5416
4900
5038
5172
5302
5428
35
5441
5l6T
5682
5798
5911
5453
5575
5694
5809
5922
5465
5587
5705
5821
5933
5478
5599
5717
5832
5944
5490
5611
5729
5843
5955
5502
5623
5740
5855
5966
5514
5635
5752
5866
5977
5527
5647
5763
5877
5988
5539
5658
5775
5888
5999
5551
5670
5786
5899
6010
"36
37
38
39
40
41
42
43
44
6021
6128
6232
6335
6435
6031
6138
6243
6345
6444
6042
6149
6253
6355
6454
6053
6160
6263
6365
6464
6064
6170
6274
6375
6474
6075
6180
6284
6385
6484
6085
6191
6294
6395
6493
6096
6201
6304
6405
6503
6107
6212
6314
6415
6513
6117
6222
6325
6425
6522
45
46
47
48
49
6532
6628
6721
6812
6902
6542
6637
6730
6821
6911
6551
6646
6739
6830
6920
6561
6656
6749
6839
6928
6571
6665
6758
6848
6937
6580
6675
6767
6857
6946
6590
6684
6776
6866
6955
6599
6693
6785
6875
6964
6609
6702
6794
6884
6972
6618
6712
6803
6893
6981
50
51
52
53
54
6990
7076
7160
7243
7324
6998
7084
7168
7251
7332
7007
7093
7177
7259
7340
7016
7101
7185
7267
7348
7024
7110
7193
7275
7356
7033
7118
7202
7284
7364
7042
7126
7210
7292
7372
7050
7135
7218
7300
7380
7059
7143
7226
7308
7388
7067
7152
7235
7316
73%
No.
1
2
3
4.
5
6
7
8
9
LOGARITHMS
431
TABLE OF MANTISSAS
No.
1
2
3
4
5
6
7
8
9
55
56
57
58
59
7404
7482
7559
7634
7709
7412
7490
7566
7642
7716
7419
7497
7574
7649
7723
7427
7505
7582
7657
7731
7435
7513
7589
7664
7738
7443
7520
7597
7672
7745
7451
7528
7604
7679
7752
7459
7536
7612
7686
7760
7466
7543
7619
7694
7767
7474
7551
7627
7701
7774
60
61
62
63
64
7782
7853
7924
7993
8062
7789
7860
7931
8000
8069
7796
7868
7938
8007
8075
7803
7875
7945
8014
8082
7810
7882
7952
8021
8089
7818
7889
7959
8028
8096
7825
7896
7966
8035
8102
7832
7903
7973
8041
8109
7839
7910
7980
8048
8116
7846
7917
7987
8055
8122
65
66
67
68
69
8129
8195
8261
8325
8388
8136
8202
8267
8331
8395
8142
8209
8274
8338
8401
8149
8215
8280
8344
8407
8156
8222
8287
8351
8414
8162
8228
8293
8357
8420
8169
8235
8299
8363
8426
8176
8241
8306
8370
8432
8182
8248
8312
8376
8439
8189
8254
8319
8382
8445
70
71
72
73
74
8451
8513
8573
8633
8692
8457
8519
8579
8639
8698
8463
8525
8585
8645
8704
8470
8531
8591
8651
8710
8476
8537
8597
8657
8716
8482_,
'8488
8549
8609
8669
8727
8494
8555
8615
8675
8733
8500
8561
8621
8681
8739
8506
8567
8627
8686
8745
854~3
8603
8663
8722
75
76
77
78
79
8751
8808-
8865
8921
8976
8756
8814
8871
8927
8982
8762
8820
8876
8932
8987
8768
8825
8882
8938
8993
8774
8831
8887
8943
8998
8779
8837
8893
8949
9004
8785
8842
8899
8954
9009
8791
8848
8904
8960
9015
8797
8854
8910
8965
9020
8802
8859
8915
8971
9025
80
81
82
83
84
9031
9085
9138
9191
9243
9036
9090
9143
9196
9248
9042
9096
9149
9201
9253
9047
9101
9154
9206
9258
9053
9106
9159
9212
9263
9058
9112
9165
9217
9269
9063
9117
9170
9222
9274
9069
9122
9175
9227
9279
9074
9128
9180
9232
9284
9079
9133
9186
9238
9289
85
86
87
88
89
9294
9345
9395
9445
9494
9299
9350
9400
9450
9499
9304
9355
9405
9455
9504
9309
9360
9410
9460
9509
9315
9365
9415
9465
9513
9320
9370
9420
9469
9518
9325
9375
9425
9474
9523
9330
9380
9430
9479
9528
9335
9385
9435
9484
9533
9340
9390
9440
9489
9538
90
91
92
93
94
9542
9590
9638
9685
9731
9547
9595
9643
9689
9736
9552
9600
9647
9694
9741
9557
9605
9652
9699
9745
9562
9609
9657
9703
9750
9566
9614
9661
9708
9754
9571
9619
9666
9713
9759
9576
9624
9671
9717
9763
9581
9628
9675
9722
9768
9586
9633
9680
9727
9773
95
96
97
98
99
9777
9823
9868
9912
9956
9782
9827
9872
9917
9961
9786
9832
9877
9921
9965
9791
9836
9881
9926
9969
9795
9841
9886
9930
9974
9800
9845
9890
9934
9978
9805
9850
9894
9939
9983
9809
9854
9899
9943
9987
9814
9859
9903
9948
9991
9818
9863
9908
9952
9996
No.
1
2
3
4
5
6
7
8
9
432 GENERAL MATHEMATICS
The preceding work may be briefly arranged as follows :
Let N= 48x55
log 48 = 1.6812
log 55 = 1.7404
Then logA 7 = 3.4216
By the table, N = 2640.
493. Logarithm of a product. The discussion and ex-
amples in Art. 492 have shown that to the problem of
multiplying two powers of 10 there corresponds the problem
of adding their logarithms (exponents). This may be stated
briefly as the first law thus: The logarithm of the product
of two numbers is the sum of the logarithms of the factors;
or, by formula, log (aby = log a -f- log b.
It is easily shown that the law also holds for any number
of factors in a product ; that is, log abc = log a + log b + log c,
and so on.
EXERCISES
1. Check by actual multiplication the logarithmic method of
finding the product of 48 and 55.
2. Find by means of logarithms the products of the follow-
ing numbers :
(a) 10 x 100 x 1000. (b) 51 x 40. (c) 83 x 6 x 2.
3. Find by using logarithms the area of a triangular garden
plot whose base is 38 ft. and whose altitude is 17 ft.
Solution. The formula for the area of any triangle is
Hence, in this case, A = 19 17.
log 19 = 1.2788
log 17 = 1.2304
Then log A = 2.5092
By the table, A = 323.
LOGARITHMS
4. Solve the following problems by means of logarithms :
4 (a) What is the area of a rectangle whose base is 70 and
whose altitude is 32 ?
(b) What is the area of a parallelogram whose base is 64 and
whose altitude is 85 ?
(c) What is the area of a square, a side of which is 29 ?
Solution. The formula for the area of a square is
A = ,s * = *-.
In this case .1 = 29 x 29 = 29 2 .
log 29 = 1.4624
log 29 = 1.4624
Therefore log A = 2.9248
and, by the table, A =841.
494. Logarithm of a power. Obviously the result in
Ex. 4, (c), would have been the same if we had multiplied
the logarithm of 29 by 2 instead of adding it to itself.
In like manner, if we want the logarithm of 50 3 , we can
obtain it either by using the logarithm of 50 three times
as an addend or by taking three times the logarithm of 50.
Thus, log 50 3 = 3 log 50 = 3 x 1.6990 = 5.0970.
This discussion illustrates a very important law ; namely,
that the logarithm of a number roiled to a power in the exponent
of the power times the logarithm of the number ; or, by for-
mula, log a" = n log a.
EXERCISES
1. Find by the preceding method the logarithms of the
following numbers: 22 2 ; II 8 ; 15 4 .
2. Find by logarithms the volume of a cubical cake of ice
one edge of which is 10 in. ; 11 in.
3. The area .1 of a circle is given by the formula A = irr,
where / is the radius of the circle. What is the area of a circle
whose radius is 6 in. ? (Use logTr = 0.4971.)
434 GENERAL MATHEMATICS
495. Logarithm of a quotient. The method of computing
by logarithms is as useful in division as in multiplication.
In order to make the method clear, let us review our two
methods of dividing one number by another.
(a) - - = 1000, by actual division.
(b) - = - 2 = 10 3 = 1000, by subtracting the exponents.
1UU 1U
Here, as in multiplication, we obtain the same result
by either method, but the second method reduces the
operation of actual -division to a simple problem of sub-
tracting exponents.
EXERCISES
1. Find the following quotients by the two methods just
discussed :
100,000 1,000,000
1000 ' 10,000
2. Divide 100 by 31.623 by using the table of Art. 489.
So "* ion - =
XOTE. The student should check this result by actual division.
3. Divide 562.34 by 31.62 by using the table of Art. 489.
We may obtain the quotient of any two numbers in
like manner by subtracting the exponent of the power of
10 equal to the divisor from the exponent of the power of
10 equal to the dividend. Keeping in mind the definition
of a logarithm it is clear that this fact may be expressed as
a law ; thus, the logarithm of the quotient of two numbers is
the logarithm of the dividend minus the logarithm of the
divisor ; or, as a formula, log IT) l9 a ~ 19 ^
LOGAEITHMS 435
EXERCISES
1. Given log 2 = 0.3010, log 3 = 0.4771, find log f ; log J
2. Find the value of the following fractions to three
significant figures by using logarithms :
x 59 x 85 . 381 x II 3
< a > 43" ~^~
. 752 x 350 71 x 48 x 253
-~ ~"
HINT. To find the logarfthm of each fraction, arid the logarithms
of the factors of the numerator and from this sum subtract the sum
of the logarithms of the factors of the denominator.
496. Position of the decimal point. Since the multipli-
cation or division of a number by 10 amounts to a moving
of the decimal point one place to the right in multiplication
and one place to the left in division, and since the multi-
plication or division of a number by 100 amounts to
moving the decimal point two places to the right or left,
and so on, the position of the decimal point in a number
whose logarithm we are seeking affects the characteristic onli/.
The truth of the foregoing statement can be seen best by
means of the table in Art. 489. In this table, for example,
! O o.25 =1.7782, or log 1.7782 = 0.25.
If we multiply both sides of this equation by 10, we get
10 1 - 2B = 17.782, or log 17.782 = 1.25.
Again, multiplying both sides of this last equation by 10,
1 02.25 = 177.82, or log 177.82 = 2.25,
and so on. The student will observe that only the integral
part of the exponent of 10 (the logarithm of the number)
is changing, and that the array of figures remains the
436 GENERAL MATHEMATICS
same even though the decimal point moves one place to
the right after each multiplication. In like manner, if we
divide both sides of the equation 10 - 25 = 1.778 by 10, and
continue the division, we obtain
loo.*- 1 = 0.1778, or log 0.1778 = 0.25 - 1;
100.25-2 = 0.01778, or log 0.01778 = 0.25 - 2,
lOo.ffi-3 = 0.001778, or log 0.001778 = 0.25 - 3,
and so on.
The logarithms 0.25-1, 0.25-2, 0.25-3, etc. are
negative quantities, but they are not in the form in which
.we usually write negative numbers. However, if we adopt
these forms, the mantissas of all our logarithms will not
only be positive but they will be the same for the same
array of figures no matter where the decimal point is found.
Thus the mantissa of log 1.778 is the same as the
mantissa of log 0.001778, as was shown above. These two
logarithms differ, therefore, only in their characteristics.
In some texts log 0.1778 is written 1.25 instead of 0.25 1.
To agree with this statement the dash above the 1 means
that only the 1 is negative. Some books prefer the form
9.25-10 instead of 0.25-1 or 1.25. The student can
easily see that 9.2510 has the same value as 0.251.
\Ve shall later see another advantage of the form 9.25 1 0.
The preceding discussion may be summarized in two
statements :
1. We agree to express the logarithm of any number in a
for tn such that its mantissa shall be positive. This can
always be done, whether the number is greater or less than
unity. In either case the positiveness or negativeness of
the number is shown entirely by means of the characteristic.
2. Two numbers containing the same succession of digits,
that ?X differing only in the position of the decimal
LOGARITHMS 437
will have logarithms that differ only in the value of the
characteristic. This explains why we called the table a
table of mantissas and why, in looking up the logarithm
of a number, we need pay no attention to the decimal
point in the number. The same table of mantissas serves
both for numbers greater and less than unity.
497. Table of characteristics. The following 'table is
given in order that the student may determine more
quickly the characteristic of the logarithm of any number
between 10 to the minus 6th power and 10 to the plus
7th power. This is about as much of a range as we
shall ever need. The table can be extended upward or
downward at will.
TABLE OF CHARACTERISTICS
10~ 6 = 0.000001 10 J =10
10-5-0.00001 io 2 =ioo
10- 4 -0.0001 10 3 = 1,000
io-* = o.ooi io 4 = 10,000
10-2-0.01.' 10 5 = 100,000
10- ! = 0.1 io 6 =1,000,000
10 = 1 IO 7 - 10,000,000
For example, if we want the logarithm of 2142, we
know that its characteristic is 3, because 2142 lies between
the 3d and 4th powers of 10. Again, if we want the
logarithm of 0.0142, we know that the characteristic of
the logarithm is 2, because 0.0142 lies between the
minus 2d and the minus 1st power of 10.
498. Interpolation. So far we have shown the student
how to find the logarithms of three-digit numbers only. In
order to be able to' find the logarithms of numbers of more
than three digits, and to find the numbers corresponding
438 GENERAL MATHEMATICS
to various logarithms which we may obtain in calculation,
it is necessary for us to learn how to make further use of
the table in Art. 492. We shall proceed to consider two
typical examples.
1. Find the logarithm of 231.6.
Solution. From the table of Art. 497 it is clear that the char-
acteristic of the logarithm is 2. To find the mantissa, we ignore the
decimal point and look in the table of Art. 492 for the mantissa
of 2316. Reading down the left-hand column of the table, headed
" No.," we find 23. The numbers in the same horizontal row with 23
are the mantissas of the logarithms 230, 231, 232, and so on.
We want to find the logarithm of 2316. We can now write
log 232 = 2.3655
log 231 = 2.3636
Tabular difference = 0.0019
Now since 231.6 is f' 6 of the way from 231 to 232, we add {'^
of the tabular difference 0.0010 to the logarithm of 231. Thus,
log 231.6 = 2.3636 + ^ x 0.0019.
Therefore log 231.6 = 2.3647.
The process of obtaining the logarithm of a number
in this way is called interpolation. The student should
practice this method by finding the logarithms of several
four-digit numbers.
2. Find the number whose logarithm is 0.3883 1.
Solution. We know at once that the number is a decimal fraction
lying between the minus 1st and the power of 10. This tells us
that the decimal point comes just before the first significant figure
in the number. If we look in the table of Art. 492 we do not find
the mantissa 3883, but we find 3892, which is a little greater, and
3874, which is a little less ; that is,
0.3892 - 1 = log 0.245.
0.3874 - 1 = log 0.244.
Since 0.3883 1 is the logarithm of the number we want, the
number lies between 0.244 and 0.245. Now 0.3883 - 1 is T 9 8 of the
LOGARITHMS . 439
way from log 0.244 to log 0.245 ; hence the number corresponding
to the logarithm 0.3883 - 1 lies ^, or i, of the way from 0.244 to
0.245. Therefore the number desired is 0.2445.
Here the process of interpolation is used on the inverse
problem, that of finding a number when its logarithm is
given.
EXERCISES
1. Find the logarithms of the following numbers: 745; 83.2;
91200; 0.567; 0.00741. (No interpolation.)
2. Find the logarithms of the following numbers : 6542 ;
783.4; 91243; 0.4826; 0.002143. (Interpolation.)
3. Find the numbers whose logarithms are 2.6075 : 1.4249;
0.3054; 0.0212-2; 0.8457-1. (No interpolation.)
4. Find the numbers whose logarithms are 2.3080; 1.936;
0.8770 ; 0.0878 - 2. (Interpolation.)
499. Extraction of roots by means of logarithms. In
Art. 460 we discussed the meaning of fractional exponents
and showed that
a* = Va ; * = Va ; a* = V ; etc.
If we assume that the theorem of Art. 494, regarding
the logarithm of a number raised to a power, holds for
fractional exponents, then
log Va = log a* = -| log a,
log v a = log a * = 1 log a,
and so on. This illustrates the truth of another theorem,
namely, that the logarithm of any root of a number is equal
to the logarithm of the number divided by the index of the root.
9
Thus, log V542 = J log 542 = ^~- = 1.3670.
Now 1.3670 is the logarithm of 23.28 -. Therefore the
square root of 542 is approximately 23.28 .
440 GENERAL MATHEMATICS
If the logarithm of the number is negative, the root
may be found as in the following examples.
Find by logarithms :
(a) V0.472. (b) ^0.472. (c) -V/O472.
Solution. Log 0.472 = 0.6739 1. Now if we attempt to take ^
of this negative logarithm, we shall obtain a fractional characteristic
that would be confusing. Therefore, in order to make it possible
to keep the mantissa positive and the characteristic (after the
division) an integer, we write
log 0,172 = 1!).<>73!> -20,
a number which the student can readily see is equal to 0.6739 1,
and which has the added advantage referred to above. We now get
(a) log Vo.472 .= * (19.6739 - 20) = 9.8369 - 10.
In like manner,
(b) log V 0.472 = i (29.6739 - 30) = 9.8913 - 10,
and () log v 0.472 = J (39.6739 - 40) = 9.9185 - 10.
In (a), above, log Vo.472 = 9.8369 -10. This means that
the characteristic is 1 and that the mantissa is 0.8369.
By reference to the table, we find that 0.8369 is the man-
tissa of the logarithm of 687. Since the characteristic is
- 1, the number lies between the minus 1st and the power
of 10 ; hence the decimal point comes just before the 6, and
V0.472 = 0.687. The student should check this result by
actually extracting the square root of 0.472 by the method
given in Art. 446.
EXERCISES
1. Find by logarithms :
V9604 ; V153.76 ; V0.000529 ; A/10648 ;
^4275 ; -v/3.375 ; ^0.001728.
2. Given a = 4.25, ft = 22.1, and c = 0.05, find by logarithms
I 72
the value of \| to three significant figures.
LOGARITHMS 441
. 182.41 ' , 3)2.42x35.1
3. Find by logarithms the value of V and -\\ ~^r^ T^
i 4.oo^ i i>. X J."
to three significant hgures.
4. The velocity v of a body that has fallen a distance of
,s- feet is given by the formula v = V2 </.->. If g = 32.16, what
is the velocity acquired by a body in falling 30 ft.?
5. If a bullet is shot upwards with an initial velocity of v,
the height .s to which it will rise is given by the equation
r = V2 gs. What must be the minimum velocity at the start
of a bullet fired upwards if it is to strike a Zeppelin 1500 ft.
high? (Assume y = 32.16.)
6. The time t of oscillation of a pendulum / centimeters long
is given by the formula t = 7r\| j- Kind the time of oscilla-
i j)oO
tion of a pendulum 78.22 centimeters long.
* 7. If the time of oscillation of a pendulum is 1 sec., what
is its length ?
*8. The area of any triangle is given by the formula
A = Vs (s a) (s b} (s c),
where a, b, and c are the sides of the triangle, and s equals
a + b 4- c
one halt the perimeter, or
&
Using the preceding formula, find by means of logarithms the
number of acres in a triangular field whose sides are 15, 38.2,
and 45.3 rods respectively. (1 A. = 160 sq. rd.)
*9. The area A of the cross section of a chimney in square
feet required to carry off the smoke is given by the formula
0.6 P
/~ '
V//
where P is the number of pounds of coal burned per hour, and
h is the height of the chimney in feet. Find out what the area
of a cross section of a chimney 80 ft. high should be to carry
off the smoke if 560 Ib. of coal are burned per hour.
442
GENERAL MATHEMATICS
* 10. The average velocity v of the piston head in inches per
second for a steam engine is given by the formula
where s denotes the distance (in inches) over which the piston
head moves, and p is the number of pounds of pressure of
steam in the cylinder. Find v if s = 30.24 in. and p = 115 Ib.
*11. In the equation x W 1J what is the value of x when
y = ? when y 1 ? when y = 2 ? when ?/= !?
*12. Fill in the following table for the equation x =10 ?/ .
y
i
2
-1
- 2
-3
i
I
i
4
5
X
i
10
100
NOTE. For y equaling ^, ^, 2^, etc., use the table of mantissas,
Art. 492.
* 13. Plot the results in the table of Ex. 12 and draw the curve,
thus obtaining the'graph for x = lO 2 ' (or y log x) (Fig. 300).
FIG. 300. GRAPH OF x =
*14. Show that the graph for x =10^ (Fig. 300) makes clear
the following principles :
(a) A negative number does not have a real number for its
logarithm.
LOGARITHMS 443
(b) The logarithm of a positive number is positive or nega-
tive according as the number is greater or less than 1.
(c) The greater the value of x, the greater its logarithm.
(d) As x gets smaller and smaller, its logarithm decreases
and becomes smaller and smaller.
*15. Find by the graph of Ex. 13 the logarithm of 2.25;
of 4.5 ; of 1.1 ; of 2.8.
*J6. Of what number is 0.35 the logarithm? 0.5? 0.42?
*17. Check your results for Exs. 15 and 16 by the results
given in the table of Art. 492.
* 500. Exponential equations. Instead of finding the log-
arithm of 1000 to the base 10, we could arrive at the same
result by solving the equation 10 r = 1000, for this equa-
tion asks the question, What power of 10 equals 1000?
In other words, What is the logarithm of 1000 to the
base 10 ? An equation like this, in which an unknown is
involved in the exponent, is called an exponential equation.
EXERCISE
Give five examples of exponential equations where 10 is
the base.
*501. Method of solving exponential equations. The sim-
plest exponential equations may be solved by inspection
just as the logarithms of many numbers can be given by
inspection. Where an exponential equation cannot be
solved readily by inspection, logarithms may be employed
to simplify the process. We will illustrate each case.
Solution I (by inspection).
(a) If 2 X = 4, then x = 2. (e) If 10?' = 100, then y - 2.
(b)If3 a; = 9, then* = 2. (f)IflO j; = 1000, then x - 3.
(c) If 2V = 8, then / = 3. (g) If 10 = 10,000, then y = 4.
(,1) Tf 3* = 81, then x = 4.
444 GENERAL MATHEMATK \S
Solution II (by using logarithms).
Solve the equation 2 X = 6 for x.
Taking the logarithms of both sides :
log 2 X = log 6,
or x log 2 = log 6.
. z = !2e = 02 = 2 58
log 2 0.3010
1 G.
The student must remember that ^ - is not equal to
6 log2
log - The first is a fraction obtained by dividing one
A
logarithm by another, and involves division ; the second
indicates that the logarithm of a fraction is to be found,
and involves subtraction.
EXERCISE
Solve the following equations for x :
(a) 2 X = 7. (b) 3 X = 5. (c) 4* = 10. (d) (1.12)* = 3.
502. Interest problems solved by logarithms. Some im-
portant problems in interest may be solved by means of
exponential equations and logarithms. The following
simple example will illustrate the principle:
In how many years will a sum of money double itself at 6%
if the interest is compounded annually ?
Solution. In one year $1 will amount to $1.06 ; in two years the
amount will be 1.06 x 1.06, or (1.06) 2 ; in three years the amount
will be (1.06) 3 ; and so on. Therefore in x years the amount of $1
will be (1.06) 3 '. Then, if the money is to double itself in x years,
the conditions of the problem will be represented by the equation
(1.06)'* = 2. Solving this equation for x, we get x = 12.3 (approx.).
Therefore a sum of money will double itself at 6% (compounded
annually) in about 12.3 yr.
LOGARITHMS 445
EXERCISES
1. Explain the solution of the problem given in Art. 502.
2. If the interest is compounded annually, in how many years
will a sum of money double itself at 3% ? 3|-% ? 4% ? 5% ?
3. In how many years will a sum of money treble itself at
4^ interest compounded annually ? semiannually ?
4. The amount of P dollars for n years at /%, compounded
annually, is given by the formula .1 =P(1 +/)". Find the
amount of $1200 for 10 yr. at 4%.
Solution. In this problem P = 1200, r = 0.04, n = 10.
^ = 1200(1 + 0.04)
The computation may be arranged as follows :
log 1200 = 3.0792
10 log 1.04 = 0.1700
log A = 3.2492
Therefore A = 1775, number of dollars in the
amount.
NOTE. As a matter of fact, this value of A is not exact, because
we are using only four-place tables. In practice the value of the
problem should determine the kind of tables used. The greater the
number of places given in tin' tables used, the greater the accuracy
of the result.
5. What will $5000 amount to in 5 yr. at 3%, interest
compounded annually ? semiannually? quarterly?
6. Approximately three hundred years ago the Dutch
purchased Manhattan Island from the Indians for $24. What
would this $24 amount to at the present time if it had been
placed on interest at 6%' and compounded annually?
7. What would he the amount of 10 at the present time
if it had been placed on 3% annual compound interest tifty
years ago ?
446 GENERAL MATHEMATICS
8. A boy deposited 300 in a savings bank on 3% interest,
the interest to be compounded annually. He forgot about his
deposit until fifteen years later, when he found the receipt
covering the original deposit. What did the 300 amount to in
the fifteen years ?
9. What sum will amount to $1600 in 10 yr. at 6%, interest
being compounded annually ?
10. What sum will amount to $ 2500 in 5 yr. at 3%, interest
being compounded annually ?
11. In how many years will $4000 amount to $8500 at 6%,
interest being compounded annually ?
12. What would be the amount to-day of 1 cent which
nineteen hundred and twenty years ago was placed on interest
at 6%, compounded annually ? Find the radius in miles of a
sphere of gold which has this value.
NOTE. A cubic foot of gold weighs 1206 pounds avoirdupois,
one pound being worth approximately $290.
The volume of a sphere is given by the formula V = 3 irr 3 , where
V is the volume and r the radius of the sphere.
No doubt the pupil is convinced of the value of log-
arithms as a labor-saving device in complicated arithmetic
computations. Since he will meet numerous opportunities
for applications, the lists of problems in the chapter are
brief, the aim being to give just enough illustrations to
make clear the meaning of the principles involved.
HISTORICAL NOTE. Logarithms were invented by John Napier
(1550-1617), baron of Merchiston in Scotland. His greatest purpose
in studying mathematics was to simplify and systematize arithmetic,
algebra, and trigonometry. The student should read about Xapier's
"rods," or "bones," which he designed to simplify multiplication and
division (Encyclopaedia Britannica, llth ed.).
It was his earnest desire to simplify the processes that led him
to invent logarithms ; and, strange as it may seem, he did not consider
a logarithm as an exponent. In his time the theory of exponents was
LOGARITHMS 447
not known. A Swiss by the name of Jobst Biirgi (1552-1632) may
have conceived the idea of logarithms as early or earlier than
Napier and quite independently of him, but he neglected to pub-
lish his results until after Napier's logarithms were known all over
Europe.
Henry Briggs (1561-1630), who, in Napier's time, was professor of
geometry in Gresham College, London, became very much interested
in Napier's work and paid him a visit. It is related that upon
Briggs's arrival he and Napier stood speechless, observing each other
for almost a quarter of an hour. At last Briggs spoke as follows :
" My lord, I have undertaken this long journey purposely to see your
person, and to know by what engine of wit or ingenuity you came
first to think of this most excellent help in astronomy, namely, the
logarithms, but, my lord, being by you found out, I w y onder nobody
found it out before, when now known it is so easy."
After this visit Briggs and Napier both seem to have seen the
usefulness of a table of logarithms to the base 10, and Briggs devoted
himself to the construction of such tables. For this reason logarithms
to the base 10 are often called Briggsian logarithms.
Abbott says that when Napoleon had a few moments for diversion,
he often spent them over a book of logarithms, which he always
found recreational.
Miller in* his " Historical Introduction to Mathematical Litera-
ture (p. 70) says : "The fact that these logarithms had to be computed
only once for all time explains their great value to the intellectual
world. It would be difficult to estimate the enormous amount of time
saved by astronomers and others through the use of logarithm tables
alone." (For further reading see Cajori's "History of Elementary
Mathematics," pp. 155-167. Consult also the New International Cy-
clopedia, which contains a great deal of excellent historical material.)
SUMMARY
503. This chapter has taught the meaning of the follow-
ing words and phrases : logarithm, characteristic, mantissa,
interpolation, and exponential equation.
504. The theory and practical value of logarithms has
been discussed in as elementary a way as possible so that
448 GEN KKA L MATHEMATK 'S
the student may be able to appreciate the value of this
powerful labor-saving device.
505. This chapter has taught the student four impor-
tant logarithmic formulas :
(a) log ab log a + log b* (c) log a u = n log a.
(b) log - = log a log b. (d) log Va =
o n
506. The position of the decimal point in any result
depends entirely upon the characteristic of the logarithm
of the number sought. Thus, two numbers having the
same succession of digits will have the same mantissa.
The mantissa of the logarithm of a number is always
positive; the characteristic may be either + or .
507. This chapter has taught methods of solving loga-
rithmic and exponential equations.
508. The student has been taught how to solve verbal
problems by means of logarithms, for example, the interest
problem.
CHAPTER XVIII
THE SLIDE RULE AND GEOMETRIC REPRESENTATION OF
LOGARITHMS PRACTICALLY APPLIED TO THE SLIDE RULE
509. General description of the slide rule. The theory of
the slide rule is based on the elementary ideas and prin-
ciples of logarithms, and anyone can learn to use the slide
rule who has the ability to read a graduated scale.
The slide rule may be used in nearly all forms of cal-
culation and is gradually coming into general use in
many different fields: to the practicing engineer and to
the student of the mathematical sciences it is invaluable ;
to the accountant and the statistician it is an instru-
ment of great service. The student will therefore find it
advisable to learn all he can about the actual use of this
important device.
In Chapter II we discussed the graphical method of
adding and subtracting line segments as an interpretation
of the addition and subtraction of numbers. Since to a
multiplication of two numbers there corresponds an addi-
tion of their logarithms, it is possible to obtain the loga-
rithm of a product graphically by adding line segments
whose lengths represent the logarithms of the factors. In
order to carry out this plan, a method of actually finding
a line segment whose length shall represent the logarithm
of a given number has been developed.
The slide rule is a mechanical device for determining
products, quotients, powers, and roots of numbers by
graphical addition and subtraction.
449
450
GENERAL MATHEMATICS
Mantissas of logarithms of numbers from 1 to 10 (which,
as we have seen, are the same for numbers from 100 to 1000
or from 1000 to 10,000) are laid off to a certain scale on
two pieces of rule (see Fig. 301) which are made to slide by
nr
FIG. 301. SLIDE KUL.K
each other so tliat the sums or differences of the logarithms
can be obtained mechanically.
The scale is numbered from 1 to 10 at the points which
mark the logarithms of the several numbers used. The com-
mon scale is 5 in. long and the common rule 10 in. long, so
that the series of logarithms is put on twice, and the number-
ing either repeated for the second set or continued from 10
A 1
2 34 68
B 1 2 34
C
D
FIG. 302
to 100. The most common form of the rule is the Mannheim
Slide Rule. On this rule, which is made essentially as shown
in Figs. 302 and 303, there are two scales A and B just alike
(A on the rule and B on the slide), and two other scales,
C and D, just alike (Z> on the rule and C on the slide).
The student will note that the distance from 1 to 2 on
scales A and B is the same as the distance from 2 to 4 and
451
from 4 to 8. This means that if we add the distance from
1 to 2 on scale B to the distance from 1 to 2 on scale A by
means of the slide B, we shall obtain the product 2x2,
or 4. In like manner, if we add the distance from 1 to
4 on scale B to the distance from 1 to 2 on scale A, we
shall obtain the product 4 x 2, or 8.
C and D differ from A and B in being graduated to a
unit twice as large as the unit to which A and B are
graduated, so that the length representing the logarithm of
a given number on C and D is twice as long as the length
A 1
D
FIG. 303
on A and B representing the. logarithm of the same number.
Therefore, any number on the lower rule or slide is oppo-
site its square on the upper rule or slide ; and if the upper
rule and slide be regarded as standard logarithmic scales,
the lower rule and slide will be standard logarithmic scales
of the squares of the numbers shown. The student can
easily understand from the preceding statement how the
square roots of numbers are found.
It should be said, however, that the values of the loga-
rithms themselves are not shown on the scales. What we
find is the numbers which correspond to the logarithms.
Each unit length on the scales (graduated lengths) repre-
sents equal parts of the logarithmic table. Thus, if the
logarithm of 10 be selected as the unit, then the logarithm
452
of 3, or 0.477, will be represented by 0.477 of that unit;
4 by 0.602 ; 5 by 0.699 ; and so on, as can be seen by
referring to the following table of corresponding values :
Number
1
2
3
4
5
6
7
8
9
10
Logarithm
0.301
0.477
0.602
0.699
0.778
0.845
0.903
0.954
1.000
The numbers between 1 and 2, 2 and 3, 3 and 4, and
so on, are represented on the logarithmic scales by inter-
mediate divisions, and the entire scale has been graduated
as closely as is possible for convenience in reading.
The preceding discussion should therefore make it clear
that at the ^y^th division along the scale on the slide rule
we should find 2 and not its logarithm, and at the ^nnj^h
division we should find 5 and not 0.699, and so on.
It is clear that this scheme eliminates entirely the
process of finding the numbers corresponding to certain
logarithms, as we had to do when we computed by means
of the table in Art. 492.
The student will observe, further, that the left index of
the scales (that is, the division marked 1) may assume
any value which is a multiple or a decimal part of 1 (for
example, 10, 100, 1000, 0.1, 0.01, 0.001, etc.), but when
these values are assumed, this same ratio must be held
throughout the entire scale. In this case the proper values
for the subsequent divisions of the scale in order would be
20, 30 ; 200, 300 ; 2000, 3000 ; 0.2, 0.3 ; 0.02, 0.03 ; 0.002,
0.003. It follows that as the value of the 1 at the begin-
ning of each scale varies any number such as 382 may
have the value 38200, 3820, 382, 38.2, 3.82, 0.382, 0.0382,
and so on.
If a number which the student has to read does not
come exactly at a graduation he must estimate the values
THE SLIDE RULE 453
as closely as possible ; for example, if a certain number were
indicated ^ of the way from 152 to 153, he would read
152.3, on the assumption, of course, that the left index of
the scale has the value 100. We shall now proceed to
show by specific examples how the slide rule is used.
510. Multiplication with the slide rule. All calculations
in multiplication, division, and proportion are worked out
on scales C and Z>, as by reason of the greater space allotted
to each of the divisions the results obtained are more
accurate. To find the product 2x3 set scale C so that
its left-hand index (the division marked 1) falls exactly
opposite the division marked 2 on scale D (see Fig. 302).
Then directly opposite the division marked 3 on scale C
we shall find on scale D the division marked 6, which is
the required product.
This process is justified by the fact that to log 2 on
scale D we add log 3 on scale (7, thus obtaining log 6
on scale D.
In general, to multiply any constant number a by another
number ft, set 1 of scale C opposite a of scale D and read the
product ab on scale D opposite b of scale C.
EXERCISES
1. Use the slide rule to find the products in the following
problems :
(a) 2 and 4 ; 2 and 5 ; 2 and 6 ; 2 and 7 ; 2 and 8 ; 2 and 9 ;
2 and 10.
(b) 3 and 2 ; 3 and 3 ; 3 and 4 ; 3 and 5 ; 3 and 6 ; 3 and 7 ;
3 and 8 ; 3 and 9 ; 3 and 10.
2. How would you find the product of 20 and 30 ?
NOTE. The Mannheim Slide Rule will enable us to secure results
correct to three significant figures, and in exceptional cases results
454 GENERAL MATHEMATICS
correct to even four significant figures may be obtained. However,
in the work of this chapter we shall be content if we make our com-
putation correct to two, or perhaps three, significant figures, because
in actual practice this is sufficient.
511. Division with the slide rule. To divide 6 by 3, set
3 of scale C opposite 6 of scale D (see Fig. 302) and read
the quotient 2 on scale D opposite 1 of scale C. This
process is justified from the fact that from log 6 we sub-
tract log 3, thus obtaining log 2 on scale D.
In general, to divide any constant number a by another
number 6, set b of scale C opposite a of scale D and read the
quotient - on scale D opposite 1 of scale C.
EXERCISE
1. Use the slide rule to find the quotients in the following
problems :
(a\ 4.6-8.10 /Vl 8 . 8 8 . 8 ( n\ 10- 10. 10- 10
W 2' 2' 2> ~2~" W f f ' * f* \- ) "3"' ~4~' ~~' TO*
512. The runner. Each slide rule is equipped with a
runner which slides along the rule in a groove and by
means of which the student is enabled to find more quickly
coincident points on the scales. It is also valuable to mark
the result of some part of a problem which contains several
computations. Thus, if
25.2 x 3.5 x 3.68
-22*-
we can compute 25.2 x 3.5 by one setting of the slide and
then bring the runner to 88.2, the result. We then bring
the index of the slide to 88.2 and multiply by 3.68 ; this
gives 324.6 (approx.). Bring the runner to this result
and divide by 22.6; the quotient should be 14.4 (approx.).
THE SLIDE RULE 455
The student can easily determine the use of the runner
in finding powers and roots after he has read the next
two articles.
513. Raising to powers with the slide rule. The student
will observe that if the logarithm of any number in the
table of Art. 492 be multiplied by 2, the logarithm thus
obtained will correspond to the square of that number.
Thus, if the logarithm of 2 (0.301) be multiplied by 2, the
result (0.602) is the logarithm of 2 2 , or 4. This is in
accord with the law of Art. 494 regarding the logarithm
of a number raised to a power.
In like manner, the logarithm of 2 multiplied by 3 is 0.903,
which is the logarithm of 2 3 , or 8, and so on. Since this same
relation holds for any number to any power, we may raise
a number to any power by using the slide rule as follows :
1. Squares of numbers.
To find the value of S 2 look for 3 on scale D and read
3? = 9 directly opposite 3 on scale A.
2. Cubes of numbers.
To find the value of 3 s , set 1 of scale B opposite 3 of
scale A and find 3 s , or 27, on scale A opposite 3 of scale C.
3. Fourth power of numbers.
To find the value of 3*, set 1 of scale C to 3 on scale D
and find 3*, or 81, on scale A opposite 3 of scale C.
4. Higher powers.
Higher powers of numbers may be found by a method
similar to the preceding, but we shall not need to discuss
these here, as most of our problems will deal with the lower
process of numbers.
514. Square roots found by means of the slide rule. To
find the square root of any number, bring the runner to
the number on scale A, and its square root will be found
456 GENERAL MATHEMATICS
at the runner on scale D exactly opposite the number.
This process is seen to be the exact inverse of finding the
square of a number.
If the number contains an odd number of digits to the
left of the decimal point, its square root will be found on
the left half of the rule ; if it contains an even number
its square root will be found on the right half. If the
number is a fraction, and contains an odd number of zeros
to the right of the decimal point, the root is on the left
half ; if it contains an even number (or no zeros) the root
is on the right half. If the student prefers he may deter-
mine the first figure of the root mentally and then find
the proper half of the rule to use by inspection.
EXERCISES
1. Find the square roots of the following numbers by means
of the slide rule and compare your results with those of the
table in Art. 449 : 169 ; 576 ; 625 ; 900 ; 2.25 ; 3.24 ; 1.96 ; 4.41.
2. What is the side of a square whose area is 784 sq. ft.?
515. Cube roots found by means of the slide rule. To
find the cube root of a number by means of the slide rule,
move the slide either from left to right or from right to
left until the same number appears on scale B opposite the
given number on scale A as appears on scale D opposite the
left or right index (division 1) on C. The number which
appears on both scales B and D is the cube root of the
given number. For example, to find the cube root of 125,
move the slide to the left till 5 on scale B appears opposite
125 of scale A, and 5 on scale D lies opposite 1 (the right
index) on scale C. Thus 5 is the cube root of 125.
A second method of finding the cube root is to invert
the slide (see Art. 517) and set 1 on scale C under the
THE SLIDE RULE 457
given number on scale A and then find the number on
scale B which lies opposite the same number on scale D.
This number will be the cube root sought. The student
should observe that the process of extracting the cube
root is the inverse of that of cubing a number.
EXERCISES
1. Find the cube roots of the following numbers by means
of the slide rule and compare your results with those of the
table in Art. 449 : 64 ; 125 ; 0.729 ; 2197 ; 0.001331.
2. Find one edge of a cube whose volume is 1728 cu. in.
516. Proportion. A great many problems in proportion
can be solved very easily by means of the slide rule.
The student will observe that if the left-hand indexes
of scales C and D coincide, the readings on the slide
are in direct proportion 1:1 to the opposite readings on
the rule.
In like manner, if the left index of C is made to coin-
cide with 2 on D, the ratio will be 1 : 2, and so on. Hence,
10 25
to find the fourth term in the proportion = , we
15 x
move the slide to the right until 10 on scale C is opposite
15 on scale D, and opposite 25 on C read off x = 37.5 on D.
In general, to find the fourth term in a proportion, set the
first term of the proportion on scale C to the second term on
scale D and find the fourth term (unknown) on scale D oppo-
site the third term on C.
517. Inverted slide. The slide may be inverted so that
scale C faces scale A. This gives inverted readings on
scale C which are the reciprocals of their coincident read-
ings on scale D, and vice versa. Thus, 3 on scale D lies
opposite 0.33 on scale (7, and vice versa.
458 GENERAL MATHEMATICS
The inverted slide is useful, as is shown in finding cube
roots and also in problems involving an inverse propor-
tion, as in the following example :
If 10 pipes can empty a cistern in 22 hr., how long will it
take 50 pipes to empty the cistern ?
Solution. Invert the slide and set 10 of scale B at 22 on scale D,
and opposite 5 on C find 4.4 (the result) on scale D.
518. Position of the decimal point. The student will be
able in most practical problems to determine the position
of the decimal point. If there is any considerable difficulty
in any later work he should consult some standard manual
on the use of the slide rule.
MISCELLANEOUS EXERCISES
Solve the following problems by means of the slide rule :
1. Find the product of 58.2 x 2.55; 33.4 x 75.6; 22.5 x
33.3 x 8.2 ; 0.12 x 0.09 x 0.003.
82.5 3.5 0.04
2. tind the following quotients: ; ; ; ;
35.3 x 75.5 . 7.2 x 83.5 x 0.09
22.8 3.6
3. Perform the indicated operations: 25 2 ; 3 3 3 ; 2 x 2 2 x 15 8 ;
12 4 ; 1.2 2 x 7.5 2 x 0.9 2 .
4. Extract the indicated roots: V2? V3; V5; V7; Vl2;
V576; VTO6; Vl37.2 ; -^8; -^15; -\/Tl2; A/64; ^1728.
5. Find the circumference of a circle whose diameter is 6;
5.5 ; 2.83. (See a slide-rule manual for a short cut.)
6. If 10 men can do a piece of work in 4 da., how long
will it take 5 men to do the work if they work at the
same rate?
7. What will be the cost of 13| ft. of rope at 3j$ per foot ?
THE SLIDE RULE 459
8. What is the volume of a cubical stone 6.3 ft. long, 4.5 ft.
wide, and 3.2 ft. high ?
9. What distance will a train travel in 10 hr. and 20 min.
at a rate of 30.5 mi. per hour ?
10. What is the interest on $5600 for 1 yr. at 3J% ?
at 5%? at 6%?
11. In how many hours will a train travel 756 mi/ if it
travels at the rate of 41.2 mi. per hour ?
12. How many miles in 2783 ft. ? in 17,822 ft. ?
13. Find the area of a circle whose diameter is 10 in.;
7.5 in.; 0.351 in. (Use A = 0.7854 d 2 .)
14. Find the volume of a cube one of whose edges is
3.57 in. ; of a cube one of whose edges is 82.1 in.
15. If a man can save $3120 in 26 mo., how many dollars
will he save at the same rate in 1 mo. ? 6 mo. ? 12 mq. ?
16. Find the mean proportional between 6 and 27.
17. If goods cost 550 a yard, at what price must they be
sold to realize 15% profit on the selling price ?
18. The formula for the area of the ring in Fig. 304 is
A = irR* Trr 2 . Since 2 r = d, this formula may be
written A- =- 1, or A= ~^J 2 ~
Using the last formula, find the area of the ring
when R = 8.5 in. and r =' 5.3 in. FIG. 304
19. Find the amount of $225 invested for 12 yr. at 6%
simple interest. (Use the formula found under logarithms or
refer to a slide-rule manual for a short cut.)
20. What force must be applied to a lever 5.2 ft. from the
fulcrum to raise a weight of 742 Ib. which is 1| ft. on the
opposite side of the fulcrum ?
460
GEXK 1 1 A L M AT H KM ATICS
21. A shaft makes 28 revolutions and is to drive another shaft
which should make 42 revolutions. The distance between their
centers is 60 in. What should be the
diameter of the gears ? (See Fig. 305.)
HINT. Refer to a slide-rule manual
or, better, develop the formula between
the diameter of the gears, the number
of revolutions, and the distance between
their centers, as follows : FIG. 305
28 circumferences of the large gear = 42 circumferences of the small
gear. Why '.'
Then 28 (60 - x) = 42 x (see Fig. 305).
28 60 = 70 x. Why ?
| = f.
Then apply the slide rule.
22. Since F = ^C + 32 is the equation representing the
relation between the Fahrenheit and centigrade thermometers,
we may compare readings of these two thermometers by the
following scheme :
c
Set 5
Below degrees centigrade
I)
to 9
Read degrees plus 32 equals Fahrenheit
W r hat then is F. when C. = 25 ? 18 ?
*23. Trigonometric applications are greatly simplified by the
slide rule. The solution of a formula
be sin^4 . .
like A = is readily obtained.
Sin A may be used directly as a factor
in performing the operation. Find the
area of the corner lot in Fig. 306.
6=84'
FIG. 306
*24. Find the value of the lot in Ex. 23 at $871.20 per acre.
NOTE. For numerous applications of the slide rule to practical
problems the student should consult a standard slide-rule manual.
THE SLIDE KULE 461
SUMMARY
519. This chapter has taught the meaning of the follow-
ing words and phrases: slide rule, runner, inverted slide.
520. The theory and practical value of the slide rule
have been discussed and illustrated so that the student
can get at least an elementary working knowledge of this
powerful labor-saving device.
521. The student has been taught how to use the slide
rule in multiplying, dividing, raising to powers, and ex-
tracting roots.
522. The student has been shown how problems in
proportion and many other verbal problems may be solved
by the slide rule.
523. The student has been referred to the slide-rule
manual for methods of solving the more difficult problems.
CHAPTER XIX
QUADRATICS; QUADRATIC FUNCTIONS; QUADRATIC EQUA-
TIONS; GRAPHS OF QUADRATIC EQUATIONS; FORMULAS
INVOLVING QUADRATIC TERMS
524. Quadratic-equation problem. An engineer increased
the speed of his train 2 mi. an hour and made a run of
180 mi. in 1 hr. less than schedule time. What was the
speed when running according to schedule ?
Solution. Let x = the ordinary rate of the train.
Then x + 2 = the rate after the increase.
180
= the schedule time.
x
180
= the time it takes after the speed is increased.
x + 2
180 180
Then = hi. A\ hv . ;
x x + 2
The L.C.D. is x (x + 2). Why ?
Multiplying through by x (x + 2) we get
180 (x + 2) = ISOx + x(x + 2).
180 x + 360 = 180 x + x 2 + 2 x.
360 = x* + 2x. Why ?
x 2 + 2x- 360 = 0. Why?
We are not able to simplify the equation o?+ 2# 360 =
further. The methods of the preceding chapters will not
solve it. In fact we appear to have come to the end of
the road. It is clear that the problem is solved if we
can find a value of x which will make the value of the
quadratic trinomial z 2 + 2 x 360 equal to zero.
462
QUADRATIC EQUATIONS
463
An equation in which the highest power of the unknown
is the second power is called a quadratic equation. Many
problems in geometry, science, and mechanics are solved
by quadratic equations. It is our purpose in this chapter
to develop the power to solve quadratic equations and to
apply quadratic methods to verbal problems. This process
will be illustrated by the solution of the given equation,
3? + 2 x - 360 = 0.
525. Quadratic function. The expression x* + 2 x 360
is a quadratic function of x, or a function of the second
degree ; with every change in the value of x the value of
the function x 2 + 2 x 360 changes. We shall get some
light on the question, What value of x will make the ex-
pression 3? + 2x 360 equal to ? by studying how the
value of the expression x 2 + 2 x 360 changes as we give
different values to x. This variation is best shown by
means of the graph.
526. Graph of a quadratic function. In order to under-
stand more about the graph of a quadratic function we
shall consider a few simple exercises.
INTRODUCTORY EXERCISES
1. Find the value of the function a- 2 + 2 a- 360 for each of
the following values of x : 0, 10, 10, - 15, 15, 20, 18, 19, 21.
2. Fill in the following table of corresponding values
for x and the function x 2 + 2 x 360 :
X
10
-10
+ 15
-15
+ 16
-16
+ 17
-17
J 2 + 2x-360
-360
-240
-280
x
+ 19
-19
+ 21
-21
+ 25
-25
+ 30
-30
x 2 + 2x-360
464
GENERAL MATHEMATICS
If we transfer the corresponding values of x and the function
x 2 + 2 x 360 from the table to a sheet of cross-section paper
f(t
\
1
1
\
I
f
V
i
\
1
f
v
i
\
1
\
\
1
f
v
1
^
f
v
1
*y
(
)
?,<
\
-i
(}
(i
1
(1
I
?
)
?,
n
'
f
\
f
^
f
V
J
s
/
s
f
s
,
/
s
^
^
FIG. 307. SHOWING THE GRAPH OF A QUADRATIC FUNCTION
so as to secure the points which correspond to these values,
we shall obtain a series of points which, when connected by
a smooth curve, will be like the curve in Fig. 307.
EXERCISES
1. From the graph (Fig. 307) determine how the value of
a-2 _j_ 2 x _ 360 changes as x changes from 25 to 20 ; from to
15 ; from to - 15 ; from - 25 to - 30 ; from + 25 to + 30.
2. What value of x will make the function or 2 + 2x 360
equal to 300? 200? 150? -250? -300?
The preceding exercises show us that the graph enables
us to see what value x must have in order that the expres-
sion a^+ 2 x 360 shall have a given value. The pupil will
recall that, in the speed problem with which we started, the
QUADRATIC EQUATIONS 465
one question we could not answer was, What value of x
will make the expression a^+ 2 x 360 equal to zero ? This
question is now easily answered. A glance at the graph
shows that the expression becomes at two places ; namely,
when a: = 18 or when a; = 20. Thus the equation a?+2x
360 = is satisfied by x = 18 or x = 20. Check these
values by substituting in the equation. Hence the speed
of the train running according to schedule was 18 mi. per
hour. We reject the 20 as meaningless. (Why ?)
527. The two solutions of a quadratic equation. In the
preceding article we rejected 20 as a meaningless solution.
Though a quadratic equation has two solutions, this does
not mean that every verbal problem that leads to a quad-
ratic equation has two solutions. The nature of the con-
ditions of the verbal problem may be such as to make one
or even both of the solutions of the quadratic impossible
or meaningless. When neither of the two solutions of the
quadratic is a solution of the problem it usually means
that the conditions of the problem are impossible, or that
the problem is erroneously stated. In fact it would be
easy to make up an arithmetic problem whose result
could not be interpreted ; for example, it would be diffi-
cult to jump out of the window 2i times. To decide
which solution, if either, meets the conditions stated in
a verbal problem, it is necessary to reread the problem,
substituting the solutions in the conditions of the prob-
lem, and to reject solutions of the equation which do
not meet the conditions.
528. How to solve a quadratic equation graphically. We
may now summarize the method of solving a quadratic
equation revealed in the discussion of Arts. 526-527 as fol-
lows : (1) Reduce the equation to the form ax 2 + bx + c = 0.
(2) Make a table showing the corresponding values of x and
460
G EN ER A L M AT 1 1 E M ATICS
the function ax 2 -f bx + c. (3) Transfer the data of the
table to squared paper and construct the curve represent-
ing the function ax 2 + bx + c. This curve shows the values
of ax 2 + bx + c which correspond to the different values
of x. (4) By inspection determine the points of the curve
where the expression is zero. The values of x for these
points are the solutions of the equation.
EXERCISES
Solve the following equations graphically, and check :
1. ic 2 - 9z-f 14 = 0.
Plot the function between the
limits to 12. This means con-
struct the table by letting x equal
the following values : 0, 1, 2, 3,
4 ... 12.
6. 4a 2
Plot from 2 to 5.
Plot from - 4 to + 2.
8. 100 x 2 - 5 x -495 = 0.
Plot from 5 to 4- 5.
9. 6a; 2 -17x = 20.
Subtract 20 from both mem-
2. z 2 -6z + 5 = 0.
Plot from - 1 to 7.
3. x 2 - 3x -10 = 0.
Plot from - 3 to 6.
4. x 2 -11 x + 24 = 0.
Plot from 1 to 10.
5. x*-llx + 25 = 0.
Plot from 1 to 10.
11. 9x 2 + 3x + 20
Subtract 2 x 2 + 2 x + 50 from both members of the equation.
529. The graph solves a family of equations. At this
point the student should note that a single graph may be
used to solve a whole family of equations. Thus, if we turn
to Fig. 307 we see that the curve for x 2 2x 360 can
bers and plot 6 x 2 -11 x- 20.
10. 2x 2 - 9 = 3x.
Subtract 3 x from both mem-
bers and plot the function
2:r 2 -9 -3z.
QUADBATIC EQUATIONS 467
be used not only to solve the equation x* + 2 x 360 = but
also to solve every equation of the type a? + 2 x 360 = c
(where c is some arithmetical number). For example, if we
ask what value of x will make x z + 2 x 360 equal to 100,
we can tell by looking at the curve that the answer is 20.5
or 22.5, and this is precisely the same as saying that the
two roots of the equation j? + 2 x 360 = 100 are 20.5
and - 22.5.
EXERCISES
Solve by the graph :
1. x 2 + 2x- 360 = 200. 6. x 2 + 2 x = 400.
2. ** + 2* - 360 =180.
Subtract 36Q
3. x- + 2 x- 360 = 400.
4. & + 2x - 360 = - 250. ? x 2 + 2x - 500 = 0.
5. x 2 + 2x- 360 = -360. Add 140. Why?
The last two exercises show that the graph solves all
equations in which x* + 2 x= c (some arithmetical number).
For we can write the given equation x 2 + 2 x 500 = in
the form x* + 2 x 360 = 140. This last form we are able
to solve at sight by the graph.
530. The parabola. The curve representing the func-
tion x*+ 2x 360 shown in Fig. 307 is called a parabola.
Study and discuss the general shape and symmetry of the
curve. Compare the curves you and your classmates have
drawn in the exercises of this chapter and see if you can
find a parabola in an earlier chapter of this text.
The graph of a quadratic function in one unknown is
a parabola. iWis a symmetrical curve. No three points of
the curve lie on a straight line. The parabola is a common
notion in physics and mechanics. Thus, the path of a
projectile (for example, a bullet) is a parabola. A knowl-
edge of the theory and application of many such curves
468 GENERAL MATHEMATICS
was of extreme importance in the recent world war. The
soldiers who had been trained in some of the more advanced
mathematical courses, especially in trigonometry and graphi-
cal work, were in demand and were given plenty of oppor-
tunity to put into practice what they had learned in school.
In plotting functions like x? + 2 x 360 we plot the
values of x along the a>axis and the corresponding values
of the function on the ?/-axis. This suggests that the
curve obtained in Fig. 307 is the graph of the equation
y = z 2 + 2 a: - 360.
It follows that whenever y is a quadratic function of x,
or when x is a, quadratic function of y, the graph of the
equation is a parabola.
EXERCISES
Graph each of the following equations :
1. y = a; 2 -4. 3. z = y 2 + 5 ?/ + 4.
2. y = x 2 + 3 x + 2. 4. x = f - 1 y + 6.
531. Maxima and minima. The theory of maxima
(greatest values) and minima (least values) of functions
has many important applications in geometry, physics, and
mechanics.
This article will present one example drawn from each
subject. A careful study of the following example will
suggest the proper method of attack.
ILLUSTRATIVE EXAMPLE
A rectangular garden is to be inclosed on ^iree sides, the
fourth side being bounded by a high wall. WlJft is the largest
garden that can be inclosed with 20 rd. of fencing ?
Solution. Let x represent the width.
Then 20 2 x represents the length,
and 20 x 2 x 2 represents the area.
469
We are now interested in a maximum (greatest possible) value
that 20 x 2 x 2 can have. By trial we obtain the corresponding
values for x and the function 20 x 2 x 2 shown
in the table below.
Common sense, the table, and the curve of
Fig. 308 show us that if the garden is made
very wide or very narrow the area is very small.
The table and the curve of Fig. 308 suggest that
50 is probably the largest area. In this case the
dimensions of the garden are 5 and 10. By
taking x first a little larger and then a little
smaller than 5, we may check our conclusion.
We can as a matter of fact, save our-
X
20z 2x 2
1
18
2
32 .
3
42
4
48
5
50
6
48
7
42
10
selves much of the labor of these computations by an alge-
braic method, which we shall present in Art. 538. At this
-/(*>
-20
FIG. 308. SHOWING THE MAXIMUM VALUE or A QUADRATIC FUNCTION
stage, however, we shall be content to plot the curves and
find the highest or the lowest point on the curve.
470 GENERAL MATHEMATICS
EXERCISES
1. If a ball is thrown upward with a velocity v , the dis-
tance d from the earth to the ball after a given time t is given
by the physics formula
d = v t- 16 1 2 .
How high will a ball rise which is thrown with an initial
velocity of 100 ft. per second ?
HINT. The formula becomes d = 100 1 16 < 2 . Plot the function
and find by inspection its greatest value.
2. Divide 10 into two parts such that their squares shall be
a minimum.
*3. Find the most advantageous length of a lever for lifting
a weight of 100 Ib. if the distance of the weight from the
fulcrum is 2 ft. and the lever weighs 4 Ib. to the foot.
532. Limitations of the graphic method of solving quad-
ratic equations. By this time the student is no doubt con-
vinced that the graphic method of solving quadratic
equations has its limitations. We may enumerate the
following: (1) The results are frequently rough approx-
imations. This is evident the moment we attack problems
of some slight difficulty. In fact the earlier problems of the
chapter are artificially built so that in all probability the
student will accidentally get an accurate result. We must
remember that the graphic method depends for its accuracy
upon the mechanical (or nonintellectual) conditions, such
as the skill of the student at this type of work, the exact-
ness of squared paper, and our ability to estimate fractional
parts of the unit. (2) Aside from the fact that the signifi-
cance of a graph is sometimes obscure, the work is a bit
cumbersome and tedious. (3) It is not economical of time,
as we shall presently show.
QUADRATIC EQUATIONS. 471
533. More powerful methods of solving quadratic equa-
tions. Because of the foregoing limitations of the graphic
method we are ready to proceed to the study of more
efficient methods. These methods rest purely on an intel-
lectual basis (that is, the accuracy is independent of
constructed figures). We shall observe that they get the
results quickly and with absolute accuracy.
534. Quadratic equations solved by factoring. The factor-
ing method may be illustrated by the following solution
of the speed problem with which we opened the chapter :
Solution. Given x 2 + 2 x - 360 = 0.
Factoring the left member,
(x + 20) (x - 18) = 0.
The preceding condition, namely, (a; + 20) (z 18) = 0, will be
satisfied either if x + 20 = or if x 18 = 0, for we learned in
Art. 236 that the product of two numbers is zero if either factor
is zero. Thus, 5 x = or x 8 = O/
Now if x + 20 = 0,
then x - 20.
And if x - 18 = 0,
then x - 18.
Hence x - + 20,
or x-- 18.
In the next solution we shall omit a considerable part of the dis-
cussion and show how the work may be arranged in a few simple
statements.
Solve z 2 - 9 x + 14 = 0.
Factoring, (x - 7) (x - 2) = 0. (1)
This equation is satisfied if x 7 = 0, (2)
or if x - 2 = 0. (3)
From equation (2) x = 7,
and from equation (3) x = 2.
The numbers satisfy the equation, consequently 2 and 7 are the
roots of the equation x' 2 9 x + 14 = 0.
472 GENERAL MATHEMATICS
EXERCISES
Solve the following quadratic equations by the method of fac-
toring and test the results by substituting in the equations :
1. x 2 - 5x + 6 = 0. 15. 20x - 51 = x 2 .
2- if -7 y +12 = 0. 16. 77+4d = d 2 .
4. x 2 - 6x-27=0. 18.10^ + 33^=7.
. 5. x 2 -2x-35 = 0. 19. 6z 2 =23z + 4.
6. cc 2 + 5x = 6. x I I
20. + =
HINT. Subtract 6 from '2 2 x
both members before apply- 3
ing the method. Why? 21> 15a; + 4 = ^'
7. ar + a; = 56. r _ jt 33
o-l
HINT. Subtract 2x from oo K ~ Q _.
&O> t-J *)U O ^~~
both members and rearrange x
terms before factoring. y 15 ?/ 2
9. ^ + 4 = 42,. " 2 + T" = 14'
10. x 2 - 85 =12 a-. 25.-^-+^ =4.
11. 2 =10^ + 24. ~ y
12. ?/? 2 91 = 6 w. 26 - 2 = "3 a + 3'
XO* i-C ~~ r~ 3C - -r*-. /y or
07 x _ i _ o
14. m 2 +112 = 23m. ' x-3 a; + 3
VERBAL PROBLEMS
Solve the following problems by the factoring method and
test the results by substituting the solution in the conditions
of the problem :
1. A crew rows across a calm lake (12 mi. long). On the
return trip it decreases the rate by 1 mi. per hour and makes the
trip in 7 hr. Find the rate of the crew both going and returning.
QUADRATIC EQUATIONS
473
2. A man drives 'a car 80 mi. out of town. On the return
trip he increases his rate 8 mi. per hour. He makes the trip
in 4^ hr. What was his rate while driving out ?
3. The base of a rectangle exceeds the altitude by 5 in. The
area equals 150 sq. in.
Find the
altitude.
base and
40
-90-2W-
T
I
40 -2W
f
W
I
90
FIG. 309
4. The base of a
triangle is 7 in. less
than the altitude. The
area equals 85 sq. in.
Find the base and
altitude.
5. A farmer is plowing a field of corn 40 rd. wide and 90 rd.
long (Fig. 309). At the end of a certain day he knows that he
has plowed five sixths of the field. How
wide a strip has he plowed around the
field?
6. A piece of tin in the form of a
square is taken to make an open box.
The box is made by cutting a 1-inch
square from each corner of the piece of
tin and folding up the sides (Fig. 310).
The box thus made contains 36 cu. in.
Find the length of the side of the original piece of tin.
535. Limitations of the factoring method. In some of the
preceding exercises the quadratic expressions were very
difficult to factor. This is usually the case when the
constant terms are large numbers. Indeed, most verbal
problems lead to quadratic equations which cannot be
solved by the factoring method. The following problem is
a simple illustration:
FIG. 310
474 GENERAL MATHEMATICS
What is the length of a side of a square (Fig. 311) whose
diagonal is 2 ft. longer than a side ?
Attempted solution by factoring method:
Let x = a side of the square.
Then x + 2 = the length of the diagonal, x
By the theorem of Pythagoras
x z + x 2 = x 2 + 4 x + 4.
Simplifying, & - 4 x - 4 = 0.
This appears to be the end of the road ;
we cannot factor x 2 4 x 4, for we cannot obtain a combination
of whole numbers or fractions whose product is 1 and whose sum
is 4. And yet we are probably convinced that such a square does
exist though we are forced to admit that the solution of the problem
by the factoring method is hopeless.
536. Solution of the quadratic equation by the method of
completing the square. If we were able to make the left
member of the equation z 2 4 a; 4 = a perfect square
without introducing the unknown (#) into the right mem-
ber, we could take the square root of each member of the
equation and thus obtain a linear equation which would
be easily solved. This is precisely the method we wish to
employ. However, we must first learn to make the left
member x 2 - 4 x 4 a perfect square.
ORAL EXERCISES
1. Find (x + 2) 2 ; (x + 3) 2 ; (x + 4) 2 ; (a; - 2) a .
2. When is a trinomial a perfect square ? (See Art. 250.)
3. Make a perfect square trinomial of the following: a- 2 6 x;
7 x
x 2 + 7 x ; x 2 + 9 x ; x 2 +
QUADRATIC EQUATIONS 475
The preceding exercises show that it is easy to complete
the square of a binomial of the form x 2 + ax, for we need
only to add the square of half the coefficient of x. Then, too,
the constant term of a trinomial can always be made to
appear in the right member of the equation, leaving the
left member in the form x 2 + ax. We now proceed , to
solve the equation x 2 4 # 4 = 0. Write the equation
/ 4\ 2
thus: X s 4 # = 4. Add (5-) or 4, to make the first
\ 2 /
side a trinomial square, and we obtain
x z _ 4 x _|_ 4 = 8.
Taking the square root of both sides and remembering
that 8 has two square roots, + V8 or V8, we get
(1)
or ar-2=-V8. (2)
From equation (1) we get x = 2 + V8, and from equation
(2) we get a:=2_-V8.
If we obtain V8 either by the arithmetical method taught
by Art. 446 or by using the table of Art. 449, the result
(accurate to three places) is 2.828. Then x= 2 + 2.828, or
4.828. Hence the side of the square whose diagonal is 2 ft.
longer than a side is 4.828 ft. We can check this result by
applying the theorem of Pythagoras.
We reject 2 V8, or 0.828, because it does not satisfy
the conditions of the problem. However, the student should
realize that 0.828 is just as much a solution of the
equation z 2 4 x + 4 = 8 as is 4.828.
The method of completing the square is further illus-
trated by the following solution of the equation
10 z 2 - 9* + 2 = 0.
476 GENERAL MATHEMATICS
Write the equation 10 x 2 - 9 x = - 2. Why '?
9 x 1
Dividing by 10, z 2 - = - - .
1U 5
Note that, the left member is now more easily completed. Why ?
x _
10
x2 _9 + ^L-.J_. Why?
10 + 400 400
Taking the square root of each member,
x - & = 5 V
Whence x = % or f .
537. Summary of the method for solving quadratic equa-
tions by the method of completing the square.
1. /Simplify the equation and reduce to the form ax* +
bx = c.
2. If the coefficient of x 2 is not 1, divide both members of
the equation by the coefficient so that the equation takes the
form x 2 +px = q.
3. Find half the coefficient of x; square the result ; add
the square to both members of the equation obtained in step 2.
This makes the left member a perfect square.
4. Express the right member in its simplest form.
5. Take the square root of both members, writing the double
sign before the square root in the right member.
6. Set the left square root equal to the positive root in the
right member of the equation in 5. Solve for the unknown.
This gives one root.
7. Repeat the process, using the negative root in 5. This
gives the second root of the equation.
8. Express the roots first in simplest form.
QUADRATIC EQUATIONS 477
EXERCISES
Solve by the method of completing the square, and check :
' 1. x a -6 = 91. 11. 4z 2 + 45z-36 = 0.
2. x 2 -8a; = 48. 12. 6z 2 + 7cc-20 = 0.
3. X 2_ x _3 = o. 13. 2 2 + 62=l.
4. y 3 _|_ 4y _)_ 3 = 0. HINT. Compute roots to the
- o i A K n nearest hundredth.
5. if + 4y 5 = 0.
6. /,* _(- 8 b - 20 = 0. 14. x 2 + 4 a = 16.
7. y 2 +14^-51 = 0. 15- z 2 = 24 + 4a:.
8. m 2 + 5 wi - 6 = 0. 16. 7 + 2x = a; 2 .
9. a; 2 -13z + 40 = 0. 17. 75-3x 2 =75.
10. x 2 + 6x + 5 = 0. 18. 19 - a = 4 a 2 .
VERBAL PROBLEMS
1. If 4 is taken from a certain number the result equals 96
divided by that number. Find the number.
2. Find the two consecutive numbers the sum of whose
squares equals 113.
3. In physics we learn that the distance in feet which a
stone thrown downward goes in a given time equals 16
multiplied by the square of the number of seconds it has fallen,
plus the product of the velocity with which it is thrown and
the number of seconds fallen ; that is, s = vt + 16 1*. Suppose
that v = 20 ft. per second and s = 1800 ft. Find the value
of t. Try to state the meaning of this problem in simple
(nontechnical) words.
4. How long will it take a baseball to fall from the top of
the Washington Monument (555 ft.) if it starts with a velocity
of 50 ft. per second ?
HINT. Solve the equation 16 t 2 + 50 t = 555.
478 GENERAL MATHEMATICS
5. How long will it take a body to fall 800 ft. if it starts at
20 ft. per second ?
6. How long will it take a bomb to fall from a Zeppelin
1000 ft. high if it starts with no initial velocity ?
7. Two trains are 175 mi. apart on perpendicular roads and
are approaching a crossing. One train runs 5 mi. an hour faster
than the other. At what rates must they run if they both
reach the crossing in 5 hr. ?
NOTE. The 175 means the distance along the track.
8. The circumference of the fore wheel of a carriage is less
by 3 ft. than the circumference of the hind wheel. In traveling
1800 ft. the fore wheel makes 30 revolutions more than the
hind wheel. Find the circumference of each
wheel.
9. A window (Fig. 312) in the form of a
rectangle surmounted by a semicircle is found
to admit the most light when the width and
height are equal. If the area of such a window p
is 175 sq. ft., what is its width ?
10. A boy has a piece of board 16 in. square. How wide a
strip must he cut from each two adjacent sides to leave a
square piece whose area is three fourths that of the original
piece ? In what form would you state your result to meet all
practical purposes ?
11. A lawn is 30 ft. by 80 ft. Two boys agree to mow
it. The first boy is to mow one half of it by cutting a
strip of uniform width around it. How wide a strip must
he cut?
12. A farmer has a field of wheat 60 rd. wide and 100 rd.
long. How wide a strip must he cut around the field in order
to have one fifth of the wheat cut ?
QUADKATiC EQUATIONS 479
13. In a circle of radius 10 in. the shortest distance from
a given point on the circumference to a given diameter is 8 in.
Find the segments into which the
perpendicular from the point divides
the diameter.
HINT. Study Fig. 313 and try to re-
call the various theorems we have proved
involving mean proportional. If you fail
to get a solution, refer to the mean pro-
portional construction (Art. 374).
14. A broker sells a number of- rail-
way shares for $600. A few days
later, the price having risen f 10 a share, lie buys for the same
sum three less shares than he sold. Find the number of
shares transferred on each day and the price paid.
15. A boy sold a bicycle for $24 and lost as many per cent
as the bicycle had cost him in dollars. Find the cost.
16. A line 20 in. long is divided into two parts, A C and CB,
so that AC is the mean proportional between CB and AB.
Find the length of A C.
*538. Maxima and minima algebraically determined. We
have seen in Art. 531 that the graph of a quadratic func-
tion may be used to determine the maximum or minimum
values of the function. This method is not as exact, how-
ever, as the algebraic method of determining maxima and
minima, and it is longer. Take, for example, the problem
of Art. 531 to find the maximum value for 20 x 2 a- 2 .
If we represent this maximum value by m, then we may
write 20 x 2 a- 2 = m. If we divide both sides of the
equation by 2 and complete the square on the left side
of the equation,
(x - 5)2= 25 -|.
480 GENEKAL MATHEMATICS
It is evident from this equation that if x is a real
number, m cannot be greater than 50. Therefore the maxi-
mum value of m, or of 20 x 2x*, is 50.
In like manner by letting m represent the minimum
value of a function we can determine when in is a mini-
mum more quickly and more accurately than we can by
the graphic method.
EXERCISES
Determine the maximum or minimum values of the follow-
ing functions :
1. 3x a -4ar-l. 3. ->x*-x-l. 5. x* - 6x + 8.
2. 2 + 2Z-2X 2 . 4. 1-x 2 . 6. 6-x-x 2 .
SUMMARY
539. This chapter has taught the meaning of the follow-
ing words: parabola, maxima and minima.
540. This chapter has taught three methods of solving
a quadratic equation of one unknown: the graphical
method, the factoring method, and the method of com-
pleting the square.
541. The graphical method proved to be the most con-
crete. It presented a clear picture of the changes hi the value
of the function which correspond to changes in the value of
the unknown. It also served as a sort of " ready reckoner."
542. The factoring method was more expedient; the
results were obtained by this method much more quickly
and with greater accuracy.
543. The method of completing the square was used
to solve quadratic equations which were not solvable by
the method of factoring.
544. Both the graphical and algebraic methods of de-
termining maxima and minima were presented.
INDEX
Absolute value, 156, 161
Acute angle, 49
Addend, 39
Addition, algebraic, 162
commutative law, 38
geometric, 36
law of, 4
of monomials, 160-163
order of terms in, 38
of polynomials, 168
of positive and negative numbers,
160-169
Adjacent angles, 60
Ahmes, 88
Ahmes Papyrus, 360
Algebra, origin of word, 8
"shorthand" of, 13
Alloy problems, 338
Altitude, 67
Angle, acute, 49
bisection of, 68
central, 52
complement of, 119
construction of, 56, 62
definition, 47
degree of, 55
of depression, 352
of elevation, 350
initial side of, 48
left side of, 114
measurement of, 54, 66
negative, 159
notation for, 50
obtuse, 49
positive, 159
Angle, reflex, 49
right, 48
right side of, 114
straight, 48, 111, 112
supplement of, 116, 118
symbols for, 48
terminal side of, 48
vertex of, 47
Angles, adjacent, 60
alternate exterior, 127
alternate interior, 124
comparison of, 59
complementary, 119
corresponding, 68
exterior, 139
geometric addition of, 61
geometric subtraction of, 61
interior, 125, 130
positive and negative, 159
size of, 48
supplementary, 116, 118
supplementary adjacent, 111,
116
vertical, 122
Arc, degree of, 65
intercepted, 52
Area, measurement of, 74
of a parallelogram, 79
practical method of estimating, 74
of a rectangle, 75
of a rhombus, 81
of a square, 78
of a trapezoid, 82
of a triangle, 81
unit of, 74
481
482
(JKNKKAL .MATIIK.MATK'S
Areas, calculating, 205
proportionality of, 341
Arithmetic average, 244, ii4o, 247
Arrangement, 167
Ascending powers, 167
Axioms, 21, 22, 37
Bar diagram, construction of, 224
interpreting, 222
Base, 102
Beam problems, 336, 385
Bearing, of a line, 353
of a point, 354
Bhaskara, 108
Binomial, cube of, 416
geometric square of, 91
square of, 92
Bisector, construction of, 66, 68
perpendicular, 66
Braces, 175
Brackets, 175
Briggs, Henry, 59, 447
Cartograms, 230
Centigrade, 289
Central tendencies, measures of,
244
Characteristic, 428
table of, 437
Circle, arc of, 52
center of, 52
circumference of, 52
construction of, 51
definition of, 51
degree of arc of, 53
diameter of, 52
quadrant, 52
radius of, 52
semicircle, 52
Class interval, 239
Class limits, 240
Coefficient, 39
Coincide, 34
Commutative law, 38, 85
Compasses, 31
measuring segment, 32, 33
Compensating errors, 255
Complement of an angle, 119
Consecutive-number problems, 13
Constant, 301
Coordinates, 265, 368
Corresponding parts, 346
Cosine, 357, 358, 361
Cube, 98
Cube root, of arithmetical num-
bers, 420
by slide rule, 456
by table, 397, 398
Cumulative errors, 255
Curve, normal distribution, 257
skewness of, 260
symmetry of, 259
Data, 214
Decagon, 44
Decimal point, logarithms, 435
Degree, of angle, 53
of arc, 53
of latitude, 53
of longitude, 53
of a number, 166
Dependence, 300
Dependent variable, 300
Depression, angle of, 352
Descartes, 108
Descending powers, 167
Difference, of monomials, 42
of two-line segments, 38
Direct variation, 308
Dissimilar terms, 7
Distance, 280
Division, checking long, 209
INDEX
483
Division, definition of, 194
law of signs in, 195
of monomial by monomial, 105
of negative numbers, 194
of polynomial by monomial, 197
of polynomial by polynomial, 207
with slide rule, 454
by zero, 211
Drawing to scale, 345-355
Elements of geometry, 88
Elevation, angle of, 350
Elimination, 373
by addition or subtraction, 374,
375
by substitution, 377
summary of methods of, 379
Equal segments, 34
Equation, checking, 6
definition and properties, 12
laws for solving, 2-5, 9
members of, 2
number satisfies, 6
quadratic, 462
root of, 7
substituting in, 6
translation of, 12
Equations, contradictory, 371
dependent, 372
equivalent, 372
exponential, 443, 448
identical, 372
inconsistent, 372
indeterminate, 370
outline of systems, 373
pupils' test of simple, 243
simultaneous linear, 367, 369
system of, 369
systems containing fractions, 381
in two unknowns, 373
Equiangular, 149
Equilateral, 44
Euclid, 88
Evaluation of formulas, 93, 279,
290
Exponent, indicates degree, 166
logarithm as an, 427
Exponents, 102, 425
fractional, 412
negative, 415
zero, 414
Extremes, 323, 328
Factors, 198
common monomial, 198
"cut and try," 199
of difference of two squares, 202
equal, 102
prime, 200
of trinomial square, 202
Fahrenheit, 289
Fallacious proof, 211
Formula, definition, 78, 273
evaluating, 93, 279, 290
interest, 273
motion problem, 279
solving, 276, 294
summary of discussion, 279
translating into, 287
work problem, 283-285
Fourth proportional, 334
construction of, 334
Frequency table, 239
Fulcrum, 336
Function, 359
defined, 299
dependence, 300
graph of, 301
graphical solution, 304
linear, 303
quadratic, 463
Functions, trigonometric, 359
484
GENERAL MATHEMATICS
Geometric problems, 376
Geometry, origin of word, 88
Graph, 215 '
of centigrade, 290
of constant-cost formula, 261
of data, 214
of Fahrenheit, 290
of functions, 301
of linear equations, 263-269
of quadratic equations, 463-469
of simultaneous linear equations,
369
of variation, 308-311
of x = 10*, 442
of y = x 2 , 391
Graphic curve, construction of, 233
interpreting, 231
Graphing, terms used in, 267
Hexagon, 44
Hipparchus, 360
Hyperbola, 311
Identity, 204
Independent variable, 300
Index, 420
Indirect measurement, 345-355
Inequality, 34
Inertia of large numbers, 255
Intercepted arc, 52
Interest formula, 273-275
involving amount, 277
Interest problem, graphical solu-
tion of, 276
solved by logarithms, 444
Interpolation, 437
Intersection, point of, 26
, Inverse variation, 309-311
Is8sceles triangle, 138
Joint variation, 312
Labor-saving devices, 424
Latitude, 63
Law of the lever, 336
Laws, 21, 38, 85, 179, 196, 336
Least common denominator, 324
Least common multiple, 11
Length, 26
measurement of, 27
units of, 28
Lever arm, 181
Line, bearing of, 353
of sight, 351
Line segments, 26
difference of two, 38
equal and unequal, 34
ratio of, 35
sum of, 36
Linear equations, definition, 266
graphic solution of simultaneous,
369
Locus, 266
Logarithm, of a power, 433
of a product, 432
of a quotient, 434
Logarithms, 424, 450, 451
applied to slide rule, 449
historical note on, 446
notation for, 427
position of decimal point, 435
Longitude, 53
Mannheim slide rule, 450, 453
Mantissas, 428
table of, 430-431
Maxima, 468-469, 480
Mean proportional, 329
Means, 323, 328
Measurement, of angles, 54
of areas, 74
errors in precise, 30
indirect, 345-355
INDEX
485
Measurement, of line segments, 27
of volumes, 99
Median, 66, 250
construction of, 67
how to determine, 251
Members of an equation, 2
Metric system, 28
advantages of, 29
historical note on, 29
Minima, 468-469, 480
Minutes of angle, 53
Mixture problems, 338, 379
Mode, 248
advantages of, 249
disadvantages of, 249, 250
Monomial, degree of, 167
factors of, 198
Monomials, division of, 195
multiplication of, 184
sum of, 177
Motion, circular, 284
Motion problems, 279, 384
graphical illustration of, 283
Multiplication, abbreviated, 95
algebraic, 89-90
by balanced bar, 180
commutative law of, 85
law of signs in, 179
of monomials, 184
of a polynomial by a monomial,
186
of positive and negative num-
bers, 178-194
several factors, 184
with slide rule, 453
special products, 192, 193
of two binomials, 190
of two polynomials, 187, 188
by zero, 183
Napier, John, 446
Negative number, 151
Newton, Isaac, 418
Normal distribution, 257, 268
Notation, for angles, 50
for logarithms, 427
for triangles, 130
Number, absolute value of, 156
algebraic, 151
degree of, 167
literal, 6
negative, 150-157, 178-180
numerical value of, 156
positive, 150-154
prime, 198
Number scale, 170
Number-relation problems, 378
Numbers, difference of algebraic,
177
ratio of, 35
representation of, 152-153
sum of algebraic, 177
Obtuse angle, 49
Octagon, 44
Opposite angles, 122
of parallelogram, 142
Order of powers, 167
Parabola, 467
Parallel lines, construction of, 69
definition, 68
Parallelepiped, 99
volume of oblique, 100
volume of rectangular, 99
Parallelogram, defined, 70
opposite angles, 142
Parenthesis, 175-176
Partial products, 89
Pentagon, 44
Perfect trinomial square, 393
Perigon, 48
480
GENERAL MATHEMATICS
Perimeter, 43
Perpendicular, 65
bisector, construction of, 66
Pictograms, 214-220
Point, bearing of a, 354
determined by, 26
Polygons, classified, 44
equilateral, 44
similar, 318
Polynomial, 40
degree of, 167
Polynomials, addition of, 168
classified, 40
division of, 208
multiplication of, 187
subtraction of, 173
Positive numbers, 151-154
Power, 103
Powers, ascending, 167
descending, 167
table of, 397, 398
of ten, 425-426
Prime number, 198
Problems, alloy, 338
beam, 337, 385
clock, 284
consecutive-number. 13
digit, 378-379
geometric, 376
interest, 273
lever, 336
mixture, 338, 379
motion, 279, 384
number-relation, 378
recreation, 386-388 ,
specific-gravity, 340
work, 285
Product, accuracy of, 93
geometric, 89
law of order, 85
monomial, 85
Product, partial, 89
of a polynomial and a monomial,
86
of powers, 103
of two polynomials, 187
Proportion, 322
beam problems, 337, 385
different arrangements of, 327
mean proportional, 329
means and extremes of, 323, 328
mixture and alloy problems, 338
specific-gravity problems, 340
test of proportionality, 324, 328
Proportional, constructing a mean,
332
fourth, 334
inversely, 309
Protractor, 54
Pyramid, frustum of, 98
triangular, 98
Pythagoras, 401
historical note on, 402
theorem of, 397, 399, 400
Quadrant, 52
Quadratic equation, 463
Quadratic equations, completing
the square, 475-477
factoring method, 471-474
graphic method, 465
two solutions of, 465
Quadratic function, 463
graph of, 463
Quadratic surd, 392, 406-407
Quadratic trinomial square, 392
Quadrilateral, 44
Quotients in per cents, 335
Radical expression, 420
Radical sign, 390
large number under, 406
INDEX
487
Radicand, 390
Rate, 280
Ratio, 35, 315, 345
trigonometric, 359
Rationalizing denominator, 409
Rectangle, 70
Reflex angle, 49
Regiomontanus, 360
Removal of parenthesis, 175-177
Rhombus, 81
Right angle, 48
Right triangle, 135
Right triangles, similar, 355
sum of acute angles in, 137
Root of an equation, 6
Roots, fractional exponents, 412
higher, 420
by logarithms, 439
table of, 397-398
Scale drawings, 345-355
Seconds, of an angle, 53
Sector, 217
Segment, line, 26
unit, 27
Semicircle, 52
Series, continuous and discrete, 237
Signs, 151
change of, 171
law of, 179, 195
minus, 151-152
plus, 152
Similar terms, 40, 42
Similar triangles, 315
construction of, 314-316
corresponding sides of, 315, 330
Similarity, 345
Simultaneous equations, 367, 369
Sine of an angle, 357
Slide rule, 449
decimal point, 458
Slide rule, index of, 452
inverted slide, 457
proportion problems, 457
raising to powers, 455
runner, 454
Solids, geometric, 98, 99
Solving equations, 6
Specific gravity, 340
Sphere, 98
Square, 71
of the difference, 192
of the sum, 192, 193
trinomial, 202
Square root, 390-397, 451
algebraic rule for finding, 396
of a fraction, 408, 409
by graphic method, 391
by logarithms, 439-440
by mean proportional, 405
memorizing, 408
of a product, 406
by ruler and compass, 404
by slide rule, 455
table of, 398
Squared paper, 32
Statistical regularity, law of,
254
Statistics, 214
defined, 238
historical note on, 270, 271
limitations of, 253
use of, 238
Steel tape, 349
Straight angle, 48
' Substituting, 6
Subtraction, algebraic, 171
graphical, 38, 170
Sum, of angles about a point,
112-113
of angles of a triangle, 131
Supplement, 116, 118
488
GENERAL MATHEMATICS
Surds, addition and subtraction of,
410
multiplication and division of , 41 1
Surveying, 58
Surveyor's chain, 349
Symmetry, 259
Systems of equations, 369
Table, of mantissas, 430-431
of roots and powers, 397-398
of trigonometric ratios, 361
Tangent of an angle, 358
Tape, 349
Terms, 7
dissimilar, 7
order of, 7
similar, 7
Test of proportionality, 324
Tetrahedron, 106
Theorem of Pythagoras, 397-399
Transit, 58
Transversal, 68
Trapezoid, 82
Triangle, altitude of, 67
area of, 81
base angles of, 138
base angles of isosceles, 138
defined, 43
equilateral, 44
exterior angles of, 139
isosceles, 138
notation for, 130
perimeter of, 43
right, 135
sides of, 43
similar right, 317
solving, 359
sum of exterior angles, 139
sum of interior angles, 131-135
trigonometric area formula, 365
Triangle, vertex, 47
vertices, 43
wooden, 136
Triangles, construction of, 142-147
similar, 314
Trigonometric ratios, 359
Trigonometry, 356
Trinomial, 40
factoring, 199
Trinomial square, 202
Turnin : tendency, 181
Unequal angles, 60
Unequal lines, 34
Units of measure, 28, 74, 99
Value, absolute, 156
Variable, 300
dependent, 300
independent, 300
Variation, direct, 305
inverse, 309
joint, 312
Vel Deity or rate, 280
Verbal problem, method of solving,
16
Vertex of an angle, 47
Vertical angles, 122
Vieta, 108
Vinculum, 175
Volume, of cube, 102
measurement of, 99
of parallelepiped, 100
unit of, 99
Work problems, 285, 383
Zero, division by, 211
multiplication by, 183
ftftftS
UNIVERSITY OF CALIFORNIA LIBRARY
Los Angeles
This book is DUE on the last date stamped below.
URL
RENEWAL
.79
FEC'D LD-URt
Form L9-Series 444
3 1 158 004ft fi
UC SOUTHERN REGIONAL LIBRARY FACILITY
A 000 984 760 9
0 comments:
Post a Comment