FIRST-YEAR MATHEMATICS
BRESXICH
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THE UNIVERSITY OF CHICAGO MATHEMATICAL SERIES
Eliakim Hastings Moore
General Editor
THE SCHOOL OF EDUCATION TEXTS AND MANUALS
George William Myers
Editor
FIRST-YEAR MATHEMATICS for SECONDARY SCHOOLS
THE UNIVERSITY OF CHICAGO PRESS CHICAGO, ILLINOIS
Agf uta THE CAMBRIDGE UNIVERSITY PRESS
LONDON AND EDINBURGH
THE MARUZEN-KABUSHIKI-KAISHA
TOKYO, OSAKA, KYOTO
KARL W. HIERSEMANN
LEIPZIG
THE BAKER & TAYLOR COMPANY
NEW YORK
RENE DESCARTES
RENE DESCARTES was born at La Haye, near Tours, March 31, 1596, and died at Stockholm, February 11, 1650. As a boy his health was delicate, and at the age of eight his father sent him to a Jesuit school at La FlSche. Finishing his education at this school in 1612, he went to Paris. Here he devoted two years to mathematical studies with Mydorge and Mersenne. In his day the only callings open to the sons of the nobility were the army and the church. He chose the former and joined the army of Prince Maurice of Orange, then at Breda, a town in Holland. He then devoted himself to the study of philosophy, science, and mathematics. Read the story of his asking Isaac Beeckman to decipher a Dutch placard for him in Bail's History of Mathematics, pp. 269-70 (5th ed.).
Descartes was a small man with a large head, projecting brow, prominent nose, and black hair that grew down nearly to his eyebrows. His voice was feeble and he was cold and selfish in disposition. He is said to have despised all learning and art unless something tangible could be gotten from them. He went to Stockholm at the invitation of the Queen of Sweden in 1649 and died there of lung trouble after a few months.
In the year 1637 he wrote a book. Discourse on Methods, which contained an appendix on geometry that constitutes his title to enduring fame. The appendix showed how to study geometrical figures by means of algebraic equations and contained the following contributions to algebra:
1. Established the custom of denoting known numbers by letters at the beginning of the alphabet and unknowns by letters at the end of the alphabet.
2. Introduced the system of indices now used in mathe- matics.
3. Contributed the earliest recognition of the advantage of taking all terms of an equation to its first member.
4. Realized the true meaning of negative numbers and used them freely.
5. Gave the rule for finding the number of positive and the number of negative roots of an equation, and this is still called Descartes' Rule.
6. Introduced indeterminate coefficients in solving equa- tions.
7. Gave the first statement of the so-called Euler Theorem connecting the faces, edges, and angles of a polyhedron.
These contributions and other minor ones give him a better right than Vieta to the cognomen "Father of modern algebra." Furthermore, his geometry is regarded as con- taining the outlines of analytical geometry. He is regarded as the originator of this branch of mathematics.
Jx^cn. Girtej-uu.
First -\fear Mathematics for Secondary Schools
BY
ERNST R. BRESLICH
Head of the. Department of Mathematics in the University High School, The University of Chicayo
THK UNIVERSITY OF CHICAGO PRESS CHICAGO, ILLINOIS
B7Z
Copyright 1906 By
George W. Myers
Copyright 1909 and 191 5 By
The University of Chicago
First Edition privately printed October 1906
Second impression published April 1907
Second Edition August 1907
Third Edition September 1909
Second Impression December 1910
Third Impression August 191 1
Fourth Impression September 191 2
Fifth Impression September 191 3
Sixth Impression October 1913
Seventh Impression September 1914
Eighth Impression November 1914
Fourth Edition September 191 5
EDUCATION DEPT.
Composed and Printed By
The University of Chicasro Press
Chicai^o, Illinois, U.S.A.
EDITORIAL PREFACE
The course of study in American high schools is in process of extensive change. The change commenced with the intro- duction of new subjects. At first science began to compete with the older subjects; then came manual training, commercial and agricultural subjects, the fine arts, and a whole series of new literary courses. In the beginning the traditional subjects saw no reason for mixing in this forward movement, and such phrases as ''regular studies," ''substantial subjects," and "serious courses" were frequently heard as evidences of the complacent satisfaction with which the well-established departments viewed the struggles for place of the newer subjects. Today, however, the teachers of mathematics and classics are less anxious than formerly to be classified apart. Even the more conservative now write books on why they do as they do and they speak with a certain vehemence which betokens anxiety. They also pre- pare many editions of their familiar type of textbook, saying of each that it is something which is both old and new. All these indications make it clear that the change in the high-school curriculum which began with the introduction of new subjects will not come to an end until many changes have been made in the traditional subjects also.
Over against the obstinate conservatism of some teachers is to be set the vigorous movement within all subjects to fit them effectively to the needs of students. The interest of today is in supervised study, in better modes of helping students to think, in economy of human energy and enthusiasm. This means inevitably a reworking of the subjects taught in the schools. It is the opportunity of this generation of teachers to work out the changes that are needed to make courses more productive for mental life and growth.
During this process of reform, mathematics has changed perhaps less than any other subject. The textbooks in algebra have modified but little their list of topics or their mode of
vii
M23265'P'
viii EDITORIAL PREFACE
exposition. Most of the later books introduce graphs and have graded their problems better and have omitted some of the intricacies which were included a generation ago. These improvements are welcome but insufl^cient, and if algebra has been conservative, what words shall one find to describe Euclid- ian geometry! Most teachers of mathematics continue to indict themselves by failing abnormally large percentages of their students; and, what is more, the extreme conservatives among these teachers regard it as a virtue that they do not bring their students to the passing level. It is useless to argue with a teacher who puts on the student body the blame when 25 per cent of them are unable to profit by contact with himself. Such a teacher has no insight into the social relations of which he is a part; he is absorbed in subject-matter or in some other considerations remote from real school life. He fails to realize the significant historical fact that the time has passed when the chief duty of the teacher is to eliminate students.
Evidences are not wanting, however, that a thorough reor- ganization of the body of secondary-school mathematics is at hand. Since the presidential address before the American Mathematical Society of one of the undersigned (cf. Science, March 13, 1903), many papers have agreed in recommending radical changes, both in organization and content of the courses and in method of instruction. The School of Education of the University of Chicago as a center for educational experimen- tation under test conditions has since 1903 been developing a solution of this difficult problem of mathematical reorganization. The solution is sought along the lines of fusion. The reasons for this type of solution as detailed in former editions of this book are generally understood, and during recent years the trend of enlightened opinion has set strongly toward the educational desirability of this type of material for beginners in mathe- matics.
The salient historical facts of the experiment of which this book is a product are as follows:
In the school year of 1903-4 a tentative program for mathe- matics along fusion lines was worked out in conferences of the
EDITORIAL PREFACE ix
teachers of mathematics of the college and the high school of the School of Education. This program was revised, and used in mimeograph form as a text for the first-year classes in 1904-5. It was again revised, remimeographed, and the improved form was the text of these classes in 1905-6. After another revision the material was published by the University High School office in the summer of 1906 under the title of First-Year Mathe- matics for Secondary Schools. The University Press soon began supplying other high schools that desired to try out the material, and a little later took over the publishing end of the enterprise. The teachers co-operating with Mr. George W. Myers in the authorship of the first edition were Messrs. William R. Wickes, Ernst R. Breslich, Harris F. MacNeish, and Ernest A. Wreidt.
To continue the fusion type of work with the second-year classes of 1906-7 the little manual. Geometric Exercises for Algebraic Solution, was compiled that year, and published by the University Press in the summer of 1907. This manual was to supplement a standard text in geometry which was then in use with second-year students.
In the summer of 1909 the University Press published the revised and completer edition of First-Year Mathematics under the same title as before, Messrs. Arnold Dresden and Ernest L. Caldwell having been added to the list of participating authors.
In the spring of 1910 the University Press published Second- Year Mathematics for Secondary Schools from manuscripts that had been in use in mimeograph form in second-year classes for over a year. This was also avowedly a tentative form of the second-year material for provisional use, though most of the critics have failed or refused to see the avowal.
First- and Second-Year Mathematics have been the only texts used in the respective years in the University High School since their publication.
The present edition of First-Year Mathematics, which is soon to be followed by a revised form of its companion, Second-Year Mathematics, bears the sole name of Mr. Breslich, head of the High School department of mathematics, as author. While both books are the natural outgrowth of the experiment begun
X EDITORIAL PREFACE
twelve years ago, Mr. Breslich has entirely recast and rewritten the texts. His earnest and untiring work on the experiment from the outset has peculiarly fitted him for the task. Authorial credit for the present form of the material is entirely due to him. In certain important respects the plan of the present edition is a distinct departure from that of former editions. The work here having been done by a single author, who has himself been teaching the material to his own classes, some very specific gains have been brought to the book, a few of which may well be enumerated :
1. Greater homogeneity of the material and closer and more persistent correlation of matter drawn from the several mathe- matical branches. This means a much smoother gradation of difficulties and a better sustained conformity to an original plan.
2. The frank and full criticism of the author's work by his colleagues, both privately and in conference, has resulted in many both major and minor alterations that make the work more easily teachable by those who are inexperienced in the use of such material. The organization of the material has thus acquired a high degree of objectivity and practicality.
3. A much more searching scrutiny of the psychology of the final organization was thus realized. This subject-matter may accordingly claim to be, not only empirically and experimentally suitable, but also to be psychologically justifiable.
4. Much greater emphasis is here placed upon experimental and inductive geometry than was done in former editions. Moreover, the inductive work shades over into deductive pro- cedures more gradually here than in the former books.
These are not the only gains of this edition over former editions, but they are the gains of greatest scope and significance. The language of the book has been carefully studied with refer- ence to its easy comprehensibihty by beginners. The exercises have been very carefully chosen, graded, and related. In short, no pains have been spared and no labor stinted to make the present edition both an easily workable text for those who try the material for the first time, and a highly profitable body of ideas for beginners.
EDITORIAL PREFACE xi
To those who may examine this book from the point of view of the critical mathematician, it ought to be said that it is designedly a pedagogical rather than a logical organization of general and fundamental mathematical notions. Rigor in the pure mathematical sense is not attempted in definitions, axioms, or principles. Insight has everywhere been the con- trolling consideration. Experimentation, intuition, and induc- tion are freely employed. It takes time to learn deduction and the approach to this goal is gradual but sure. Clearness and comprehensibility under normal classroom conditions and in public-school environments have been guiding motives. Austere ideals of abstract rigor have been sacrificed. Still it is believed that as much rigor is demanded from stage to stage of the develop- ment as can profitably be attempted by the present-day thirteen- or fourteen-year-old boy or girl.
As to incommensurables, little more is attempted than to make clear to the student that they exist and the nature of the difficulty they present. His mind will be opened toward them and he will be ready for fuller ideas in good time.
The School of Education and the Department of Mathe- matics of the University of Chicago share the settled conviction of Mr. Breslich and his colleagues that this new book will con- tribute to the solution of the problems which confront the mathematical sciences in their efforts to be a vital part of the new course of study in American high schools.
Eliakim H. Moore George W. Myers Charles H. Judd
AUTHOR'S PREFACE
In planning the work of the first year, the following facts have been kept in mind :
1 . Each of the various divisions of secondary mathematics — algebra, geometry, and trigonometry — includes simple prin- ciples relatively easy to master, and also difficult, complex principles. The simpler principles are best suited for beginners, and may therefore be brought together in an introductory course which leads up to more complex aspects of these various branches of mathematical science.
2. Because they make the acquaintance of only one of the three subjects during the first year, many students fail to get an insight into secondary mathematics and are discouraged from continuing the study. Thus it is commonly the case that the student is brought into contact only with algebra in the first year. When he finds algebra very difficult he frequently misses the opportunity to discover that he can be successful in geometry. If an introductory course can be formulated in which algebra and geometry are taught together, success in one field will arouse an interest and enthusiasm which will encourage the student to attack the other with increased vigor. The result will be a gain of mathematical power and no loss in general training.
3. The relationship between algebra and geometry has long been recognized by technical students of mathematics, quite apart from any consideration of the desirability of teaching them together. Algebra and geometry supplement each other. Both are used to express facts about quantity; e.g., the graph and the formula both express the law of a group of numerical facts. Both give these facts in a form easily taken in by the trained eye; both state the facts in generalized form and thus make the deduction of any number of particular instances possible. By correlating the two related forms of thought in a
XIV AUTHOR'S PREFACE
single course of instruction the student's comprehension of quantity is at the same time simplified and deepened; simpli- fied, because the double method of attack makes it easier to overcome difficulties; deepened, because of the more enduring impression made upon the mind. In the course presented in this book, geometry is used throughout the book to illustrate algebraic processes, while algebra carries on the reasoning in the compact and abstract symbols which generalize quantita- tive facts in a degree which is impossible in graphic expression.
4. The number of mathematics courses required for gradua- tion from the high school is constantly being reduced. A student taking only one year of mathematics will, under ordinary circumstances, for this reason fail to come into contact with that very important body of geometrical ideas necessary to increase his understanding of his space environment.
5. A student will be most interested in subjects in which practical values are most clearly exhibited. If instruction in various branches of mathematics is given in the introductory course, the student will see the usefulness of various modes of treatment of the facts of quantity. He is made to realize the value of algebra by explicit references whenever the superiority of algebraic over geometric methods appears. Up to this point only the relation of algebra to geometry has been commented on. It is also true that the fundamental notions of trigonometry, which are commonly kept from the student until the third or fourth year of the high school, appeal to him because of their usefulness as tools in problem-solving. Hence, these notions introduced at an early stage are presented in a way not difficult for the beginner. Practical apphcations, especially to surveying, have therefore in the following pages received considerable attention.
When the various branches of mathematics are treated as separate subjects, there is a tendency for each to take on the rigid form of the final science. This tends inevitably to a certain formalism in mode of presentation. Such formalism is not the best method for the beginner. Correlation helps to avoid excessive formalism. Rigor is not carried beyond the under-
AUTHOR'S PREFACE XV
standing of the pupil. Indeed, it must be said that when rigor is attempted beyond the comprehension of the student it is only apparent. For correlated mathematics it is relatively easy to adopt a method of approach which is largely inductive. In geometry the peculiar properties of the appropriate figures are studied and the results are then combined into a theorem. This brings about an easier and a much better understanding than a beginner can obtain from a logical proof. Not until toward the end of the geometry of the first year does the demon- stration take the form of a logical proof. Axioms usually assumed to be self-evident are in the following pages illustrated in order to make their meaning apparent and vital. Algebra is introduced as a natural means of expressing facts about number and gradually becomes a symboHc language especially well adapted to stating the conditions of a problem in a natural and helpful way. The growing difficulty and complexity of problems then lead to the necessity of learning how to manipu- late algebraic symbols. The symbolism of algebra thus becomes a highly clarifying instrument of problem-analysis and problem- solving. The laws of algebra are carefully illustrated, thus avoiding the danger of symbol-juggling without insight into the real meaning.
There are certain processes which belong together logically but which should be separated in treatment because they make difficulties for the beginner. Hence, wherever the processes are not needed as instruments of instruction they are taught separately; e.g., the meaning of positive and negative numbers, the laws of signs, and the operations with positive and negative numbers are not studied until the pupil has become thoroughly familiar with unsigned literal numbers and with the operations and laws of such literal numbers. The fusion plan makes pos- sible a wide choice of process for the particular difficulty in hand and thus very materially facilitates conformity to the peda- gogical dictum, "One major difficulty at a time."
Until recently the character of secondary texts has been nearly uniform. An attempt to reorganize traditional material will not bring the best results unless this material is presented in a form
XVI AUTHOR'S PREFACE
in which even the inexperienced teacher can use it successfully. In making the experiment from which the text resulted, the author has intentionally gone about the work deliberately. He has watched for several years the difficulties encountered by new teachers coming into the mathematics department of the University High School and he has also had the good fortune 'of getting the frank and most helpful criticism of the material from teachers who have been in the department for some time. In addition to this, he has had the advice and criticism of 'several professors of the College of Education, who, from their interest in the work as an educational experiment, have made a detailed study and criticism of the material. The result of it all is that the book has become easily teachable.
One of the aims of every high-school teacher is to teach his pupils to work independently and to be able to use their books. The author shares this aim with his colleagues and has accord- ingly made a careful study of the difficulties met by high-school pupils in preparing their work. The organization of the ma- terial is such that a student will be aided in distinguishing the essential from the less important. Summaries given at the end of each chapter will be helpful in reviewing the work of the chapter. The list of typical problems in chapter XIX will make it possible for the student to review the whole course with increasing interest, because the ground is here covered differ- ently from the way in which it was gone over the first time.
, The chief gains of this text over the traditional treatment may be summarized as follows:
The student receives a broader mathematical preparation. At the end of a year he will know a number of important geo- metrical facts; he will know enough algebra to manipulate formulas and to solve equations in one or more unknowns. He has learned to use both algebraic and geometric methods of solving problems.
The program is richer in content. There is much dissatis- faction on the part of parents and students with a course in mathematics offering only one subject during the whole year. Tlie teaching of several subjects makes it possible to bring in
AUTHOR'S PREFACE xvii
practical problems from each, with the result that the course will lose in formalism and be better suited for beginners. The student sees the superiority of algebraic methods over geometric methods and of trigonometric methods over both, and comes to appreciate the value of the course and is more likely to decide to go on with more advanced work.
This results in an increased measure of economy of time, because progress is continuous; whatever is learned is kept available, and topics commonly treated in two or more mathe- matical branches are here treated once for all.
The teacher will receive new suggestions as to methods of teaching. For in the attempt to open up new fields and to treat traditional material partly according to new methods it became necessary to organize this material with unusual care and to supply a large number of illustrative examples and suggestions as to the aim and methods involved. The new material also makes it much easier for the teacher to elicit greater spontaneity of effort from pupils.
The author desires to render full acknowledgment to Pro- fessor Charles H. Judd for his interest and very substantial aid in the way of suggestions and criticisms, and to Principal Franklin W. Johnson, whose encouragement, assistance, and con- tinued interest have made this educational experiment possible. He is also indebted to his colleagues in the department of mathematics, Messrs. Wilham D. Reeve, Raleigh Schorhng, and Horace C. Wright, who have read all of the chapters in detail and whose constructive criticisms have been most valuable.
The portraits appearing as inserts, with the exception of that of John Wallis, have been taken from Philosophical Portrait Series, published by The Open Court Publishing Co., Chicago.
Ernst R. Breslich
CONTENTS
CHAPTER PAGK
I. The Straight Line 1
Measurement of Line-Segments 1
Ways of Expressing Facts about Quantity . . 6
IL Addition and Subtraction 15
Graphical Addition and Subtraction . . . . 15
Perimeters 21
Algebraic Addition and Subtraction . . . . 23
III. The Equation 30
Use of Axioms in Solving Equations . . . 33
Problems to Be Solved by the Aid of the Equation 36
IV. Angles 43
Classification of Angles 43
The Measurement of Angles 45
The Use of the Protractor in Measuring Angles . 48
The Sum of the Angles of a Triangle .... 50
The Sum of the Exterior Angles of a Triangle . 53
To Draw an Angle Equal to a Given Angle . 56
V. Areas and Volumes. Multiplication . . 62
The Area of a Square 62
Area of a Rectangle 66
Volume of Cube and Parallelopiped .... 68
Graphing Equations 69
Multiplication of Monomials 70
Addition of Monomials 73
Multiplication of a Polynomial by a Monomial . 74
Multiplying Polynomials by Polynomials . . 77
Area of Parallelogram and Triangle .... 79
VI. Angle-Pairs 83
Adjacent Angles 83
The Sum of the Adjacent Angles about a Point on
One Side of a Straight Line 85
xix
XX CONTENTS
CHAPTER PAGE
The Sum of the Angles at a Point .... 86
Supplementary Angles 87
Complementary Angles 90
Opposite Angles 93
The Acute Angles of a Right Triangle . . 95 Angle-Pairs Formed by Two Lines Intersected by a
Third . 96
VII. Parallel Lines. Lines and Planes in Space . 99
Parallel Lines 99
Angles of the Parallelogram and Trapezoid . . 104
Models of Geometrical Solids 108
VIII. Measurement of Lines in Space. Similar
Figures 112
Drawing to Scale 112
Ratio . . .118
Similar Figures 120
Problems in Similar Figures 123
IX. Ratio. Variation. Proportion .... 127
Trigonometric Ratios 127
Ratio .130
Direct Variation 136
Inverse Variation 140
Proportion . . . . . . 142
Proportionality of Areas . . . . . . 147
X. Congruence of Triangles . . . . 150
Congruence 150
The Isosceles and the Equilateral Triangle . .154
The Right Triangle 160
XI. Constructions. Symmetry. Circle . . . 164 The Fundamental Constructions Summarized and
Proved 164
Apphcations of the Fundamental Constructions . 166
Symmetry . 169
The Circle 172
CONTENTS ■ XXI
CHAPTER PAGE
XII. Positive and Negative Numbers. The Laws of
Signs 178
Uses of Positive and Negative Numbers . . . 178
Graphing Data . . . . . • • • 181
Addition of Positive and Negative Numbers . . 184
Subtraction of Positive and Negative Numbers . 188
' Law of Signs in Multiplication 19 1
Multiplication by Zero 195
Product of Several Factors 195
Law of Signs for Division . . . . . 196
XIII. Addition and Subtraction 201
Review of the Laws of Addition . . . .201
Addition of Monomials 202
Addition of Polynomials 203
Subtraction of Monomials 206
Subtraction of Polynomials 208
Removal of Parentheses 210
XIV. Multiplication and Division . . . . .213
MultipHcation of Monomials 213
Multiplication of Polynomials by Monomials . . 215
MultipHcation of Polynomials by Polynomials . 216 Multiplication of Arithmetical Numbers . . .218
Division of Monomials 220
Reduction of Quotients 221
Monomial Factors 222
Reduction of Quotients 224
Division of Polynomials 225
XV. Special Products. Factoring. Quadratic
Equations 230
The Square of a Binomial 230
Factoring Trinomial Squares . " . . . . 232 Product of the Sum of Two Numbers by Their
Difference 233
Factoring the Difference of Two Squares . . 234 The Product of Two Binomials of the Form
(ax+6)(cx+ri) 236
xxil CONTENTS
CHAPTER PAGE
Factoring Trinomials of the Form ax'^-\-bx-\-c . 236
The Theorem of Pythagoras 238
Square Root of Arithmetical Numbers . . .241
Quadratic Equations 245
Quadratic Equations Solved by Factoring . 246 Quadratic Equations Solved by Completing the
Square 248"
XVI. Problems Leading to Equations of the First
Degree in One Unknown 252
Solution of Problems and Equations . . . 252
Geometric Problems .256
Problems Involving Number Relations . . . 258
Motion Problems 260
Clock Problems 267
Problems on Percentage and Interest . . . 269
Mixture Problems 272
Lever Problems 273
XVII. Linear Equations Containing Two or More
Unknown Numbers 278
A System of Two Linear Equations .... 278
Graphical Method of Solving a System of Equations 279
Algebraic Solution of Equations in Two Unknowns 283
Geometric Problems 286
Motion Problems 288
Miscellaneous Problems 289
Fractional Equations 291
Systems of Three or More Linear Equations . . 292
XVIII. The Formula 297
The Formula as a General Rule .... 297
Evaluation of Formulas 302
Expressing One of the Letters of a Formula in
Terms of the Others 305
XIX. Review and Supplementary Questions and
Problems 308
STUDY HELPS FOR STUDENTS^
The habits of study formed in school are of greater impor- tance than the subjects mastered. The following suggestions, if carefully followed, will help you make your mind an efficient tool. Your daily aim should be to learn your lesson in less time, or to learn it better in the same time.
1. Make out a definite daily program, arranging for a definite time for the study of mathematics. You will thus form the habit of concentrating your thoughts on the subject at that time.
2. Provide yourself with the material the lesson requires; have on hand textbook, notebook, ruler, compass, special paper needed, etc. When writing, be sure to have the light from the left side.
3. Understand the lesson assignment. Learn to take notes on the suggestions given by the teacher when the lesson is assigned. Take down accurately the assignment and any references given. Pick out the important topics of the lesson before beginning your study.
4. Learn to use your textbook, as it will help you to use other books. Therefore understand the purpose of such devices as index, footnotes, etc., and use them freely.
5. Do not lose time getting ready for study. Sit down and begin to work at once. Concentrate on your work, i.e., put your mind on it and let nothing disturb you. Have the will to learn.
'^ These study helps are taken from Study Helps for Students in the University High School. They have been found to be very valuable to students in learning how to study and to teachers in training students how to study effectively.
XXIV STUDY HELPS FOR STUDENTS
6. As a rule it is best to go over the lesson quickly, then to go over it again carefully; e.g., before beginning to solve a problem read it through and be sure you understand what is given and what is to be proved. Keep these two things clearly in mind while you are working on the problem.
7. Do individual study. Learn to form your own judgments, to work your own problems. Individual study is honest study.
8. Try to put the facts you are learning into practical use if possible. Apply them to present-day conditions. Illus- trate them in terms familiar to you.
9. Take an interest in the subject. Read the corresponding literature in your school library. Talk to your parents about your school work. Discuss with them points that interest you.
10. Review your lessons frequently. If there were points you did not understand, the review will help you to master them.
11. Prepare each lesson every day. The habit of meeting each requirement punctually is of extreme importance.
EUCLID OF ALEXANDRIA
THE dates of Euclid's birth and death are unknown. He lived and taught mathematics at Alexandria in Egypt from 306 to 283 B.C. He had probably studied in Plato's school at Athens and in Aristotle's school at Stagira, both in Greece. Being well versed in both mathematics and Aristotelian logic, when he became head of the mathematical school at Alexandria he undertook to cast all Greek mathematics into the form of syllogistic reasoning. The result was his Elements of Geometry, which has become the most cele- brated mathematical text ever written. The Greeks had always regarded the proofs of theorems as a real part of geometrical study. Most other mathematical peoples included only the results and conclusions. The form Euclid gave to these proofs was so .excellent that his Elements soon replaced all other texts of his time, gave him the nickname of ''The Author of the Elements," and for nearly 2,500 years has made his name a synonym for his science. In England boys are even today said to be studying Euclid when it is meant they are studying geometry.
The form of our modern texts in geometry is nearly the same as Euclid gave his text. Euclid's text is at least the basis of our American texts. You will be interested to read the story of Euclid's ordering his slave to give a student a coin for studying geometry, in order to show the contempt he had for those who studied geometry for gain; also the story of his telling the young Ptolemy ''There is no royal road to geometry." Reason's highway is the only road we know of which leads to a knowledge of geometry, even today. Remember that if the road seems rough and steep at places, there are charming views at the top.
Read Euclid's biography in some encyclopedia or history of mathematics. While reading think how long men and boys have been trying to learn how to prove theorems as Euclid proved them three cen- turies before the birth of Christ. A study that has stood about the same for so long is not likely to ]:>e much different during your lifetime.
CHAPTER I
THE STRAIGHT LINE
Measurement of Line-Segments
1. Straight lines. Lay a ruler on a sheet of notebook paper and pass the point of a sharp pencil along the edge of the ruler. The drawing obtained is what in ordinary- speech is called a straight line.
We use the same term when we refer to a tight telegraph wire, a stretched string, a ray of sunlight passing into a dark room through a small opening, etc.
Give other examples of straight lines that can be seen in the classroom.
2. Geometric lines. The different examples in § 1 all have characteristics which differ, but they have one char- acteristic in common, namely, their length. When we speak in exact scientific terms we use the word ''line" to refer to the length of each of the above, not to their width and thickness. Thus, geometric lines have length only, but not width nor thickness.
3. Points. A particle of dust in the air, a small chalk dot on the blackboard, the sharp end of a needle, are examples of points as we ordinarily use that word. Again, for purposes of scientific exactness we must neglect the whiteness of the crayon point, the grayness of the dust point, and the spread-outness, however small, of each of the objects. In science a point is merely the position, but not length, breadth, nor thickness.
4. Notation for points. The position of a point on a straight line is indicated by a small cross-line or dot.
I
iJ/ ; ; ; ; MlrOT-y'EAR MATHEMATICS
Thus in Fig. 1 three points are marked. They are named
A, B, and C. It is customary in
— ' ' ' science to use capital letters in naming
Fig. 1 points.
5. Line-segments. The portion of a Unc bounded by two pomts on the Une is a line-segment, or, briefly, a segment, as A B or BC, Fig. 1.
6. Measurement of length. One edge of a ruler is
graduated in inches.
Place the edge of ^ ^
your ruler along the
line-segment A B, Fig. 2, and count the number of inches contained 'm. AB.
When we compare the length of a line-segment, as A B, Fig. 2, with such a well-known and well-established segment as an inch, or a yard, we are measuring the line-segment; or, put in general terms, to measure a line- segment is to find out how many times it contains another line-segment, called the unit-segment.
7. Units of length. Most of the civilized nations have derived a unit of length from the length of the human foot. The result was that they differed in the standard length. A commission appointed by the National Assembly of France devised a standard length which was adopted by France in 1793 and is now used very generally in scientific work in all countries. This stand- ard unit is called the meter. It is a bar of platinum, equal approximately to 1.1 yards, which is approxi- mately one ten-millionth of the distance from the North Pole to the equator, measured along a meridian. The meter is divided into 1,000 equal parts, called millimeters. Ten millimeters make a centimeter, ten
THE STRAIGHT LINE 3
centimeters make a decimeter, and ten decimeters make a meter.
EXERCISES
1. Using a ruler whose edge is graduated in centimeters,
measure AB, Fig. 3, in centi-
meters. ^ ^
Fig. 3
2. With a ruler measure
in centimeters the edge of a page of this textbook.
8. Compass. A compass, Fig. 4, is an instrument used for measuring. Because a good compass is accurate, it is useful in many forms of exact work. One who expects to do exact work should learn to use it freely.
9. Use of the compass. The compass may
be used to measure line-segments. Instead
of laying the ruler directly on AB, Fig. 2,
Fig. 4 ^^ was done in §6, the sharp points of the
compass are placed at A and B. Then the
points are placed on the marks of the ruler and the
number of inches or centimeters between them is
counted.
EXERCISES
1. With the compass measure AB, Fig. 2, in inches.
2. With the compass measure AB, Fig. 3, in centimeters.
3. Measure the length of the page of this book with the compass.
If an inch is not contained exactly, leaving a remainder less than an inch, count the number of eighths or sixteenths of an inch left over.
4. Mark two points on paper and estimate the distance between them. Test the correctness of the estimate by meas- uring the same distance with the compass.
FIRST-YEAR MATHEMATICS
10. Use of squared paper for measuring. The fol- lowing exercises explain the use of squared paper in measuring line-segments.
EXERCISES
1. Measure the length of AB, Fig. 5, in centimeters, using squared paper.
Place the points of the compass at A and B. Then place the
points on one of the heavy lines
"^ ^ of the squared paper, as at C and
D. Each side of a large square being 1 cm. long, count the num- ber of centimeters in CD and esti- mate the remainder to tenths of a centimeter. Express the length of CD in the form of a decimal.
— |
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1 1 1 |
1 i |
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1 1 |
1 1 |
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i 1 |
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1 |
1 1 1 |
1 |
1 1 |
Fig. 5
2. Draw a line-segment and measure its length in centimeters. Express the result in the form of a decimal fraction,
3. Draw a line-segment and measure its length approximately to two decimal places.
To find the length of a segment to two decimal places use as a unit a segment equal to ten times the length of the side of a small square on centimeter paper. Then AB, Fig. 6, is equal to 1 and CD is equal to .1. Why?
Imagine CD divided into 10 equal parts. Then .1 of CD equals .01 of AB.
Line-segment AB, Fig. 7, equals 2.35. Why?
X |
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Fig. 6
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Fig. 7
4. Measure to two decimal places the length of segments CD and EF, Fig. 7.
5. Draw several line- segments and using compass and squared paper measure them to two decimal places.
THE STRAIGHT LINE
6. Measure to two decimal places the segments AB, BC, and CA, Fig. 8.
7. Draw a triangle and measure the three sides to two decimal places.
B
Fig. 8
11. Equal segments.
If a segment can be placed upon another so that the end-points of one coincide with (exactly fit upon) the end-points of the other, the segments
are said to be equal. In Fig. 9 seg- ments a and h are equal. This is read a equals h and is written a = h.
Draw two equal line-segments.
Fig. 9
12. Unequal segments. If a segment b when placed
upon a segment a covers only a part
~ , of a, 6 and a are said to be unequal.
c This is written a=^b (read a is not
equal to b).
In Fig. 10, a is less than b and The first is written a<b and the
Fig. 10
greater than c. second a>c.
13. Notation for line-segments. In the preceding paragraphs line-segments have been denoted in two ways : (1) by marking two of the points of the segment by capital letters, (2) by marking the segment with a small letter placed near the middle of the segment. In the second case the small letter generally stands for the length of the seg- ment and is therefore a number that can be found by measuring the segment.
6
FIRST-YEAR MATHEMATICS
14. Literal numbers. A number denoted by a letter is a literal number.
In Fig. 11 find, by measuring, the numbers denoted by a, b, and c.
15. Representation of num- bers. Numbers may be repre- sented either by line-segments or by letters. The numbers that the line-segments represent are found
Fig. 11
by measuring the segments.
EXERCISES
1. Represent the following numbers by segments: 3, 14, 4.5,7.8,2.47, 1.64, .32.
2. Draw several line-segments and find, by measuring, the numbers represented by them.
16. Graphical representation. A line-segment used to denote a number is called a graph of the number.
The diagram below represents graphically the amounts in millions spent in 1909 by the people of the United States for certain necessities and luxuries.
0 25 50 75* 100 125 ISO 175 200 22$ 250 275 300 325 350
Newspapers and Periodicals
Druggists' Preparations
Pianos and Organs Jewelry
III 1 1 |
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Fig. 12
17. Ways of expressing facts about quantity. Graphs are an effective help when used to illustrate numerical
THE STRAIGHT LINE 7
statistics and scientific data. There are various ways in which the same facts about quantity can be expressed.
First, we may take the exact figures from the census reports about the population of the United States from 1790 to 1910. This gives the following table of numbers:
Year
Population Year
Population
1790 3,929,214
1800 5,308,483
1810 7,239,881
1820 9,638,453
1830 12,860,702
1840 17,063,353
1850 '. 23,191,876
Second, we may represent these facts graphically by line-segments. Thus, Fig. 13 shows at a glance the whole
1860 31,443,321
1870 38,558,371
1880 50,155,783
1890 62,947,714
1900 75,994,575
1910 91,972,266
ruv |
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-A |
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A |
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c — c — 0 |
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3 C 3-a |
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c — r — a |
3 C •.- 0 a- 0 |
c 0-0 0|— a |
3 J |
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Fig. 13
group of facts giveil in the preceding table. One can also see the change in the population without knowing the exact values of the numbers represented by the line- segments. This is brought out even better by joining the end-points of the vertical segments. We may even pre- dict what the population will be in 1920. This mode of
8
T^IRST-YEAR MATHEMATICS
expressing quantity is especially good when we wish to compare magnitudes with each other; e.g., national debts, navies of various countries, consumption of coal, etc.
Compare the two methods above as to their advantages and disadvantages.
Third, we may represent facts by literal numbers, as may be seen from the following illustration :
A train traveling in equal time-intervals over equal distances is said to have uniform motion. The distance passed over in the unit of time (e.g., an hour) is called the velocity of the train. Thus, the velocity of a train is 30 mi. per hour, if the train travels 30 mi. every hour. It follows that the train travels in 2 hours a distance of 2X30 mi., in 3 hours 3X30 mi., in 4 hours 4X30 mi., etc., and in t hours a distance of ^X30 miles. In general, distance = tiyneXvelocity^ or mlQiier^
tX'
Thus, tXv represents the following facts, if tJ = 30
Time in hours . . |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
etc. |
||||||||||||
Distance in miles |
30 |
60 |
9C |
120 |
150 |
180 |
210 1 |
240 |
270 |
30( |
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1 |
3 |
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300 |
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10 |
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120 |
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w
(B)
(C)
Fig. 14
The same facts may be represented graphically, as in Fig. 14. Notice that the graph in Fig. 14 is regular, in fact it is a
THE STRAIGHT LINE
straight line. This enables us to read from the figure the distances passed over during a given number of hours.
EXERCISES
1. Tell from the graph the number of miles the train travels in 15 hours; in 12 hours.
2. A train travels at the rate of 20 mi. per hour. Represent in a table, as in (B) p. 8, the distance passed over in the first ten hours. Represent the same facts graphically, as in (C).
18. Graphing data. Make the graphs for the following and tell what the graphs show.
EXERCISES
1. Graph these average heights of boys and girls:
Age |
2 |
4 |
6 |
8 |
10 |
12 |
14 |
16 |
18 |
20 |
Boys Girls |
1.6 ft. |
2.6 |
3.0 |
3.5 |
4.0 |
4.8 |
5.2 |
5.5 |
5.6 |
5.7 |
1.6 ft. |
2.6 |
3.0 |
3.5 |
3.9 |
4.5 |
4.8 |
5.2 |
5.3 |
5.4 |
Mark off on squared paper the heights, as in Fig. 15, call- ing the vertical side of a large square one foot. Connect the points as shown. The broken dotted line so obtained is the graph of the heights of girls, the full line is the graph of the heights of boys.
At what age do boys grow most rapidly ?
The answer may be seen from the table or from the graph.
2. The populations, in mil- lions, of the United States for each ten years, beginning 1790, are: 3.9,5.3, 7.2,9.6, 12.9, 17.1, 23.2,31.4, 38.6, 50.2,62.9,76.0,92.0. Make the graph.
r-bi |
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d-'X. |
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It /L |
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X |
X- |
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x_i |
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2_L4 6 8 10 |
tl2l 14._16i_ia 20 |
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Mill |
Fig. 15
10
FIRST-YEAR MATHEMATICS
3. The monthly average rainfall or snowfall, in inches, at a certain place for 30 years is given below.
Jan 2.8 May 3.59 Sept 2.91
Feb 2.30 June 3.79 Oct *2.63
Mar 2.56 July 3.61 Nov 2.66
Apr 2.70 Aug 2.83 Dec 2.71
Graph these data, letting twice the side of a large square represent one inch, and tell what the connecting line shows.
4. In a certain school the eyes of the pupils of all classes were tested two or three times each year for sixteen years. The following table gives the percentage of short-sighted pupils for each grade and the increase from one grade to the next. Graph the data. When is the percentage of increase lowest? When is it highest? Point out striking peculiarities of the graph.
Grade |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
Percent- age . . . |
2.8 |
4.6 |
7.8 |
11.7 |
12.1 |
15.3 |
17.0 |
22.5 |
29.7 |
36.0 |
41.7 |
47.7 |
Increase |
1.8 |
3.2 |
3.9 |
.4 |
3.2 |
1.7 |
4.5 |
7.2 |
6.3 |
5.7 |
6.0 |
5. The following table gives the population, in millions, of the United States in 1910, by sex, race, nativity, and parentage. Draw the graphs.
Native White Native Parentage |
Native White Foreign Parentage |
Foreign White |
Negro |
Indian |
Japa- nese |
Chi- nese |
|
Male. . . |
25.2 |
9.4 |
7.5 |
4.9 |
.13 |
.063 |
.067 |
Female. |
24.3 |
9.5 |
5.8 |
4.9 |
.13 |
.009 |
.005 |
in latitude 38°: 9.7 |
10.8 |
12.0 |
13.7 12.5 11.2 |
10.5 |
9.5 |
in latitude 45°: 9.1 |
10.4 |
11.9 |
14.1 12.6 11.1 |
9.6 |
8.8 |
THE STRAIGHT LINE 11
6. Graph these average lengths of day from sunrise to sun- set in latitude 42°.
Hours Hours Hours
Jan. 16 9.5 May 16 14.5 Sept. 15 12.5
Feb. 15 10.5 June 15. .. .15.0 Oct. 16 11.2
Mar. 16 11.9 July 16. .. .14.9 Nov. 15 9.6
Apr. 15 13.3 Aug. 16 13.9 Dec. 16 9.1
7. Using the same sheet, scale, and dates as in problem 6, graph the average days' lengths
13.3 14.4 14.9 14.6 13.5 14.9 15.6 15.3
What differences in the change of the day's length in differ- ent latitudes do the three graphs show ?
8. A boy saved two dimes a week, placing them in a savings box. Tabulate, as in (B), p. 8, the contents (number of dimes) for each of the first ten weeks. Represent these facts graphically. Represent the same facts in letters, as in (A), p. 8, knowing that the contents c is always 2 times the corresponding number of weeks w.
19. To test a ruler. To make sure that lines drawn with his ruler are straight, a carpenter makes the following test: He sights along the edge, making the end-points appear to fall together. If all other points in the edge fall together with the end-points, the edge of the ruler is straight.
A ruler may also be tested as follows : Mark two points on a sheet of paper. Lay the ruler on the paper so that the edge passes through the two points. Draw a line- segment through these points. Then lay the ruler on the opposite side of this line, making the edge pass through the
12 FIRST-YEAR MATHEMATICS
two points. If the edge falls exactly along the line, the ruler is straight.
Test your ruler in both ways.
20. Coinciding lines. From § 19 it is seen that two straight lines fall together if two points of one fall on two
points of the other. The lines
^^y:^^^^^^^y^ ^^® said to coincide. Fig. 16
^A^ — ~-_,^^^ shows that any number of lines
Fig. 16 can be drawn from one point to
another, but only one of them is
a straight line; i.e., Through two points one and only one
straight line can he drawn. The two points are said to
determine the straight line.
21. To produce a segment. Since only one straight line can be drawn between two points, a segment may be extended by moving the straight-edge ^ ^ along the segment. "^^ ' ' *"
The segment, Fig. 17, has been extended in the directions indicated by the arrows; it is said to be produced.
Mark three points A, B, and C not on the same straight line. Draw line-segments AB, BC, and CA. Show by meas- uring AB, BC, and CA that CA is shorter than the path CBA.
22. The shortest path between two points is the straight line-segment joining the points. Even when the exact scientific statement of this principle is not known to us, our ordinary behavior conforms to it; e.g., when hurrying from one place to another we choose the straight line as the shortest path between them. Apparently the fact is known even to animals; e.g., the dog when called by his master runs to him in a straight line.
THE STRAIGHT LINE 13
23. Axiom. Statements like those printed in italics in §§20 and 22 when assumed to be true are called axioms. Usually axioms are statements so simple that they seem evident. Axioms that belong especially to geometry are sometimes called postulates. For example, some writers would call the statements in §§20 and 22 postulates. In this book they will be referred to as axioms 1 and 2 respectively.
24. Point of intersection. Through a given point any number of straight lines can be drawn. When two straight lines have only one point in common, they are said to intersect and the point is called the point of inter- section.
EXERCISES
1. Through a given point A draw several straight lines.
2. If A is the point of intersection of two segments, Fig. 18, show that any point B on one of the segments cannot lie /^ on the other.
^ ,„ 25. Axiom 3. Two straight lines
Fig. 18 . . . . ^
can intersect m only one point.
For, if they had two points in common, they would coincide (§20).
Summary
26. This chapter has taught the meaning of the follow- ing terms and phrases: straight line, line-segment, point, measurement of length, unit of length, equal and unequal segments, literal number, to produce a segment, coinciding lines, axiom, point of intersection.
27. The use of the following instruments has been learned: the ruler, for drawing and measuring straight
14 FIRST-YEAR MATHEMATICS
line-segments; the compass, for laying off line-segments; and squared paper, for measuring segments and graphing data.
28. The following symbols are used: =, meaning is equal to; > for is greater than; < for is less than; ^ for is not equal to; X , meaning multiplied by.
29. .Three ways of representing statistics and scientific data are studied:
(1) tabulating them.
(2) representing them graphically by means of line- segments.
(3) representing facts by means of Uteral numbers.
30. Axioms.
1. Through two points one and only one straight line can he drawn.
2. The straight line-segment joining two points is the shortest path between them.
3. Two straight lines can intersect in only one point.
FRANCOIS VIET A
T^RANgOIS VIETA, better known by his Latin '*' name, Franciscus Vieta, a French mathema- tician of great originaUty, was born at Fontenay in Poitou in 1540 and died at Paris in 1603. In his book entitled Isagoge in artem analyticam, pubUshed in 1591, he introduced so many new and important algebraic ideas that he has been called the ''Father of modern algebra." He used vowels for unknown numbers and con- sonants for knowns. He discussed cubic and quartic equations, devised an excellent method of finding the approximate roots of numerical equations, and solved a 45th-degree equation that was famous for its difficulty in his age. He used trigonometrical ideas in solving it. He displayed an ingenuity that was almost uncanny in de- ciphering the Spanish secret signal code, by which he was able to read all the Spanish military dis- patches that fell into the hands of the French, to the great profit of his king. His mathematical knowledge made him of great service to his country in his public life. He studied mathe- matics as a pastime and for the love of it.
Besides the Isagoge, he wrote several other important mathematical works. Read his biography in some encyclopedia or history of mathematics.
CHAPTER II
ADDITION AND SUBTRACTION
Graphical Addition and Subtraction
31. Number scale. Beginning from a point 0, Fig. 19, a unit-segment is laid off repeatedly, forming a geometric picture of the series of numbers 1, 2, 3, etc. According to this arrangement, to every
number 1, 2, 3 corre- ^ i 2 s \ s e 7 ^^
sponds a point on the line OA, Fig. 19
the number indicating how
many units the corresponding point is from 0. A series
of numbers, as 1, 2, 3 thus arranged, is a number
scale. Such scales are found on the meterstick, the thermometer, the engineer's steel tape, etc.
32. Sum. If two line-segments, 2 cm. and 1 cm. long, respectively, are placed upon an indefinite straight line, as
AB, Fig. 20, adjoining each
^ 2 ^ ;■— ^ ^ other and having only one
Fig. 20 end-point in common, the
segment AD thus formed
is the sum of 2 cm. and 1 cm. This is written 2 + 1=3.
EXERCISES
1. By use of line-segments ^ ^
find the sum of 4 and 3; 2.2
and 3.1; 6 and 1.4. c — d
2. Let AB and CD, Fig. 21, ^ « , ^ n
be any two line-segments. De- ^
noting their lengths by a and h respectively, find the sum a+h, as EH.
The problem shows how a letter may be used to denote different segments. One advantage of expressing lengths by letters is that
15
16 FIRST-YEAR MATHEMATICS
we can find the sum before we know the numbers for which the letters stand.
3. Denoting the length of EH, Fig. 21, by c, show by meas- uring AB, CD, and EH that A c E B a-\-b = c.
^^^- 22 4. Show by measuring that
the whole segment AB, Fig. 22, equals the sum of all its parts AC, CE, and EB.
Problem 4 illustrates the following axioms :
33. Axiom 4. The whole is equal to the sum of all its parts.
34. Axiom 5. The whole is greater than any of its parts.
EXERCISES
1. Draw the segments a = 3. 4 cm., 6 = 2.3 cm., c = 1.5cm. Draw the sum a-]-b-{-c.
2. A, B, C, and D are four ^ b c ~d consecutive points on a straight Fig. 23
line. Fig. 23, such that AB = CD. Show by measuring that AC = BD.
3. Show without measuring that AC = BD, Fig. 23.
Since it is given that AB = CD, Fig. 23 and since BC = BC,
it follows that AB-{-BC = CD+BC
or AC = BD (Axiom 4)
In the third step of the solution of problem 3 the fol- lowing axiom is used: "* ^ 35. Axiom 6. If the same number
be added to equals the sums are equal.
G H 1. Let AB and CD, Fig. 24, be two
Fig. 24 equal segments. Also let EF and GH
ADDITION AND SUBTRACTION 17
be equal. Find the sums AB+EF and CD+GH. Show by measuring that these sums are equal.
2. Letting a, b, c, d be the lengths of four segments such that a = b and c = d
show by measuring that a-\-c = b+d
Problem 2 illustrates the following axiom :
36. Axiom 7. Equals added to equals give equal sums . (addition axiom).
37. Axioms 4-7, although so far related only to line- segments, hold equally for all other things. Thus, the number of pupils in a class is greater than the number of pupils in one row. What axiom is used? If the same number of apples is placed in each of two boxes already containing an equal number, the boxes will again contain an equal number of apples. Why? If the number of boys a in one classroom equals the number of boys b in another, and if the number of girls c in the first room equals the num- ber of girls d in the second, then by axiom 7, a-{-c = b-\-d; i.e., the number of pupils in both rooms is the same.
38. Difference. Place a line-segment, as A B, Fig. 25, upon another, as CD, so that the two
have one end-point in common. The a b
amount ED by which CD is greater ^ ^ ^
then A jB is the difference between CD ^ig. 25 and AB. This may be stated as fol- lows: CD minus AB equals ED, or in written form: CD-AB = ED.
39. Symbols. The symbols + and — used in the pre- ceding pages have the same meaning as in arithmetic.
18 FIRST-YEAR MATHEMATICS
They are read plus and minus and indicate the operations of addition and subtraction.
EXERCISES
1. Subtract by means of line-segments (graphically) 3.5 from 6.
As in Fig. 25, draw CD = 6 cm., AB=3.5 cm., and measure ED.
2. Subtract, graphically, a number, as s, from another larger number m. Denoting the difference between m and s by d, show- by measuring that m — s = d.
40. Equation. The statements a-{-h = c, m — s = d, CD — AB = ED, etc., express equalities and are called equations.
EXERCISES
1. A man had 22 acres and sold 8 acres. He had I acres left. Express this in equation form.
2. A man had 22 acres and sold s acres. How many had he left ? Give answer in the form of an equation.
3. A man possessed p acres and sold s acres. How many had he left?
Express by equations the relations of the numbers in the following problems :
4. A boy has m marbles and buys b more. He has then M marbles.
5. There are b boys and g girls in a class of p pupils.
6. A boy earns c cents a day for d days. He then has C cents.
7. A profit of p dollars added to a capital of c dollars amounts to a dollars.
8. An article bought for b dollars is sold for s dollars at a loss of I dollars.
9. Express a number 5 greater than n; c greater than n; 7 less than 6; x less than n.
ADDITION AND SUBTRACTION 19
10. The difference of two numbers is 110. If the smaller is
s, what is the larger ?
I I
11. The difference of two nmnbers ^ y ^^ ^ ^ is d. If the larger is I, what is the
smaller ? b
12. In Fig. 26, AB = CD. Show by measuring ^/ W AB, CD, and CB that AB-CB = CD-CB. a^ ^c
T^"tp 97
13. In Fig. 27 AB = BC
and AD = CE. Show by measuring
thsit BD = BE; i.e., AB-AD = BC-CE
Problems 12 and 13 illustrate the following axiom:
41. Axiom 8. // the same number or equal numbers be subtracted from equals the differences are equal (subtraction axiom).
Give the reasons for the following conclusions :
1. a = 9, 6 = 3, therefore a+6=12
2. m=10, s = 4, therefore m — s = 6
3. a = x, b = y, therefore a-\-b = x-\-y
4. m = r, s = t, therefore m — s = r — t
42. Axiom 9. The sums obtained by adding unequals to equals are unequal in the same order as the unequal addends.
E.g., if AB>CD Or, with numbers, 18> 12
and if EF=GH and 10 = 10
show that AB-^EF>CD-\-GH Hence, 18+10>12+10
43. Axiom 10. The sum obtained by adding unequals to unequals in the same order are unequal in the same order.
E.g., if AB>CD Or, with numbers, 7>2
and if EF>GH and 8>3
show that AB+EF>CD-\-GH Hence, 7+8>2+3
20 FIRST-YEAR MATHEMATICS
44. Axiom 11. The differences obtained by subtracting unequals from equals are unequal in the order opposite to that of the subtrahends.
E.g., if AB = CD Or, with numbers, 15 = 15
and if EF<GH and 3<10
Bhovf\h2itAB-EF>CD-GH But 15 -3> 15 -10
45. Historical note. About 300 b.c. the mathema- tician EucUd of Alexandria wrote a mathematical text called The Elements, in which he added, subtracted, and multiplied by use of line-segments. The first to use algebraic addition, subtraction, and multiplication seems to have been Diophantus, about 250 a.d.* The Italian Pacioli in 1494 was the first to give rules for all processes of addition, subtraction, multiplication, and division.
It is the purpose of the remainder of this chapter to study some of the laws of algebraic addition and sub- traction.
46. Multiples. The line-segment A B, Fig. 28, is laid off
repeatedly on the segment CD,
^ ^ making CE=EF = FG, etc.
^ E 'f~g S ^ ^ Denoting the length oi AB Fig. 28 by I, then CE = l, CF = 2l,
CG = 3l, etc. The segments CF, CG, CH, etc., are called multiples of AB.
47. Graphical addition and subtraction. Literal num- bers may be added and subtracted by means of line- segments.
* See Cajori, History of Mathematics, pp. 35 and 74, or Ball's History of Mathematics, pp. 52 and 103.
ADDITION AND SUBTRACTION 21
EXERCISES
1. Let a and h denote two numbers. Find the sum 3a +26.
Represent a and h by line- a a a b b
segments. Since 3a = a+a+a, ' ' ^ ^ '
and 26 = 6+6 lay off a on a Ai sa-^-ab IB
straight line three times and Fiq, 29
then 6 twice. Then the distance
AB, Fig. 29, from the starting-point to the end-point is 3a +26.
2. If a = 1 and 6 = 2, find graphically the sum 3a +26. Let a be represented by 1 cm., 6 by 2 cm.
3. Find graphically Sx-\-y — 2z, x, y, and z denoting three numbers.
4. Find graphically 3x-\-y-2z, if a: = 4.4, ?/ = 2. 3, 2=1.2.
5. If w = 2. 1 find graphically 3m+5m.
Perimeters
48. Triangle. Three points A, B, C, Fig. 30, and the three segments a, h, and c joining them, form a triangle. A triangle has three vertices (corners) A, B, and C and three sides a, 6, and c.
49. Perimeter. The sum of the three sides of a triangle, as a+6+c, Fig. 30, is the perimeter of the tri- angle.
EXERCISES
1. A lot has the form of an equal-sided (equi- lateral) triangle. Fig. 31, each side being x rods long. How many rods of fence will be needed to
inclose it? pj 3^
2. What is the sum of the sides of a triangle. Fig. 32, whose sides are 2x ft., 2x ft., and 3x ft. long?
3. What is the perimeter of a tri- angle whose sides are 2a, 5a, and 6a?
Fig. 30
22 FIRST-YEAR MATHEMATICS
50. Polygon. The figure ABCDEF, Fig. 33, formed
by joining the points A, B, C, D, E, and F by Une-segments, is a polygon. The word '^polygon" comes from the Greek and means many-cornered.
Polygons having 4, 5, 6 n sides
are called quadrilateral, pentagon, hex- agon rj-gon, respectively. The
sum of the sides of a polygon is its perimeter.
51. Equilateral. A polygon having all sides equal is an equilateral polygon.
EXERCISES
Of what kind of polygons may the following equations express the perimeters, p?
p = 7x 9. p = 12x
p = 8x 10. p = 20x
p = 9x 11. p = nx
p = 10x 12. p^ax
52. Value. A value of a letter is a number for which the letter stands. A letter may have a value even before the value is numerically determined.
Thus, in exercises 1-4, § 51, what values has x in order that the perimeter in each case be 60 ?
EXERCISES
1. Determine the value of x in equations 8-10, § 51, which makes the perimeter in each case 60.
2. Find the values of the perimeters in exercises 1-12, § 51, for a; = 2 in.; x = 3 ft.; x = 6 yards.
p = 3x = x+x+x = 2-|-2+2=6, etc.
3. Show, by a sketch, polygons whose perimeters, p, are given by 7) = 6.r+4; p = 4x-\-l6; p = Sx-\-Sy; p = 4x-^2y; p = r)x-\r^y.
1. |
p = Sx |
5. |
2. |
p = ix |
6. |
3. |
p = 5x |
7. |
4. |
p = Qx |
8. |
ADDITION AND SUBTRACTION 23
4. Find the values of the perimeters in exercise 3 for x = 5, y = 2; x = 12, 2/ = 4; x = l, ?/ = 7.
5. Show that 4s has the same value as 4Xs. E.g., lets = 3, then4s = 3+3+3+3=4X3.
6. Show that 8p = 8Xp.
53. Monomials. Numbers like 4s, 8p, 5c, etc., are monomials, or terms.
54. Coefficient.J The arithmetical factors 2, 3, 4, etc., in 2x, Sx, 4:X, etc., are the coefficients of the literal factor x. When no coefficient is written, as in x, a, n, the coefficient is understood to be 1. Thus, a means la, n means In.
55. Similar and dissimilarnumbers. Monomials having the same (common) literal factor are called similar num- bers, as 56 and 86, but 46 and 4a are dissimilar numbers.
56. Poljntiomial. An algebraic number consisting of two or more monomials, as 4a+10, 5a: -[-6a — 6, is a poly- nomial. 'Tolynomial" means many-termed.
57. Binomial. Trinomial. Polynomials of two terms are called binomials, as 2a +36. Polynomials of three terms are trinomials.
Algebraic Addition and Subtraction
58. Addition of similar monomials. The sum of simi- lar numbers, as 4a and 3a, can be simplified (reduced) according to a law which may be illustrated as follows :
Reduce to the simplest form the sum 4a+3a. Since 4a+3a = (a+a+a+a) + (a+a+a)* =7a, it follows that 4a+3a = 7a.
t The term "coefficient" was first used by the French mathe- matician Vieta (1540-1603) in a pamphlet on calculation published in 1591.
*The symbols ( ) are used to show the parts that come from 4a and 3a, respectively.
24
FIRST-YEAR MATHEMATICS
Thus, the sum of similar monomials is a monomial having the coefficient equal to the sum of the coefficients of the given monomials and having the same literal factor as the given monomials.
Using this law, reduce to the simplest form each of the follow- ing sums: 5x+Qx', 18n+5/i+4n; a+3a+10a; 12s+6s+s+16s.
59. The advantage of adding numbers according to the law of § 58 may be seen from the solution of the following problem :
The tickets for a football game are sold at 25 cents by John, Henry, Kenneth, William, and James. They report sales as follows: John sold 56 tickets, Henry 75, Kenneth 27, William 83, James 69. At the gate, 123 tickets are sold. Find the total receipts.
Solution I: John, 56X25c.=$ 14.00
Henry, 75X25c.= 18.75 Kenneth, 27X25c.= 6.75 William, 83X25c.= 20.75 James, 69X25c.= 17.25 Gate, 123 X 25c. = 30.75
Total receipts = |
$108.25 |
|
Solution II: |
John, 56 tickets. |
56 X 25c. |
Henry, 75 " |
75X25C. |
|
Kenneth, 27 " |
27x25c. |
|
William, 83 " |
83X25C. |
|
James, 69 " |
69X25C. |
|
Gate, 123 " |
123X25C. |
433 tickets, 433 X25c. =$108.25
Solution II is the simpler. Because the different terms to be added have the common factor 25, they are added by prefixing the sum of the coefficients to the common factor.
ADDITION AND SUBTRACTION 25
EXERCISES
1. The tickets being sold at x cents, John sells 60 tickets,' Henry 78, Kenneth 45, William 36, and James 84. At the gate 137 tickets were sold. Find the total receipts.
Total receipts: 60x +78x+45a;+36a;+84a; + 137a; = 440a;.
Again the terms have a common factor x and the sum was simplified by prefixing the sum of the coefficients to the common factor.
Is solution I possible in this case? Give a reason for your answer.
2. Is it possible to reduce to a simpler form the sum of 10 apples, 5 pears, and 4 plums ? Give a reason for your answer.
3. Which of these sums can be reduced ?
4a;+2, Sy+2y, 5a+2a4-6
4. The length of the school hall is I feet. I go through the hall 6 times on Monday, 8 times on Tuesday, 4 times on Wednesday, 6 times on Thursday, and 10 times on Friday. How many feet do I travel along the hall during the week ?
5. The running track in the playground is y yards. While in training, I run around it 6 times on Monday, 8 times on Tuesday, 10 times on Wednesday, 12 times on Thursday, and 14 times on Friday. How many yards do I run during the week?
6. Add as indicated:
(1) ix-\-20x-\-7x-\-nx (4) ^6+f6+i6
(2) A2j+ly+5y-\-2y+Sy (5) {5z-i-z)-{-{4z+2z)
(3) 10m+4m+2m+2n+3n (6) 12.5c+1.2c+4c
7. Two trains leave a station at the same time traveling in opposite directions at the rate of m miles per hour. How far apart are two towns, if the trains reach them in 8 and 12 hours respectively ?
26 FIRST-YEAR MATHEMATICS
60. Subtraction of similar monomials. The law for simplifying the difference of similar monomials is similar to the law for addition and may be illustrated as follows:
Subtract 4a from la.
7a=a-\-a-\-a-\-a-\-a-\-a-\rci
and 4a = a+«+a+a
Subtracting equals from equals, 7a — 4a = a +a +a = 3a
Thus, the difference of similar monomials is a monomial having the coefficient equal to the difference of the coefficients of the given monomials and having the same literal factor.
EXERCISES
1. Reduce to the simplest form each of the following differ- ences:
10e-7e; 13a-5a; 14w-2n; l^w-^w, 3.48fir- .25fir; 1.04x— .08a:; |m — im; fp — 4p.
2. Reduce the following sums and differences to their simplest forms:
(1) 5a+7a-4a (4) \h-\-lh-\h
5a+7a-4a (5+7-4)a = 8a
(2) Ux-x+Zx (5) (82 -2) + (52 -22)
(3) 13.5c+2.4c-c (6) 15x+(6x-4a;)
61. Commutative law. The following problems illus- trate the commutative law:
1. John has two kinds of marbles, 8 of one and 3 of the other. How many has he in all ?
To find the number of marbles he has he may add cither the 3 marbles to the 8 or the 8 to the 3, thus: 8+3=3+8.
2. Show by adding line-segments , ^ _,_3— ^ ^j^^^ 8+3 = 3+8 (Fig. 34).
I — -5--» 8 i
A i I I I I 1 I B ^- Sbow as in problems 1 and 2
Fig. 34 that 4+3 = 3+4.
4. Show that a+6 = 6+a. 5. Show that a+6+c = c+a+6 = 6+c+a.
ADDITION AND SUBTRACTION 27
Problems 1 to 4 show that the value of a sum remains unchanged by changing the order of the addends. This is called the Commutative Law of Addition.
Add in the most advantageous way, making use of the commutative law:
875+316+25; 9,993+4,287+7
62. Parentheses. The symbols (),[],! \, called parenthesis, bracket, and brace, respectively, are used to inclose numbers. Sometimes one of the symbols ;s in- closed within another, thus:
8+4+7+2 = [(8+4)+7]+2 and 8+4+7+2+3 = j[(8+4)+7]+2j+3
EXERCISES
1. A boy has three kinds of marbles, 4 of one kind, 7 of another, and 5 of a third. How many marbles has he ?
To find the sum, the first kind may be added to the second and the result to the third; i.e., 4+7+5 = (4+7) +5 = 16. Or he may add the marbles of the second and third kinds and these to the
first kind, giving the equation
, .iH-^ 1 5. , 4 + (7+5)=16. It follows that
,,,,,,. (4+7)+5 = 4 + (7+5). Why?
I j^—4. -2 + 5 1 2. Show by means of line-
FiG. 35 segments (Fig. 35) that
(4+7)+5 = 4+(7+5).
3. Show by means of segments that (a+6)+c = a+(&+c).
4. Show by means of segments that
a+6+c+d = a+(6+c)+d=(a+6) + (c+d).
Problems 1 to 4 illustrate the following law:
63. Associative Law of Addition. The sum of several numbers is the same in whatever way two or more of the numbers are combined into a sum before adding in the rest.
28 FIRST-YEAR MATHEMATICS
EXERCISES Using the commutative and associative laws the following sums are to be found in the most advantageous way:
1. (4356+ 1483) + (4356 -1483)
2. (34+ 128) +66
3. 381 + (436+19) Simplify the following:
4. x+\7x-{-{ix-2x)-\-iSx-x)\
5. [15a-(3a+2a)] + [(8a-a)-4a]
6. 3m+[m+(7m— 4m)+3m] — (5m— 4m)
7. Sy-2y+[\l2y-8y-{2y-\-y)-{-7y\-y]
Summary
64. This chapter has taught the meaning of the fol- lowing terms: number scale, sum, difference, equation, triangle, perimeter, polygon, quadrilateral, pentagon, hexagon, n-gon, equilateral polygon, value of a letter, monomial, coefficient, similar and dissimilar numbers, polynomial, binomial, trinomial.
65. The following symbols have been introduced: + for addition, — for subtraction, the parenthesis ( ), the bracket [ ], and the brace \ \.
66. Axioms.
4. The whole is equal to the sum of all its parts.
5. The whole is greater than any of its parts.
6. // the same number be added to equals the sums are equal.
7. Equals added to equals give equal sums.
8. // the same number or equal numbers be subtracted from equals the differences are equal.
ADDITION AND SUBTRACTION 29
9. The sums obtained by adding unequals to equals are unequal in the same order as the unequal addends.
10. The sums obtained by adding unequals to unequals in the same order are unequal in the same order.
11. The differences obtained by subtracting unequals from equals are unequal in the order opposite to that of the subtrahends.
67. Laws.
1. The sum (difference) of similar monomials is a monomial having the coefficient equal to the sum (difference) of the coefficients of the given monomials and having the same literal factor.
2. The value of a sum remains unchanged by changing the order of the addends (Commutative Law of Addition).
3. The sum of several numbers is the same in whatever way two or more of the numbers are combined into a sum before adding in the rest (Associative Law of Addition).
68. Algebraic numbers may be added or subtracted graphically.
CHAPTER III
THE EQUATION Use of Axioms in Solving Equations
69. Uses of the equation. The equation was used in the first two chapters to state the equahty of two numbers or of two geometric magnitudes and to express verbal statements in brief form. Thus, the statement that the distance passed over by a body moving with uniform velocity is obtained by multiplying the velocity by the time, takes the simple form d = vXt. Or, denoting the minuend by m, the subtrahend by s, and the difference by d, the statement that the difference equals the minuend less the subtrahend is expressed in equation form hy d = m—s. In mathematics the equation is of great importance, espe- cially as a tool for solving problems, for the statement of a problem in most cases takes the form of an equation.
In making a study of the equation we must begin with some very simple problems in order that we may clearly understand the new laws to be developed. If these laws are mastered in connection with simple cases, it will be easy to apply them later to more comphcated and difficult cases.
A bag of grain of unknown weight, w ounces, together with an 8-oz. weight just balances an 18-oz. weight. How much does the bag of grain weigh?
The problem may be stated in an equation, thus:
w;+8 = 18. Findw;.
Suppose 8 oz. to be taken from each pan, giving
u.
The bag of grain weighs 10 oz.
Fig. 36 Showing w-\-S
18
30
THE EQUATION 31
70. Member of an equation. An equation, as iy+8 = 18, may be regarded as an expression of balance between the numbers on the two sides of the equahty sign. The number to the left of the equality sign is the left side, or left member of the equation, the number to the right is the right side, or right member.
Thus, in the equation a+5 = 7, a+5is the left member and 7 is the right member.
EXERCISES
1. State in words problems in weighing expressed by the following equations:
w;+5 = 7; iy+2 = 4; 3+w; = 8; l0 = 5-\-w
2. Show how by the aid of the balance the value of w in each of the equations of problem 1 can be found.
71. Determining the value of the unknown number.
An equation, as w-\-S = lS, may be regarded as stating the question: What number added to 8 gives 18? It has been shown that the answer may be found by interpreting the equation as a problem in weighing and then taking 8 oz. from both pans of the balance. Just as the scales will balance if the same number of weights are taken from each pan^ we may subtract the same number from both sides of an equation and get another equation. The work of finding the unknown number this way may be arranged in written
form thus:
Let w;+8 = 18
8= 8
Then w =10
For, if the same number be subtracted from equal num- bers, the remainders are equal (subtraction axiom, § 41).
To test the correctness of the result, replace the un- known number in the original equation by 10, obtaining
32 FIRST-YEAR MATHEMATICS
10+8 = 18. Since both members of the equation reduce to the same number, the result ly = 10 is correct.
72. Substitution. When a number is put in place of a literal number, it is said to be substituted for the literal number.
73. Satisfying an equation. 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, 2 satisfies the equation a; +4 =6. Why?
74. Root. A number that satisfies an equation is a root of the equation.
75. Solving an equation. The process of finding the values of the unknown number or numbers which satisfy an equation is called solving the equation.
76. Check. A test, or a check, of the correctness of the result obtained by solving the equation can be made by substituting in the original equation the result in place of the unknown number. If the equation is satis- fied, the result is a root of the equation.
EXERCISES
Solve the following equations and problems and check the results:
1. 2-hn=17
Let 2+n = 17 2 =2 Then n = 15, by subtracting 2 from both members
Check: 2+15 = 17, by substitution Hence, 17 = 17
2. ii;+2 = 10 4. Z+ 7 = 18 6. 8+?;= 15
3. a;+4 = 36 5. rf+10=14 7. 13 = s+3
THE EQUATION
33
8. Two equal, but unknown, weights, together with a 1-lb. weight just balance a 16-lb. and a 1-lb. weight together. How heavy is each un- known weight?
In the form of an equation, the problem is stated thus:
2w;+l = 17. Findw;.
Suppose 1 lb. to be taken from each pan, giving
2w = lQ Then w = S
Fig. 37. — An equa- tion is an expression of balance of values.
9. A man finds that three bags of shot, of equal, but unknown, weights, together with two 2-lb. weights on the left scale-pan and a 12-lb. and a 4-lb. weight on the right scale-pan just bal- ance. Find the weight of each bag of shot. In the language of an equation we may
write
3w;+4 = 16
Taking 4 lb. from each pan, we have
Fig. 38.— If the weight in one pan is changed, the weight in the other must be changed correspondingly.
3w = 12 Then w= 4:
Each bag of shot weighs 4 lb.
10. Show how the values of w in the following equations may be found with the balance: 2w;+6=16; 3w;+7 = 19; 3w;+ 10 = 28; 5iy+7 = 27
77. Use of axioms in solving equations. The equations in problems 9 and 10 may be solved by use of axioms as
follows :
3i(;+4 = 16
Let
4= 4
12, by subtracting 4 from both sides
12
— , by dividing both sides by 3.
Then 3m; Sw 3 w = 4, by reducing the fractions
State the axiom used in obtaining the equation dw = 12.
34 FIRST-YEAR MATHEMATICS
In obtaining the equation w = 4: from 3w = 12 the fol- lowing axiom was used:
78. Axiom 12. If equals be divided by equal numbers (excluding division by 0) the quotients are equal (division axiom). E.g., if a segment AB = CD, then ^AB=^CD. Or, since 8+4 = 2 + 10, it follows that J of (8+4) equals J of (2+10).
Solve by use of axioms the equations in problem 10, § 76, stating the axiom used in each step of the solutions.
79. The equation n— 2 = 5 may be regarded as stating the question : What number diminished by 2 gives 5 f The number is 7 and may be found by adding 2 to both sides of the equation.
In written form the work of solving the equation is arranged as follows :
Let n-2 = 5
2 = 2 Then n+2— 2=2+5, for equals added to equals give equal sums.
Or, n = 7
EXERCISES
State each of the following equations in the form of a question. Then solve the equations, giving in each step the axiom used, and check.
1. |
X — |
8= 3 |
8. |
2a;+ 6 = |
16 |
15. |
Qx -7 =74 |
2. |
X — |
5= 7 |
9. |
Sx+ 7.= |
19 |
16. |
4m+3.2=15.2 |
3. |
n — |
2= 4 |
10. |
3^+ 6 = |
27 |
17. |
5n -1.4= 8.6 |
4. |
10 — |
7 = 14 |
11. |
2t -11= |
21 |
18. |
3a; +2 =2x+5 |
5. |
X — |
10= 1 |
12. |
4a;- 5 = |
23 |
19. |
5x -4 =4x+7 |
6. |
10- |
x= 3 |
13. |
9A;-15 = |
93 |
20. |
2?/ +3 =3?/-5 |
7. |
2x- |
1= 9 |
14. |
9a:+ 8 = |
116 |
21. |
7s -2 =6s+8 |
THE EQUATION 35
80. Axiom 13. If equals he multiplied by the same num- ber or equal numbers, the products are equal (multiplication axiom) .
E.g., since 3i = 1 and 2 =2 it follows that 2X3| =2Xi
Axiom 13 may be used to remove the fractions in an equation. To illustrate this,
Let ix = 9 Multiplying both members by 2, 2X^x= 2X9 (by axiom 13) Reducing to simplest form, re = 18
EXERCISES |
||
Solve the following equations: |
||
1.H |
3. 1=3 |
5. ^=10 |
2j=9 |
-h^ |
6. 1=15 |
81. Problems to be solved by arithmetic or algebra. Many problems may be solved either by arithmetic or by the use of the equation. When the solution of a problem is made by the use of the equation, it is commonly called an algebraic solution.
1. Divide a pole 20 ft. long into two parts so that one part shall be 4 times as long as the other.
Arithmetic Solution
The shorter part is a certain length.
The longer part is 4 times this length.
The whole pole is then 5 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 4X4 ft., or 16 ft.
Hence, the parts are 4 ft. and 16 ft. long.
36 FIRST-YEAR MATHEMATICS
Algebraic Solution
Let n be the number of feet in the shorter part. Then 4n is the number of feet in the longer part, and n+4n, or 5n=20 n= 4 4n = 16
Hence, the parts are 4 ft. and 16 ft. long.
2. A farmer wishes to inclose a rectangular pen with 80 ft. of wire fencing. He wishes the pen to be 3 times as long as it is wide. How long shall he make each side ?
Algebraic Solution
Let X be the number of feet in the smaller side. Then Sx is the number of feet in the longer side,
and X +3x, or 4a:, is the number of feet half-way round the pen .
4x = 40
a; = 10
3a: = 30
Hence, the sides are 10 ft. and 30 ft. long.
Problems to Be Solved by the Aid of the Equation
82. Important steps in the algebraic solution of prob- lems. The solutions of the foregoing problems 1 and 2 illustrate the following important steps :
a) In every problem certain facts are given, or known, and others are to be determined.
b) In solving the problem one of the unknown numbers is denoted by a symbol, as x.
c) Then all the given facts are expressed in algebraic language, the number x being used as if it were known.
d) Two different expressions for (i.e., denoting) the same number are equated (joined by the sign = of equality) .
THE EQUATION 37
e) The solution of the equation thus obtained gives the required values of the unknown number.
/) The correctness of these values is tested by sub- stituting them in the conditions of the problem. The result is correct if these conditions are satisfied.
EXERCISES
83. Geometric problems. The following problems con- tain geometric relations and are to be solved by the aid of the equation: ^
1. The perimeter of an equilateral quadrilateral is 48 ft. Find a side.
Denoting the side by s the perimeter may be represented by 4s or by 48. Hence, 4s = 48.
2. The perimeter of an equilateral hexagon (6-side) is 186. Find a side.
3. The perimeter of an equilateral decagon (10-side) is 285. Find a side.
4. The perimeter of an equilateral dodecagon (12-side) is 264. Find a side.
5. The opposite sides of the quadrilat- eral, Fig. 39, are equal. The perimeter is 432 yd. and the length is 26 yd. greater ' wTW than the width. Find the dimensions. Fig. 39
6. A Maypole 22 ft. high breaks into two
pieces so that the top piece, hanging beside the lower piece, lacks 6 ft. of reaching the ground. How long is each piece? Draw a figure.
7. Draw a figure representing a regulation football field, the length of which is 663 yd. greater than the width, the sum of the length and width being 1633 yards.
8. Make a sketch of a rectangle, one side of which is 12 ft. longer than 3 times as long as the other, the perimeter being 80 feet.
38 FIRST-YEAR MATHEMATICS
9. A garden is 5 times as long is it is wide. It takes 120 yd. of fence to inclose it. Find the length and width.
10. A side of a lot 126 ft. long is to be divided into two parts so that one is 6 times as long as the other. Find the length of the short part.
84. Problems expressing number relations. The fol- lowing problems give training in expressing given number relations in algebraic form :
1. If 4 times a number is decreased by 5 the result is 35. Find the number.
2. If i% of a number is increased by 12 the result is 20. Find the number.
3. If 12 is subtracted from 2 times a number the result is the same as the number increased by J of itself. Find the number.
4. Six times a number increased by 2 of itself equals 11. Find the number.
5. Four times a number increased by i of itself is the same as twice the number increased by 11. Find the number.
6. John was asked to think of a number, to treble it, and to add 4. He obtained 25 as a result. What was the original number ?
85. Consecutive number problems.
1. Find two consecutive numbers whose sum is 203. Let X be one number and x-\-l the other.
2. Find three consecutive numbers whose sum is 474.
3. Find two consecutive odd numbers whose sum is 204. Let X be one number and x+2 the other.
4. Find two consecutive even numbers whose sum is 378.
5. Find three consecutive even numbers whose sum is 372.
THE EQUATION 39
86. Miscellaneous problems.
1. James has 3 times as much money as Charles and 4 times as much as William. All together they have 57 cents. How many cents has each ?
2. A farmer has 3 times as many sheep as his neighbor. Having sold 22 sheep, he has the same number as his neighbor. How many did each have before the sale?
3. Two men divide $5,247 between them. One receives $324 more than twice as much as the other.. How much does each receive ?
4. A purse of $75 was divided between two persons. One received $27 more than the other. How much did each receive ?
5. Three men. A, B, and C, wish to divide 1,584 shares of stock among themselves so that A shall have 25 more than B and C shall have 50 more than B. How many shares must each receive ?
6. At an election in the Freshman class of a certain high school 84 votes were cast. Candidates A and B each received a certain number of votes; candidates C and D each received
3 times as many as candidate A; and candidate E received 4 times as many as D plus 4. How many votes did each candi- date receive ?
7. A man divides his 160-acre farm as follows: He takes a certain number of acres for lots, 4 times as much for pasture,
4 times as much for corn as for pasture, 2 as much for wheat as for corn, and 15 acres for meadow. How many acres does he assign to each purpose ?
8. The manager of a high-school football team received from the manager of a visiting team a check for $63 for expenses. Cashing the check he obtained $2 and $5 bills, the same number of each. How many bills of each kind were there ?
9. A box of oranges was sold for $3 . 60, giving a profit equal to -5- of the cost. What was the cost ?
40 FIRST-YEAR MATHEMATICS
10. Two partners in business make a profit of $5,248. Of this amount they decide to give $325 to charity. The remainder is to be divided between them so that one receives twice as much as the other. How much does each receive ?
11. The annual income of a family is divided as follows: One-tenth is used for clothing, one-third for groceries and meat, milk, and help, and one-fifth for rent. This leaves $1,320 for other expenses, and for the savings account. How much is the income ?
12. The manager of a business receives a salary of $2,000 a year and 1 per cent of the year's profit. In one year the profit of the business was 10 per cent of an income of $700,000. What was the manager's income that year ?
13. A boy has $4 in his bank and saves 40 cents each week. His brother has $16 and draws 10 cents each week. After how many weeks will they have equal amounts in the bank?
14. A bicyclist plans to take a trip of 148 mi. in 6 days. He intends to ride a certain number of miles on Monday, | as many on Tuesday, as many on Wednesday, f as many on Thursday, and 2 as many on Friday. Having followed this schedule he finds it necessary to ride 20 mi. on Saturday. How many miles did he ride each day ?
15. After 8 years a father will be 4 times as old as his son is now. How old is the father, the present age of the son being 12 years ?
16. Show how to divide a sum of $1,278 among three persons in such a way that the share of the first shall be 3 times that of the second, and the share of the second twice that of the third.
17. Divide $260 into two parts such that one is $24 more than the other.
THE EQUATION 41
EXERCISES
87. Solve the following equations:
^' 2^4 3
Solution: Multiplying both sides of the equation by the least common multiple of the denominators, i.e., by 12: 12y 12?/^ 12X1 2 "^ 4 3
Then 6?/ +3?/ = 4, by reducing the fractions. This clears the
equation of fractions
9y = 4, by combining like terms
4 y—Q, by dividing both sides of the equation by 9
Check the result.
2. 252-17 = 113
3. 28a;+14 = 158
4. 28x-9 = 251
5. 20a;+2a:-18a: = 22
6. 17?/ -3?/+ 16?/ =105
7. 17s+7s-13s = 88
8. 16^+2«-13^ = 22i
9. 321a:-109a;+8a: = 22
10. 404?/ -304?/+ 12^ = 560
11. 3.4x-1.2x+4.8x = 70
12. 3.5a;+7.6a:-8.6:c=15
13. 5.8?/-3.9?/+12.6?/ = 58
14. 6s-3.5s+5.5s = 68
15. 6.82s+1.18s-3.54s = 42
16. 8a;-4.5x+5.2x = 87
17. 16.5a:+15.8-2.3x = 186.2
18. 6.15?/-1.65?/+7.8 = 57.3
19. 8?/+6.875+2?/=46.875
20. 2+5.372-8.73 = 61.34
21. 13^-8.75^+6.87 = 57.87
22. 15x+3.73a;-9.23 = 65.69
23. 3x+7x-\-15x-2x+5 = 74. 38. f:c+Ja;-|a: = 18
24. 2x+7x-3x-6 = 24 39. 6a;-fa:-ia;+7 = 129
25. |
ix=Q |
26. |
ix = 3 |
27. |
h=s |
28. |
|a; = 20 |
29. |
^» |
30. |
X |
31. |
^- |
32. |
x+h = Q |
33. |
x-ix = 7 |
34. |
i+j-' |
35. |
l-h' |
36. |
H- |
37. |
H-« |
42 FIRST-YEAR MATHEMATICS
Summary
88. This chapter has taught the meaning of each of the following terms: members of the equation, substitution, satisfying the equation, root of the equation, solving and checking equations.
89. In solving equations the following axioms are used : 6, 7. If the same number or equal numbers be added to
equals the sums are equal (addition axiom).
8. 7/ the same number or equal numbers be subtracted from equals the remainders are equal (subtraction axiom) .
12. If equals be divided by equal numbers {excluding division by 0) the quotients are equal (division axiom).
IS. If equals be multiplied by the same number or equal numbers the products are equal (multiplication axiom).
90. In many problems the algebraic solution is simpler than the arithmetical solution.
91. In solving verbal problems algebraically it is helpful to observe the following steps :
Denote by a letter the unknown number called for in the problem.
Express the given data in algebraic form.
Obtain the equation by equating two expressions de- noting the same number.
Solve the equation and check the results.
92. An equation containing fractions may be cleared of fractions by multiplying every term in the equation by the least common multiple of the denominators and then reducing all fractions to the simplest form.
93. Some geometric problems may be solved by the aid pf the equation.
T H A L E S OF MILETUS
T HALES was born at Miletus in Asia Minor about 640 B.C. and died there about 542 b.c. He was probably of Phoenician parentage. In early life he was a merchant and traveled in Egypt in the pur- suit of business. While there he studied Egyptian science, and it is said he amazed the king by determin- ing the height of a pyramid by measuring its shadow. On his return home he abandoned business, opened at Miletus the first Greek school of philosophy, about 600 B.C., and devoted his life thenceforth to scientific and philosophical pursuits. He became famous as an astronomer, a mathematician, and a philosopher. He was the first to study mathematics scientifically. In proving mathematical truths he used the super- posing of figures and other reasoning methods. He was the originator of the mathematical method of 'Hhinking it out."
His school lasted over a hundred years after his death. The Pythagorean school grew out of it. The geometry of Thales' school consisted of half a dozen of our most elementary theorems, but it is worth while noting that he insisted that they be proved by sound reasoning.
Students will be interested to look up in some bio- graphical dictionary or history of mathematics the story of his cornering the olive presses to illustrate his shrewdness, the story of his trouble with the badly behaved donkey to illustrate his sagacity, the story of his tumbling into a ditch to show his absent-mindedness, and the story of his prediction of the eclipse which later gave him a place among the Seven Sages of Greece.
Gow's History of Greek Mathematics gives a good account of Thales. See also Smith's Dictionary of Greek and Roman Biography.
CHAPTER IV
ANGLES
Classification of Angles
94. Angle. Clockwise and counter-clockwise direction of turning. What part of a complete turn does the long hand of a clock make (Fig. 40) when it rotates (turns) about the center post from 12 to 3? 12 to 6? 12 to 9? 12 to 4? 12 to 8? 12 to 2? 12 to 10? If a line, as OA, Fig. 41, be imagined to rotate in a plane about a fixed point, 0, in the direction indicated by the arrowheads, until it reaches the posi- ^^^' ^^
tion OB, it is said to turn through an angle, a. The same angle may be formed by rotating OB about 0 until it
takes the position OA. In the latter case the direction of turning is clockwise; in the first case it is counter-clockwise, i.e., opposite to the direction in which the hands of a clock move. The amount of rotation needed to bring OA to the position 05 is the angle formed by OA and OB. The word ''angle" comes from the Latin angulus, meaning corner.
Fig. 41
95. Symbols for angle.
angles it is ^ .
The symbol for angle is Z ; for
43
44 FIRST-YEAR MATHEMATICS
96. Notation for angles. An angle may be denoted by a small letter written within the angle, as a in Fig. 41. Often it is convenient to denote an angle by the point of intersection of the lines forming the angle prefixed by the angle symbol, as Z 0. Sometimes three letters are used, as ZAOB or Z BOA, the first and last letters denoting points on the lines forming the angle, the middle letter denoting the point of intersection.
Fig. 42 ^'^' Sides. The Unes AO and BO,
Fig. 42, forming the angle AOB, are the sides of ZAOB.
98. Vertex. The point of intersection of the sides of an angle is the vertex of the angle. ''Vertex" is a Latin word, meaning turning-point.
99. Right angle, straight angle, perigon. If a line rotating in a plane about one of its points makes i of a complete turn, the angle formed is a right angle, Fig. 43. If the line makes a half turn, the angle is a straight angle,
rr\
a
Fig. 43 Fig. 44 Fig. 45
Fig. 44. Thus, a straight angle is an angle whose sides lie in the same straight line on opposite sides of the vertex. If the line makes a complete turn, the angle is a perigon, Fig. 45.
1. Draw a right angle; a straight angle; a perigon.
2. Draw an angle made by a line which has rotated f of a complete turn; ^ of a complete turn.
3. Show that a perigon is equal to two straight angles or four right angles.
ANGLES
45
100. Acute and obtuse angles. An angle less than a right angle is an acute (sharp) angle, Fig. 46. An angle greater than a right angle and less than a straight angle is an obtuse (blunt angle, Fig. 47).
1. Draw an acute angle; an obtuse angle. 2. Point out a number of right angles in the classroom.
Fig. 46 Acute angle
Fig. 47 angle.
-Obtuse
3. In the letter A, Fig. 48, point out the acute angles; the obtuse angles.
Fig. 48
The Measurement of Angles
101. Comparison of angles. Angles are compared by placing their planes one over the other so that the vertex and a side of one angle coincide (fit) with the vertex and a side of the other. If the other sides of the angles coincide, the angles are equal. If the other sides do not coincide, the angles are unequal, the smaller angle being the one* whose second side falls within the other angle.
1. Compare angles A and B, Fig. 49; B and C.
Fig. 49
To place angle B on A make a trace of Z J5 on thin paper and fit this on Z A.
2. Draw two angles that appear to you to be equal and com- pare them as in problem 1.
46
FIRST-YEAR MATHEMATICS
Fig. 50
102. Size of an angle. Since an angle is forined by rotating a line about a fixed point, the size of the angle
depends entirely upon the amount of turning and not upon the length of the sides. Thus, Z P is greater than Z 0, Fig. 50, although the sides of Z 0 are longer than the sides of Z P.
103. Graphical addition and subtraction of angles. If
one angle is placed adjacent to another, so that the vertex and one side coincide, the angles are said to be added. Thus,
AAOC=AAOB+ABOC,
Fig. 51. To sub- j,^^ ,,
tract one angle
from another, place the first on the second so that they have the vertex and one side in common. Thus, AAOC=AAOB-ABOC, Fig. 52. 1. Draw two unequal angles and find
Fig. 52
the sum; the difference.
Use thin paper, as in problem 1, § 101.
2. Let X, y, and z de- note three angles so that x>y,y>z- Dmwx-^-y-^-z; x—y-\-z; x-\-y — z.
104. Degree, min- ute, second. The right angle XOY, Fig. 53, is divided into 90 equal o
Fig. 53
ANGLES
47
parts. Each of these is a degree (1°). The degree is used as angle unit. A degree can be divided into 60 equal parts, called minutes (10 . Each minute contains 60 equal parts, called seconds (V). The Greek mathe- matician Hypsicles (180 b.c.) was the first to divide an angle into degrees.*
EXERCISES
1. What part of a perigon is a degree ? Of a straight angle ? Of a right angle ?
2. Read the following: 13°24'3.5".
3. How many seconds are contained in an angle of 20°14'22"?
4. How many degrees are in ZXOA, Fig. 53? In ZXOBf In ZAOB? In ZXOC? In zAOCf
5. How many degrees are contained in the angle made by the long hand of the clock when it has rotated from 12 to 3? 12 to 6? 12 to 9? 12 to 4? 12 to 8? 12 to 2? 12 to 10?
6. How many degrees, right angles, straight angles, are in the angle formed by the hands of a clock at 9 o'clock? At 6 o'clock? At 4 o'clock? At 2 o'clock?
7. Draw freehand angles containing 45°, 60°, 30°, 90°, 180°, 360°.
8. How many degrees are in two straight angles ? In four right angles ? In I of a right angle ? In } of a right angle? In g- of a right angle? In r right angles ?
105. Circle. Center. Opening the compass and placing the sharp point at a point P, Fig. 54,- the pencil can be made to draw a curved line, called a circle. Thus, the circle is a closed curved line, all points of which lie in the
* See Ball, p. 85.
Fig. 54
48
FIRST-YEAR MATHEMATIQS
same plane and are equally distant from a fixed point within the curved line. The fixed point is the center of the circle.
Some writers use the word ''circumference" in the sense in which we use ''circle."
The Use of the Protractor in Measuring Angles
106. The protractor. The protractor, Fig. 55, is an instrument for measuring angles. The circular rim is divided into 180 equal parts. If straight lines were
drawn from the center, 0, to each mark on the rim, 180 equal angles would be formed at 0, each of which would be one degree (1°).
107. Arc-degree. A part of the circle, Fig. 55, included between two consecutive marks is an arc-degree.
1. Suppose a pointer to rotate about the fixed point, 0, Fig. 55, in the direction indicated by the arrow. In rotating from the position OX to the position OZ, through how many right angles would it turn? Through how many straight p,ngles ? Through how rnany degrees ?
ANGLES
49
2. Find the number of degrees and the number of right angles in each of the following angles of Fig. 55: XOZ; AOZ; COZ; XOA; XOB; XOY; XOC; XOD; AOB; AOY; BOC; BOD.
108. Radius. A line-segment drawn from the center to a point on the circle is a radius. Thus, OX is the radius of the circle XYZ, Fig. 55.
109. Arc. A part of a circle is an arc. An angle whose vertex is at the center of a circle intercepts the arc cut off by its sides.
110. Semicircle. Quadrant. An arc equal to one- half of a circle is a semicircle. An arc equal to one-fourth of a circle is a quadrant.
Fig. 56
111. To measure an angle. To measure an angle, as XOB, Fig. 56, by means of a protractor, place the pro- tractor on the angle, making the center of the circle fall on the vertex 0 of the angle, and making the line from the center to the zero-reading fall along one side, OX. of the angle. Then read along the circle from the zero-readir)g to the point where the side OB cuts the circle. Thus, Z XOB = 30°.
50
FIRST-YEAR MATHEMATICS
EXERCISES
1 . Draw an angle . By use of the protractor find the number of
degrees in the angle.
2. Draw a tri- angle. How many angles does it have? Read each angle in three letters.
Fig. 57
3. Fill out the table, Fig. 58, with reference to the triangle ABC, Fig. 57.
4. Draw a triangle. Fill out a table like Fig. 58 with refer- ence to this triangle.
Angle |
Classify Angle lAcute, Obtuse, Etc.) |
Amount of Rotation (Complete, Half, Less than Half, Etc.) |
No. of Degrees (Estimated) |
No. of Degrees (Measured) |
Error |
a |
|||||
b |
|||||
c |
|||||
Sum |
Fig. 58
The Sum of the Angles of a Triangle
112. The sum of the angles of a triangle. From exercises 3 and 4 of § III, it is seen that the sum of the angles of a triangle is a straight angle, or 180°.
113. Theorem. It can be demonstrated (proved) by geometry that the sum of the angles of any triangle is a straight angle.* A statement to be proved is called a theorem.
* The first mathematician to prove this theorem probably was Thales (640 B.C.-550 B.C.). Ball, pp. 16-17.
ANGLES
51
EXERCISES
1. Draw a triangle. Tear off the corners and place the angles adjacent to each other as in Fig. 59. What seems to be the sum of the angles of the triangle ?
Fig. 59
2. That the sum of the angles of a triangle is 180° can be shown by rotating a stick or a pencil successively through the angles, as follows: Draw a triangle, Fig. 60. Place a pencil or stick in position 1 and note the direction it is pointing. Rotate the pencil through angle x. Then move it along AB io position 2. Turn the pencil through angle y and move it along BC to position 3. Turn it through angle z to position 4. The pencil has now rotated through an Fig. 60
amount equal to x-\-y-\-z. Note the
direction the pencil is pointing in the last position. Through what part of a complete turn has it rotated? Through how many right angles? Through how many degrees?
3. State by an equation the number of degrees in the sum of the angles x, y, and s of a triangle.
By means of this equation it is possible to find an angle of a triangle when the other two are known. This is of great importance to surveyors, as it enables them to find all angles of a triangle although they may be able to measure directly only two angles.
114. Problems. Solve the problems below, observ- ing the following steps: Make a sketch of the triangle, denoting the angles as given in the problem. Use the
52 FIRST-YEAR MATHEMATICS
theorem of § 112 to obtain the equation. Solve the equa- tion and find the values of the angles.
1. The angles of a triangle are Sx, x, and dx. Find the numerical values.
2. Find the value of each angle of a triangle, the first angle of which is twice the second, the third being 3 times the first.
3. Find the angles of a triangle if the first angle is 6 times the second, and the third is 2 of the first.
4. The three angles of a triangle are equal. Find them.
5. One angle of a triangle is 27°. The second angle is 27° larger than the third. How large is each angle ?
6. One angle of a triangle is | as large as another. The third is 3 times as large as the first. How large is each angle ?
7. Find the angles of a triangle if the first is 3 of the second, and the third is 2 of the first.
8. Find the angles of a triangle if the first angle is 18° more than the second, and the third is 12° less than the second.
9. The difference between two angles of a triangle is 20°, and the third angle is 36°. Find the unknown angles.
10. Find the angles of a triangle if the first is 25° more than the second, and the third is 3 times the first.
11. Find the angles of a triangle if the first angle is double the second, and the third is 3 times the first, less 9°.
12. Find the angles of a triangle if the first is 4 of the second, and the third is 7 of the first, plus 18°.
13. Find the angles of a triangle if the first is 3| times the second, minus 8°, and the third is j of the second.
14. Find the angles of a triangle if the first is 6 times the second, plus 18°, and the third is 2 of the first, minus 7°.
15. Show that a triangle cannot contain more than one right angle; not more than one obtuse angle.
ANGLES 53
16. Two angles, a and h, of one triangle are equal respectively to the angles r and s of another triangle. Show that the third angle, c, of the first triangle equals the third angle, t, of the other.
The Sum of the Exterior Angles of a Triangle
115. Exterior angles. If a side as A 5 of a triangle ABC, Fig. 61, is extended the angle s, formed by the consecutive side BC and /
the extension of A 5, is an exterior angle.
1. How many exterior angles are there at each corner?
2. How many exterior angles has a tri- angle ?
3. Draw a triangle. Prolong one side at
each vertex and measure the three exterior angles formed. What is the sum of the exterior angles, if one angle is taken at each vertex?
4. In Fig. 62, find the sum of the three exterior angles by
rotating a pencil as indicated.
5. Show that the sum of the exterior angles of a triangle, taking one at each vertex, is 360°.
Suggestion: How many degrees are in the sum x+w, Fig. 61 ?
In z-\-tl In y+s? Fig. 62 Show that
(x+w^) + (3+0 + (2/+s)=3Xl80°. = 540°. This equation may be written:
ix+y-{-z) + {w-\-s+t) =540°. Why?
But x±y±z =180°. Why?
Therefore, w+s-\-t =360°. Why?
54
FIRST-YEAR MATHEMATICS
Fig. 63
116. Interior angles. The angles of a triangle are called interior angles when contrasted with the exterior angles.
117. Proof. Reasoning like that of problem 5, § 115, is very common in geometry. Such reasoning is called proving, or proof.
EXERCISES
1 . The three exterior angles of a triangle are equal. Find the value of each exterior and interior angle.
2. Find the unknown interior and exterior angles of the triangles of Figs. 63 and 64.
3. Draw freehand the following angles: 30 135°, and by means of a protractor test the accuracy of your
drawings.
4. In the mariner's compass, Fig. 65, state what angles are formed by these directions : E and SE; W and NE; N and E; NW and SE; S and NW; WSW and E; ENE and WNW.
5. Divide an angle into two equal parts, first by estimating, then by means of the protractor.
Fig. 65 6. Fold a piece of paper, Fig. 66,
as in Fig. 67. Fold again, making
Unfold the paper. Fig. 68. The
Fig. 64 45°, 60°, 90°,
the edge OA fall along OB.
'<%'
Fig. 67
D
Fig. 68
ANGLES 55
creases are seen to form four angles x, y, z, w. Why are these angles equal ? Why are they right angles ?
118. Problem 6 illustrates the theorem that all right angles are equal.
1. Show by measuring with the protractor that a/i exterior angle of a triangle equals the sum of the two remote interior angles.
2. Show by measuring that the two exterior angles at the same vertex of a triangle are equal.
119. Classification of triangles as to angles. A triangle all of whose angles are acute is an acute triangle. A triangle one of whose angles is obtuse is an obtuse triangle. A triangle one of whose angles is a right angle is a right triangle. An equiangular triangle is a triangle whose angles are equal.
Show that if one angle of a triangle is a right angle the other two are acute angles.
120. Measurement of angles out of doors. To meas- ure angles in the open, e.g., the angle formed by two roads, an angle-measurer can be made as follows:
On a drawing board, Fig. 69, tack a protractor. Place the board upon a table or fasten it to a tripod and place the center of the circle of the protractor as nearly as possible pj^ gg
over the vertex of the angle
to be measured. Bring the table into horizontal position. Use a ruler, with a pin stuck in it at each end, to sight in the direction of each side of the angle and each time note the reading on the protractor. The difference between these readings is the number of degrees in the angle.
56
FIRST-YEAR MATHEMATICS
When accuracy is important, as in astronomy and surveying, better instruments are used (see the engineer's transit, Fig. 70). The principal parts of these instru- ments are the graduated circles for taking the readings for the sides of the angles, and the small telescope for sighting in the direction of the sides.
To Draw an Angle Equal to a Given Angle
121. By use of protractor to draw an angle of given size. Let it be required to draw an angle of 60°.
Draw an indefinite line as AB, Fig. 71. Place the protractor with the center of the circle on A and with the zero reading on AB. At the 60° mark place a point C. Remove the protractor and draw AC. Angle BAC is the required angle.
1. With a protractor draw the following angles: 80°; 92|°; 170°; 36°; 91°; 160°.
2. Which of the angles in problem 1 are acute ? Obtuse ?
Fig. 70.— Engi neer's transit.
122. Oblique angles, oblique lines, perpendicular lines.
Acute and obtuse angles are oblique angles. The sides of oblique angles are said to be oblique to each other;
ANGLES 57
the sides of a right angle are said to be perpendicular to each other.
123. By use of the protractor to draw an angle equal to a given angle.
Measure the given angle with the protractor and construct an angle containing the same number of degrees, using the method of § 121.
An angle may be constructed equal to a given angle by use of ruler and compass only. This construction is a consequence of the following relation between angles, whose vertices are at the center of a circle (central angles), and their intercepted arcs.
124. Theorem: Equal central angles in the same or equal circles intercept equal arcs.
For angle A, Fig. 72, may be placed upon angle B, making ZA coincide with /.B. (Why fs this possible ?) Then circle Fig. 72
A will coincide with circle B.
(Why ?) This will make C fall on E and D on F. Therefore arc CD coincides with arc EF.
125. Theorem: In the same or equal circles equal arcs are intercepted by equal central angles.
For, circle A, Fig. 72, may be made to coincide with circle B so that arc CD coincides with arc EF. (Why can this be done?) Then ZA must coincide with ZB. (Why ?)
126. The theorems in §§124 and 125 explain why the protractor may be used to measure angles. For every angle-degree at the center intercepts an arc-degree on the rim of the protractor. Every angle therefore contains as many angle-degrees as there are arc-degrees in the intercepted arc. This may be expressed briefly thus: A central angle is measured by the intercepted arc.
58 FIRST-YEAR MATHEMATICS
127. Problem: At a given point on a given line to construct an angle equal to a given angle*
Let ZABC, Fig. 73, be the given angle and let D be the point on the given line EF.
c
K
Fig. 73
Using B and D as centers and the same radius draw arcs GH and IK.
With a radius equal to GH and / as center draw an arc meeting IK at L.
Draw line DL.
Angle IDL is the required angle.
Check the correctness of the construction by measuring AGBH and IDL with the protractor,
EXERCISES
1. Draw a right angle by using a right tri- angle, Fig. 74.
2. Draw a right angle by using the protractor (see §121).
3. Draw a right angle by using ruler and compass.
Draw a line-segment, as AB, Fig. 75. yfF
Using a point C on AB and a convenient radius, draw arcs of a circle meeting AB in the points D and E.
With D and E as centers and a con- venient radius draw arcs meeting at F. ^^^- '^ Draw CF. Angle BCF is the required angle.
* Oenopides of Chios (500 B.C.-430 b.c.) has the reputation of being the first to solve this problem (see Cajori, p. 19).
ANGLES 59
4. Draw an angle equal to the sum of two given angles, using the protractor.
Measure the given angles with a protractor and draw an angle containing as many degrees as are contained in the given angles together.
5. Using ruler and compass draw an angle equal to the sum of two given angles.
Let A ABC and DEF, Fig. 76, be the given angles whose sum is to be constructed.
^.
H
Construction: Draw a line-segment, as GH.
With B, E, and G as centers and the same radius
draw arcs AC, DF, and IL. Draw arc IK = a.rc AC and arc KL — arc FD. Angle IGL is the required sum. Proof: ZABC=ZIGK. Why?
ZDEF=ZKGL. Why? ZABC+ZDEF=ZIGK+ZKGL. Why? ZABC+ZDEF= ZIGL. Why?
6. Let X be a given angle. Draw 2a;; 3a;; using the methods of problems 4 and 5.
7. Let X, y, and z be three angles. Construct x-\ry-]rz.
8. Draw the differences of two given unequal angles, using the protractor.
9. Using ruler and compass draw an angle equal to the difference of two given unequal angles.
Proceed as in problem 5, but lay off KL from K in the direction opposite to KL, i.e., on KI .
10. Of the given angles x, y, z, angle x is greater than y and y is greater than z. Draw an angle equal to x — y-\-z; x-\-y — z.
60 FIRST-YEAR MATHEMATICS
11. Using a protractor, divide an angle into 2, 3, 4 , etc.,
equal parts.
Measure the angle and take 2, i, i , etc., of the numeri- cal measure.
12. Using ruler and compass divide an angle into two equal parts.
Let ZABC, Fig. 77, be the given angle.
With B as center and any radius draw an arc cutting BA and BC at D and E.
With D and E as centers and a con- venient radius, draw arcs meeting at F.
Draw BF.
BF divides ZABC into two equal parts.
128. Bisecting a magnitude. To divide a magnitude into two equal parts is to bisect it.
1. Two ports A and B are 15 miles apart. They contain cannon of 11-mile range. Make a figure showing the space these guns can shell.
2. Bisect a straight angle, using the method of problem 12, § 127. Compare the construction with that of problem 3, § 127.
3. Using the construction of problem 2, construct a perpen- dicular to a given line at a given point in it.
Summary
129. In this chapter the meaning of the following terms was taught: angle; acute, right, obtuse, straight, and oblique angle; clockwise and counter-clockwise direction of turning; perigon; sides and vertex of an angle; equal and unequal angles; size of an angle; angle- degree and arc-degree; minutes, seconds; circle; center and radius of a circle; arc of a circle; quadrant and semicircle; intercepted arcs; theorem; proof; interior and exterior angles of a triangle; acute, obtuse, right, oblique, and equiangular triangle; to bisect a magnitude.
ANGLES 61
130. The following symbols have been introduced: Z for angle, A for angles, (°) for degree, {') for minute, {") for second.
131. Angles may be denoted by a single small letter written within the angle, or by a capital letter written outside of the angle at the vertex, or by three letters, the first and last denoting points on the sides and the middle letter denoting the vertex.
132. The truth of the following theorems has been shown :
1. The sum of the angles of a triangle is a straight angle.
2. The sum of the exterior angles of a triangle is 360°.
3. An exterior angle of a triangle equals the sum of the two remote interior angles.
4. Equal central angles in the same or equal circles inter- cept equal arcs and equal arcs are intercepted hy equal central angles.
5. A central angle is measured hy the intercepted arc.
133. The protractor is an instrument for measuring angles and for drawing angles equal to given angles.
134. Ruler and compasses were used to add and sub- tract given angles; to draw an angle equal to a given angle; to bisect an angle; to construct a perpendicular to a given line at a given point in the line.
CHAPTER V
AREAS AND VOLUMES. MULTIPLICATION
The Area of a Square
135. Original meaning of geometry. When land is to be sold or divided, it becomes necessary to measure it. The Greek writer Herodotus tells that in Egypt originally the land had been divided equally among the Egyptians. Whenever the river Nile washed away part of the land, the Egyptians actually had to rediscover their lands after the flood, and in order to adjust the taxes overseers were sent out by the king to measure out by how much the land had become smaller. Thus, measuring was the beginning of geometry.
The word ^'geometry" comes from the Greek words ge, meaning earth, and metron, meaning measure.
136. Kinds of quadrilaterals. Pieces of land, fields, pastures, lots, etc., often have the form of polygons. The following are the most familiar quadrilaterals.
\ |
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/ / |
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/ \ Trapezoid |
\ \ |
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Parallelogram |
Rectangle |
Fi( |
Square 3.78 |
Rhombus |
EXERCISES
1. What quadrilaterals in Fig. 78 are equilateral? Equi- angular ?
2. What quadrilaterals contain right angles ?
3. In what respect does the square differ from the rectangle ? From the parallelogram ? From the rhombus ?
62
AREAS AND VOLUMES. MULTIPLICATION 63
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Fig. 79
4. In what respect does the rhombus differ from the parallelo- gram?
137. Area of a square the length of whose side is a whole number. To measure a square place it upon squared paper, as A BCD,
Fig. 79, and count the num- ber of square centimeters con- tained in it.
This may be done as follows: Suppose the side AB to be 3 cm. long. The square can then be divided into 3 vertical rectangular strips, as AEFD. Each strip, being
3 cm. high, can be divided into 3 equal sq. cm., as HGFD. There- fore ABCD contains 3X3 sq. cm., or 9 sq. cm.
The result, 9, is the area of ABCD. One square centimeter, as HGFD, is the unit of area.
1. Find in a similar way the area of a square whose side is
4 cm. long.
2. Find the area of a square whose side is 6 cm. long.
The foregoing problems show that the number of square units contained in a square may be obtained by multiply- ing the number of units of length in one side by itself. This may be stated briefly trhus:
The area of a square is equal to a side multiplied by itself.
Since the length (base) of a square is equal to the width (altitude), the area A of a square whose side is a units long is given by the equation A=aXa. This may be expressed in another form, thus: A equals a-square, or in symbols, A = a^, a^ meaning aXa.
138. Formula. In the two preceding paragraphs two methods of finding the area of a square are given. First,
64 FIRST-YEAR MATHEMATICS
the geometric method. A drawing of the square is made and the number of unit squares contained in it is counted. Second, the algebraic method. The length of the side is substituted in the general expression A=a^. The advan- tage of the last method over the first is apparent. The equation A=a^ is called a formula for finding the area of a square.
Find by means of the formula the areas of squares whose sides have the following lengths : 4; 6; 10; 100; x; I; s.
139. Units of area. The standard units of area are the square meter (sq. m.), square decimeter (sq. dm.), square centimeter (sq. cm.), square millimeter (sq. mm.), square foot (sq. ft.), square inch (sq. in.), and square yard (sq. yd.). Their sides are respectively of the following lengths: 1 m., 1 dm., 1 cm., 1 mm., 1 ft., 1 in., and 1 yd.
The area of land is expressed in acres, an acre being a quantity of land containing 4,840 square yards.
Show that 1 sq. m. = 100 sq. dm.; 1 sq. dm. = 100 sq. cm.; 1 sq. cm. = 100 sq. mm.
140. Area of a square the length of whose side is a fraction. In § 137 it was shown that the area of a square the length of whose side is a whole number (i.e., an integer) may be found from the formula A =a^. It will be shown that the same formula may be used to compute the area of a square the length of whose side is a fraction.
EXERCISES
1. Find the area of a square whose side is 2\ cm.
Draw a square on a line-segment 2\ cm. long and divide it as in Fig. 80. Find the sum of the parts of the large square.
2. Verify the result of exercise 1 by letting a = 22 in the formula A = a^.
AREAS AND VOLUMES. MULTIPLICATION
65
2 |
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Fig. 80
3. Find the area of a square whose side is 1.5 m.
By taking a smaller unit of length the side of this square may be expressed as an integral number: since 1 m. = 10 dm., 1.5 m. = 15 dm. Using the formula A=a'^, we have ^ = (15)^ sq. dm.= 225 sq. dm. = 2. 25 sq. m., since y^Q sq. m. = l sq. dm.
4. Verify the result of exercise 1 by solving it by the method of exercise 3.
Change a = 2| cm. to
a =2. 5 cm. =25 mm.
5. Find the area of a square whose side is 2f m. Change 2 J m. to 2.75 m. =275 cm. Then apply A =o?.
6. In the formula A = a? let a = 2f m. and show that the result obtained agrees with that of exercise 5.
Exercises 1-6 shov^ that when the length of the side of a square is fractional, as a = 1 J m., this fraction may be changed to a decimal, a = 1 . 25 m., which may be expressed as an integral number in terms of a smaller unit of length, a = 125 cm. By means of the formula A=o? the area of the given square may then be expressed in terms of a smaller unit square, A = 15,625 sq. cm. Since 1 sq. cm. = To^.Vcnr sq. m., the area may now be expressed in terms of the original square unit. Thus, A =1.5625 sq. m. However, this result agrees with the result obtained by let- ting a = 1 J- m. in A =o?. For, A = (llY sq. m. = f f sq. m. = 1 . 5625 square meters.
It follows that the formula A=a^ may be applied to squares the lengths of whose sides are fractional or integral numbers.
This result may be expressed in the form of the follow- ing theorem:
Theorem: The area of a square is equal to the square of one side. A = a^
66 FIRST-YEAR MATHEMATICS
Area of a Rectangle
141. Formula for the area of a rectangle. The area of a rectangle may be expressed by a formula found in a manner similar to finding the formula for the square.
EXERCISES
1. Denoting the base of a rectangle by b and the altitude by h, draw on squared paper a rectangle for which 6 = 3 cm. and h = 4 cm. Find the area by counting the number of square centi- meters in the rectangle.
2. Draw rectangles for the following pairs of values of b and h : 6 = 3.4 cm., h = 2cm.; 6 = 4.2 cm., h = 2 cm,, and find the areas by counting the number of square centimeters.
3. By the method of § 141 find the area of rectangles having a = 2.8 m., h = 3.6 m., a = 4.8 m., h = l.S meters.
4. Show that the results of exercises 1-3 may be obtained from the following theorem:
Theorem: The area of a rectangle equals the product of the base by the altitude.
A = bXh
EXERCISES
1. Find the perimeter and area of a square whose side is 4 ft.; a; ft.; .06 cm.; 2.48 meters.
2. The perimeter of a square floor is 64 ft.; 4:X ft; 8a ft. Find the area.
3. How many feet of wire are needed to fence in a square piece of ground whose area is 2,500 sq. ft. ? a^ sq. m. ? x^ square yards ?
4. The base of a rectangle is 8 times as long as the altitude. The sum of the base and altitude is 16 inches. Find the sides and area of the rectangle.
AREAS AND VOLUMES. MULTIPLICATION
67
5. Show by an equation the other dimension of a rectangle, having one side equal to 9 ft. and an area equal to 18 sq. ft.
" « « u 9 ft. " « « " " 15 sq. ft.
" " " " " 6 yd. " " " " " 12 sq. yd. " " " " 4 yd. " " " " " A sq. yd.
6. The frame of a picture of rectangular form of dimensions 22 in. by 17 in. is 2^ in. wide. Find the area of the frame and the outer perimeter.
7. A mantel is 36 in. high and 42 in. wide. The grate is 30 in. high and 30 in. wide. Find the area of the mantel and the number of square tiles contained in it if each tile is 3 in. long.
8. How many square tiles of 8 in. length are needed to make a walk 28 ft. long and 3 ft. wide ?
9. In a rectangular garden 25 ft. wide and 95 ft. long a 3-ft. wide walk is laid along the whole edge. The midpoints of the long sides are joined by a 2-ft. path. How many square feet are left for the garden ?
Cube and Parallelepiped
142. Cube. A solid like the one represented in Fig. 81 is called a cube. A cube has six /aces, all of which are squares. Two faces come together in an edge. Thus, there are 12 edges in the cube. The cube has 8 corners.
A
143. Parallelepiped. The solid
represented in Fig. 82 is y\q. 81
a rectangular parallele- piped. The faces of a rectangular paral- lelopiped are rectangles.
How many faces has a parallelopiped ? How many edges ? How many corners ?
z.
Fig. 82
68
FIRST-YEAR MATHEMATICS
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Volume of Cube and Rectangular Parallelopiped
144. Unit of volume. To measure a solid a cube is. used whose edge equals the unit-length. The number of times this cube is contained in a given solid is called the volume of the solid, the cube being the unit volume.
145. To find the volume of a parallelopiped. Let
Fig. 83 represent a rectangular parallelopiped 5 cm. long, 4 cm. deep, and 3 cm. high. Since the face on which the figure stands is a rectangle 5 cm. long and 4 cm. wide,
a layer of 5X4 unit cubes may be placed on it. Since the solid is 3 cm. high, it contains three such layers and therefore 3X5X4 unit cubes. Thus, the parallelo- piped contains as many units of volume as the number of units obtained by multiplying the length by the width and their product by the height.
This is usually expressed in the form of a theorem, thus : Theorem: The volume of a parallelopiped equals the product of the length hy the height by the width. V=lXhXw
146. Volume of a cube. The volume of a cube is com- puted in the same way as for the parallelopiped. Since the edges of a cube are all equal, the theorem of § 145 takes the following form:
The volume of a cube whose edge is e units long is given by the formula V = eXeXe.
The formula V = eXeXe may be written briefly V = e^, read V equals e-cuhe, e^ meaning eXeXe.
In words, this formula is expressed as follows :
Theorem : The volume of a cube is equal to the cube of one edge.
AREAS AND VOLUMES. MULTIPLICATION
69
147. Graphing equations. la chapter I, § 17, three ways of representing data were shown. Let us apply them to the following problem:
A number of rectangles are 3 in. wide and their lengths are 4", 5", 6", 7", 8", 9", 10^', 11", 12", 13". Calculate the areas of these rectangles.
The first mode of representing the areas for various lengths is to tabulate for each value of the length, /, the corresponding value of the area, A [see Table (I) in Fig. 84].
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Fig. 84
Second, the same data may be represented graphically.
From the point 0, Fig. 84, lay off the length horizontally and from the end-point of the length lay off the area vertically. Then join the end-points of the vertical lines by straight lines.
Third, the equation A=3Z represents the data of the table and graph. The straight line PQ, Fig. 84, is said to be the graph of A =3/.
1. From the graph. Fig. 84, find the area when Z= 1, 2, 3, 14.
70 FIRST-YEAR MATHEMATICS
2. Graph the following equations:
1. A = U 3. L = 2c+1 5. C = w^
2. D = t-\-l 4. 7j = 2x+S 6. y = x^+3
148. Linear equations. Equations whose graphs are straight lines are linear equations.
Multiplication of Monomials
149. S3mibols for multiplication. The formulas A = bXh, V = lXhXw, A=o?, F = e^ suggest that the products of numbers may be represented geometrically; e.g., the product of any two numbers may be expressed graphically by a rectangle whose dimensions are equal to the given numbers.
The product of two equal numbers, or of three equal numbers, may be represented by the area of a square or the volume of a cube, respectively. Hence, the notation a-square and e-cube. The product of four or more equal numbers cannot be expressed geometrically. However, in algebra the process is extended. Thus, we write aXa = a^, aXaXa = a^, aXaXaXa = a'^, read a-fourth, aXaXaXaXa = a^, read a-fifth, etc.
The product of two different literal numbers, as x and y, is shown by writing the letters side by side, as xy, with no sign between them. We are familiar with the form xXy from arithmetic. The form xy is most used in algebra. It is often convenient to use the form X • y.
150. Exponent. Base. Power. In x~, t^,x^ x^
the 2, 3, 4 n are called exponents of x.
What is the meaning of x-fifth? .r-sixth? ar-seventh? a;-tenth ? a;-nth ?
Write these numbers in symbols.
AREAS AND VOLUMES. MULTIPLICATION
71
The exponent of a number is the small figure or letter written to the right and a little above the number symbol, to denote the number of equal factors in a product. In 6^, meaning the product 6X6X6, the 3 is the exponent, the number 6 is the base, and the product, 6^, is the power. Thus 216 (=6^) is the third power of 6. (Why?)
When no exponent is written, as in ax, the exponent is understood to be 1 for each letter, as though the number were written a^x^.
Notice that 4x means 4 • x, or that x is to be used as an addend 4 times, while x"^ means x-x- x-x, or that x is to be used as a factor 4 times, and similarly for the other forms, as dx and x^, 5x and x^, etc.
EXERCISES
1. Letting a; = 6, give the meaning and value of each of these numbers.
1. 2a; 4. x^ 7. 5a; 10. Sx^
2. a;2 5. 4a; 8. x^ 11. 5x^
3. 3a; 6. x* 9. 2a;2 12. 2x*
2. Show from Fig. 85, (1) that the perimeter, p, of a square of 3a; is given by p = 12a;; (2) that the area A is given
by 9a;2.
3. Express by equations 'the perimeter and area of the rectangle, Fig. 86.
4. Show from a figure the perimeter and area of a rectangle 3a by 5a.
5. Write the following products in
x^
Fig. 85
xz
f j; X
Fig. 86
briefest form:
a; • a; • a;
3j
If
72 FIRST-YEAR MATHEMATICS
6. Find the values of the monomials: 2^; IP; {^y- 34^ (1.2)2; 7\
7. Letting a = 4, x = S, 6 = 1, y = 2, find the values of the polynomials :
x2+2x-hl; x''-\-Sxy+if; a^-2a''b-\-3ab^-¥
8. Two messengers leave a town at the same time, traveling in opposite directions. They travel a and 6 mi. an hour respec- tively. How far apart are they after t hours?
9. Find the values of the following numbers, if x = 4; 2x^; {2xy; Sx'; {SxY
10. Letting a = 2, 6 = l,c = 5, (^ = 3, e = 4, find the value of ab-\-bc-j-cd-\-de _ a^-\-b^+c''-\-d^-\-e'^ ^ a^-{-Sa%-^3ab'^+¥ a-\-b-\-c-{-d ' a+6+c+^ ' a+b
151. Product of powers having the same base. The product of two powers having the same base can be simplified. Thus, x^ • x^=xx ■ xxx = x^; x^ • x =xxx • x=x'^.
Give orally the products of the following pairs of factors in briefest form:
1. |
42.42 |
2. |
8-83 |
3. |
102 . 104 |
4. |
a2.a |
5. |
123 . 124 |
6. |
a^.a^ |
7. |
a -a^ |
9. |
62. 6^ |
10. |
c-c^ |
11. |
c'.c' |
12. |
x'-x' |
13. |
a 'X |
14. |
b'C |
15. |
b-b |
16. |
a -a^ |
17. |
6-6=^ |
18. |
K'-K |
19. |
K'^-K'^ |
20. |
a-x^ |
21. |
a2 -x |
22. |
g-t' |
23. |
a?'b |
24. |
x'^-if |
152. Comutative law (law of order). To illustrate the equation 5X2=2X5 draw a rec- tangle 5 units long and 2 units wide, Fig. 87. The area is 10 sq . units = 5 X 2 sq . units. A rec- tangle 2 units long and 5 units Fig. 87 wide differs from the rectangle
. i 1 |
AREAS AND VOLUMES. MULTIPLICATION 73
in Fig. 87 only in position. Its area is 2X5 sq. units. Since 5X2 and 2X5 represent the area of the same rec- tangle in different positions it follows that 5X2=2X5.
1. Show that 7X8 = 8X7.
2. Show that a6 = 6a.
These problems illustrate that in algebra, as in arith- metic, the factors of a product may be changed in order without changing the value of the product. This is the Commutative Law of Multiplication.
Simplify the following products, doing all you can mentally:
1. 4:xy 'bxhf
4:xy • 5x^y^ = 4 • 5 • x • x^ • 2/ * 2/^ (Why ?) = 20xV
2. 462c. 56V 5. (2xijy 8. 2.3-5-2.5.4
3. abc'a%c^-2abc^ 6. {2m^n^Y 9. 3 -5 -4.4 -4
4. 4ci3.5ci2.2a 7. {Za^xY 10. {Zax) {^o?x) {SaH'')
Addition of Monomials
153. Coefficient. The coefficient of any factor in a term is the product of all the other factors of the term. Thus, in ^iaxy the coefficient of axy is 4, of xy is 4a, of ax is
4?/. In -^ the coefficient of c is - , of b, is ^ , etc. When
o o o
the coefficient of the term is spoken of, the arithmetical factor is usually meant. It is common to say the coeffi-
bc . \ cient of the term ^axy is 4; of ^ is ^, etc. If no coefficient
o o
is written, as in a, x^, 1 is understood to be the coefficient, as though la, la;^ were written.
154. Similar terms. Terms which have a common factor are said to be similar with respect to that factor and are called similar or like terms, as 2x'^y'^, I2x^y^, Sx^y^.
74 FIRST-YEAR MATHEMATICS
155. Dissimilar terms. Terms which have no com- mon factor are called dissimilar or unlike terms, as 4a2 and 3c62.
EXERCISES
In each of the following polynomials point out with respect to what factor the terms are similar, state in each case the coefficient of the common factor, and then reduce to the simplest form. Do all you can without writing down your work.
1. ^a%-\-a%+5a%-{-Sa%
The common factor is a^b. The coefficients are 4, 1,5, 3. Hence, 4a%+a%-\-5a^b+Sa% = {4: + l+5+3)a% = I3a%.
2. 4x+7x-f 20x+35a:
3. ax-\-25x-\-bx+^Qx
4. ax-\-bx-{-cx-\-dx
5. Spqh+lApqh-^12pqh
6. 3a%-\-5a'^b-\-7a'bj-Sa%
7. 2ax-\-Sxa+7xa+5ax
8. 3pg2+6«g24-8rg2+l2s^2
9. 4:axz-\-7cxz-\-5dxz-\-9exz
10. abm-{-pmq-\-xmy-\-'mdc
11. 3(a+fe)+4(a+6) + 12(a+6)
12. 8(x2+?/2) + 10(x2+?/2) + 12(^:2+^2)
13. SUpr-q'')-\-5Hpr-q')+^Mpr-f) Multiplication of a Polynomial by a Monomial
156. Graphical multiplication of a sum by a monomial,
Solve the following problems: ^ ^ ^
1. Express by an equation the area of a rectangle of dimension 6 and a; +3,
6X
X E 3 B
Fig. 88. YiG, 88
The area of ABCD = Q{x-^^). Why? (1)
The parenthesis ( ) means that the number within is to be multiplied by 6.
AREAS AND VOLUMES. MtTLTIPLICATION 75
Dividing the rectangle ABCD into two rectangles by drawing line EF, we have
ABCD = AEFD -{-EBCF. Why ? (2)
But AEFD=Qx. Why ? (3)
and EBCF^IS. Why? (4)
Substituting for ABCD, AEFD, and EBCF their equals obtained from equations (1), (3), and (4), equation (2) takes the form 6(a:+3) =6x+18.
2. Show from Fig. 89 that a{m-{-n) = am-\-an. "
3. Show from a figure that ^ ^
d{a+b-{-c)=da-\-db-\-dc
The equations in problems 2 and 3 exemplify the axiom : A whole equals the sum of all its parts.
157. Partial products. The products am and an, problem 2, are the partial products of a{m-\-n)
What are the partial products of d(a-\-b-{-c)?
158. Product of a pol3momial by a monomial. Prob- lems 1, 2, and 3, § 156, illustrate the following prin- ciple :
A polynomial is multiplied by a monomial by multiplying each term of the polynomial by the monomial and then adding the partial products.
1. Multiply as indicated 2(3a:+4?/); {x+2y)U; 3a+46-2c; 56-4rc+2?/.
2. Simplify by carrying out the multiplications before the additions and subtractions n{a-\-b)—b; {x-\-y)m — ym; 3.x2+4a:?y2a:+10.
3. Letting a = 4, 6 = 1, c = 3, find the values of the following numbers: (a+6)c; 2(a+6).-c; 4a+2(6+c); 8rtH-(36+4c); 3a(2a6+8-5c); 25c3; a^; (2a-h3)5+(46+5)2.
159. Graphical multiplication of a difference by a monomial. The number {a—b)c means that b is
a D I
a-h
76 FIRST-YEAR MATHEMATICS
subtracted from a and the difference multiplied by c. The F E product (a—b)c may be repre-
sented geometrically, thus: c
In Fig. 90, ABCD = ABEF-DCEF
ABEF = ac ^ « ^ DCEF = hc
Fig. 90 ABCD = ia-b)c
Substituting in the first equation for ABCD, ABEF, and DCEF their equals {a — b)c, ac, and be, we have {a — b)c = ac — bc.
EXERCISES
1. State the principle illustrated by the preceding problem.
2. Express the following products as polynomials: (2x-37/)a; 2{^x-&y)\ 2a; (4a +56 -3c); {a'-2ab^-b'')a.
3. Multiply as indicated: a%z^'¥; Sm'^n'^mn^; {2x-\-^y)2; (4x'^-2x-\-3)2x; xY'^^^', ^rs^t-Sst^; 3xix-7j); 4:y{x+y-2z).
4. Simplify 5{2x-3y)+2{2x-y)-\-{x-4y)2x.
5. Give the factors of the following products: ab-\-bc; xy+xz; 2a-\-m; 2xy+10x; 3x-Qy; 2ax-8ay; A^xa"" - 12xb' ; Sxa%+9xy%.
6. The width of a rectangle is x — S, the length x, and the perimeter 66 yards. Find the width and the length in yards.
7. The area of a square equals that of a rectangle whose sides are 4 cm. less and 8 cm. greater respectively than the side of the square. Find the area of the rectangle.
8. After 12 years a man will be twice as old as he was 12 years ago. What is his age ?
9. Solve the following equations:
1. 3(a;+ 4) = 22+a;
2. 9(a:+35) = 5(2a;+45)
3. 3(a:+15)+5 = 2(2x+9)H-4(x+3)
ac |
ad |
be |
M |
AREAS AND VOLUMES. MULTIPLICATION 77
4. M = 8
6. -g-+l--j^+2
Multiplying Polynomials by Polynomials
160. Graphical multiplication. The product of a-\-h by c-\-d is written {a-{-b){c-\-d). It may be represented geometrically by a rectangle of dimen- sions a+fe and c-\-d, Fig. 91.
The rectangle ABCD is composed of four rectangles ac, be, ad, and bd. b
Therefore, ^ " ^' "' ^
Fig 91
{a-^b){c-{-d)=ac+bc-\-ad-\-bd,
both sides of the equation representing the area of rectangle ABCD.
The product {a—b){c-{-d) is represented by a rectangle having the dimensions (a—b) and (c+d), Fig. 92.
^ J j^ The rectangle A J5J^i^ = ac+ ac?.
T-' ' ""i Subtracting from this the rectangles be
and bd, we obtain the rectangle ABCD.
Therefore, (a — b){c-\-d)=ae-\-ad — bc — bd, both sides of the equation representing the A c G d B area of rectangle ABCD. Fig 92
Show from a rectangle that
{a-\-b-\-c){m-\-n) = am-\-hm-{-cm-{-an-{-hn-\-cn
161. Multiplication of poljmomials. In § 160 the
following principle is illustrated:
A polynomial is multiplied by a polynomial by multiply- ing each term of one polynomial by every term of the other and adding the partial products.
)\ |
be |
bd |
|
D |
H |
||
i-b |
78 FIRST-YEAR MATHEMATICS
EXERCISES
1. Using the principle of § 161 express the following products as polynomials:
1. {x-\-y)(a-\-b) 4. {r+s)(a-\-x)
2. {x+y){m+n) 5. 12(7+6-4)
3. {a+x){b+y) 6. 2a(3a2-4a+2)
2. Solve the equation :
(s+2)(s+5) = (s+l)(s+3)-f22
3. Find for x = 3, y = 2 the value of (Sx-2ij)(2x-i-3y); (2^2 -5a; +6) (5a; -3); (7x^y-Sxif){x^y-y^x)
4. By means of figures express as polynomials the following squares:
1. {a+b){a-\-b) or {a+by 3. {x-\-yy
2. (c+dy 4. (m-hn)2
5. Draw squares whose areas are expressed by the following trinomials:
1. a^-\-2ax-\-x^ 3. k''-\-2kb+b^ 5. x^+6x-{-9
2. fe2+26c+c2 4. s2+2s^+^2 6^ ^2+80+ 16
6. What are the factors of the trinomials in problem 5 ?
7. The area of a rectangle is 96 sq. yd., the base is (8+ a;) yd., and the altitude 8 yards. Find the base.
8. One side of a rectangle is 5 yd., the other is 7 yards. By how much must the longer side be increased so that the area is- if times as large as before ?
162. Check. Problems in multiplication may be checked by substituting in the problem and in the product convenient values for the letters. Both should reduce to the same number.
EXERCISES
Multiply as indicated and check. Use pencil only when needed.
AREAS AND VOLUMES. MULTIPLICATION 79
1. Sx{x^+^y+2y^)
3x (a;3 +4x1/ +2i/2) = 3x4 + 12x21/ +6x1/2 Check: Let x = l, y=2
3x(x3+4xt/+2i/2) =3(1+8+8) =51
Sx* + 12x^y +6x1/2 = 3 +24 +24 = 51
3. (a2+2a+l)(a-3)
4. {x^-\-xy+7f)(x'^-xy+y')
5. (p+?+r)2
6. (.4a;-.3i/+.7^)(10^+20i/+302)
7. ( . 4a+ . 36)2+ ( . 6a+ . 26)^+ ( . 2a+ . 36) ( . 5a+ . 16)
8. (2x3-7x+3)(2a;+5)
9. (a3-32/+5)(a2+10) 10. (p*+2p2+7p)(2p-l)
Area of Parallelogram and Triangle
163. Area of parallelogram. To find the area of a parallelogram, as A BCD, Fig. 93, ^
draw line BE at right angles to AB, Jy ^fyc:
dividing the parallelogram into two Z y
parts, the triangle BEC and the
quadrilateral ABED. The triangle
is then cut off and moved to the left until BC coincides
with AD, forming a rectangle as ABEF, Fig. 94.
Then rectangle AB^F = parallelo-
h\
F E
gram A BCD. Why? But rectangle
^ J, ^ ABEF = b'h. Why? Therefore, par-
FiG. 94 allelogramABCZ) = 6 • /i. Why? This
gives the following theorem: Theorem: The area of a parallelogram is equal to the product of the base by the altitude.
P=bh Find the area of a parallelogram (1) if 6 = 28, and /i = 19
(2) if 6 = 10.3, and /i=14. 6
80 FIRST-YEAR MATHEMATICS
164. Area of a triangle. A triangle, as ABC, Fig. 95, is one-half of a parallelogram obtained by drawing the
lines DC and AD. Since the tri-
^^ -? angle and the parallelogram have
/|/i^^^^^^^ / the same base BA and equal alti-
^Z__j — _ -^^A tudes h it follows that the area of
Fig. 95 ABC is equal to one-half of the
product of A 5 by /i. (Why?) This leads to the following theorem :
Theorem : The area of a triangle is equal to one-half of the product of the base by the altitude.
T=hb'h
1. Find the area of a triangle
(1) if 6 =12 ft., and /i = 16 ft.
(2) if 6 = 8.2 ft., and /i = 7. 78 ft.
2. Show by an equation the other dimensions of a triangle having
a base = 6 ft. and an area = 24 sq. ft. " " =6 ft. " " " =9 sq.ft. an altitude = 4 yd. '' " " =16sq. yd. " =aft. « " " =a sq.ft. " =cin. " " " =6sq. in.
165. Quotient. The quotient of x divided by y is
written - or x^y and is read ^^x over i/" and ^^x divided
y ^
by ?/" respectively.
166. The area of a trapezoid. To determine the area of a trapezoid, as A BCD, Fig. 96, draw the line-segment BD, dividing the trapezoid into two n a c triangles.
'1^
Triangle A2)B=i/i6. Why? X ^!
6 B
Triangle Z)5C=|;ia. Why? ^ ^^^^^
AREAS AND VOLUMES. MULTIPLICATION
81
Therefore, trapezoid ABCD= ^hb -\-iha. Why? or trapezoid ABCD=ih{b+a). Why?
CaUing a and b the bases of the trapezoid this result expresses the following theorem:
Theorem: The area of a trapezoid is equal to one-half the product of the altitude by the sum of the bases.
The altitudes, bases, and areas of parallelograms are as given below. Find the value of the unknown dimension.
Altitude |
Base |
Area |
3a:+2 16 2.r + l 10 |
8 5x-7 15 3a: +2 |
56 288 105 110 |
Summary
167. In this chapter the meaning of the following terms was taught: parallelogram, rectangle, square, trapezoid, rhombus; area of a figure, unit of area; formula; cube, parallelopiped; volume, unit of volume; exponent, power, base; coefficient; similar and dissimilar terms; partial products; quotient.
168. The following symbols were introduced: the exponent, to indicate the number of equal factors. To indicate the product of unequal factors use the symbols X, •, or no symbol at all. Division is indicated by the symbol -i- or by placing one letter or figure (number) over another with a line between.
169. Equations may be graphed. Equations whose graphs are straight lines are linear equations.
170. The product of two numbers may be represented geometrically as the area of a rectangle.
82 FIRST-YEAR MATHEMATICS
171. Problems in multiplication may be checked by substituting convenient values for the letters in the prob- lem and in the result.
172. The following algebraic principles were taught :
1. Commutative law. The factors of a product may he changed in order without changing the value of the product,
2. A polynomial is multiplied by a monomial by multi- plying each term of the polynomial by the monomial and then adding' the partial products.
3. 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.
173. The following theorems and formulas have been obtained :
1. The area of a square is equal to the square of one side.
2. The area of a rectangle is equal to the product of the base by the altitude.
3. The volume of a rectangular parallelopiped is equal to the product of the length by the height by the width.
4. The volume of a cube is equal to the cube of an edge.
5. The area of a parallelogram is equal to the product of the base by the altitude.
6. The area of a triangle is equal to one-half of the product of the base by the altitude.
7. The area of a trapezoid is equal to one-half the product of the altitude by the sum of the bases.
JOHN WALLIS
JOHN W A L L I S
JOHN WALLIS was born at Ashford in England, November 22, 1616, and died at Oxford, October 28, 1703. During a school holiday, when he was only fifteen years old, he was struck with curiosity at seeing the odd symbols and signs of an arithmetic in the hands of his brother. He borrowed the book and in a fort- night, with his brother's assistance, he had mastered the book. He studied to become a physician, and he was the first to maintain that the blood circulates in the human body. His main interest however was in mathematics.
Wallis became professor of geometry at Oxford in 1649 and lived there thenceforth until his death. Besides many mathematical works, he wrote on theology, logic, philosophy, and he devised a system for teaching deaf-mutes. His genius thus revealed itself in many ways.
He wrote an arithmetic and an algebra that were long the standard texts. Newton is said to have learned algebra from WalHs' text. In his algebra he introduced into mathematics the symbol ll for parallelism. He was one of the pioneers in the field of calculus and he developed the theory of interpolation very fully. He discovered
2-2-4-4-6-6-8-8
TT
= 2
1 3 3 5 5-7-7-9.
He was familiar with all the mathematics of his day, added greatly to the subjects of higher arithmetic and algebra, and did much toward laying the founda- tions in English science for what mathematicians now call analysis.
Read the life of Wallis in Ball's History of Mathe- matics, pp. 288-93 (5th ed.).
CHAPTER VI
ANGLE-PAIRS
Adjacent Angles
174. Adjacent angles. Two angles that have the same vertex and a common side between them are adjacent angles. The sides which are not common are the exterior sides.
EXERCISES
Give reason for
Fig. 97
1. Are b and d, Fig. 97, adjacent angles? your answer. Are a and b adjacent angles? b and c?
2. Read the exterior sides oi Ax and y, Fig. 98; the common side.
3. Draw two adjacent acute angles whose sum is a right angle.
4. Draw three acute angles whose sum is a right angle.
5. Draw two adjacent obtuse angles whose sum is 3 right angles.
6. Draw two adjacent angles, one obtuse and the other acute, whose sum is 2 right angles.
7. With a protractor draw adjacent angles of 75° and 85°; of 103° and 57°; of 3U° and 2^°. In each case check the work by finding the sum arithmetically and then measuring the sum with a protractor.
8. Draw two intersecting straight lines making a pair of adjacent angles equal and show by measuring that the angles are right angles.
83
84
FIRST-YEAR MATHEMATICS
175. Perpendicular lines. If two straight lines inter- sect making a pair of adjacent angles equal, each line is perpendicular to the other.
1. In Fig. 99 which lines are perpendicular? Point out the equal adjacent angles.
2. Draw two lines perpendicular to each other, using only the ruler.
/
Fig. 99
176. Theorem. At a given point in a given line only one perpendicular can he drawn to the line.
For, if two perpendiculars to AC, Fig. 100, were drawn at B, the two right angles DBC and EEC would be unequal. This contradicts the A — theorem that all right angles are equal and is therefore impossible.
B Fig. 100
177. Draw two adjacent angles of 56° and 124°; of 19^ and 160^; of 92° and 88°. With a ruler, or straight- edge, see if the exterior sides of each pair of angles form a straight line. What term is applied to each angle-sum?
This problem leads to the following the- orem:
2j Theorem: If the
sum of two adjacent angles is a straight angle, the exterior sides are in the same straight line.
178. From a point C on a straight Hne AB, Fig. 101, draw three lines as in the figure. By estimating find the
c Fig. 101
ANGLE-PAIRS
85
number of degrees in each angle. Then measure each angle with the protractor. Fill out the table in Fig. 102. What seems to be the sum of all the angles about a point on one side of a straight line? This illustrates the theorem of § 179 below.
... Angle |
a |
b \ c |
d |
Sum |
Estimate |
40 |
1 |
||
Measure |
39 |
Fig. 102
The Sum of the Adjacent Angles about a Point on One Side of a Straight Line
179. Theorem: The sum of the adjacent angles about a point, on one side of a straight line, is a straight angle, or 180°.
EXERCISES
1. Find the number of degrees in each angle of Fig. 103. We may write
9x+a;+(37-2x) + (5x-26) = 180. Why? Changing the order of the terms,
9x-\-x-\-5x- |
-2a;+37-26 = 180. |
What law is used ? |
13x+ll = 180. |
Why? |
|
13.x = 169. |
Why? |
|
x = 13. |
Why? |
|
Therefore |
9x = 117 37-2x = ll 5x-26 = 39 |
Check: x+9x-{-'S7 -2x+5x■ 2. With a protractor make a drawing of Fig. 103.
3. All the angular space about a point in a plane, on one side of a straight line, is divided into angles represented by the
180
5X-26
Fig. 103
86 FIRST-YEAR MATHEMATICS
following expressions. Find x and each angle in degrees. Draw figures for exercises 1, 2, and 3.
1. X, 5x, lx — 2
2. Sx, 48 -3a;, 5a; -22, 4a; -14
3. 25f +5a;, 8x+8|, 3a;, 9^^
4. 3a;, 2(a;+9), x, 42-a;
5. 2a;, 2(a;-10), a;- 18, 3(36-a;)
6. 3(a;-3), a;+33, 2(41-a;)
7. 2.8X+39.33, 1.2a;-32.09, a;+7. 16
8. 6.93a;, 4.82a;, 1.27a;+5.09, 138. 91 -9. 02a;
9. la;, 88+ia;
10. fa;+10, 86-|a;
11. 4a;+14, 97-fa;
The Sum of the Angles at a Point
180. The sum of the angles just covering the angular space about a point may be found from the following problem.
Draw four lines from a point, Fig. 104. Find the number of degrees in each angle, first, by estimating, then by measuring with a protractor. What seems to be the sum of the angles that just fill the angular space about a point ? This may be stated in the form Fig. 104 of a theorem as follows:
Theorem; The sum of all the angles at a point just covering the angular space about the point is a perigon, or 360\
ANGLE-PAIRS
87
EXERCISES
1. Find by solving an equation the number of degrees in each angle of Fig. 105.
All the angular space about a point in a plane is divided into angles repre- sented by the following expressions; find X and each angle in degrees. With pro- tractor draw figures for exercises 2, 3, and 4. Fig. 105
5X+27
2. 2x, X, 4a: -1-40, 180- |
-3x |
||
3. x,SQ-\-5x,Sx-9 |
|||
4. 3a:, 5a:, 5a:+45, 27- |
-X |
||
5. 15a:+16j, 37j-2a: |
, 8x-9 |
||
6. 3a:, 5a: 4-261, 2a:, 9x4-143 |
3 5 |
||
7. 7x4-24, 14x4-531, |
120f- |
■3x |
|
Solve the following equations: |
|||
8. 15-6^4-8^ = 25 9. 6^-74-4^=13 |
15. |
^+8-^=10 |
|
10. 7!/ -37/+ 107/ = 39 11. 2(x-3)4-12 = 18 |
16. |
i-t-« |
|
12. 8+5(s4-7) = 63 |
17. |
f-15+| = 52 3.7Z-3.6-2.9^ = 2.4 |
|
13. 18r-hl3-10r=75 |
18. |
||
14. 1+5 = 9 |
19. |
2x+4(x+10)+3x=130 |
Supplementary Angles
181. Supplementary angles. Two angles whose sum is a straight angle (180°) are supplementary angles. Each angle is said to be the supplement of the other.
If the supplementary angles are adjacent they are called supplementary adjacent angles.
88 FIRST-YEAR MATHEMATICS
EXERCISES
1. Draw two supplementary adjacent angles.
2. Are 50° and 130° supplementary? 37° and 133°? 60° and 120°? 90° and 90°?
3. How many degrees are there in the supplement of an angle of 45° ? Of 120|° ? Of 90° ? Of a° ? Of x° ?
4. Write the supplement of a°; of b°; of 5d°; of ^ .
5. If angles of 120° and a° are supplementary, what does a represent ?
6. If a;°+80° = 180°, what is the supplement of x° ? What is the value of a; ?
7. In the equation, a°-i-6°=180°, what is the supplement of a°? Of6°? Why?
8. State by an equation that the following pairs of angles are supplementary:
(1) x° and 60° (4) 50° and a;°+70°
(2) 70° and y° (5) 2a;°+3° and 27a:°-2°
(3) 6° and c° (6) lx° and V-a^°+112|°
9. The supplement of a; +3 degrees is 2a; +27 degrees. Find X, x-\-3, and 2a:+27.
We may write X +3 +2rc +27 = 180. Why? (1)
Combining like terms Sx +30 = 180 (2)
Subtracting 30 3a; = 150. What axiom is used? (3)
Dividing by 3 a; = 50. What axiom ? (4)
Whence a:+3=53. Why? (5)
2a;+27 = 127 (6) Check: a;+3+2a;+27 = 180
10. a;° is the supplement of a:°+84°. Find the angles.
11. One of two supplementary angles is 98° larger than the other. Find the angles.
12. One of two supplementary angles is 27° smaller than the other. Find the angles.
ANGLE-PAIRS 89
13. One of two supplementary angles is 3^ times as large as the other. Find the angles.
14. The difference of two supplementary angles is 110°. Find them.
Let x° be one angle and a;°+110° the other.
15. Find two supplementary angles whose difference is 21°; 36§°; 73i°; d°.
16. The difference between an angle and its supplement is 37°. Find the angle.
17. How many degrees are there in the angle, x°, if it is the supplement of 5a;° ? Of 1x° ? Of 3|a;° ?
18. How many degrees are there in an angle that is the supple- ment of 4 times itself ? Of 8 times itself ? Of 10 times itself ? Of 2\ times itself ? Of f of itself ? Of i of itself ?
19. Express in algebraic language:
(1) the double of an angle, x
(2) 15° added to 3 times the angle
(3) 29° subtracted from 6 times the angle
(4) 4 times the sum of the angle and 13°
(5) Two-thirds of the sum of the angle and 17°
20. Express in algebraic symbols:
(1) the supplement of an angle, x
(2) 5 times the supplement
(3) 3 times the supplement
(4) 14° added to 3 times the supplement
(5) 16° subtracted from 3 times the supplement
(6) the supplement increased by 10°
(7) the supplement diminished by 18°
(8) the supplement divided by 4
(9) one-third of the supplement
(10) 17° added to the supplement
(11) 20° added to one-third of the supplement
(12) 19° subtracted from f of the supplement
90 FIRST-YEAR MATHEMATICS
21. If an angle is doubled and its supplement is increased by 20°, the sum of the new angles thus obtained is 280°. Find the two supplementary angles.
Let X be one angle, and 180— x the other; then by the conditions of the problem 2x+(180-a;+20) =280.
22. If an angle is trebled, and its supplement is diminished by 112°, the sum of the angles obtained is 168°. Find the supplementary angles.
23. The sum of an angle and \ of its supplement is 90°. Find the angle.
24. If an angle is increased by 12°, and its supplement is divided by 5, the sum of the angles obtained is 80°. Find the supplementary angles.
25. If 20° is added to 5 times an angle, and 15° is subtracted from 2 times the supplement of the angle, the sum of the angles obtained is 401°. Find the supplementary angles.
Solve the following equations :
26. 2(x-5)+|(x+l)=3 29 a:+5 2x-5 ^_^
27. 5(x-5) + (x+3)=2 ' 3 5
15z '^n 6(^-2) , a:+3
28. i|^- 20+ (2;+ 12) = 9 ^"- 3 "^ 4 ~'^~^
31. Find x and each of the following supplementary angle- pairs :
^+2+172i and ^^-^ ; x-^ and -jH-90
32. Draw a figure showing that the supplements oj equal angles are equal.
Complementary Angles
182. Complementary angles. Two angles whose sum is a right angle are complementary angles. Each angle is said to be the complement of the other.
ANGLE-PAIRS 91
EXERCISES
I. What is the complement of a, Fig. 106? Of 6?
2. Draw two adjacent complementary angles. May either angle be obtuse? Point out two perpendicular lines.
3. Show whether 22° and 68° are comple- mentary; 43f ° and 46|°; 89|° and f °.
~ ~~ 4. What is the complement of 60°? Of 30°?
5d° 3s°+2f°
5. Write the complement of d°; of 3c°; of -^ ; of ^ — ;
of 7(a+b) degrees; of dx"^ degrees; of 7y^ degrees; of Sx'^ — 5y^ degrees.
6. If angles of 40° and d° are complementary, how many degrees does d stand for?
7. If i/°+70° = 90°, what is the complement of ?/? Why? What is the value oi y?
8. In the equation, c°-fcP = 90°, what is the complement of c°? Oid°? Why?
9. State by equations that the following pairs of angles are complementary:
(1) y° and 50° (2) 30° and z° (3) w° and x°
(4) a°-|-30°anda°-20° (5) 2x°+7° and 5x°-2°
(6) 3 (a: +7) degrees and 5(2x — 8) degrees
(7) |x— 15j degrees and 26|^a;4-43f degrees
10. x° is the complement of x°+48°. Find the angles.
II. One of two complementary angles is 24° larger than the other. Find the angles.
12. One of two complementary angles is 28° smaller than the other. Find the angles.
13. How many degrees are there in the angle x, which is the complement of 4:X ? Of 6a: ? Of 5^x ?
92 FIRST-YEAR MATHEMATICS
14. How many degrees are there in an angle that is the com- plement of 3 times itself ? Of 7 times itself ? Of 6 times itseK ? Of Si times itself ? Of i of itself ? Of i of itself ?
15. The difference of two complementary angles is 83°. Find them.
16. Find two complementary angles whose difference is 21°; 36i^°; 73i°; d°.
17. The difference between an angle and its complement is 27°. Find the angle.
18. If an angle is doubled, and its complement is increased by 40°, the sum of the new angles thus obtained is 160°. Find the complementary angles.
19. If an angle is trebled, and its complement is diminished by 40°, the sum of the angles obtained is 130°. Find the com- plementary angles.
20. The sum of an angle and § oi the complement is 75°. Find the angle.
21. If an angle is increased by 15°, and the complement is divided by 3, the sum of the angles obtained is 75°. Find the complementary angles.
22. If 20° is added to 3 times an angle, and 6° is subtracted from 3 of the complement, the sum of the angles obtained is 102°. Find the complementary angles.
Solve the following equations :
23. y +88+^ =180
24. f+10-g=15
25. Draw a figure showing that the complements of equal angles are equal.
ANGLE-PAIRS Opposite Angles
93
183. Opposite angles. Two angles having a common vertex, and having sides in the same straight line, but in opposite directions, are called opposite or vertical angles (as X and z. Fig. 107).
Fig. 107
a protractor.
EXERCISES
1. Read both pairs of opposite angles in Fig. 107.
2. On tracing paper make a trace of angles y and z (Fig. 107). Put this trace on angles x and w and see whether z coincides with (fits on) x, and y with w. How do the opposite angles com- pare in size ?
3. Test your conclusion in exercise 2 by drawing two intersecting straight lines and measuring both pairs of opposite angles with
4. Show that in Fig. 107 x+y=l80 Show that y-^z= ISO Show that x-\-y = y-\-z Then x = z. Why?
5. Show, as in exercise 4, that y = w.
Exercises 4 and 5 show the truth of the following theorem :
Theorem : // two lines intersect, the opposite angles are equal.
EXERCISES
1. Two intersecting straight lines form four angles a, b, c, and d. If a = 40°20'10'', how large are b, c, and d?
94
FIRST-YEAR MATHEMATICS
2. Show that the bisectors of two adjacent supplementary angles are perpendicular to each other.
Why?
Proof: |
a+6+c+d = 180°, Fig. 108, |
a = 6. Why? |
|
c=d. Why? |
|
Therefore |
6+6+c+c = 180. Why? |
\ |
26+2c = 180. Why? |
\ |
fe+c = 90. Why? |
Fig. 108
3. Find x and the four angles made by two intersecting straight lines, if two opposite angles (Fig. 109) are 3a; +37 and
bx+1.
Since the given angles are opposite Fig. 109 angles,
5x+7 = 3a;+37. Why? (1)
Subtracting 7 5x = 3a;+30. What axiom? (2)
Subtracting 3a; 2a; = 30. What axiom? (3)
Dividing by 2 a; = 15. What axiom? (4)
5a;+ 7 = 5-15+ 7 = 82
3a;+37 = 3-15+37 = 82 Each of the other two angles is 180 - 82 = 98. Why ? Check: 82+98+82+98=360
4. Find x and each of the following pairs of opposite angles made by two intersecting straight lines. Draw figures for exercises (1) and (2).
and 4a; +87 anda;+103
and |a;+24|-
and 1tt^+57
and fa; +42 28 and x
(1) 7a;+27
(2) 3a; -17
(3) |a;+16i
(4) 2^\x-n (5) (6)
-3^+3^
^x+^x- Zx
(7) 5a;+
(8)
Tx 4"
and y+ 130
2a; and -^+66
ANGLE-PAIRS |
|
(9) |
l+l and 1+18 |
(10) |
f-f and|+8f |
5. Solve the following equations: |
|
(1) |
1+16+1-14 = 7 |
(2) |
1-15+1=8 |
95
The Acute Angles of a Right Triangle
184. Theorem: The acute angles of a right triangle are complementary angles. Show that this is true.
Find the values of the acute angles of a right triangle
if one angle is 3 times the other; 5 times the other; 3 of the
other; l\ of the other; 6 more than 7 times the other; \ of the other diminished by 33.
185. One of the acute angles of a right triangle is twice as large as the other. Find the acute angles. Draw a right triangle containing these angles.
If the side opposite the 90° angle is twice as long as the side oppo- site the 30° angle the construction is correct.
This problem illustrates the following theorem:
Theorem: In a right triangle whose acute angles are 30° and 60° the side opposite the 90° angle (hypotenuse) is twice as long as the side opposite the 30° angle.
The acute angles of a right triangle are given equal. Find the number of degrees in each angle. Make a drawing of the triangle. How do the lengths of the sides about the right angle compare ?
186. Isosceles triangle. A triangle having two sides equal is an isosceles triangle.
96
FIRST-YEAR MATHEMATICS
EXERCISES
1. In Fig. 110 A ABC, ADC, and BDC are right triangles. Show that X and a are complements of the same angle and
therefore equal.
Show that y and b are equal angles. ^
2. Solve the following equations:
X
■5- — . - - A
3a:-2
(1) 9+1
(2)
12+x = 9 3
1 =
(3) 9x-g-l = 7f
(4) ^f +15-f^ = 28i
3. The following angle pairs are acute angles of a right triangle. Find x and the angles.
X . ^ , 87 5a: X . , a; . 35
;-\-2x and -^ — n
8+^ and J2+-2
Angle-Pairs Formed by Two Lines Intersected by a Third
187. When two lines are intersected by a third line (transversal), eight angles are formed, Fig. 111.
1. Make a list of the supplementary adjacent angles and for each pair state the equation expressing the relation between the angles.
^^* 2. Make a list of the opposite angles and
state the equation for each pair.
188. When two lines, Fig. 112, arc cut by a transversal,
a and e the angles of the angle-pairs
b and / d and h c and g^
are called J> corresponding
angles,
angles c, d, e, f are called interior angles, angles a, b, g, h are called exterior angles,
* ANGLE-PAIRS 97
the angles of f d and e 1 are called interior angles on the the angle-pairs } c and / J same side of the transversal,
,, , -. f , , .1 on opposite sides of the trans-
the angles 01 a and / , n i ix
,, , . < , > versal, are called alternate
the angle-pairs c and e \ . ^ . ,
\ J interior angles,
,, , r. f 7 , , 1 on opposite sides of the trans-
the angles of o and h \ , n i i^
^, 1 . ^ ^ r versal, are called alternate
the angle-pairs a and q \ _^ . ,
[ ^ J exterior angles.
EXERCISES
1. If a = e prove that c = e.
2. If a = e prove that h and e are supplements.
3. If a = e prove that h and a are supplements.
4. li c = e prove that c and / are supplementary angles.
5. If c = e prove that a = g.
6. If c = e prove that/ =d.
7. If c+f= 180° prove that b =/.
8. If c+f= 180° prove that /= d.
9. If c-\-f= 180° prove that a+f= 180°. 10. If c4-/= 180° prove that c+/i= 180°.
Summary
189. This chapter has taught the following terms: adjacent angles, opposite angles; perpendicular lines; supplementary angles; complementary angles; isosceles triangle; corresponding angles formed by two lines cut by a transversal; alternate interior angles, alternate exterior angles, interior angles on the same side of the transversal; hypotenuse of a right triangle.
98 FIRST-YEAR MATHEMATICS '
190. The truth of the following theorems has been shown :*
1. At a given point in a given line only one perpen- dicular can be drawn to that line.
2. If the sum of two adjacent angles is a straight angle the exterior sides are in the same straight line.
3. The sum of the adjacent angles at a point on one side of a straight line is a straight angle.
4. The sum of all the angles at a given point covering the angular space about the point is a perigon.
5. Supplements of equal angles are equal.
6. Complements of equal angles are equal.
7. If two lines intersect, the opposite angles are equal.
8. The acute angles of a right triangle are complementary.
9. In a right triangle whose acute angles are 30° and 60° the hypotenuse is twice as long as the side opposite the 30° angle.
* These theorems were probably all first proved by the School of Pythagoras, founded at Croton in Southern Italy about 529 b.c. See Ball, p. 19.
PYTHAGORAS
PYTHAGORAS
PYTHAGORAS was born at Samos about 569 b.c. of Phoenician parents and died, probably at Metapontum, in Southern Italy, about 500 b.c. He was primarily a moral reformer and philosopher, but he was celebrated also as a mystic, a geometer, an arithmetician, and as a teacher of astronomy, mechanics, and music. His system of morals and his philosophy were founded on mathematics. He is said to have been the first to employ the word mathematics. The meaning he gave it was what we understand by general science. With him geometry meant about what high-school people today mean by mathematics.
He divided his mathematics into numbers absolute or arithmetic, numbers applied or music, magnitudes at rest or geometry, and magnitudes in motion or astronomy. His suc- cessors for many years regarded this as the necessary and sufficient course of study for a liberal education. It is the origin of the famous "quadrivium" that constituted higher education for 2,000 years.
After completing his studies near his home under Pherecydes of Syros and Anaximander, the latter a disciple and successor of Thales of Miletus, Pythagoras traveled extensively, studying mathematics in Egypt, Chaldea, and Asia Minor. Returning from his travels he settled at Samos and gave lectures with indifferent success until some time near 529 b.c, when he migrated to Tarentum. After a brief stay here he removed to Croton in Southern Italy, where he opened his famous school of philosophy and mathematics in 529 b.c. Here his school was attended by enthusiastic audiences.
He divided his hearers into probationers and Pythagoreans. The probationers were much the larger group. He formed the Pythagoreans into a brotherhood, like a modern fraternity. All possessions were to be held in common and all discoveries were to be referred to the founder. The chief mathematical discoveries were revealed only to the Pythagoreans. Read in some history of mathematics the story of the drowning of a Pythagorean for divulging a mathematical discovery and claiming it as his own, and other more significant facts about this secret order and its wonderful founder. The leading teachings of the brotherhood were self-command, temperance, purity, and obedience. Its secrecy and strict discipline soon gave the society such power in the state as to arouse the jealousy and hatred of certain influential classes in that democratic community. Instigated by his political opponents, a mob murdered many of Pythagoras' followers and finally, after his flight, probably to Metapontum, mm-dered the leader himself. After the death of their leader the Pythagoreans were dispersed over Southern Italy, Sicily, and the Grecian peninsula. They renounced secrecy, opened schools at divers centers, and they and their disciples continued pubhcly to teach Pythagorean doctrines for a hundred years after the death of Pythagoras.
Pythagoras' geometry consisted of the substance of what is contained in the first book of our school geometries about triangles, parallels, and parallelograms, together with some few isolated theorems about irrational magnitudes. His reasoning was often not rigorous, e.g., he sometimes assumed the converse to be true without a proof. His most original work was in the theory of numbers, called by the Greeks Arithmetica. Pythag- oras left no books or other writings, so that what we know of him is traditional. He himself believed, not in publicity, but in secrecy.
CHAPTER VII
PARALLEL LINES. LINES AND PLANES IN SPACE
Parallel Lines
191. Parallel lines. Two lines arc said to be parallel if they lie in the same plane but do not meet however far extended.
EXERCISES
1. Point out the parallel edges of a table top, of a sheet of notebook paper, of a rectangular box, of a cube. Give other examples of parallel lines.
2. Show by measuring that the oppo- t^ ^ ^ ^^ site sides of a rectangle are everywhere equally distant, i.e., EF = GH = KL, Fig. 113. ^ E G K
3. Extend the opposite sides of a Fig. 113 carefully drawn rectangle and show that
the extensions are everywhere equally distant.
192. The last two problems illustrate the following fact:
Parallel lines are everywhere equally distant.
193. Symbol for parallelism.* The symbol for parallel- ism is IL Thus, the statement AB is parallel to CD is
written A J5 II CD.
194. Let A, Fig. 114, be a
point not on the straight line B C.
Let AD be a straight line through
• A, intersecting BC at D. If AD
* The symbol II for parallelism was introduced by John Wallis. 99
100
FIRST-YEAR MATHEMATICS
is made to rotate about A, point D moves along BC, taking positions as E, F, G, etc. We will assume that there is a position of the rotating line, as AK, such that it does not intersect BC, i.e., when the rotating line is parallel to BC. Further, we will assume that this is the only position in which the rotating line does not meet BC. Thus, when it has passed the position AK hj the least amount, it will intersect BC to the left of D. These assumptions are stated in the form of the following axiom : Axiom: One and only one parallel can he drawn to a line from a point outside the line.
EXERCISES
1. Draw a line, as AB, Fig. 115, and place along this line one side PQ of a triangle T, cut from paper or wood. Along the
second side, PR, of the triangle draw CD. Move triangle T, letting PQ move along BA until it takes the position PiQi. Along PiZ^i draw line CiDi. Notice that the size oi Ax has remained un- changed. Show by measuring, as in § 191, that any two points on CD are equally distant from Why?
Fig. 115
CiDi. Thus, CD II CiZ)i
2. Draw a line, as AB, Fig. 116. At any two points of ^45, as P and Pi,
draw with the pro- tractor two
equal an- ^j^ ^^g
gles, X and
xi. Show as in exercise 1 that CDWEF.
3. Draw a line, as AB, Fig. 117. With the compass draw /.x= Axu Show that CDWEF.
PARALLEL LINES ' 'lOl ^' I
195. Exercises 1, 2, and 3 show how to draw parallel lines by aid of the triangle, protractor, and compass, respectively. All three ways are based upon the following theorem :
Theorem: // two lines are cut by a transversal making the corresponding angles equal the lines are parallel.
EXERCISES
1. To a given straight line, I, draw a parallel line passing through a point, A, not on I.
First method: Move a triangle along the edge of a ruler.
Second method: Draw a line through A intersecting I at B. Construct an angle at A equal to the angle formed at B, making A A and B equal corresponding angles.
2. Show that two lines perpendicular to the same line are parallel.
Use the theorem of § 195.
3. Prove the following theorem: Two lines are parallel if two alternate interior angles formed with a transversal are equal.
First show algebraically that the corresponding angles are equal if the alternate interior angles are equal. Then use the theorem of § 195.
4. Prove as in exercise 3 that two lines are parallel if the interior angles on the same side formed with a transversal are supplementary.
5. Prove that two lines parallel to the same line are parallel to each other; i.e., if AB and CD, Fig. 118, are parallel to EF, then prove ABWCD.
Proof: li AB and CD were
B
not parallel, they would inter- c B
sect if far enough extended, as
is indicated by the dotted lines. ^ ^
Then from the point of inter- Fig. 118
section K there would be two
lines KA and KC parallel to EF. Since this is impossible (why ?),
AB is parallel to CD,
l\)2 ' '' "- -Wlist-YEAR MATHEMATICS
6. State the conditions which will show that two lines are parallel to each other.
7. Draw two parallel straight lines and a transversal, as in Fig. 119. Measure and compare the corresponding angles a and b; c and d; e and /; g and h.
8. Draw two lines that are not par- allel, and a transversal. Measure and compare the corresponding angles.
196. Exercises 7 and 8 illustrate the following theorem:
Theorem: // two parallel lines are cut by a transversal the corresponding angles are equal.
EXERCISES
1. Two parallels and a transversal make angles designated as shown in Figs. 120, 121. Find the value of x and of all the 8 angles in each figure.
Use theorem of § 196.
(79 -t- Sx)" \^
Fig. 120 Fig. 121
2. Prove that if two parallels are cut by a transversal the alternate interior angles are equal.
First show by § 196 that the corresponding angles are equal, i.e., a = e, Fig. 122. Then prove that the alternate interior angles are equal, i.e., that e = c.
3. Prove that if two parallels are cut by a transversal the interior angles on the same side are supplementary.
Proof: In Fig. 122, a = e. Why? a+d = 180. Why? There- fore e+d= 180. Why?
PARALLEL LINES
103
4. In Fig. 122 lines AB and CD are given parallel State which angles are equal and which angles are supple- mentary.
5. With two parallels and a trans- versal the alternate interior angles are 7(x+l)° and (181-2a:)°. Find x and all the 8 angles.
6. With two parallels and a trans- versal the alternate interior angles are
(3x-5)° and 5(x-7)°. Find x and all the 8 angles.
7. Let ZA, Fig. 123, be a given angle. Through a point P
draw two lines parallel respectively to the sides of ZA.
Prove that a = b.
Proof: a = d. Why? d = b. Why? Therefore a = b.
8. Prove that a = c, Fig. 123.
9. Prove that a and e, Fig. 123, are supplementary.
Fig. 123
197. Exercises 8 and 9 show that if two angles have their sides parallel respectively they are either equal or supple- mentary.
Prove this theorem for the angle-pairs in Fig. 124.
Z.
Fig. 124
104
FIRST-YEAR MATHEMATICS
198. Using Fig. 125, in which DE is parallel to AC, prove that the sum of the interior angles of a triangle is two right angles.
Proof: ai+6+Ci = 180. Why?
ax =a. Why?
j^ut c ^^ Ci =c. Why?
Fig. 125 Therefore a-\-h+c = 180. Why ?
■E
199. Using Fig. 126 prove that an exterior angle of a triangle equals the sum of the two remote interior angles.
Proof: a=ai. |
Wliy? |
|
b = bi. |
Why? |
/ |
a+6+c = 180. |
Why? |
/^V |
ai+6i+c = 180. Therefore a+6+c = ai+6i+c. or a+6 = ai+6i. |
Why? Why? Why? |
|
Fig. 126 |
EXERCISES
1. Prove that the sum of the interior angles of a quadrilateral is 360°. (See Fig. 127.)
2. Find the angles of a quadrilateral if each angle is 30° greater than the consecu- tive angle.
3. The angles of a quadrilateral are to each other as 1:2:3:4. How large is each ?
Let the first angle be denoted by a, the second by 2a, etc.
Fig. 127
Angles of the Parallelogram and Trapezoid
200. Parallelogram. A quadrilateral whose opposite sides are parallel is a parallelogram.
PARALLEL LINES 105
201. Consecutive and opposite angles of a quadrilateral.
The angle-pairs x and y, y and s, s and t, t and X, Fig. 128, are consecutive angles. i^ The pairs x and s, y and t are opposite
^^S^^^- '"fig. 128
EXERCISES
1. With a ruler and protractor draw a parallelogram having two consecutive sides 3 in. and 5 in. respectively and the angle included between them equal to 60°.
^'A__ 2. Prove that the consecutive angles
I of a parallelogram are supplementary.
/ (See Fig. 129.)
; 3. Prove that the opposite angles of
! ' a parallelogram are equal.
Fig. 129
4. Find the angles of a parallelo- gram if one angle is 3 times as large as the consecutive angle.
5. The difference of two consecutive angles of a parallelogram is 20°. Find the values of all the angles of the parallelogram.
6. Two consecutive angles of a parallelogram are so related that 3 times one angle diminished by the other is equal to 30°. Find the values of both angles.
202. Trapezoid. A quadrilateral having only one pair of parallel sides is a trapezoid.
EXERCISES
1. In the trapezoid, Fig. 130, prove that x and y are supple- mentary angles.
2. Prove that the sum of the interior /^ ~\
angles of a trapezoid is four right / \
angles. ^ ^
, Fig. 130
3. In the trapezoid, Fig. 130, C is f
times as large as B and D is 4 times as large as A. How large is each ?
106
FIRST-YEAR MATHEMATICS
203. Solid. Surface. The cube, Fig. 131, and the rectangular parallelepiped. Fig. 132, are solids. Other
j
/
Cube Fig. 131
Rectangular Parallelepiped Fig. 132
Prism Fig. 133
Pyramid Fig. 134
soHds that are frequently found are the prism. Fig. 133, the pyramid. Fig. 134, the cylinder. Fig. 135, the cone. Fig. 136, and the sphere. Fig. 137.
Cylinder Fig. 135
Sphere Fig. 137
A solid consists of matter. It may be hard or soft, heavy or light. It has color. It fills a limited portion of space and is separated from the surrounding space by its surface. In geometry when we study a solid we are interested only in the form of the solid, its size and its shape. We disregard the color, weight, density, etc., and think only of the limited portion of space which it occupies. Such a solid is a geometric solid.
Which of the solids. Fig. 131-137, have flat surfaces ? Which have curved surfaces ?
204. Plane. A flat surface is called a plane surface or a plane.
Point out plane surfaces in the classroom. Point out sur- faces that are not plane.
LINES AND PLANES 107
To test whether a surface is plane a straight edge is appHed to it. If in every position the straight edge coincides entirely with the surface it is said to be a plane surface.
Since only one straight line can be drawn through two points it follows that a straight line two of whose points lie in a plane lies entirely in the plane.
How may a carpenter making a plane surface determine whether it is a plane ?
205. In § 191 it was seen that two lines in the same plane are parallel if they do not meet, however far ex- tended. When two lines in space do not meet they are not necessarily parallel.
Thus in Fig. 138 lines AiDi and AB do not meet, yet they are not parallel lines.
Point out lines in the classroom which do not meet and are not parallel; others which are parallel.
206. Parallel planes. Two planes which do not meet however far extended are parallel planes.
Thus, the planes of faces A BCD and AiBiCiDi in Fig. 138 are parallel.
Point out parallel planes in the classroom.
207. Parallel lines and planes. No line drawn in plane AiBiCiDi, Fig. 138, can meet plane ABCD. (Why?) A line which does not meet a plane however far extended is parallel to the plane.
In the classroom point out lines which are parallel to planes.
Line BBi, Fig. 138, is perpendicular to BC and BA. B Bi is also perpendicular to any line through B drawn in
108
FIRST-YEAR MATHEMATICS
plane A BCD. This may be verified by means of a try- square on a wooden model of a cube. BBi is said to be perpendicular to plane A BCD.
Point out in the classroom lines that are perpendicular to planes; e.g., one edge of a door is perpendicular to the plane of the floor. (Why?)
208. Lines perpendicular to a plane. If a line inter- sects a plane and is perpendicular to any line in that plane passing through the point of intersection it is said to be perpendicular to the plane . The symbol for expressing that one line is perpendicular to another is _l . Thus, the state- ment AB is perpendicular to CD is written AB±CD.
Models of Geometrical Solids
209. The cube. The cube may be constructed from a figure like Fig. 139. On cardboard draw the figure. 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. Measure the edge of the cube and compute the area of the whole surface. Find the volume of this cube.
Fig. 139
210. The parallelopiped. A
rectangular parallelopiped may be constructed from a figure like Fig. 140. Make the model. Compute the volume of the solid and the area of the sur- face.
Fig. 140
LINES AND PLANES
109
211. The prism. Construct a prism from a drawing like that in Fig. 141. Find the area of the entire surface.
To find the area of the hexagon divide each into 6 equilateral triangles. Meas- ure the base and altitude of one triangle and compute the area.
To draw the hexagons, draw a circle, step around it, using the radius for a step, Fig. 142, and connect consecutive marks.
212. The pyramid.
To construct models of pyramids use Figs. 143 and 144.
Fig. 141
213. The cone. Cut out two unequal circles, Fig. 145. Place the circles so that they touch each other, making point C fall on D. Then turn the circles, keeping them always touching each other, until C meets circle B at D'. This will make arc DED' equal to the length of circle A. Cut out angle DBD' and use the remainder for the curved surface of the cone. Use circle A as base.
Fig. 142
Fig. 143
Fig. 144
Fig. 145
no FIRST-YEAR MATHEMATICS
Summary
214. The chapter has taught the meamng of the following terms: parallel Imes; opposite and consecutive angles of a quadrilateral; solid, prism, pyramid, cone, cylinder, sphere; plane, parallel planes; lines parallel or perpendicular to a plane.
215. The symbol for parallelism is !!. The symbol for perpendicularity is ± .
216. The following theorems and axioms have been taught :
1. Parallel lines are everywhere equidistant.
2. One and only one parallel can he drawn to a line from a point not on the line.
3. 7/ two lines are cut by a transversal making the corre- sponding angles or alternate interior angles equal the lines are parallel.
4. If two lines are cut by a transversal making the interior angles on the same side supplementary they are parallel.
5. Two lines perpendicular to the same line are parallel.
6. Two lines parallel to the same line are parallel.
7. If two parallel lines are cut by a transversal the alternate interior angles are equal, the corresponding angles are equal, and the interior angles on the same side are supplementary.
8. If two angles have their sides parallel respectively they are either equal or supplementary.
9. The exterior angle of a triangle is equal to the sum of the two remote interior angles.
10. The opposite angles of a parallelogram are equal.
LINES AND PLANES 111
1 1 . The consecutive angles of a paralleloynmi are supple- mentary.
217. The following construction was taught: Through a given point outside of a given line to draw
a line parallel to the given line.
218. It was seen how models can be made of the following geometrical solids: cube, rectangular parallelo- piped, prism, pyramid, and cone.
CHAPTER VIII
MEASUREMENT OF LINES IN SPACE. SIMILAR FIGURES
Drawing to Scale
219. Indirect measurement. In the preceding chap- ters lengths of distances and sizes of angles were deter- mined by direct measurement with ruler, compass, and protractor. However, in many cases problems call for the lengths of distances which cannot be measured di- rectly, e.g., the distance across a river, the height of a tree, etc. The following problems will show that when it is impossible to determine distances or angles by direct measurement other related lines or angles may be meas- ured which will enable us to determine the required parts. This is called indirect measurement.
A man starting from a point 0, Fig. 146, walks 1.15 mi. south, then 2.2 mi. east, and then 1 . 75 mi. north. How far is he from the starting-point ?
Letting 2 cm. on squared paper repre- sent 1 mi., make a drawing of the meas- ured distances, as Fig. 146. Then meas- ure CO in the drawing. This is the required length.
220. Drawing to scale. The distances of the last problem as graphed in Fig. 146 are said to be drawn to
112
~ |
1 1 |
- |
~ |
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2 |
0- |
r-i |
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' |
||||||||||||||||||||||||||
^ |
.— |
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^ |
" |
■' |
||||||||||||||||||||||||
^ |
||||||||||||||||||||||||||
r |
^ |
r- |
||||||||||||||||||||||||
\^ |
-* |
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^ |
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14 |
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>^ |
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i> |
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4 |
? |
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_ |
_ |
Fig. 146
MEASUREMENT OF LINES
113
the scale: 2 cm. to 1 mi. should be indicated.
On all scale drawings the scale
EXERCISES
1. Draw to the scale 2 cm. to 1 mi. the following distances: 2.34 mi.; 1.06 mi.; 3.90 mi.; 0.15 mi.; 2.63 miles.
2. A man starting from a point S walks 45 yd. east and then 60 yd. north. Find the direct distance from the stopping- point to the starting-point.
Make a drawing to the scale 1 cm. to 10 yards.
221. Surveying instruments. The
instruments commonly used by surveyors for measuring distances are the surveyor's chain, Fig. 147, and tape, Fig. 148. Angles are measured with the transit (p. 56).
The chain is made
up of a succession of
links having loops at
the ends, which are
connected by rings.
In measuring distances, chaining
pins. Fig. 149, are used to mark the
end-point of the chain or
tape.
Fig. 147.— Surveyor's chain.
Fig. 148.— Steel Tape
EXERCISES
1. A man walks 80 yd. south, then 144 yd. east, and then 120 yd. north. Find the direct distance from starting-point.
Use the scale 1 cm. to 12 yards.
2. Two men start from the same point. One walks 5 mi. west, and then 3 mi. north; the other walks 4 mi. south, and then 5 mi. east. How far apart are they?
Fig. 149. — Chaining pins.
114
FIRST-YEAR MATHEMATICS
3. Draw to the scale 1 cm. to 2 ft. a plan of a room 24 ft. by 18 ft. and find the distance diagonally across the floor.
4. Draw a plan of a rectangular field 16 rods long and 12 rods wide, using the scale 1 in. to 4 rods, and find the distance diagonally across the field.
5. Draw to the scale 1 cm. to 10 ft. a plan of the end of a house, Fig. 150, and find the height of the top of the roof from the ground.
6. The Union Station in Washington, D.C., is 760 ft. long and 343 ft. wide. The main waiting-room covers an area of 220 X 130 sq. ft. The station exceeds the United States Capitol in dimensions, the latter being 751 ft. long and 350 ft. wide. Draw rectangles whose dimen- sions are scale drawings of the dimensions of both buildings and the waiting-room.
7. A railroad surveyor wishes to measure across the swamp AB, Fig. 151. He measures the distance from a tree A to a stone C and finds it to be 165 feet. The distance from a tree at B to the stone C is 150 feet. Find the distance in feet across the swamp, the angle at C being 80°.
8. To measure the width ^C of a stream. Fig. 152, without crossing it, an engineer lays off a line, BC, on one side
of the river, and measures the angles at B and C. Draw a triangle to scale from the data in the figure and determine the width of the river.
9. A boy wishes to determine the height, HK, Fig. 153, of a factory chimney. He places the angle-measurer first at B and then at A and measures the angles x and y. The angle- measurer lies on a box, or tripod, 3^ ft. from the ground. A and
Fig. 152
MEASUREMENT OF LINES
115
B arc two points in line with the chimney and 50 ft. apart. What is the height of the chimney if the ground is level and if a; = 63° and ?/ = 33|°?
10. The line AB, Fig. 154, passes through a building. Explain how to find by means of a scale drawing the distance y ■ m between A and B,
/
A
Bn
Fig. 153
Fig. 154
222. Angle of elevation. A telescope is pointed hori- zontally in the direction EH, Fig. 155, toward the tower HT, and the farther end is then raised (elevated) until the telescope points to the top, T, of the tower. The angle HET through which the telescope turned is the angle of elevation of the top of the tower, from the point of horizLfaiii'^» — T observation. Fig. 155
EXERCISES
1. From point A, Fig. 153, what is the angle of elevation of the top of the chimney? From point B?
2. When the angle of elevation of the sun is 25°, a building casts a shadow 90 ft. long, on level ground. Find the height of the building.
3. Find the angle of elevation of the sun when a tree 40 ft. high casts, on level ground, a shadow 60 ft. long.
4. The angle of elevation of the top of a tree is 38°, the observer standing 20 yd. from the tree. How high is the tree ?
5. On the top of a tower stands a flagstaff. At a point A, on level ground, 50 ft. from the base of the tower, the angle of ele- vation of the top of the flagstaff is 35°. At the same point A
116 FIRST-YEAR MATHEMATICS
the angle of elevation of the top of the tower is 20°. Find the length of the flagstaff.
A-
jj±oHzonianineT__ 223. Angle of depression. A
telescope at 7", on the top of a cliff (Fig. 156), is pointed hori- zontally, and then the farther end Fig. 156 ^^ lowered (depressed) until the
telescope points to the buoy at B. The angle HTB through which the telescope turned is the angle of depression of the buoy from the point T.
EXERCISES
1. If the height of the cliff, Fig. 156, is 100 ft., and the angle of depression of the buoy, as seen from T, is 40°, what is the distance of the buoy from the bottom of the cliff ?
2. A boat passes a tower on which is a searchlight 120 ft. above sea-level. Find the angle through which the beam of light must be depressed from the horizontal so that it may shine directly on the boat when it is 400 ft. from the base of the tower.
3. From the top of a cliff 150 ft. high the angle of depression of a boat is 25°. How far is the boat from the top of the cliff ?
4. From a lighthouse, situated on a rock, the angle of depres- sion of a ship is 12°, and from the top of the rock it is 8°. The height of the lighthouse above the rock is 45 feet. Find the distance of the ship from the rock.
224. Bearing of a line. The angle which a line makes with the north-south line is the bearing of the line.
The bearings of the arrows in Fig. 157 are read 30° east of north, 20° west of south, 60° west of north, 75° east of south. These may be written briefly: N 30°E, S20°W, N60°W, S75°E.
MEASUREMENT OF LINES
117
EXERCISES
1. With a ruler and protractor draw lines having the follow- ing bearings: 45° east of north (northeast); 45° west of south
N
N
W-
W-
i
N
N
60'
-E W-
-E W-
75°
Fig. 157
(southwest); 65° east of south; 65° east of north; of south; 40° west of north.
2. Write in abbreviated form the bearings of the lines in exercise 1.
225. Bearing of a point. An observer standing at a point A is looking in the direction of another point B. The bear- ing of the line from A to B with refer- ence to the north-south line through the point of observation A is called the bearing of the point B from A.
Thus, in Fig. 158, the bearing of B from A is N 50°E ; the bearing of A from B is S 50°W.
Why are angles NAB and SBA, Fig. 158, equal ?
EXERCISES
1. The bearing of a fort B from A, both on the seacoast, is N65°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 7 mi. apart. Find the distance from each fort to the vessel.
Make a drawing to the scale 1 cm. to 2 meters. To get the direc- tions of the lines draw north-south lines first through A, and through B when the position of B is determined.
118 FIRST-YEAR MATHEMATICS
2. Point Q is 6 . 4 mi. east and 9 . 8 mi. north of P. Find the distance from P to Q. What angle does PQ make with the north-south line through P f
3. The view from a battery at B, Fig. 159, to the enemy's fort at F is obstructed by a hill H. A point P is found near the bottom of the hill from which F is observed to bear 4 mi. northeast. P is 6.25 mi. northwest of B. Find the distance and bearing of F from B.
4. A man wishes to measure the width of a river without crossing it. The river flows due west. Standing at A, on the bank, he observes a tree on the other bank in the direction N20°E. He walks Fig. 159 along the bank 50 rods east to B and there
observes the tree in the direction N60°W. Find the width of the river.
Ratio
226. Ratio of numbers. The quotient found by dividing 6 by 3 is the ratio of 6 to 3. This is sometimes
written 6:3, or 6/3. For the present purpose ^ will be
considered the most convenient form. Thus the ratio
of 3 to 4 is ^; of a to 6 is ^ • The ratio of any number
to another number is the quotient found by dividing the first number by the second.
Any fraction may be regarded as an expression of the ratio of the numerator to the denominator. A whole number expresses the ratio of itself to 1.
227. Ratio of line-segments. If a line-segment is 2 in. long and another 3 in., the first is f of the second. The number f is the ratio of the two line-segments.
What is the ratio of two line-segments of length 5 in. and 7 inches?
MEASUREMENT OF LINES 119
In practice, to find the ratio of two line-segments each is measured in a convenient unit, as inch or centimeter, and one measure is then divided by the other. Thus, the ratio of two line-segments is the ratio of the numerical measures of the segments, both being measured with the same unit.
228. Notation for ratio of segments. According to
§ 227 the ratio of two line-segments, as AB and CD,
-n- ^nr^ mesiSUTe oi AB -,, ,, . .i. .
Fig. 160, means tttt. > with the agreement that
^ measure of CD ' ^
a common unit is used in measuring. a b
A R This is usually written briefly 77^ . ^ ^
, ^ CD Fig. 160
EXERCISES
AB
1. Using 2 cm. as a unit find the ratio ^^, Fig. 160, and
express the result as a decimal fraction.
2. Two line-segments are measured in yards and are found to be 7 yd. and 3 yd. long respectively. Find the ratio.
3. What are the measures of the segments in exercise 2 in feet ? What is the ratio ?
4. Draw two line-segments and measure them in centimeters and in inches. In each case find the ratio expressed as a decimal fraction. How do the results compare ?
229. Exercises 2, 3, and 4 illustrate the fact that a change in the unit of measure does not change the ratio.
EXERCISES
2
1. Draw two line-segments having the ratio -. How many
2 pairs of line-segments have the ratio - ?
3 5
2. Give several pairs of numbers having the ratio p.
3. Show that Sx and 7x represent all number pairs whose ratio is ~.
120 FIRST- YEAR MATHEMATICS
4. Divide 85 into two parts in the ratio 2:3. Let the parts be denoted by 2x and Sx.
5. Divide 84 into three parts which are as 3:4:5.
5
6. Two numbers are in the ratio „. If 12 is subtracted
o 3 from each, the differences are in the ratio -. What are the
4 numbers ?
7. What number added to 12 and subtracted from 30 gives
5 results that are in the ratio 777 ?
1^ 2i
8. The ratio of two Une-segments is -j . The longer line is
34 30 centimeters. Find the shorter line.
230. Ratio of angles and polygons. The ratio of two angles or of two polygons is the ratio of their numerical measures.
EXERCISES
1. Draw two angles and measure each with a protractor.
Find the ratio of the angles.
-R-
R,
Fig. 161
2. Rectangles R and Ri, Fig. 161, are divided into equal parts. Find the ratio.
3. The angles of a triangle are in the ratio 1:2:3. Find them.
4. The angles covering the plane around a point are in the ratio 2:3:4. Find them.
5. The ratio of 2 times an angle to 8 times the supplement is 2' Find the angles.
6. If 6° be taken from an angle and added to its complement
2 the ratio of the angles thus formed is ^. Find the angles. .
Similar Figures
231. Construction of triangles. In §§ 219-225 it was shown how unknown distances and angles could be deter- mined by indirect measurement. In the following sections
MEASUREMENT OF LINES
121
an algebraic method of finding distances will be developed which is more advantageous than the geometric method of using scale drawings.
EXERCISES
1. Draw a triangle whose angles are respectively 35°, 65°, and 80° and measure each side.
Draw a line-segment of convenient length, as AB, Fig. 162. At A draw on AB an angle of 35°. At B draw an angle of 65°. Check the accuracy of the construction by measuring angle C. Measure to two decimal places AB, AC, and BC.
2. Starting from a length of AB different from that in Fig. 162, construct a second tri- angle whose angles are 35°, 65°, and 80° respec- tively and measure its sides.
3. Find the ratio, to two decimal places, of each side of the triangle in exercise 1 to the corresponding side of the triangle in exercise 2. How do your results compare with the results obtained by other members of your class ?
4. How do the triangles constructed in exercises 1 and 2 compare as to shape? Are these triangles of the same shape as the triangles made by the other members of the class ? Are the triangles of the same size ?
5. Draw two triangles of different sizes in each of which the angles are 30°, 60°, and 90°. Compare the triangles as to shape. Measure all sides of the triangles. Find the ratios of the sides of one triangle to the corresponding sides of the other.
6. Draw two triangles of different sizes each of which has angles equal to 90°, 25°, and 65° respectively. Compare the
-p- |
^ T |
V x' |
i" ^q4- |
A |
^ ^ 5 i |
y^\ |
v^ ^ |
n ^ V- |
O^ S |
Y ^X- |
^^ \ |
.^ \ |
^ A -iv |
^^f it «s |
^2! _ _ ji -jii^ |
4 2\M- -^' |
Fig 162
122 FIRST-YEAR MATHEMATICS
triangles as to size and shape. Compare the ratios of the corresponding sides.
7. Draw a triangle. Draw another having the angles equal respectively to the angles of the first triangle. Are the triangles of the same shape? Are they necessarily of the same size? Compare the ratios of the corresponding sides.
232. Similar triangles. Triangles having the same shape are called similar triangles. Similar triangles are not necessarily of the same size.
233. Exercises 1-7, §231, illustrate the following theorem :
Theorem: If the angles of one triangle are respectively equal to the angles of another the triangles are similar.
It was also seen that if the corresponding angles of two triangles are equal, it follows that the ratios of the corresponding sides are equal.
This explains the following definition of similar tri- angles and polygons.
234. Similar polygons. If two polygons have their corresponding angles equal and the ratios of the corre- sponding sides equal they are called similar polygons.
Similar figures are of frequent occurrence, e.g., two squares, two circles, two equilateral triangles, a figure and its scale drawing, a photograph and an enlarged or reduced picture, and so on.
Similar triangles may be regarded as the same triangle magnified or minified to a definite scale, or both may be regarded as scale drawings of the same triangle to different scales.
Similarity of figures is indicated by the symbol ^. Thus, AXYZ - AXiFiZi means that AXFZ and XiYiZi are similar.
MEASUREMENT OF LINES
123
235. Problems in similar figures. The fact that the ratios of the corresponding sides of similar polygons are equal suggests an algebraic method of finding distances.
EXERCISES
1. To measure the height of a tree.
The tree, the length of the shadow, and the sun's rays passing over the top of the tree form a triangle, Fig. 163. The shadow is measured and is found to be 80 ft. long.
At the same time the
shadow of a vertical
stick 3 ft. high is
measured and found to
be 8 ft. long, Fig. 164. The stick, the shadow,
and the sun's rays form a triangle similar to the Why? Letting h denote the height of the tree,
Fig. 163
Fig. 164
first triangle, we have
80
^8
Why'
Whence, h = m*
2. To measure the height of a flagpole a boy holds a pencil AB, Fig. 165, in vertical position and 2 ft. from his eye so that
Fig. 165
it covers the pole. If the pole is 360 ft. away and if the pencil is 6 in. long, how high is the pole ?
* According to a report by Plutarch this method of finding dis- tances was devised by Thales (640 B.C.-550 b.c.) who used it to find the height of a pyramid. (See Ball, p. 16.)
124 FIRST-YEAR MATHEMATICS
3. The gables of a house and a porch have the same shape. The sides of the porch-gable are 7 ft., 7 ft., and 10 feet. The longest side of the house-gable is 25 feet. How long are the other two sides of the house-gable ?
4. The sides of a triangle are 8, 10, and 13. The shortest side of a similar triangle is 11. Find the other sides.
5. The sides of a triangle are 4.6 cm., 5.4 cm., and 6 centi- meters. The corresponding sides of a similar triangle are a; cm., y cm., and 15 centimeters. Find x and y.
6. The sides of a triangle are 1, 2, and 3, and the longest side of a similar triangle is 20. Find the other sides of the second triangle.
7. Two rectangular flower beds are to be made of the same shape, but of different size. One is to be 3 ft. wide and 5 ft. long; the other 12 ft. wide. How long should it be ?
8. A city block and a lot within the block have the same shape. The lot is 100 ft. by 150 ft. and the block is 300 ft. wide. How long is it ?
9. Prove that if a line is drawn parallel to one side of a
given triangle, meeting the other two sides, ^ a triangle is formed similar to the given
n/X.^ triangle.
A^ — ^B . Show that the angles of triangle ABC,
Fig. 166 Fig- 166, are equal to the corresponding
angles of triangle DEC. Then apply §233.
10. In Fig. 166 let AC =21, BC=S5, ^e DC=S. Find^C. ''f^^^
11. InFig.l661etA(7=3+a:,i)C = 3, fy loo \^^
EC = 4:, and EC=l, Find x and the / A^ ^T^r^
length of AD. ck^,:^^--------^D
12. To find the distance across a Fig. 167 lake CD, Fig. 167, triangle CEB is
drawn and EC, AE, and AB are measured. Find CD.
MEASUREMENT OF LINES
125
13. Line AB, Fig. 168, is 20 ft. long, having the ratio f. How long is each part?
Draw AC±AB and 2 units long. Draw BD-LAB and equal to 3. Draw CD. Show that triangles AEC and EBD are similar and determine x by means of an equation.
14. Divide a line 24
7 parts having the ratio ^;
Divide it into parts
into
ft. long 4 6 7' 5"
15. Draw a triangle. Draw a second triangle whose sides are respectively twice as long as the sides of the first.
Let ABC, Fig. 169, be the first triangle. On an indefinite line lay off AiBi equal to twice AB. With Ai as center and radius
-^.Cr
twice as long as AC strike an arc, as at Ci. With Bi as center and radius twice as long as BC strike an arc meeting the first arc. Draw AiCi and BiCi. Then AiBiCi is the required triangle.
16. Compare the triangles constructed in exercise 15 as to shape. Are they similar? Measure the corresponding angles. What are the ratios of the corresponding sides ?
17. Draw a triangle with sides 3 times as long as the corre- sponding sides of another triangle. How do the ratios of the corresponding sides compare? Are the triangles similar? Measure the corresponding angles.
236. Exercises 15 to 17 illustrate the following theorem: Theorem: Two triangles are similar if the ratios of the' corresponding sides are equal.
126 FIRST-YEAR MATHEMATICS
Summary
237. This chapter has taught the meaning of the following terms: indirect measurement, drawing to scale; angle of elevation, angle of depression; bearing of a line, of a point; ratio of numbers, of line-segments, of angles, of polygons; similar figures. The symbol ^ is used to indicate similarity.
238. The following instruments have been shown to be used by surveyors and engineers: the transit, sur- veyor's chain, surveyor's tape, chaining pins.
239. The truth of the following theorems has been shown :
1. Two triangles are similar if the angles of one are respectively equal to the angles of the other.
2. Two triangles are similar if the ratios of the corre- sponding sides are equal.
240. Two ways of finding distances by indirect meas- urement have been used:
1. The required distance is found by means of a scale drawing obtained from measurement of lines and angles related to the required distance.
2. The required distance is found by solving an equa- tion obtained from the similarity of two triangles, one of which contains the required distance as a side.
241. Similar polygons are defined in geometry as polygons whose corresponding angles are equal and whose corresponding sides have equal ratios.
AKClllMEDEti
A R C H 1 M E D P: S
ARCHIMEDES was born at Syracuse on the island of Sicily 287 b.c. and died there 212 b.c. It is said that he was related to the royal family of Syracuse. He studied mathematics under Conon at the University of Alexandria in Egypt. His great mechanical ingenuity was often called into the service of his government. He held it to be beneath the dignity of a scientist to apply his science to practical use; nevertheless he was the inventor of numerous practical devices and mechanical contrivances. Read about his detecting the fraudulent goldsmith, his invention of burning- glasses, his lever device for launching ships, and his device for pumping the water out of ships and even of inundated fields, etc., in Ball's History of Mathematics.
It was on the occasion of launching one of the king's large new ships that he remarked that he could move the earth if he but had a fulcrum to place his lever on.
He wrote many mathematical and scientific works, including important contributions to almost every field of science then known. He did especially valuable work in plane and solid geometry. In a book on the Circle he showed for the first time that the ratio of the length of a circumfer- ence to its diameter is between 3f and S^J-. He also worked out in another place the relations as to volume and surface of the cone, cylinder, and sphere. He regarded his discoveries on the round bodies as so important and so beautiful that he requested that the figure of a sphere inscribed in a cylinder be carved on his tombstone.
His contributions to pure and applied science were so numerous and so important that he is often referred to as "The Newton of antiquity." See whether you think the title appropriate after reading in Ball or elsewhere what both he and Newton did for science.
In his Sand Counter, Archimedes undertook to calculate and to express in numbers the number of grains of sand it would take to fill the universe. It is thought that the reason he did this was to show his scientific countrymen that there could be devised a great deal more powerful way of writing numbers than the way the Alexandrian scholars were teaching people to write them.
CHAPTER IX
RATIO. VARIATION. PROPORTION
Trigonometric Ratios
242. In chapter VIII the importance of ratio* in deter- mining distances has been recognized. What follows here will illustrate further the use of ratios in the solution of problems.
I wish to know the height of a recently constructed chimney, AB, Fig. 170. I ^^^^
have determined that I am ^^^'
about 250 ft. from the chim- ney and that the angle of ^" Tsd'" elevation of the top is ap- Fig. 170 proximately 30°.
The problem is easily solved by means of a scale drawing.
Let A^iBiCi, Fig. 171, be a scale drawing of AABC.
Fig. 171
Then AA.BiCi^ AABC.
Hence,
h AiBi
2.50
''A,C,
or h=250-
A,B, AiBx
Why? Why?
The last equation shows that if the ratio ^^^ were known, h could be found by multiplying this ratio by 250.
* Ratio and proportion is one of the oldest topics of mathe- matics. It was employed by the Egyptians, Babylonians, Chinese, Greeks, Arabs, and mediaeval Europeans. The equation has largely supplanted the proportion in modern mathematics.
127
128 FIRST-YEAR MATHEMATICS
A 7? However, the ratio . ^ can be obtained from aw?/ right triangle one
of whose acute angles is 30°. For, let AAiBiCi , Fig. 171, and ^ABC, Fig 170, represent any two right triangles having ZC= ZCi=30°.
AB AC Then, since these two triangles are similar, -r-^- =-r-7T •
AiBi Aid
Multiplying both sides of this equation by AiBi we have
ACXAiBr
AB =
AiCi
Dividing both sides by AC, we have ~Tr< = -rjr •
Akj A\\^i
Thus, the problem of finding AB may be solved as follows:
From any right triangle containing an angle of 30°, determine the
A H ratio -r^ . Then find h by substituting the result in the equation Aid
. = 250.^.
243. Table of tangents. Denoting the angle of elevation by C, Fig. 172, the height by h, and the distance from the point of observation to the chimney by /, the equation
may be used as a formula for computing h. The value
AB of the ratio -j-^ may be computed for various values of
ZC and tabulated. Any height may then be found by measuring the angle of elevation C and the distance /,
AB
and then multiplying I by the value of the ratio -j-^
AB
corresponding to C in the table. The ratio -j-^ is called
the tangent of ZC and the table is the table of tangents.
RATIO
129
TABLE OF TANGENTS OF ANGLES FROM 1° TO 89=
1 |
bC < |
3.1 bO |
.2 tsC |
1 |
1 < |
r |
1 |
§ Eh |
|
1° |
.02 |
19° |
.34 |
37° |
.75 |
55° |
1.43 |
73° |
3.27 |
2 |
.03 |
20 |
.36 |
38 |
.78 |
56 |
1.48 |
74 |
3.49 |
3 |
.05 |
21 |
.38 |
39 |
.81 |
57 |
1.54 |
75 |
3.73 |
4 |
. .07 |
22 |
.40 |
40 |
.84 |
58 |
1.60 |
76 |
4.01 |
5 |
.09 |
23 |
.42 |
41 |
.87 |
59 |
1.66 |
77 |
4.33 |
6 |
.10 |
24 |
.44 |
42 |
.90 |
60 |
1.73 |
78 |
4.70 |
7 |
.12 |
25 |
.47 |
43 |
.93 |
61 |
1.80 |
79 |
5.14 |
8 |
.14 |
26 |
.49 |
44 |
.96 |
62 |
1.88 |
80 |
5.67 |
9 |
.16 |
27 |
.51 |
45 |
1.00 |
63 |
1.96 |
81 |
6.31 |
10 |
.18 |
28 |
.53 |
46 |
1.03 |
64 |
2.05 |
82 |
7.11 |
11 |
.19 |
29 |
.55 |
47 |
1.07 |
65 |
2.14 |
83 |
8.14 |
12 |
.21 |
30 |
.58 |
48 |
1.11 |
66 |
2.25 |
84 |
9.51 |
13 |
.23 |
31 |
.60 |
49 |
1.15 |
67 |
2.36 |
85 |
11.43 |
14 |
.25 |
32 |
.62 |
50 |
1.19 |
68 |
2.47 |
86 |
14.30 |
15 |
.27 |
33 |
.65 |
51 |
1.23 |
69 |
2.60 |
87 |
19.08 |
16 |
.29 |
34 |
.67 |
52 |
1.28 |
70 |
2.75 |
88 |
28.64 |
17 |
.31 |
35 |
.70 |
53 |
1.33 |
71 |
2.90 |
89 |
57.29 |
18 |
.32 |
36 |
.73 |
54 |
1.37 |
72 |
3.08 |
EXERCISES
Solve the following problems :
1. The rope of a flagpole, BC, Fig. 173, is stretched out so that it touches the ground at a point 20 ft. from the foot of the pole, and makes an angle of 73° with the ground. Find the height of the flag- pole.
RC
h = 20-jYi by the formula
;i=20(3.27), from the table
Hence,
20' C
Fig. 173
2. A balloon is anchored to the ground at a point A by a rope, making an angle of 67° with the ground. The point C on the ground directly under the balloon is 139 ft. from A. Assuming the rope-line to be straight, find the height of the balloon.
130 FIRST-YEAR MATHEMATICS
3. The angle of elevation of an aeroplane at a point A on level ground is 60°. The point C on the ground directly under the aeroplane is 300 yd. from A. Find the height of the aeroplane.
4. The angle of elevation of the top of a tower is 27° and the distance from the foot of the tower is 259 feet. Find the height of the tower.
5. A vertical pole 8 ft. long casts on level ground a shadow
9 ft. long. Find the angle of elevation of the sun.
Since the ratio -j-^ = - = . 89 approximately,
Fig. 174, the table of tangents shows the cor- responding value of angle C to be approxi- mately 42°.
Fig. 174 6. A tree is broken by the wind into two
parts forming a right triangle with the ground. The upper part makes an angle of 40° with the ground. The top of the tree is 48 ft. from the foot, the dis- tance being measured along the ground. How high was the tree?
7. From the roof of a building the angle of depression of a point directly opposite on the other side of the street is 74°. The street being 45 ft. wide, find the height of the building.
Ratio
244. On squared paper draw a triangle, as ABC, Fig. 175. Through a point on A C, as D, draw line DE parallel to ^ J5. Find
the ratios j^ and -^ to two deci- mal places. ""' pj^ J75 How do the ratios compare ?
RATIO 131
This problem illustrates the following theorem: , Theorem : A line drawn parallel to one side of a triangle
divides the other two sides into corresponding parts having
equal ratios*
EXERCISES
1. Find EB, Fig. 175, if DC = 8, DA = 2.5, and CJ5;=10.2.
2. Find EB, if DC = 4, AD= .27, and CE=li DC.
3. Find EB, if DC = a, AD = b, and C^ = c.
4. If AD = DC, Fig. 175, show that BE = EC.
5. State exercise 4 in the form of a theroem.
245. Draw a triangle, as ABC, Fig. 176. Bisect
AC AD
angle ACB and find the ratios -^^ and -^ . How do
they compare? Express the result as a theorem and compare your statement with the fol- lowing theorem: c
Theorem : A line bisecting an angle of a triangle divides the side opposite
that angle into parts whose ratio is ^^ p"
equal to the ratio of the other two sides. Fig, 176
EXERCISES
1. If in Fig. 176 AC = S", BC = Q'\ and if AD = 2" less than DB, find the lengths of AD and DB, using the foregoing theorem :
2. Find the segments of AB, Fig. 176, if AB = 8, AC = 4, and CB = 7.Q.
* Eudoxus (408-355 b.c.) gave the first satisfactory proof of this theorem. Archimedes (287-212 b.c.) extended it to apply even if the parallel cuts the extensions of the sides.
132
FIRST-YEAR MATHEMATICS
2^6. Draw a triangle, as ABC, Fig. 177, measure AB, and divide it into 7 equal parts. Lay off BD = | A B, thus dividing 2
Fig. 177
AB at D in the ratio
divide CB in the ratio
Similarly
2
7'
Draw
DE. Measure angles BDE and DAC.
If the construction is well done, A BDE and DAC will be found to be equal.
Show that DE is parallel to AC.
This problem illustrates the following theorem:
Theorem: If two sides of a triangle are divided into parts having the same ratio, the line joining the points of division is parallel to the third side of the triangle.
EXERCISES
1. In triangle ABC, Fig. 177, find EC, if AD = i, DB^G, and 5^= 10.
2. The distance AB across a swamp is to be found. Fig. 178.
A point C is found in the same line with A and B.
At C and B lines CD and BE are drawn perpendicular to CB and the line AD is drawn. The measured values of CB, DE, and EA are 160 ft., 175 ft., and 642 ft. respectively. Find the dis- tance AB.
3. Divide a line-segment into two equal parts.
_F Let AB, Fig. 179, be the given
7 7^ segment.
}y^^ / Draw AC through A making an acute
"'^>.^/ angle with AB.
^"--^^ On AC lay off AD = DE.
^C Join E to B.
^^«- 1^^ Draw/)/?' 11^5.
Show that AF = FB. (See exercise 4, § 244.)
^•^
RATIO 133
4. Divide a line-segment into two parts whose ratio is |. Let AB, Fig. 180, be the given seg- ment. . F
^-r:: 7 yB
Draw AC making an acute angle *-^v' ^
with.4B. ^ "^^ /
On AC lay off AD = 2 units and iS^^^,.
Z)^ = 3units. ""^^c
Join ^ to i?. Fig. 180
Through D draw DF li EB.
Show that ^ = ^.
3 17
5. Divide a line-segment into parts having the ratio t; ^i q.
247. Use of compass in finding ratios of line-segments.
Let AB and CD, Fig. 181, be two segments whose ratio is to be found.
^2b
„ 27 27 221
■Z.D
Fig. 181
Let us assume that AB and CD contain a common unit of measure. (It will be shown later that there are line-segments that have no common unit of measure.) To find the common unit, proceed as follows:
Lay off the smaller segment CD on the larger AB as often as possible, leaving a remainder EB, which is less than CD.
Lay off EB on CD, leaving a remainder FD, which is less than EB.
Lay off FD on EB, leaving a remainder GB.
Lay off GB on FD, leaving no remainder.
The last remainder, GB, is a common unit of measure of AB and CD.
Using GB as unit, show that AB = SQ, and CD =27.
Therefore ^=11 .
248. Greatest common divisor. The method of find- ing the ratio of two line-segments by means of the com- pass as given in § 247 is not as practical as the method of chapter VIII in which squared paper or the marked ruler
134 FIRST-YEAR MATHEMATICS
was used for measuring. However, the method of § 247 is valuable for other purposes; e.g., it serves to illustrate a method of finding the greatest common divisor (g.c.d.) of arithmetic and algebraic numbers, so impor- tant in reducing ratios and fractions to lowest terms. Find the greatest common divisor of 86 and 27. Compare the following solution step for step with the solution in §247.
27)86(3 Divide the smaller number, 27, into the
81 larger, 86, leaving a remainder 5.
5)27(5 Divide this remainder into the smaller
25 given number, leaving a remainder 2.
2)5(2 Divide the last remainder into the pre-
4 ceding divisor, leaving a remainder 1.
1)2(2 Divide the last remainder into the pre-
2 ceding divisor, leaving no remainder.
The last divisor, 1, is the greatest common divisor of 86 and 27.
249. Prime numbers. When two numbers have no common divisor except unity they are said to be prime to each other. A number which has no divisor except itself and unity is a prime number.
EXERCISES
1. Find the g.c.d. of the following pairs of numbers:
1. 73 and 16 5. 263 and 765
2. 1,155 and 52 6. 350 and 425
3. 174 and 273 7. 3,542 and 5,016
4. 174 and 275 8. 3,795 and 2,865
2. Find geometrically, i.e., by means of line-segments and the use of the compass, the greatest common divisor of 73 and 16; of 70 and 36; of 45 and 42.
250. Reduction of ratios. In algebra, as in arithmetic, fractions are reduced to lowest terms either by dividing both numerator and denominator by all factors common to them, or by dividing both terms by their g.c.d.
RATIO 135
EXERCISES Reduce to lowest terms.
1. I 15. ^' 28. ^^
15 966 mnx
1 = 2_J ^ 2 First find g.c.d. a{a-\-h)
15 5-3 5 then reduce. 29. ^r^\jy\
14 ^^- 5,117 30.
X'Z
66 ^'- 1752 31
22 ^^ ^03^ 28m3n2A;
14m^^ '^ 18. -^ oo 12«6c^
36 1,254 32.
192 or: «6a;
4a36
12 1^ 3Q^^3^
18 1^-2,845 ^ ^'U^^
Q ^A 20. — o. m^3p^r
24 ^' 660 34.
p^qhnnr
7. 1? 21. -^ o. 15«^^c'
30 2,310 35.
30a46c
g^ 20 22. ?^ ^^ 12a262c2^
36 18x?/ 36.
24 24a;'!/ _ 0 • 4/ • x • a: • y _ 4x2
48 l8^2- 0.3^.^.y -3^ 3^
'^ 23. - 2^x22^
ioo "" ■ 38.3^
102 24. ^
60a6c2(i2
18605/ 12aa:?/''^
'2
7a%'x
25. ^ ^^ 15^V^
3xV
12- 240 ■ «^7 ^•
666 26 ^ 41 257/iW
909 * a;22 41. ^^a^^s
14. -^ 27. ^ 42. '^'''^''''
1,728 ' a;?/2w; 56a562c3a;?/
136
FIRST-YEAR MATHEMATICS Direct Variation
251. Function. If a man walks at the rate of 2 mi. an hour, the distance d he walks in a given time t is found from the equation d = 2t.
The following table gives the distance passed over in 1 hr., 2 hr., 3 hr This table shows that a change
Number of hours |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
Distance passed over . . . |
2 |
4 |
6 |
8 |
10 |
12 |
14 |
16 |
in the time causes a change in the distance passed over. The distance, is said to depend upon the time, or the distance is said to be a function of the time. The time and distance are variables, the rate is a constant.
Dependence of one magnitude upon another is met frequently. Thus, the amount a man earns in a given time depends upon the number of days he works; the price paid for a piece of goods depends upon the number of yards; the greater a man's property is the higher will be his taxes; the weight of a piece of iron depends upon its size, etc. In geometry the area of a square depends upon the length of the side; the area of a circle depends upon the length of the radius; the perimeter of an equilateral triangle depends upon the side, etc.
252. Constant. A number that has a fixed value is a constant.
253. Variable. A number that changes, assuming a series of values, is a variable.
EXERCISES
1. In the table above find the ratio of each distance to the corresponding time. How do the ratios compare? It is seen
VARIATION 137
that although the distance and time vary the ratio does not change (remains constant).
2. A man earns $20 per week. Show that the amount he earns in a given time depends upon the time. Show that as the time changes, the amount changes, but that the ratio of the amount to the time remains constant.
3. Write the equation for the weight w in pounds, and the cost c of flour at 3^ cents a pound. Find the values of c as li^ takes the integral values from 1 to 10.
What is the ratio of the values of w to the corresponding values of c ?
4. Tabulate the areas of rectangles with the altitude 5 and bases 1, 2, 3 . . 6. How does the area vary as the base varies ?
Write the equation for the area A and the base b of rectangles with altitude 5. Graph the equation.
Does a change in value of b change the value oi A?
A Does it change the value of the ratio y f
254. Direct variation. When two numbers vary so that one depends upon the other for its value, leaving the ratio of any value of one to the corresponding value of the other constant, then one is said to vary directly as the other, or to be directly proportional to the other.
Thus, the number x is said to vary directly as y, if
the ratio - remains constant, x and y both changing, or
y X
varying. The equation ^=c expresses algebraically, and
is equivalent to, the statement that x varies directly as y.
EXERCISES
Express the following statements by means of equations:
1. A man's pay p is directly proportional to the number of days t.
By definition j=k, a constant.
138 FIRST-YEAR MATHEMATICS
2. The weight of a steel rail varies directly as the length.
3. The weight of a mass of silver varies directly as the volume.
4. If a body moves at a uniform rate, the distance varies directly as the time.
5. The area of an equilateral triangle varies directly as the square of the side.
Solve the following problems :
6. The area of a rectangle varies directly as the base if the altitude remains constant; and when the area is 27, the base is 3. What is the constant ratio of the area to the base ?
Since j-=k , —=k or k = 9
Therefore i- = 9
0
7. When the base of the rectangle in problem 6 is 8, what is the area ?
8. The turning-tendency caused by a weight moved along a bar, or lever, varies directly as the lever-arm. The turning- tendency is 20 when the arm is 5. Find the turning-tendency when the arm is 7.
9. The length of a circle varies directly as the diameter. The constant ratio of the circle to the diameter is 3 . 14, approxi- mately. Write the equation for the circle c and the diameter d. When the circle is 157, what is the diameter ?
10. The distance d through which a body falls from rest varies directly as the square of the time i in which it falls; and a body is observed to fall 400 ft. in 5 seconds. What is the constant ratio of d\,o i^l
Write the equation for d and i.
How far does a body fall in 1 second ? In 2 seconds ? In 3 seconds ?
11. a: varies directly as y, and when a: = 20, ?/ = 4. Find the value of X when y=l7.
12. If z varies as x, and 2 = 48 when x = 4, find z when .r= 11.
VARIATION 139
13. Cut from a cardboard circles with diameters of various lengths. By rolling the circles along a straight line determine the lengths of the circles. Determine the ratio of each circle to the diameter. Compare yoifr result with that obtained by other members of the class.
Does the length of the circle vary as the diameter? Give reasons.
14. 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 160 ft. a second 5 seconds after it begins to fall. What will be the speed attained in 8 seconds ?
15. The distance passed over by a body moving at a constant rate varies directly as the time. Find the rate of a train which travels, at uniform rate, the distance of 225 mi. in 6 hours.
16. A stone fell from a building 560 ft. high. In how many seconds did it reach the ground? (See problem 10.)
17. The time t (number of seconds) of oscillation of a pendu- lum varies directly as the square root of the length /. A pen- dulum 39.2 in. long makes one oscillation in 1 second. Find the length of a pendulum which makes an oscillation in 2 seconds.
18. Show that the area A of a square varies directly as the square of the side s.
19. The area of the equilateral triangle varies directly as the square of the side. The area of an equilateral triangle whose side is 2 equals ^S. What is the area of an equilateral triangle whose side is 4 ?
20. The altitude of an equilateral triangle varies directly as the side. The altitude of an equilateral triangle whose side is 2/3 is equal to 3. Find the altitude of an equilateral triangle
4 whose side is 77^ •
21. The simple interest on an investment varies directly as the time. If the interest for 6 years on a sum of money is $200, what will be the interest for 8 years and 3 months ?
140 FIRST-YEAR MATHEMATICS '
22. The surface of a sphere varies as the square of the radius. The surface of a sphere with radius 7 in. is 616 sq. in. What is the radius of a sphere twice as large in surface ?
Inverse Variation
255. Inverse variation defined. In §§ 251-254 certain magnitudes were found to depend upon each other so that an increase in one caused an increase in the other, or a decrease in one caused the other to decrease. How- ever, some magnitudes depend each upon each other so that an increase in one causes the other to decrease; e.g., an increase in the number of men doing a piece of work causes a decrease in the time it takes to do the work. If 8 men do a piece of work in 12 days, 16 men will need only 6 days to do it. The number of men is said to vary inversely as the number of days, or to be inversely pro- portional to the number of days.
Suppose a man wishes to fence in a rectangular piece of ground containing 140 square feet. If he makes it 14 ft. long, the width must be 10 feet. If he makes it 20 ft. long, he can allow only 7 ft. for the width, etc. Thus, the length and width must vary in a way which leaves the area, i.e., their product, constant. This leads to the following definition: A number x varies inversely as y if the product xy remains constant as both x and y vary, i.e., if xy=k, a constant.
EXERCISES
1. Express the following statements by means of equations: (1). The apparent size of an object varies inversely as its distance.
(2). The time needed to go a certain distance varies in- versely as the rate of travel.
VARIATION
141
(3). The force of gravity varies inversely as the square of the distance.
(4). The heat of a stove varies inversely as the square of the distance from it.
2. In 7/ = - or 2j
- show that X varies inversely as y.
3. The variable y varies inversely as x. When a: = l, y = 2. Write the equation for x and y and find the value of the con- stant product. Find the value of y when x = S.
4. The number of men doing a piece of work varies inversely as the time. If twelve men can do a piece of work in 28 days, in how many days can 3 men do the same ?
5. When gas in a cylinder is put under pressure, the volume is reduced as the pressure is increased. It is found in physics that the volume varies inversely as the pressure. The volume of a gas is 4 cubic cm. when the pressure is 3 pounds. What is the volume under a pressure of 6 pounds ?
256. Graphing direct variation. The equation c = 3 . 14c? states that the length of the circle varies as the diameter (see problem 13, § 254). Complete the following table for corresponding values of c and d.
d. .. |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
c . . . |
0 |
3.1 |
6.3 |
9.4 |
Represent the equation c = 3.14d graphically as in Fig. 182.
Graph the laws of variation in problems 8, 10, 11, and 14, §254.
257. Graphing inverse variation. A distance of 42 mi. is passed over by a
25 20 |
||||||||
y |
||||||||
y |
y |
|||||||
10 |
X |
y |
||||||
-/ |
/ |
y |
||||||
/ |
X |
s h r. Fig. 182
142
FIRST-YEAR MATHEMATICS
train at various rates. Find the time it takes the train to travel this distance if it moves uniformly at the rate of 45 mi. per hour; 40 mi. per hour; 35 mi. per hour, etc.
42 Letting t denote the time and r the rate, then t = —
r
expresses the conditions of the problem. Thus, the time
varies inversely as the rate. Why ?
42
The facts expressed by the equation t=— as t and r
take different values |
may |
be tabulated as |
follows : |
||||||
r |
45 |
40 |
35 |
30 |
25 |
20 |
15 |
10 |
5 |
t |
0.9 |
1.0 |
1.2 |
1.4 |
1.7 |
2.1 |
2.8 |
4.2 |
8.4 |
y 8 7 6 5 h 3 1 |
|||||||||
1 |
\ |
||||||||
E-, |
\ |
||||||||
\ |
|||||||||
\ |
|||||||||
\ |
|||||||||
'^ |
= |
||||||||
Ra |
^L |
ine. |
0 5 10 15 20 25 30 35 UO 1,5
Fig. 183
The same facts are given graph- ically in Fig. 183. State from the graph how the time varies as the rate varies from a very small to a very large value.
Graph the following equations: xy = S', a;?/ = 24; a;?/ = 30.
Proportion
258. Proportion. Problems in similar triangles have been solved by means of equations of equal ratios, as X 5
3 27*
An equation of two equal ratios is a proportion.
mu 4 2 1 3 a c „ ,. .
Thus, ^ = o> ^^Wh^'tl ^^^ proportions. A propor- tion as T = ;7 may be read *'a over h equals c over d," or
''a is to 6 as c is to d." The proportion is sometimes
written in the form a'.h—cd.
2 5 . . , 2
Is the statement ^=To a proportion tion ? Give reasons for your answers.
Is y = 25 a propor-
PROPORTION 143
259. Means and extremes. The first and last terms in a proportion are called the extremes; the second and
a c third, the means. Thus, mr = -^,a and d are the extremes,
h and c the means.
EXERCISES
1. Compare the product of the extremes with the product
■ .u .- 2 6 3 6 20 10 12 4
oi the means m the proportions: K^TEJy — T75"o~=~r'~Q'~T"
What do you find true of the products ?
2. Make up several proportions and compare the product of the means with the product of the extremes.
Exercises 1 and 2 illustrate the following theorem: Theorem: In a proportion the product of the means is
equal to the product of the extremes.
This theorem is a convenient test of proportionality,
as it is usually simpler to find the products than to reduce
the ratios to lowest terms.
Proof: The equation r =-^ may be cleared of fractions by multi- plying both ratios by hd and then reducing them to lowest terms. It follows that ad = be.
EXERCISES AND PROBLEMS
Solve the following problems:
1. Divide $2,400 into two parts having the ratio 2:1.
TT- 1 ■ .u .■ 2,400-a: 2
1^ md X m the equation = - .
X 1
2. If 5 and 3 are added to a certain number and 3 and 4 subtracted from it the resulting four numbers are in proportion. Find the number.
3. Two numbers are to each other as 3 to 7 and their sum is 50. Find the numbers.
144 FIRST-YEAR MATHEMATICS
4. If 3 is subtracted from a number, 6 added to 2 times the number, 12 subtracted from 2 times the number, and 6 subtracted from 4 times the number, the resulting numbers are in propor- tion. What is the number?
5. The dimensions of a rectangle are 7 and 4 and the length of a similar rectangle is 8. What is the width ?
6. Find the value of x in the following proportions. Check the results by substituting in the original equation.
9. 10. 11.
1. |
X 3 160" 5 |
|
2. |
4 132 x~ 3 |
|
3. |
18 X 810 3 |
|
4. |
200 -a: X |
7 ^18 |
5. 5±E=2+x j2.
3—x 4— X l-3a -6a
2a 4a- 1
a:- 12 2 l-3a:~l
13.
14.
x-\-l |
3a:+2 |
10 " |
■ 28 |
X-4: |
x-8 |
a;-13 |
x-U |
2-x 3-x~ |
S-x 1-x |
x+5 |
x-S |
x-^S- |
'x-4: |
4x-3 2a;+3 |
4X-V2 ~2x+6 |
x-o |
x-\-4 |
4 " |
' 16 |
x+2 |
2x-3 |
x-2 |
'2x4-1 |
7. The shadows of a tree and a 5-ft. rod are 96 ft. and 8 ft. respectively. Find the height of the tree.
8. If 2" on a map represent 21 mi., what distance on the map represents 50 miles ?
9. Solve the following equations:
v^j-\^ 1+3^2-3
\~y-\ . X X
y-l 1-4 1+4
PROPORTION 145
Alloy Problems
Solve the following problems :
1. In an alloy of silver and copper weighing 90 oz., there are 6 oz. of copper. Find how much silver must be added so that 10 oz. of the new alloy shall contain f oz. of copper.
Let X be the number of ounces of silver added.
Then 90+^ is the number of oz. in the new alloy.
Since (90+.t) oz. of the new alloy contain 6 oz. of copper, 90 +a:
6
oz. contam 1 oz. of copper.
Similarly, since 10 oz. of the new alloy contain i oz. of copper, — oz. contain 1 oz. of copper.
„, . 90 +a: 10
Therefore —x — = —
b 2
i(90+x)=6 10
36+^ = 60 5
x=m
2. If 80 lb. of sea- water contain 4 lb. of salt, how much fresh water must be added to make a new solution of which 45 lb. contain | lb. of salt ?
3. In a mass of alloy for watch-cases, which contains 60 oz., there are 20 oz. of gold. How much copper must be added so that in a case weighing 2 oz. there will be 2 oz. of gold ?
4. In an alloy weighing 80 grams, there are 34 grams of gold. How much nickel must be added so that a ring made from the new alloy and weighing l| grams shall contain 4 gram of gold ?
5. In an alloy weighing a oz. there are h oz. of gold. How much of another metal must be added so that a portion weighing c oz. shall contain d oz. of gold?
6. Gun metal is composed of tin and copper. 4,100 lb. of gun metal of a certain grade contain 3,444 lb. of copper. How much tin must be added so that 2,100 lb. shall contain 1,722 1b. of copper?
146 FIRST-YEAR MATHEMATICS
Lever Problems
Two weights, Wi and w-2, Fig. 184, will balance on a beam that
lies across a stick when the distances,
riFT] [nq c/i and c?2, of weights from the stick
J ""^ Zi are in the inverse ratio of their
a, 0,2 ,
Fig. 184 weights; i.e., when -r = — .
1. Find di, if (1) ^2=18 ft., u'2 = 60 lb., w^i = 501b.
(2) ^2 = 27 in., u'2 = 36 lb., w)i = 241b.
2. Find 6^2, if (1) (/i = 40in., w;2=16 lb., w;i=18 1b.
(2) di = 25m., 'M;2 = 3.8 1b., 'w;i = 2.85 1b.
3. Find wh, if (1) di = 2.5 in., ^2 = 7.5 ft., w;2=10.51b.
(2) rfi = 6.6ft., (i2 = 9.9ft., u'2=171b.
4. Find ^('2, if (1) rfi = 3.5 ft., ^2 = 8.5 ft., u'i = 301b.
Mixture Problems
1. What per cent of evaporation must take place from a 6 per cent solution of salt and water (salt-water of which 6 per cent by weight is salt) to make the remaining portion of the mixture an 8 per cent solution?
Let X be the number of per cent evaporated. Then 100 — x is the number of per cent remaining. Since 6 per cent of the first solution is salt, \ o o ' 1^0 is the amount of salt.
Since 8 per cent of the new solution is salt, i oo(100— ^) is also the amount of salt.
Therefore ^ g o . 100 = lU (100 -x) 600 = 800 -8x a: = 25
2. What per cent of evaporation must take place from a 90 per cent solution to produce a 95 per cent solution ?
3. A physician having a 6 per cent solution of a certain kind of medicine wishes to dilute it to a 3^ per cent solution. What per cent of water must be added ?
4. A druggist has a 95 per cent solution. What must he do to change it to an 80 per cent solution required in a prescription ?
PROPORTION 147
Proportionality of Areas Prove the following theorems :
260. Theorem : The areas of two rectangles are to each other as the products of their dimensions.
Proof: Denoting the areas of the rectangles by R and Ri, and their dimensions by b, h, and 61, hi, we have
R =bh. Why? Ri=bihi. Why?
Therefore ^=^- Why?
Hi Oirli
261. Theorem: // two rectangles have equal bases, they are to each other as the altitudes.
For, -j^ = 7-7- . Reducing the ratio t-t to lowest terms, Ri bihi ^ bihi '
R _ 0h _h Ri~^i~hi
262. Theorem: // two rectangles have equal altitudes they are to each other as the bases. Prove.
EXERCISES
1, The equation 2 = 5 expresses the relation between the ratio of the areas and the ratio of the altitudes of two rectangles. Find the altitude x.
2. The area of a rectangle is 80 sq. ft., and the base is 10 yards. What is the area of a rectangle having the same altitude and a base equal to 24 yards ?
Prove the following theorems:
263. Theorem: The areas of parallelograms are to each other as the products of the bases and altitudes.
148 FIRST-YEAR MATHEMATICS
264. Theorem: The areas of two triangles are to each other as the products of the bases and altitudes.
265. Theorem : The areas of two parallelograms having equal bases are to each other as the altitudes.
266. Theorem : The areas of two triangles having equal bases are to each other as the altitudes.
Summary
267. The following new terms have been taught in this chapter: prime numbers, reduction of fractions to lowest terms; greatest common divisor; direct and inverse variation, variables, constant, function; proportion; means and extremes. The symbol i expresses square root.
268. The truth of the following theorems* has been shown:
1. A line parallel to one side of a triangle divides the other two sides into corresponding parts having equal ratios.
2. A line bisecting an angle of a triangle divides the side opposite that angle into parts whose ratio is equal to the ratio of the other two sides.
3. // two sides of a triangle are divided into parts having the same ratio, the line joining the points of division is parallel to the third side of the triangle.
4i. In a proportion the product of the means equals the product of the extremes.
5. Two rectangles, two parallelograms, or two triangles are to each other as the products of the bases and altitudes.
6. Two rectangles, two parallelograms, or two triangles having equal bases are to each other as the altitudes.
* Most of these theorems were first proved by the Pythagorean School from 500 B.C. to 400 B.C.
PROPORTION 149
7. Two rectangles, two parallelograms, or two triangles having equal altitudes are to each other as the bases.
269. A common unit of two line-segments may be found by use of the compass. The process of finding the great- est common divisor of two numbers is similar to the process of finding the greatest common unit of two line-segments.
270. Ratios and fractions may be reduced to lowest terms by dividing numerator and denominator by all factors common to them.
271. The statements x varies directly as y and x varies inversely as y are equivalent to the statements x=ky and xy=k respectively. Both equations have been represented graphically.
272. Many problems, as alloy problems, mixture problems, and lever problems, lead to proportions.
CHAPTER X
CONGRUENCE OF TRIANGLES
Congruence
273. Congruent figures.* In chapter VIII problems in finding distances were solved by indirect measurement. A drawing was made of the same shape as the figure con- taining the distance to be determined, and the corre- sponding distance in the drawing was then measured, or found algebraically by means of an equation. It will be shown in this chapter how to determine a distance from a figure having the same size and shape as the figure con- taining the unknown distance. Figures having the same size and the same shape are called congruent figures.
One of the problems of this chapter is to find out under what conditions two triangles are necessarily congruent.
EXERCISES
1. To construct a triangle having given two sides and the angle between them.
Construction: Suppose the given sides to be 3 in. and 4 in. long respectively and the given angle to be 60°.
On a line AB, Fig. 185, lay off AC equal to the 4-in. segment.
On AC at il construct Z DAC equal to the given angle.
On AD lay off AE equal to the 3-in. segment.
Draw EC. Then AAEC is the required triangle.
*The principle of congruence was employed somewhat by the School of Thales and very extensively by the School of Pythagoras. Pythagoras systematized it and gave it the dignity of a mathemati- cal principle.
150
CONGRUENCE OF TRIANGLES
151
2. Cut out the triangle in exercise 1 and see if it can be made to coincide with the triangles constructed by other mem- bers of the class. ,
4 in.
A 4 m
Fig. 185
3. Draw a triangle. Construct a second triangle having two sides and the included angle equal respectively to two sides and the included angle of the first triangle. See if the two triangles can be made to coincide.
Exercises 1, 2, and 3 illustrate the following theorem: 274. Theorem: Two triangles are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other.
The truth of this statement may be established by reasoning as follows:
Let ABC and AiBiCi, Fig. 186, represent two tri- angles having b = bi, c = ci, ZA = /.Ai.
Imagine AAiBiCi placed on A ABC.
Then ZAi can be made to fit on Z A, since Z^ is given equal to ZAi.
Side Ci can be made to coincide with side c, thus making Bi fall on B. Why ?
Side bi will coincide with b, making Ci fall on C. Why ?
Then ai and a must coincide. For only one straight line can be drawn between two points.
Therefore triangles AiBiCi and ABC coincide throughout and are congruent.
P'l
B,
Fig. 186
152
FIRST-YEAR MATHEMATICS
In Fig. 187 the inaccessible distance AB across a lake is to be determined.
A point C is selected from which A and B can be seen.
Line AC is drawn and extended to Ai making AC = A,C.
Similarly BiC is made equal to BC. Then BiAi is drawn.
n^ i,j^
' ' Show that triangles AyBiC and ABC are con-
rlG. lo7 .
gruent.
Hence, AB is determined by measuring AiBi. Why ?
EXERCISES
1. To construct a triangle having given two angles and the side included between their vertices.
Construction: Suppose the angles to be 35° and 50°, and the side to be 4 in. long.
Draw a line, as AB, Fig. 188.
On AB lay off AC equal to the given 4-in. segment.
On AC at A construct angle DAC equal to the 50° angle.
On AC at C construct an angle equal to the 35° angle.
A ACE is the required triangle. Why ?
D
A I U in.
Fig. 188
2. Cut out the triangle constructed in exercise 1 and see if it can be made to coincide with the triangles constructed by other members of the class.
3. Draw a triangle. Construct another triangle having two angles and the side between their vertices equal respectively to the corresponding parts of the first triangle. See if the triangles can be made to coincide.
CONGRUENCE OF TRIANGLES
153
Exercises 1, 2, and 3 illustrate the following theorem:
275. Theorem: Two triangles are congruent if two angles and the side included between their vertices in one triangle are equal respectively to the corresponding parts in the other.
For, let any two triangles as ABC and AiBiCi, Fig. 189, have c = ci, ZA = ZAi, ZB=ZBi.
Imagine AAiBiCi placed upon AABC.
Then AiBi can be made to coincide with AB. Why?
Angle Ai must then coincide with angled. Why? ^^«- ^^^
Angle Bi must coincide with angle B. Why?
Point Ci must therefore lie on lines AC and BC.
Thus, Ci must fall on C, the only point common to ^C and BC.
Therefore triangles ABC and AiBiCi coincide and are congruent.
276. Methods of proof: The theorems of §§274 and 275* were proved by placing one figure over the other and then showing that one fits exactly upon the other. This method of proof is known as the method by super- position. It is one of several kinds of proof used in
geometry.
In Fig. 190 the distances of two points A and B from a lighthouse S are to be determined.
On AB, at A, angle BASi is constructed equal to angle BAS, and angle ABSi equal to angle ABS. ASi is measured and found to be 2,680 ft. BSi is found to be 3,420 ft. What are the lengths of AS and BS f Give reasons for your answer.
* The theorem of § 275 was proved for the first time by Thales, who used it to find the distance of a ship at sea from the shore. (See Ball, p. 15.)
154
FIRST-YEAR MATHEMATICS
277. Notation for corresponding parts. Correspond- ing sides and angles of congruent figures may be denoted in various ways. The subscripts in Fig. 191 indicate that point Ci corresponds to point C, Ai to A, and Bi to B. Similarly side ai corresponds to side a, bi to b, and Ci to c.
One arc in an ^ ^c- angle means that
the angle corre- sponds to the angle in the con- FiG. 191 gruent figure
marked with one arc. Angles marked with two arcs, three arcs, etc., are corresponding angles.
Frequently the symbol (0 is used to indicate cor- respondence. Thus A' (A-prime) corresponds to A, Fig. 192, B' to B, and C to C.
Correspondence of sides is also indicated by one, two, or more strokes across the side, as in Fig. 192.
B A
Fig. 192
278. Symbol for congruence. The symbol for con- gruence is ^ , the ^ indicating similarity and = equality of the figures. Thus, the statement AABC and A'B'C are congruent is written AABC^AA'B'C, meaning of the same size and shape.
The Isosceles and the Equilateral Triangle
279. Conditions under which triangles are congruent.
The theorems in §§ 274 and 275 suggest a method of
CONGRUENCE OF TRIANGLES 155
proving triangles congruent, which is more advantageous than the method of superposition. According to these theorems to prove two triangles congruent it is not neces- sary to know that all parts of one triangle are equal to the corresponding parts of the other, but it is sufficient to show that one of the following two conditions is satisfied :
1. That the triangles have two sides and the included angle equal respectively, briefly expressed by ''side, angle, side" or (s.a.s.).
2. That the triangles have two angles and the side included between their vertices equal respectively (a.s.a.)-
The application of this method of proof will be illus- trated by proving some of the properties of the isosceles triangle.
280. Theorem : The base angles of an isosceles triangle are equal*
Let AABC, Fig. 193, be isosceles, i.e., let C
a = b. It is to be proved that ZA= ZB.
Proof: Draw the helping Hne CD bisecting ZC, i.e., making x = y.
Prove that AADC^ABDC. (s.a.s.)
It follows that ZA = ZB, because correspond- ing parts of congruent triangles are equal.
281. Congruent-triangle method. The method of proving two line-segments, or two angles, equal by show- ing that they are corresponding parts of congruent tri- angles is called the congruent-triangle method.
* This theorem was first proved by Thales. Ball, p. 1 5, indicates the method of proof supposed to have been used by him.
/ |
\ |
1 Fig. |
193 |
156 FIRST-YEAR MATHEMATICS
EXERCISES
Prove the following:
1. Theorem: An equilateral triangle is equiangular. Apply the theorem in § 280.
2. The bisector of the vertex angle of an isosceles triangle bisects the base and is perpendicular to it.
Use the congruent-triangle method.
3. Theorem: All points on the perpendicular bisector of a line-segment are equidistant from the end-points of the segment.
4. If a line bisects an angle of a triangle and is perpendicular to the opposite side, the triangle is isosceles.
5. Ttieorem: // two angles of a triangle are equal, the triangle is isosceles.
Let A and B be the equal angles, Fig. 194. c
Draw the helping line CD ± AB, i.e., mak- ing a; =?/. \ Since two angles of A A DC are equal to two
\ angles of ABDC, it follows that m = n. (See
Aj> p. 53, problem 16.) FiG^ 194 ^^^^^ AADC^ ABDC. (a.s.a.)
Then AC = BC. Why?
6. An equiangular triangle is equilateral. Apply the theorem in exercise 5.
7. If the perpendicular bisector of one side of a triangle passes through the opposite vertex, the triangle is isosceles.
8. Theorem: // two sides of a triangle are unequal the angles opposite them are unequal, the greater angle lying opposite the greater side.
Let CB, Fig. 195, be greater than CA. C
Lay off on CB the segment CD = CA. X n.
Draw AD and through B draw BE WAD. / \
a^rh = c. Why? A^^;;^^— '^-^D
Hence a+6=d. Why? ^^^ ^g^
But d<d-\-e. Why?
Therefore a+fe<d+e. Why?
Since a = e, it follows that hKd. For, equals subtracted
from unequals give unequals in the same order as the minuends.
CONGRUENCE OF TRIANGLES 157
9. Theorem: // two angles of a triangle are unequal the sides opposite them are unequal, the greater side lying opposite the greater angle.
Through A draw AD making x = y, C
Fig. 196.
Then AD = DB. Why?
Since AD+DOCA, it follows that
DB+DOCA. Why? ^^-^ ^^-^b
Thus, BOCA. Fig. 196
282. Construction of a triangle all of whose sides are given.
EXERCISES
1. Given the three sides of a triangle, to construct the triangle. Construction: Suppose the sides to be 2 in., 3 in., and 4 in. long respectively.
D,
2in.
Sin.
4 tn.
Draw a line, as AB, Fig. 197.
On AB lay off AC equal to the 4-in. segment.
With A as center and with the 3-in. segment as radius draw an arc at D.
With B as center and with the 2-in. segment as radius draw an arc intersecting the first arc at D.
Join the intersection point D to A and C.
A ADC is the required triangle.
2. Cut out the triangle constructed in exercise 1 and see if it can be made to coincide with the triangles constructed by other members of the class.
Exercises 1 and 2 illustrate the fact that two triangles are congruent if their sides are equal respectively.
3. By taking various lengths of the given segments, e.g., by making the sum of two segments equal to or less than the third,
158
FIRST-YEAR MATHEMATICS
show that the construction in exercise 1 is possible only when the sum of any two of the given segments is greater than the third.
4. Construct an equilateral triangle on a side 2 in. long.
5. Draw a triangle having sides of 2 in., f in. and 1 in.
6. With crayon and string or with a blackboard compass draw on the blackboard a triangle having sides 6 in., 8 in., and 10 in. long.
7. Describe how with a 100-ft. steel tape stakes may be set in the ground to be the corners of a triangle of sides 30', 50', and 60' long.
8. Given the base and one of the equal sides of an isosceles triangle, to construct the triangle.
283. Theorem: // the three sides of one triangle are equal, respectively, to the three sides of another triangle, the triangles are congruent, (s.s.s.)
For, let ABC and AiBiCi, Fig. 198, represent two triangles having AB = A,B,, BC = B,C,, and CA = CiAt.
c
Fig. 198
Imagine AAiBiCi to be placed adjacent to A ABC so that AiBi coincides with AB and that Ci and C he on opposite sides of AB. Draw CCi
Then ACi=AC. Why? Kence x = y. Why?
Since BCi =BC, it follows that xi =yi. Why ?
Therefore x+Xi=y-\-yi, or ZC= ZCi. Why?
Prove AABC^AABCi. (s.a.s.) Hence AABC^AA.BiC Why?
NICCOLO FONTANA
N I C C Q L Q F Q N T A N A
NICCOLO FONTANA, nicknamed Tartaglia (Stammerer), was born at Brescia in Italy about 1500, and died at Venice, December 14, 1557. In 1512 his native city was captured by the French army, and during the sack of the city his father was killed and he himself so severely injured as to produce the stoppage in his speech from which he was nicknamed and from which he never recovered. His mother was too poor to pay for more than fifteen days of schooling for him, and she could not afford even to buy paper for him to use in private study. In spite of all these obstacles he rose by dint of his own efforts to be the ablest and best known algebraist of his time.
Mathematical contests were the vogue in Tartaglia's day. When a mathematician made a discovery he challenged some well-known mathematician to a contest. Each contestant proposed the same number of problems for the other to solve in a specified time. The one who solved the largest number in the time set was declared the victor. It was thus he won his title as a mathematician. Tartaglia engaged in two such contests, winning both times. See an account of these contests in Ball's History of Mathematics, pp. 218 and 223 (5th ed.).
The scientific service for which Tartagalia is best known today is his discovery of the solution of the cubic equation, about 1530. He began public by lecturing on mathematics in Verona, and in 1535 he was ap- pointed to a chair of mathematics in Venice. He wrote three important works on mathematics, besides publishing an edition of Euclid in 1543 and an edition of Archimedes the same year. His Inventioni, pub- lished in Venice in 1546, contains his solution of the cubic equation, which had been published by Cardan as his own the year before.
CONGRUENCE OF TRIANGLES
159
EXERCISES
Prove the following :
1. Theorem: If a "point is equidistant from the end-points of a line-segment, it is on the perpendicular bisector of the segment.
Let C, Fig. 199, be equidistant from A and B, i.e., let AC =BC.
Draw the helping line CD from C to the mid- point D of AB.
Frove A ADC ^ABDC. (s.s.s.)
Thenx = y amd CD ^AB. A'
Therefore CD is the perpendicular bisector oi AB.
2. Theorem: // each of two points of one line is equally distant from two points of another line the lines are perpendicular.
For, by exercise 1 each of the given points A and B, Fig. 200, lies on the perpendicular bisector of the second line CD.
Since only one straight line can be drawn through two points, the line AB and the per- pendicular bisector of CD must be the same line.
D
Fig. 199
C4-
^D
Fig. 200
284. Locus of points. In exercise 1, § 283, it was shown that every point equidistant from the end-points of a segment lies on the perpendicular bisector of the segment. In exercise 3, § 281, it was shown that every point on the perpendicular bisector of a line-segment is equidistant from the end-points of the segment.
From these two theorems it follows that the perpen- dicular bisector of a segment is the place, or locus, of all points equidistant from its end-points.
EXERCISES
1. What is the locus of all points in a plane which are at a distance of 10 ft. from a given point P in the plane ? At a distance a from the given point ?
160
FIRST-YEAR MATHEMATICS
2. What is the locus of all points in space known to have a given distance from a given point ?
3. What is the locus of all points in a plane at a given distance from a given line ?
4. What is the locus of all points in a plane at equal distances from two parallel lines ?
5. What is the locus of all points in space having a given distance from a given line ?
6. What is the locus of all points in space having a distance of 2 ft. from the floor of the room ?
7. What is the locus of all points in space known to be at equal distances from two given points ?
The Right Triangle
285. Theorem: Two right triangles are congruent if the hypotenuse and a side of one are equal respectively to the hypotenuse and a side of the other.
Let ABC and AiBiCi, Fig. 201, be two right triangles having the hypotenuse (side opposite the right angle) AC equal to AiCi and J5C = J5iCi.
Place AAiBiCi adjacent to AABC, making BiCi coincide with the equal side BC, so that A and Ai lie on opposite sides of BC.
Then ABAi is a straight line and the figure ACAi a triangle. Why?
Show that A AC A I is an isosceles triangle. Why? (a.s.a.) Why?
Fig. 201
Hence ZA = ZAi.
Prove AABC^AAiBC: Then AA^B^Ci^AABC.
EXERCISES
Prove the following:
1. Two right triangles are congruent if the two sides about the right angle of one are respectively equal to the two sides about the right angle of the other.
CONGRUENCE OF TRIANGLES 161
2. Two right triangles are congruent if the hypotenuse and an acute angle of one are respectively equal to the hypotenuse and an acute angle of the other.
3. Theorem: The shortest distance from a point to a line is the perpendicular from the point to the line.
For, let AB, Fig. 202, be perpendicular to DE and let AC be any other line from A to DE.
In A ABC Z C is acute. , Why ? Therefore Z C is less than ZjB. Why? Hence AC>AB. Why? /
4. Oblique lines drawn to a line from a point d c b e on a perpendicular to the line and making equal Fig. 202 angles with the perpendicular are equal.
5. Oblique lines drawn from a point on a perpendicular to a line and making equal angles with the line are equal. -
Summary
286. The following new terms have been taught in this chapter: congruent figures, corresponding parts of congruent figures; locus of points; methods of proof.
287. The symbol ^ indicates congruence, i.e., can be made to coincide.
I. Problems in finding unknown distances and angles can be solved by construction of congruent triangles.
289. The following constructions of triangles were taught :
1. When two sides and the included angle are given.
2. When two angles and the side between the vertices are given.
3. When the three sides are given.
290. Two methods of proof have been taught :
1. The superposition method.
2. The congruent-triangle method.
162 FIRST-YEAR MATHEMATICS
291. The following theorems state the conditions that determine the congruence of triangles :
1. Two triangles are congruent if they have two sides and the included angle equal respectively.
2. Two triangles are congruent if they have two angles and the side between their vertices equal respectively.
3. Two triangles are congruent if the corresponding sides are equal.
4. Two right triangles are congruent if the hypotenuse and a side of one are respectively equal to the hypotenuse and a side of the other.
292. The following theorems on the equilateral triangle were proved:
1 . An equilateral triangle is equiangular.
2. An equiangular triangle is equilateral.
293. The following theorems on the isosceles triangle were proved:
1. The base angles of an isosceles triangle are equal.
2. If two angles of a triangle are equal the triangle is isosceles.
3. The bisector of the vertex angle of an isosceles triangle bisects the base and is perpendicular to it.
4. 7/ a line bisects an angle of a triangle and is per- pendicular to the opposite side the triangle is isosceles.
5. If the perpendicular bisector of one side of a triangle passes through the opposite vertex the triangle is isosceles.
294. The following theorems on inequalities were proved :
1. If two sides of a triangle are unequal the angles opposite them are unequal, the greater angle lying opposite to the greater side.
CONGRUENCE OF TRIANGLES 163
2. // two angles of a triangle are unequal the sides opposite them are unequal, the greater side lying opposite the greater angle.
295. The following theorems on perpendicular lines were proved:
1. All points on the perpendicular bisector of a segment are equally distant from the end-points of the segment.
2. If a point is equidistant from the end-points of a line- segment it is on the perpendicular bisector of the segment.
3. The perpendicular bisector of a line-segment is the locus of all points equidistant from its end-points.
4. If two points of a line are each equidistant from two points of another line the lines are perpendicular.
5. The shortest distance from a point to a line is the perpendicular from the point to the line.
6. Oblique lines drawn to a line from a point on a per- pendicular to the line and making equal angles with the perpendicular are equal.
7. Oblique lines drawn from a point on a perpendicular to a line and making equal angles with the line are equal.
CHAPTER XI
CONSTRUCTIONS. SYMMETRY. CIRCLE
The Fundamental Constructions
1. To bisect an angle.
296. In the preceding chapters several construction problems were taught without proof of their correctness. These constructions will now be summarized and proofs will be given.
EXERCISES
(See § 127, exercise 12.)
Make the construction indicated
in Fig. 203.
To prove this construction, draw
helping Hnes AD and BD.
Prove ABDC^ AADC. (s.s.s.) Then Z DCB = Z DC A . Why ?
2. At a given point on a given line to construct a perpendicular to the line. (See § 127, exercise 3; § 128, exercise 3.)
Make the construction indicated in Fig. 204.
According to this construction F and C are each equidistant from D and E.
Hence, FC is the perpendicular ^ - bisector of DE. (Why?) It follows that FC A. AB ate.
><F
C E
Fig. 204
3. Show that. on^y one perpendicular can be drawn to a line at a given point on the line. (See § 176.)
4. To bisect a given line-segment.
With A as center and a convenient radius draw arcs as at C and D, Fig. 205.
164
CONSTRUCTIONS
165
With B as center and the same radius draw arcs intersecting the first two arcs.
Draw CD intersecting ^5 at ^. Than AE=^EB. >r^
VoT, according to the construction C and D are each equidistant from A and B, and hence CD is the perpendicular bisector of AB. Why ? J
5. To construct the perpendicular bisector of a • E line-segment.
Construction and proof are the same as in exercise 4. "^^kp
6. From a point outmde of a line to draw Fig. 205
a perpendicular to the line.
The directions for this construction c \ are the same as in exercise 2.
I Proof: Since C and F, Fig. 206, are
equidistant from D and E, by construc- tion, CF is the perpendicular bisector of DE. Why?
Therefore CG±AB.
Fig. 20G
7. Only one perpendicular can draw 71 from a
be
point to a line.
For, if PD and CE, Fig. 207, were both perpendicular to AB, two angles of AEDC would be right angles.
This is impossible, since the sum of the three angles of triangle EDC would then be greater than 180.
8. At a given point on a given line to draw a line making ari angle with the given line equal to a given angle. * (See § 125.)
E D
Fig. 207
Fig. 208
Draw the helping lines CA and GF, Fig. 208. Prove ACBA^AGDF. (s.s.s.) Hence, ZABC = ZFDG.
* This construction was first made by Oenopides of Chios (500- 4.30 B.C.). See Ball, p. 30.
166 FIRST-YEAR MATHEMATICS
297. The fundamental constructions in § 296 are now to be used in more difficult constructions. Let it be understood that the unmarked ruler and compass* are the only instruments to be used in these problems.
Applications of the Fundamental Constructions
EXERCISES
1 . Through a given point outside of a given line to draw a line parallel to the given line. (§ 194, exercise 3.)
2. Construct a triangle having given two sides and the included angle. ( § 273, exercise 1.)
3. Construct a triangle having given two angles and the side between their vertices. (§ 274, exercise 1.)
4. Construct a triangle having given the three sides. (§ 282, exercise 1.)
5. Construct an angle equal to 60°.
Proceed as in the construction of an equilateral triangle; show that one of the angles of the triangle is 60°.
6. Construct the following angles: 30°, 15°, 120°.
7. Construct angles of 90°, 45°, 22°30'.
8. Construct angles of 135°, 75°165°.
9. Construct a triangle having given two sides and the angle opposite one of them.
Let m and n, Fig. 209, be the given sides and let Z.K ha the given angle.
* The Greek philosopher Plato (429-384 b.c.) taught geometry as a basis for the study of philosophy. As a convenient means of defining the field of elementary geometry he decided that the un- marked straightedge and the compass should be the only instru- ments to be used in a construction. Soo Ball, p. 43, or Cnjori, p. 31.
CONSTRUCTIONS
167
Construction: Draw a line of convenient length, as AB. Lay- off on AB a segment equal to m.
At A on AB construct an angle equal to Z K.
With radius equal to n and center at C draw an arc meeting AD atE.
Draw CE.
A ACE is the required triangle.
Discussion: In this construction problem a solution is not always possible.
For, if n is sufficiently small, the arc and line AD do not inter- sect. Therefore no triangle exists containing the given parts. This will always be the case when n is less than the perpendicular CF, Fig. 210, from C to AD.
E^D
When n = CF, Fig. 211, the arc will just touch AD at F and AACF is the required triangle.
When n>CF and n<m, Fig. 212, show that the arc intersects AD in two points, as E and Ei and that the two triangles ACE and AC El satisfy the conditions of the problem.
When n = m, Fig. 213, and when n>m, E/
Fig. 214, show that there will be only one solu- tion of the problem, i.e., A ACE.
298. The constructions in exercise 9 in which only one triangle is possible, i.e., those of Figs. 211, 213, and 214, illus- trate the following theorems of congruence :
1. Two right triangles are con- gruent if they have the hypotenuse and one of the other sides equal respectively. (Fig. 211.)
Fig. 214
168 FIRST-YEAR MATHEMATICS
2. Two isosceles triangles are congruent if the equal sides and the base angle of one are respectively equal to the corresponding parts of the other. (Fig. 213.)
3. Two triangles are congruent if two sides and the angle opposite the greater side are respectively equal. (Fig. 214.)
EXERCISES
1. To divide a right angle into three equal parts.*
With A as center, Fig. 215, and a con- venient radius draw an indefinite arc intersect- ing the sides of the right angle at B and C.
With B as center and the same radius draw an arc at D.
With D as center and radius equal to CD draw an arc at E.
Lines AD and AE trisect the angle BAC. For, ZBAD = m°. Why? Therefore ZD AC = 30°. Why? Show that Z DAE = 30°.
2. To construct a right triangle having given the hypotenuse and one of the other sides.
3. To construct a right triangle having given the hypotenuse and one acute angle.
* The problem of dividing a right angle into three equal parts was solved very early. However, the general problem, i.e., to trisect any angle by use of unmarked straightedge and compass, presented great difficulties. It became one of the famous problems of geome- try and much intellectual energy has been expended on these prob- lems. However, this has not been wasted, as many discoveries were made by mathematicians in the attempt to obtain a solution. It has been proved that the trisection of the general angle with ruler and compass is impossible. (See Cajori's History of Mathematics, p. 24.)
SYMMETRY 169
4. Construct an isosceles triangle having given one of the base angles and the altitude.
5. Construct a right triangle whose acute angles are 60° and 30°.
Draw an equilateral triangle and divide it into two right tri- angles by means of the altitude.
How does the length of the hypotenuse in this right triangle compare with that of the side opposite the 30° angle ?
6. Construct ain equilateral triangle having given the altitude.
Symmetry
299. Symmetry. If one hand is held in front of a plane mirror, the image obtained in the mirror is of the same size and shape as the other hand. The hand and the image are said to be symmetric with respect to the plane of the mirror.
Other illustrations of symmetry are: a pair of gloves, the two doors of an automobile opposite to each other, the two posts at an entrance to a park, a printed page and the type-page from which it was made, etc.
300. Symmetry of a single body or figure. The
human head is symmetric with respect to a plane passing midway between the eyes. The cube is symmetric with respect to several planes; indicate the position of some of these planes of symmetry. How many planes of sym- metry can be passed through a sphere? Give other examples of symmetric solids.
301. Symmetry of plane figures. If a figure is drawn in ink on paper and if the paper is folded before the ink is dry, an image of the figure is obtained. The image and the original figure are symmetric.
170
FIRST-YEAR MATHEMATICS
EXERCISES
1. Construct a figure symmetric to line AB, Fig. 216, with respect to the li;Qe CD.
From A draw AAi±CD, making AE = EAu
Similarly draw BBi±CD, making BF = FBi.
Draw AiBi.
Then AiBi and AB are symmetric with respect to CD.
2. Construct a triangle symmetric to triangle ABC, Fig. 217, with respect to the line DE.
As in exercise 1 draw lines symmetric to AB, BC, and CA.
3. Construct a figure symmetric to ^-BCD^/^G^F, a
Fig. 218. ^
Fig. 217
302. Axis of symmetry. Two
figures are symmetric with respect to a line if the line is the perpendicular bisector of all line-segments joining the corresponding parts of the figures. The line is called the axis of Fig. 218 symmetry.
EXERCISES
1 . Show that two plane figures are congruent if they are symmetric with respect to an axis of symmetry.
Use superposition to show that every point of one figure can be made to coincide with the corresponding point of the other.
2. Draw the axis of symmetry of an isosceles triangle.
3. Show thaf the bisector of the vertex angle of an isosceles triangle is the axis of symmetry. Show how some of the properties of an isosceles triangle stated in § 293 follow from the symmetry of the figure.
SYMMETRY
171
4. How many axes of symmetry has an equilateral triangle ? A square ? A rectangle ?
5. Show how the principle of axial symmetry stated in exercise 1 may be used in designing, pattern-making, printing, building, tailoring, etc.
303. Some geometric facts are easily established from the symmetry of the figures. Show that the following are true:
1 . Any point on the perpendicular bisector of a line-segment is equidistant from the end-points.
Let CD, Fig. 219, be the perpendicular bisector ^c
of AB.
Since CD is the axis of symmetry oi AB, /. AEC ^ can be made to coincide with Z BEC.
Hence AC coincides with BC.
2. Any point not on the perpendicular bisec- ^ . tor of a line-segment is not equally distant from Fig. 219
the end-points.
Let ED, Fig. 220, be the perpendicular bisector of AB.
Rotate AGF about ED as an axis. Then AFC takes the form of the broken line BFC, and AC = AF+FC=BF'\-FC>BC. Why?
Fig. 220 3. Any point on the bisector of an angle
is equidistant from the sides.
Let CE be the bisector of ZGCF, Fig. 221.
Let AD±CF and AB-lCG.
The bisector CE is the axis of symme- try of the angle, since FC can be made to coincide with GC by revolving ZGCA about CA as'^an axis.
Then AB must coincide with AD, since only one perpendicular can be drawn from a point to a line.
172
FIRST-YEAR MATHEMATICS
4. Any point not on the bisector of an angle is not equi- distant from the sides of the angle.
Let CE be the bisector of A BCD, Fig. 222.
Let AD±CD and AB±CB.
Rotating Z FCB about CE as an axis, hne AFB takes the form of the broken hne AFG.
But AF-\-FG>AG. Why?
and AG>AD. Why?
Therefore AF-\-FG> AD. Why?
304. The last two problems combined express the following theorem:
Theorem : The bisector of an angle is the locus of points within the angle which are equidis- tant from the sides.
The rectangle CDEF, Fig. 223, represents the top of a biUiard table. Find the point P on the cushion FE to which a ball at A, cued at the cen- ter, must be directed to strike a ball atB.
Locate a point Bi, symmetric to B with respect to FE. point of intersection P of ABi with FE is the required point.
For, it is learned from physics that if a perpendicular is drawn to FE at P and if the angle of incidence, a, equals the angle of reflection, b, a ball directed from A striking at P will rebound and strike B.
To prove that a = h, show that a = ai, b = hi, and ai = 6i.
Fig. 223
The
The Circle
305. The circle and circular arc have been used in the construction of many figures. Before taking up the follow- ing constructions of this chapter it is necessary to become acquainted with some of the properties of the circle.
THE CIRCLE 173
306. Tangent. A line that touches a circle in only one point, however far produced, is a tangent.
Line AB, Fig. 224, is tangent to circle Date.
307. Contact point. The point common to a circle and a tangent
is called the point of tangency, or point of contact.
308. Theorem: The radius drawn to the point of con- tact of a tangent is perpendicular to the tangent.
For, any point E of the tangent, not the contact point. Fig. 225, lies outside of the circle.
Hence, line AE must be longer than a radius, and AE>AB.
Thus, ^jB is the shortest of all lines drawn from the center A to CD. It is therefore perpendicular to CD.
309. Theorem: A line perpendicular to a radius at the outer end-point is tangent to the circle.
For, if AB, Fig. 225, is perpendicular to CD, then it is shorter than any other line, as AE, drawn from A to CD.
Hence, AE is longer than a radius and E must be outside of the circle.
Since E is any point on CD, not B, it follows that B is the only point on CD and on the circle, and that CD is tangent to the circle.
310. Regular polygon. A polygon that is equilateral and equiangular is a regular polygon.
311. Inscribed polygon. A polygon whose vertices lie on a circle is an inscribed polygon. The circle is cir- cumscribed about the polygon.
174 FIRST-YEAR MATHEMATICS
312. Circumscribed polygon. A polygon whose sides are tangent to a circle is a circumscribed polygon. The circle is inscribed in the polygon.
EXERCISES
1. At a given point on a circle to construct the tangent to the circle.
Draw the radius to the given point and erect a perpendicular to it at that point.
2. To find the center of a given circle.
Draw a chord, as AB, Fig. 226. Draw the perpendicular bisector CD of the chord AB.
CD must pass through the center. Why ?
Bisect CD at E.
E is the center of the circle.
3. Draw a circle passing through two given points.
How many circles can be drawn through those points ?
4. Circumscribe a circle about a triangle. Let ABC be the given triangle, Fig. 227. Draw the perpendicular bisectors of two sides,
a.s AB and CD. Both perpendiculars must con- tain the center of the circle. Why ?
Hence, they must intersect at the center of the circumscribed circle.
Let P be the point of intersection. Since P is equally distant from A and B and from C and B, it must be equally distant from A and C
Hence, a circle with P as center and radius PC must pass through A, B, and C.
5. Show that the perpendicular bisector oi AC passes through F, Fig. 227.
THE CIRCLE
175
6. To inscribe a circle in a triangle. Let ABC, Fig. 228, be the given triangle. Draw the bisectors of two angles, as A and B.
With their point of intersection, P as center and a perpendicular from P to AB as radius draw a circle. This is the required circle.
7. Prove that the circle con- structed in exercise 6 is tangent to the sides of triangle ABC.
Show that P is equally distant from BA and BC.
Show that P is equally distant from AB and AC.
Then P is equally distant from CA and CB.
Hence, a circle with P as center and radius PD will be tangent to the sides of^AABC. (§ 309.)
8. Show that the bisector of angle C, Fig. 228, passes through P.
9. To inscribe a square in a circle.
Draw two diameters perpendicular to each other, as AB and CD, Fig. 229.
Join the successive points A, C, B, and D.
The quadrilateral ACBD is the required square.
10. Prove that ACBD constructed in Fig. 229 is a square. Prove AAED, DEB, BEC, AEC congruent, (s.a.s.) Hence, AD = DB=BC = CA. Why?
Prove that AEAD, ADE, EDB, DBE, etc., are 45° angles. Prove that ADAC, ADB, DBC, BCA are right angles.
11. To circumscribe a square about a circle.
Draw tangents at A, C, B, and D, Fig. 229. The quadrilateral formed is the required square.
176 FIRST-YEAR MATHEMATICS
12. To inscribe a regular hexagon in a circle. Let circle A, Fig. 230, be the given circle.
With any point on the circle, as B, as center ^^_^^ and radius equal to AB draw an arc at C. With
^ I ^\z> C as center and the same radius draw an arc at D.
Similarly draw arcs at E, F, and G. "^ I ^c Join the consecutive points B, C, D, etc.
^^^^^ The polygon BCDEFG is the required hexagon.
Fig. 230 13. Prove that the hexagon constructed in
exercise 12 is regular.
Prove ABAC, CAD GAB equilateral.
Hence, BC = CD = DE, etc.
Prove that A BCD, CDE, etc., are equal to 120° and therefore equal to each other.
Summary
313. This chapter has taught the meaning of the following terms: symmetry, plane of symmetry, axis of symmetry; tangent to a circle, contact point; regular polygon; inscribed and circumscribed polygons.
314. Proofs were given for the fundamental con- structions taught in the preceding chapters.
315. The fundamental constructions were applied in a number of more complicated construction problems.
316. Symmetry of figures suggests the solution of problems of construction and proofs of theorems.
317. The following locus theorem has been proved: The locus of points within an angle which are equally
distant from the sides is the bisector of the angle.
318. The following theorems were proved :
1. A line perpendicular to a radius at the outer end- point is tangent to the circle.
THE CIRCLE 177
2. The radius drawn to the point of contact of a tangent is perpendicular to the tangent.
319. The following problems of construction were taught :
1. To find the center of a given circle.
2. To inscribe a circle in a triangle.
3. To circumscribe a circle about a triangle.
4. To inscribe a square in a circle.
5. To inscribe a regular hexagon in a circle.
CHAPTER XII
POSITIVE AND NEGATIVE NUMBERS. THE LAWS OF SIGNS
Uses of Positive and Negative Numbers
320. In the preceding chapters problems in geometry have been solved by geometric and by algebraic methods. It is the aim of this and the following chapters to develop skill in adding, subtracting, multiplying, and dividing algebraic numbers, to learn more about the solution of equations, and to make use of algebra, not only in the solution of geometric problems, but also in solving prob- lems arising outside of the field of geometry.
Two trains A and B leave • 90 »
1 60 1.
Columbus, Ohio, at 8:00 a.m.
Columbus
Two hours later A is 60 mi.
and B is 90 mi. from Columbus. ^^^' ^^^
How far is B from A ?
Show from Figs. 231
90
and 232 that there are a Coiumbus b
two answers to this pj^ 232
question, according as the trains travel in the same or the opposite direction.
321. Notation for directed line-segments. It is cus- tomary to distinguish line-segments when measured in opposite directions by the positive or plus {-\-) sign and the negative or minus (— ) sign, the "plus sign indicating direction to the right (or upward), the minus sign indicat- ing direction to the left (or downward). Moreover, the
178
POSITIVE AND NEGATIVE NUMBERS 179
plus or minus sign is prefixed to the number expressing the measure (length) of a directed segment. Thus, in Fig. 231,OA = +60, 05= +90. But in Fig. 232 0^1= -60 and OB = +90.
322. Number-scale. With the agreement of §321 it is possible to express both length and direction of line- segments by numbers. In Fig. 233 these numbers are represented geometrically on a straight line with reference to a point 0 in that line.
H 1 1 1 1 1 1 1 >
.8 .7 -6 -5 -i. -3 -2 -1 0 +i f2 +3 +4 +5 +6 +7 +«
Fig. 233
This arrangement is called the number-scale, or the algebraic scale.
323. Positive and negative numbers. A number pre- ceded by a plus sign is a positive number. A number pre- ceded by a minus sign is a negative number. The plus sign need not always be written. Thus, when no sign is prefixed to a number it is understood to be a positive number. The negative sign is never omitted.
324. Absolute value. The value of a number regard- less of sign is called the absolute value, or the numerical value. Thus, the absolute value of +4 is 4, of -7 is 7.
325. Positive and negative angles. By rotating line AB, Fig. 234, around A until it takes the position AC, the angle BAC is pj^ 234 formed. By rotating AB in the
opposite direction, angle BA Ci is formed. To distinguish between these directions one angle may be denoted by the
180 FIRST-YEAR MATHEMATICS
plus sign, whereupon the other is denoted by the minus sign. It is customary to consider an angle positive when it is formed by rotating a Hne counter-clockwise, and negative when it is formed by clockwise rotation.*
EXERCISES
1. Make drawings on the following angles, always starting from a line in horizontal position, as AB, Fig. 234: +90°, +45°, + 160°, +270°, -45°, -30°, -180°, -270°, -360°.
2. A line rotates through an angle of +75° and then through an angle of —40°. Give the size and direction of the angle formed by the first and the last positions of the line.
326. Positive and negative temperature. Temperature readings are taken with reference to a definite point on the thermometer, called the zero point. A temperature reading above the zero point is plus; below, minus.
Give the meaning of the following temperature readings: -4,-2, 0, +3, +8, +8, +4, +3, +1, 0,-1,-2.
EXERCISES
In arithmetic the + sign always denotes addition, the — sign, subtraction.
The following problems illustrate some of the uses of positive and negative numbers in algebra.
1. If a man's debts be indicated by the minus sign and his possessions by the plus sign, what is the condition of a man's affairs if his debts and possessions are given by +$1,200 and -$1,000? By +$73 and -$50? -$75 and +$60? -$300 and +$1,000?
* Euler (1707-1783) was the first to apply positive and negative signs to angles. This he did in a book called Introductio, published in 1748. Gauss (1777-1855) completed the modern science of positive and negative angles.
POSITIVE AND NEGATIVE NUMBERS 181
2. A bicyclist starts from a point and rides 18 mi. due north- ward (+18 mi.), then 10 mi. due southward ( — 10 mi.). How far is he from the starting-point ?
3. How far and in what direction from the starting-point is a traveler who goes eastward (+) or westward ( — ) as shown by these pairs of numbers: +16 mi. then —6 mi.? —20 mi. then +28 mi.? -18 mi. then +18 mi.? +100 mi. then +50 miles?
4. Denoting latitude north of the equator by the plus sign and latitude south by the minus sign, give the meaning of the following latitudes: +28°, +2°, -18°, +12°, -10°.
5. A boy starts with no money. If he earns 50 cents (+50 cents) and spend 40 cents ( — 40 cents) , how much money has he then?
6. What does the minus sign denote if the plus sign denotes above? forward? upward? to the right? east? north? gain? possession ? income ? addition ? increase ?
7. The statistics below give the value in millions of dollars of the excess of merchandise imported into or exported from the United States to the Philippine Islands for the years 1892-1912. Denoting an excess of exports by + , of imports by — , read the following: -6.2, -9, -6.8, -4.6, -4.8, -4.3, -3.7, -4, -3.3, -.4, -1.3, -7.3, -7.2, -6.4, -6.9, -2.8, +1.3, + 1.7, -.5, +2.3, +.5.
Graphing Data
327. Positive and negative quantities may be repre- sented graphically.
EXERCISES
1. On a winter day the thermometer was read at 6:00 a.m. and every hour afterward until 5:00 p.m. The hourly read- ings were -10°, -8°, -7°, -5°, 0°, +2°, +8°, +10°, +10°,
182
FIRST-YEAR MATHEMATICS
+5°, 0°, —5°. Mark off these readings on squared paper (Fig. 235). Connect the points thus obtained as shown in the figure. This forms a broken line, called a temperature-line.
From the temperature-line one may obtain some information regarding the changes of the temperature: When was it coldest? Warmest? When was the change most rapid?
^10'
t5'
/ |
■\ |
|||||||
/ |
1 1 1 |
\ |
||||||
, |
a%. |
7 S |
;/ |
0 il |
is \ p. |
^k |
,\ |
i— |
-^ |
^ |
Time |
Tempera- ture |
6:00 A.M. |
-10° |
7:00 |
- 8° |
8:00 |
- 7° |
9:00 |
- 5° |
10:00 |
0° |
11:00 |
+ 2° |
12:00 |
+ 8° |
1:00 P.M. |
+ 10° |
2:00 |
+ 10° |
3:00 |
+ 5° |
4:00 |
0° |
5:00 |
-5° |
Fig. 235
When was the change a rise ? etc. A more accurate picture of the changes in temperature could be obtained by taking the readings oftener than every hour.
2. On squared paper draw a line to show the following hourly readings, beginning at 8:00 a.m.: +2°, -2°, -4°, -2°, +2°, +4°, +4°, +8°, +10°.
3. The average monthly temperatures for a northern town
are
Jan. - 4° Feb. - 7° Mar. +14° Apr. +26°
May +42° June +52° July +62° Aug. +60°
Sept. +48° Oct. +37° Nov. +25° Dec. + 2°
Draw the temperature-line.
4. The daily average temperatures for 14 days at a certain place were +8°, 0°, -10°, +12°, -6°, +14°, +15°, +2°, -5°, + 15°, +20°, 0°, 0°, +10°. Graph these readings.
5. A ship's latitude from week to week was +42°, +38°, +30°, +20°, +12°, +2°, -1°, -6°, -3°, +12°. Graph these latitudes and tell when the ship crossed the equator.
POSITIVE AND NEGATIVE NUMBERS
183
6. The following table gives the lowest, highest, and average temperature for Chicago taken for 38 years previous to Decem- ber 31, 1911.
Jan. |
Feb. |
Mar. |
Apr. |
May |
June |
July |
Aug. |
Sept. |
Oct. |
Nov. |
Dec. |
-20 |
-21 |
-12 |
17 |
27 |
40 |
50 |
49 |
32 |
14 |
- 2 |
-23 |
64 |
63 |
81 |
88 |
94 |
98 |
103 |
98 |
98 |
87 |
75 |
68 |
24 |
25 |
34 |
46 |
56 |
66 |
72 |
71 |
65 |
53 |
39 |
29 |
Graph the three temperature-lines on the same sheet of squared paper.
7. The following numbers indicate in millions of dollars the value of the excess of exports from the United States or imports into the United States from Austria-Hungary for the years 1892-1912. An excess of exports is positive, an excess of imports, negative:
-6.19, -9.48, -6.37, -4.38, -5.20, -4.13, +.98 + .83, -2.03, -2.84, -3.98, -3.41, -2.14, +1.07, + 1.02, -.87, +.75, -1.21, -2.45, +2.56, +5.67.
Make the graph.
Historical Note. — The earliest instances of the regular use of the signs + and — occur in the fifteenth century. John Widmann, of Eger, born about 1460, and probably a physician, wrote a Mercantile Arithmetic. In this book these signs are used merely as marks signifying excess or deficiency.* The French mathe- matician Vieta (1540-1603) seems to have been the first to make regular use of the + and — signs as shorthand symbols for addition and subtraction.! Descartes (1596- 1650) showed how to represent positive and negative numbers along a line.
* See Ball, p. 206. t See Cajori, p. 150.
184 FIRST-YEAR MATHEMATICS
Addition of Positive and Negative Numbers
328. Graphical addition. Addition of positive and negative numbers may be performed graphically.
EXERCISES
1. Find graphically the sum of (+5) and (—2).
Let (+5) be a distance traveled eastward and (—2) a distance traveled westward. Find the distance and direction of the stopping- point from the starting-point of an K +5 ^ automobile going first (+5) mi.,
I . I I T ■ then (—2) miles. This distance
^ "^ ^ is considered the sum of the dis-
FiG. 236 tances passed over.
Starting from O, Fig. 236, lay off five units to the right to point A . From A lay off two units to the left to point B. The position of B is three units to the right of 0, i.e., (+3). This may be expressed in symbols by the equa- tion (+5) + (-2) = (+3).
2. Find graphically the following sums: (4-6) + (+2); (-6) + (+2); (+6) + (-2); (-6) + (-2); (+15)H-(-10); (+15) + (-20); (-18) + (+24); (-12) + (-9).
329. Positive and negative numbers may be added by letting the + and — signs denote two opposite senses, as gain and loss, income and expenditure, etc.
EXERCISES
1. Find the sum of (+25) and (-31).
Let (+25) denote a gain of $25 and (-31) a loss of $31; the final standing, a lo^s of $6, is the sum. Thus,
(+25) + (-31) = (-6)
2. In a way similar to that of exercise 1 find the following sums:
+ 15 -15 +15 -15 +38 -38 +38 -38 + 8-8-8+8 +19 -19 -19 -19
POSITIVE AND NEGATIVE NUMBERS 185
330. Algebraic addition. Show from examination of the results of exercise 2 that positive and negative num- bers may be added according to the following laws :
1. To add two numbers having like signs, find the sum of their absolute values and prefix to this sum the common sign.
2. To add a positive and a negative number find the difference of their absolute values and prefix to it the sign of the number having the greater absolute value.
EXERCISES
Solve the following exercises with the aid of positive and negative numbers. In finding the sums use the laws of § 330. Then verify the results by the method of § 329.
1. At noon a thermometer read 3° below 0°. In the evening it was 8° warmer. How many degress did the thermometer read in the evening ?
2. A boatman rows, at a rate that would carry him 3 mi. an hour through still water, down a river whose current is 2 mi. an hour. What is his rate per hour? What is his rate per hour if he rows up the river ?
3. A motor boat, having a speed that would make it go 12 mi. an hour in still water, is going down a river whose current is 2.4 mi. an hour. How fast does the boat move? How fast can the boat move upstream ?
4. If a man's property is worth $3,600 and his debts amount to $1,400, what is his financial standing?
5. A toy balloon pulls upward with a force of 9 oz. If a weight of 6 oz. is attached, will the balloon rise or fall? With what force?
6. A balloon pulls upward on a stone, weighing 6 oz., with a force of 8 oz. What is the sum of the forces ?
186 FIRST-YEAR MATHEMATICS
331. Addition of three or more numbers. The follow- ing problems call for the addition of three or more positive and negative numbers.
EXERCISES
1. John received $4 from his father and $3.50 from his mother. He paid $3 to the grocer and $3.75 to the hardware man. How much money had he left ?
(+4) + (+3.50) + (-3) + (-3.75) = + .7o
2. Translate the following into problems like exercise 1 and solve each :
(+4) + (3.50) + (-3.75) + (-3) (-f4) + (-3) + (3.50) + (-3.75) (-|-3.50)H-(-3) + (+4) + (-3.75)
3. How do the results in exercises 1 and 2 compare ?
4. Interchange in every possible way the addends in (-|-8) + ( — 6)H-(+4) and in each case find the value of the sum.
332. Commutative law. Exercises 2, 3, and 4 show that the commutative law holds for positive and negative numbers; i.e., The value of a sum of positive and negative numbers is the same whatever the order of the addends.
EXERCISES
1. An elevator starts at a certain floor, goes up 65 ft., down 91 ft., up 52 ft., down 13 ft., up 65 ft., and stops. How far and in what direction is the stopping-point from the starting-point ?
2. A vessel starting in latitude +20° sails +13° in latitude, then -60°, then +40°, then -10°. What is its latitude after the sailings ?
3. What is the latitude of a ship starting in latitude —50° after these changes of latitude: +10°, -5°, +18°, -7°, +38°, -12°, +60°?
4. What is the most advantageous way of adding several positive and negative numbers ?
POSITIVE AND NEGATIVE NUMBERS
187
5. A certain wholesale house owes various factories the following amounts: $475.50, $240.00, $638.50. Several mer- chants owe the wholesale firm these sums: $360.20, $159.45, $520 . 70. The firm has on hand $1,254 . 00. What is its finan- cial standing ?
6. Find the following sums :
+50 +35 -45
+25 -38 -20
-18 +24 +60
- 6 -15 +55
+75 -236
+ 13 +780
-86 - 95
+ 8 +45
+Sx -14a
— 6x —46a
-Ax +77a
+7a: — 5a
7. Show that the sum of two numbers having unlike signs but the same absolute value is zero.
8. Find the value of (-3x) + (-2a;) + (+6.c) + (+10a;) -\-{ — ox) and test the result by substituting a; = 2 in the given sum and in the result.
9. The heat of a metal was increased by 20.4°. The metal was then cooled 4° and finally heated 2° more. What is the change in the original temperature ?
10. Add the following, doing all you can orally:
1. +5 -3 |
5.-7 + 10 |
9. + 6 + 10 |
2. +8 -5 |
6. —2 +5 |
10. - 8 - 4 |
3. + 7 -10 |
7. -16 + 4 |
11. + 3 + 10 |
4. +6 -8 |
8. -3 + 1 |
12. - 7 - 2 |
188 FIRST-YEAR MATHEMATICS
13. |
-8a |
23. |
-6| |
33. |
-3.2s |
+4a |
24. |
+81 +7i |
34. |
-6.8s |
|
14. |
+7ia; |
||||
-Vdx |
25. |
-7i -121 |
35. |
-^\x |
|
15. |
-12m |
+3262C |
|||
+ 16m |
26. |
- 2i |
36. |
-2862c |
|
16. |
+24n |
-18^ |
-\-\^vHf |
||
- 6n |
+26i |
-24.vhf |
|||
17. |
+f ■ |
27. |
-2.12 |
37. |
+ Sax |
+1 |
28. |
-1.88 |
38. |
— Qax |
|
18. |
+3.16 |
-l2xHj |
|||
29. |
-4.08 |
39. |
- Ix^y |
||
19. |
-n\ |
-68mr3 |
|||
-1 +3^ |
30. |
+23f |
40. |
-75mr3 |
|
20. |
-6.69 |
-^cd? |
|||
-2i -5i |
31. |
+8.04 |
41. |
+3|crf2 |
|
21. |
+ 8.95 |
+4.5a:2y/2 |
|||
-2f +4f |
32. |
-11.25 |
42. |
-Sxhjz |
|
22. |
-16r |
-\-2Aa¥c |
|||
-6| |
+ 18r |
-Q.2ab'c |
Subtraction of Positive and Negative Numbers
333. Graphical subtraction. Positive and negative numbers may be subtracted graphically.
POSITIVE AND NEGATIVE NUMBERS 189
EXERCISES
1. Subtract (+6) from (+a).
C5 ^±±2 vD , (+a). ^
A-
-^B
}. — itb)—
(+a) EH G F
Fig. 237 Fig. 238
Let AB, Fig. 237, represent in magnitude and direction the number (+a) and let CD represent (+6). To subtract graphically (+6) from (+a) means to lay off the minuend (+«) on an indefinite line, as EF, Fig. 238, in its own direction, i.e., to point G; and to lay off the subtrahend (+6) from G in the direction opposite to that of (+6), to point H.
EH is the difference between (+a) and (+6).
Thus, EH ^i+a)-i-\-b).
2. Show that EH, Fig. 238, may be constructed by adding (-ha)and(-6),i.e., that(+a)-(+6) = (+a) + (-6).
3. Show graphically that the following expressions are equal:
(+5)-(+3)and(+5) + (-3);(+8)-(+10)and(+8) + (-10); (+6)-(+7)and(+6) + (-7)
4. Subtract ( — 6) from (+a).
(+a 1 ->\ (-6) —4
(-6) I rj—T H A > 1 iZ?
Fig. 239
On AB, Fig. 239, lay off the minuend (+«) in its own dh'cction to C. From C lay off the subtrahend ( — 6) in the direction opposite to that of (—b) to point Z).
AD is the difference between (+a) and ( — 6).
Thus, ^D = (+a)-(-6).
5. Show that AC, Fig. 239, may be constructed by adding (+a) and (+6), i.e., that (+a)-(-6) = (+a) + (+6).
6. Show graphically that the following expressions are equal: (+6)- (-3)and(+6) + (+3); (+4)-(-5) and (+4) + (+5); (+8)-(-8) and (+8) + (4-8); (+7)-(-4) and (+7) + (+4)
190 FIRST-YEAR MATHEMATICS
7. Show graphically that (-4) -(+6) = (-4) + (-6)
(-8)-(-4) = (-8) + (+4); (-7)-(+2) = (-7) + (-2)
(-3)-(-4) = (-3) + (+4); (-a)-(+fe) = (-a) + (-6) (-a)-(-6) = (-a) + (+6)
334. Algebraic subtraction. The exercises of §333 show that numbers may be subtracted according to the following law:
To subtract a number change the sign of the subtrahend and add the result to the minuend.
Arithmetic numbers may be subtracted by this law by considering them as positive numbers. Thus, 7-5=(+7)-(+5)=(+7) + (-5)=2. In arithmetic a number can be subtracted only from a larger number. In algebra subtraction is always possible, e.g.,
7-10 = (+7)-(+10) = (+7) + (-10) = -3.
EXERCISES
Subtrac,t the lower from the upper number of the following:
1. |
+ 19 -10 |
6. |
5 — 8 1 |
11. |
- 9c -15c |
2. |
-66 -25 |
7. |
+ 1 2 — 3^ |
12. |
+762 -962 |
3. |
-75 +25 |
8. |
-A +1 |
13. |
-8m3 +3m3 |
4. |
+ 8 + 10 |
9. |
+ v^ |
14. |
+a — a |
5. |
+'l +f |
10. |
-13c3 + 8c3 |
15. |
+6.75ci3 -7.25a3 |
POSITIVE AND NEGATIVE NUMBERS 191
16. |
-3.16c2 -0.89c2 • |
19. 20. 21. |
-4.76« +9.67^ |
22. 23. 24. |
+ .80m -\-lA2m |
17. |
+6(x+2) -9(c-s) +3(c-s) |
+ .82a2 -3.75a2 |
-3.26s +7. 49s |
||
18. |
-0.75c -0.90c |
+2.037/ -5.247/ |
Law of Signs in Multiplication
335. Graphical multiplication. The absolute value and sign of the product of two numbers may be deter- mined geometrically.
EXERCISES
1. Find the product of (+4) by (+3).
Since (+3) (+4) is the same as (3) (+4) it follows that (+3) (+4) equals (+4) + (+4) + (+4) = ( + 12). Geometrically this means that to multiply
(+4) by (+3) is to lay +, j _^, . a., _j
off (+4) three times in p-fj- ^ h—^-^i^-^
its own direction, o\ ( + /^;
Fig. 240. Thus, Fig. 240
(+3) (+4) = ( + 12).
2. Find the product of (-4) by (+3).
Since (+3) (-4) is the same as (3) (-4), (+3) (-4) = (-4) + (-4) + (-4) = (-12). Make a drawing for (+3)(-4), i.e., lay off ( — 4) three times in its ow/i direction. Thus, (+3) (—4) = (-12).
3. Find the product of (+4) by (-3).
Assuming that the commutative law holds for positive and nega- tive numbers, i.e., that the value of a product is not changed by changing the order of the factors, it follows that (-3) (+4) = (+4) (—3) = ( — 12), according to exercise 2. The same result is obtained if we agree to mean by (—3) (+4) that (+4) is to be laid off three times in the direction opposite to its own direction. Thus, ( —3) (+4) = — 12. Make a drawing for this product.
192 FIRST-YEAR MATHEMATICS
4. Find the product of (-4) by (-3).
According to the agreements made in exercises 1, 2, and 3, (—3) ( — 4) will be understood to mean that (—4) is to be laid off three times in the direction opposite to that of (—4). Thus, (-3) (-4) = ( + 12).
336. Law of signs for multiplication. Exercises 1^, § 335; illustrate the following law for determining the sign of a product:
The product of two factors having like signs is positive.
The product of two factors having unlike signs is negative.
Find the value of the following products, (1) using the law of signs, (2) geometrically:
(-2) (-3); (+3) (-2); (-2) (+4); (+2) (+8); (-3) (-5); (+3) (-5); (-2) (+5); (-2) (-5); (-3) (+6)
337. Turning-tendency. The laws of signs may be illustrated with the bar, Fig. 241. A light bar supplied
with equally spaced pegs is '^'^'^tr^"" balanced about its middle
00 point M. With a number
1 Mry of equal weights, w, the 'u u u i^i-~k^ ^- ^' "-' ^-"^^^ following experiments are
yljplry performed:
/ / 1. Hang a weight of 2w on
Fig. 241 the peg l\. This weight tends
to turn the bar. Note the num- ber of weights that must be attached to the hook H to balance this turning-tendency. Now hang the same weight 2w on peg U and measure its turning-tendency by attaching to the hook, H, weights sufficient to balance the bar.
2. In a similar manner find the turning-tendency caused by the weight 2w on the peg U', on U] on U.
A large number of experiments like 1 and 2 have shown the following:
POSITIVE AND NEGATIVE NUMBERS 193
1. The turning-tendency, or leverage, varies as the dis- tance of the peg from the turning-point, M.
2. The turning-tendency is equal to the product of the weight by the distance of the peg, where the weight hangs, from the turning-point.
The same two facts are true on the right side as on the left, but the bar turns, or tends to turn, in the opposite direction.
338. Direction of turning. The following method has been agreed upon to distinguish between the two directions of turning. When the bar turns, or tends to turn, with the hands of the clock it is said to turn clockwise; if it turns, or tends to turn, against the hands of the clock, it is said to turn counter-clockwise. Weights attached to the peg are downward-pulling weights, and are designated by the — sign. Weights attached at H pull upward on the bar and are designated by the + sign.
339. Lever-arm. 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 ri be repre- sented by +1, then the distance from M to r^ will be -|-4; from M to k by —3, and so on.
340. Multiplication of positive and negative numbers.
By means of the apparatus, Fig. 241, the product of positive and negative numbers is now to be found.
EXERCISES
1. Find the product (+4) (+3).
Whatever device may be used for finding this product, the result must agree with the result from arithmetic; i.e., it must be ( + 12).
194
FIRST-YEAR MATHEMATICS
4J
Fig. 242
\^-
+^ —
Let (+4) represent an arm 4 pegs to the right and (+3) a force puUing upward on the bar. The direction of turning is counter-clockwise. If it is agreed to call this the pZws direction then the turning-tendency,
T = (+4)(+3) = ( + 12), agrees with the arithmetical product 4X3.
2. Find by means of the bar the following products:
(+2)(+4); (+10)(+6); (+a)(+6).
3. Find the product (+4)(-3). »-
This means that at the fourth peg to the right a downward force of 3 weights is attached, Fig. 243. The direction of turning is clockwise and will be called minus, since it is opposite to the positive direction.
Hence, T = (+4)(-3) = (-12).
4. Show by means of the bar that (-4)(+3) = (- 12).
5. Show by means of the bar that (-4)(-3) = (+12).
6. Compare the result of exercises 1, 3, 4, and 5 with the sign law (§336).
7. Using the sign law find the products of the following, doing all you can mentally.
Fig. 243
1. |
(+5)(+3) |
13. |
(-|)(+f) |
25. |
(-c)(-pr) |
2. |
(-5)(+3) |
14. |
(+!)(+!) |
26. |
(-3|)(+a:?/) |
3. |
(-5)(-3) |
15. |
(-I)(-A) |
27. |
{-h\){-xz) |
4. |
(+5)(-3) |
16. |
(+tV)(-I) |
28. |
(-5|a)(+6) |
5. |
(+3)(+6) |
17. |
(_6f)(-6f) |
29. |
{-2\c){-2\d) |
6. |
(+2)(-2) |
18. |
(+6f)(-6f) |
30. |
(+7ia:)(-7iy) |
7. |
(-6)(-2) |
19. |
(-6f)(+6f) |
31. |
(+5)(-sr/2) |
"8. |
(-3)(-4) |
20. |
(+6f)(+6f) |
32. |
(-7)(-^2/) |
9. |
(-4)(+2) |
21. |
(+3)(-a) |
33. |
(-10) ( + 2^2^) |
10. |
2.-3 |
22. |
{-Z){-x) |
34. |
(+12)(-a2/3) |
11. |
-3.-4 |
23. |
(-7)(+r.) |
35. |
(-16)(+a6c2) |
12. |
(f)(+J) |
24. |
(-«)(+pg) |
36. |
(+6.)(-a) |
POSITIVE AND NEGATIVE NUMBERS 195
Multiplication by Zero
341. The product 4X0 means 0+0+0+0 = 0. In general aXO = 0. Why? Since aXO = OXa, it follows that OXa = 0. V^hy? This illustrates the following theorem : The value of a product is zero if one of the factors is zero.
EXERCISES
1. Showby meansof abar that aXO = 0; OXa = 0.
2. The dimensions of a rectangle are a and b. Keeping b the same, bisect a and form a rectangle of dimensions ^ and b.
Bisect o and form a rectangle of dimensions t and 6. What are the areas of the rectangles formed ?
3. Find the areas of rectangles having the following dimen-
sions: c > ^; T-^ , o; oo > ^5 ^^ ^^^ ^^^^ ^^ *^® rectangle
becomes smaller and smaller approaching nearer to zero, how does the area change? Show from this that 0X6 = 0.
4. Show by means of the bar or by means of the rectangle that 0X0 = 0.
Product of Several Factors
342. The product of several factors is obtained by multiplying the first factor by the second, the result by the third, etc. Often, however, it is more advantageous to rearrange the factors before multiplying.
EXERCISES
1. Find the values of the following products:
(+3). 4. (-6). 5- (-2); 3-5- (-2.8) • (-1.2);
(-1) • (Y) (-1) (f).
2. Find the values of the following powers: ( — 1)^, ( — 1)^ (-1)^ (-2)^ (-2)3, (-2)^ (-2)^; (-3)2, (-3)^, (-3)^
3. Find the value of 2x3-4x2-5x+3 for a; = -3.
196 FIRST-YEAR MATHEMATICS
4. Find the value of x^—Sx^y-\-4:Xi/ — 5i/iorx=—2.
5. Find the value of 5x^-{-Qx'^-\-Sx-\-4: for a: = 10.
6. Compare the values of the following: ( — 2)^ and —2^; (-2)4and-24; (-a;)3and -a;^; (-a;)4and -a;^.
7. What is the sign of the product of 7 factors, 4 of which arc positive and 3 of which are negative ?
8. What powers of negative numbers are positive? Nega- tive?
Law of Signs for Division
343. Division. Dividend. Divisor. Quotient. To
divide a number 12 by 3 means to find the number which multiplied by 3 gives 12. Thus, the product 12 and one factor 3 being given, the process of finding the other factor is division. Division of 12 by 3 is indicated in the fol- lowing ways: y , 12:3, 12h-3, or 12/3.
Division in algebra has the same meaning as in arith- metic : the product of two factors and either factor being given, the operation of finding the other factor is called division. The given product is the dividend, the given factor the divisor, the result of the division the quotient.
EXERCISES
1. Since -10=(-f2) (-5), what must (-10)-^(+2), or
+ 2 ^^ • - 5 •
2. Answer the following questions, giving reasons for your answers:
1. (-hl2)-(+3)=? 5. (-12)-(-3)=?
2. (-hl2)^(+4)=? 6. (-12)H-(+4)=?
3. (-12)4-(-h3)=? 7. (+12)-^(-3)=?
4. (-12)-^(-4)=? 8. (-hl2)-^(-4)=?
POSITIVE AND NEGATIVE NUMBERS 197
344. According to the definition of division :
( -\-ab) - ( +a) = ( +6) for ( +a) ( +6) = ( +ab) (4_«5)^(_a) = (-6) " {-a)i-b)={-\-ab) (_a5)^(+a)=(-6) " (+a) (-5)=(-a6) . (_a6)-(-a)=(+6) '' {-a) (+b) ={-ab)
Examine these equations and state the sign of the quotient:
1. If the signs of dividend and divisor are alike (i.e., both + , or both — ) .
2. If the signs of dividend and divisor are unlike (i.e., one — and the other +).
State the law of signs for division and compare the statement with the following:
345. Law of signs in division : // dividend and divisor have lil^e signs the quotient is positive; if dividend and divisor have unlike signs the quotient is negative.
EXERCISES
1. Using this law give answers to the following questions: 1.
2. -^=? 6. -^-^=? 10. -^=?
3.
4.
2. On a winter day the thermometer read —4° in the morn- ing, — 1° at noon, and —6° in the evening. What was the average temperature ?
To find the average (mean) temperature divide the sum of the readings by the number of readings.
3. Define quotient and show that a number divided by itself gives the quotient 1.
4. Show that a number divided by 1 gives that number as quotient.
+ 18_, + 2 • |
5. |
+ 18_, + 9 • |
9. |
xy_ y |
+ 18 ^ - 2 • |
6. |
+ 18 ^ - 9 • |
10. |
-xy y |
+ 2 • |
7. |
+ 9 • |
11. |
-xy — X |
-18=. |
8. |
~'l=' |
12. |
-xy |
- 2 |
— 9 |
- y |
198
FIRST-YEAR MATHEMATICS
5. Show that 0 divided by any other number gives 0.
346. Division by 0. The quotients ^, ? -^, ~
etc., have no meaning, for a number multiplied by 0 cannot give 1, 2, —5, a, etc.
The quotient ^ is undetermined, as every number
multiplied by 0 gives 0.
Therefore it is assumed that in all quotients hereafter the divisor is not zero nor equal to zero.
1. Solve the
1. 0^+14 =
2. -4x = S
3. 3-ha;=-
2. Find the mentally:
1. (+1)
2. (-1) ■
3. (+1) ■
4. (+4)
5. (+4)
6. (+f) •
7. (-1)
8. (+f)
9. (-1)
10. i-h
11. (-1)
12. (+f)
13. (-?)
14. i-h
EXERCISES
following equations:
10
4. fx =
1 0
— -^' 5. x-^|=-f
'6. 2^^1 = 1
quotients of the following, doing all you can
(+1) (-1) (+1) (-f) (-1) (+1) (-1) (-1) (+1) (-f) (+f) (-*) (+1) (-1)
16. |
(-2a) - |
17. |
(-2a) -^ |
18. |
(-56) -^ |
19. |
(+12:r)-^ |
20. |
(-a) -^ |
21. |
(+a) - |
22. |
(-a3) - |
23. |
(+a^) - |
24. |
(-«') - |
25. |
(-6a3)- |
26. |
(+4a6)4- |
27. |
i+Aab)^ |
28. |
(Sax)^ |
29. |
(-2xy)-^ |
30. |
(+6ir)H- |
(+2)
(-a)
(+5)
(-4ir)
(-ia)
(-H
(+a)
(+a=^) (+3a)
(-?>)
(-2a)
i+Sx)
■(+^2/) (-3lr)
POSITIVE AND NEGATIVE NUMBERS 199.
31. i+av') ~i-av) |
38. (-7f7/.) -^(-11^2/) |
32. i-av') -r-+v') |
39. (-1617/2=^) -(+4|2/^) |
33. i-abc) -^(-a) |
40. (+6.82a2)^(-31a) |
34. (H-a6c) -^(+ac) |
41. -3(a+fe2)^(a+6) |
35. i-abc') -^(-ac) |
42. (-72^8610) --(-Gasfeio) |
36. i-avr) -^(+ar) |
43. -96x'z^ -^-12xz* |
37. {+7iax)-^{-\-2Sa) |
44. -36m3n4-c2^+6m%3 |
3. Find the values of the following expressions: |
|
1. (-\-Sx*)^(-x')^{- |
■x); (-3a:). (-4x3)- (-6^2); |
(5a:3-3x2)-^2x; |
forx=-2. |
2."^"^Sora-6,6 = |
= 5. |
^ -Ux\x'-^x-\-4) |
Pnr -r = _ 9 |
7(-x)3 ^"^ - -
Summary 347. The chapter has taught the following :
1. Positive and negative numbers may be used to distinguish between the opposite senses, as upward and downward, to the right and to the left, gain and loss, etc.
2. Statistics containing positive and negative numbers may be represented graphically.
3. Positive and negative numbers may be added and subtracted graphically.
4. The sum of two numbers having like signs is the sum of the absolute values with the common sign prefixed.
5. The sum of two numbers having unlike sign's is the difference of the absolute values prefixed by the sign of the number having the greater absolute value.
6. The value of a sum is not changed by changing the order of the addends.
200 FIRST-YEAR MATHEMATICS
7. To subtract a number change the sign of the subtrahend and add the result to the minuend.
8. The product of two numbers having like signs is positive; of two numbers having unlike signs, negative.
9. The turning-tendency caused by a force acting upon a bar is equal to the product of the force by the lever-arm.
10. The value of a product is zero if one of the factors is zero.
11. Factors may be changed in order before finding the product.
12. The quotient of two numbers having like signs is positive; of two numbers having unlike signs, negative.
13. Numbers may be multiplied graphically or a turning-bar may be used to explain multiplication of positive and negative numbers.
BLAISE PASCAL
BLAISE P A S C A L
BLAISE PASCAL, a natural but somewhat erratic genius, was born at Clermont, France, on June 19, 1623, and died at Paris, August 19, 1662. He had displayed exceptional ability by the age of eight, and, despite the dis- couragements of his father and his teacher, became greatly interested in geometry at twelve years of age. Deprived of books on geometry, he discovered for himself many of the properties of figures. Seeing the boy's determination to study geometry, his father gave him a copy of Euclid's Elements, .which he mastered in a few weeks.
At the age of fourteen Pascal was admitted to the weekly scientific meetings of the French geometricians; at sixteen he wrote an essay of marked originality on conic sections, and at eighteen he constructed an important calculating machine. Thereafter he studied for a time ex- perimental science, then religion, then returned again to mathematics. He formulated a new theorem of conies, still known as ''Pascal's theorem," and invented and employed his arith- metical triangle for figurate numbers from which the coefficients of the expansion of a binomial are obtained. He laid down the foundations of the theory of probability, did much work on the cycloid, and exerted himself on the theory of indivisibles. He is said to have worn himself out completely through excessive hard work, so that he died of old age at the age of thirty-nine. See an account of his life and work in some history of mathematics.
CHAPTER XIII ADDITION AND SUBTRACTION
Review of the Laws of Addition
348. The laws of algebra which in chapter II were shown to hold for addition of literal numbers hold also for positive and negative numbers.
1. Commutative law. Show graphically that
(+8) + (-3) = (-3) + (+8). This illustrates the commutative law : The value of a sum remains unchanged by changing the order of the addends.
2. Associative law. Show graphically that
(-5)4-(+3) + (-2) = [(-5)+3]+(-2) = (-5)-f[(+3) + (-2)] This problem illustrates the associative law : In adding several numbers the sum is the same in what- ever way two or more of the numbers are combined into a sum before adding in the rest.
349. Similar terms. Terms which have a common factor are said to be similar with respect to that factor and are called similar terms. Thus,
tx^y^, 12x^y^, —Sx^y^ a,re simWsir terms. Why? Also 1 5a26c^ 10a26c^ -Sa^bc\ Why?
350. Dissimilar terms. Terms which have no com- mon factor are called dissimilar terms, as 4a2 and — 3c6^
201
202 FIRST-YEAR MATHEMATICS
EXERCISES
Point out with respect to what factor the following terms are similar and give the coefficients of the common factor :
1. 4x, -7x, 20a;, -35a;
2. ax, —25a:, —bx, 46a;
3. —Spqh, —14:pqh, —12pqh
4. Sa%, -5a%, -i-7a%, -Sa%
5. 2aa;, 3aa;, —lax, —5ax
6. -Spq\ ^tq\ -Skq\ \2sq^
Addition of Monomials
351. Express the sums of the following numbers in the form of a polynomial and combine like terms:
EXERCISES
1. Aa'h, -a%, -da%, a%, -Sa%
The common factor is a%. The coefl&cients of a% are 4, —1, — 5, 1, —3. The sum of the coefficients is —4. Hence, the required sum is —4:a%.
2. lOxy^, —4iXy^, —2x\f, -\-xy^, —5xy^
The common factor is xy^. The coefficients of xy^ are 10, —4, —2, +1, —5. The sum of the coefficients is 0. Hence, the required sum is 0. Why?
3. +15a, -7a, +18a
4. -18a;2, -12a;2, +15a;2, -3a;2
5. +2Ub, -\-Siah, -4^a6, -b^ab
6. -\-27abc, -S5abc, +10a6c, -2abc
7. 3(a+fe), -4(a+6), l2{a-{-b)
8. -8(a;2-f-2/2), -24(a;2+2/2), 17(a;2+7/2)
9. -3f(pr-g2), +5f (pr-g2), _4yV(pr-g2)
10. 18(mp-3s)2, -15(mp-3s)2, -37(mp-3s)2, 14(mp-3s)2
11. a{x-\-y-\-z), -b{x-\-y-\-z), -c{x-\-y-\-z), d{x-\-y+z)
ADDITION AND SUBTRACTION 203
12. 15ax2, -76^2, Sdx"^, -hex"
13. h8\ -I2xhj, +7zhj, -ZsH
14. -27fa6, +lS\cd, lh\ah, Ufcd
15. 5ax, —3x,x
16. 9a262, -3c3^7^ 4:a%', -4cY, -Sa%'
17. 3mp2, -8mp^, -{-5a^x, -Sa'^x, -^mp'-, 2a^x
18. -4s«, -^, 3«, +5si
19. ar'^p, —r^p, —br^p, pr^, —cpr-
20. 27(a+6)2, +4c, + 15a, —12c, -15(a+6)2, -8a
21. a(a-b), b{a-b), -c{a-b)
22. a^{a+b), -2ab{a^b), ¥{a+b)
23. -6i(a+6), i(a2+62), -^(a^+fe^), +8f (a+Z>)
24. r^(r-{-p), —Sr^(r-{-p), 3r(r+p), r+p
Addition of Pol5niomials
352. The law for adding polynomials may be seen from the following two problems:
1. The main stairway in a school consists of four flights of stairs,, having a, b, c, and d steps, respectively.
If a pupil goes up and down 8 times in one day, how many steps does he take ?
If a pupil goes up and down tlie stairs 5 times on Monday, 6 times on Tuesday, 4 times on Wednesday, 5 times on Thursday, and 4 times on Friday, how many steps does he take on the stairway every week ?
On Monday : 10a + 106 + 10c + lOd On Tuesday: 12a + 126 + 12c + 12d On Wednesday: 8a + 86+ 8c + Sd On Thursday : 10a + 106 + 10c + lOd On Friday: 8o+ 86+ 8c+ 8c?
48a+486+48c+48d Thus, these polynomials arc added by adding similar terms.
204 FIRST-YEAR MATHEMATICS
2. Add: +27x^-loxy-\-18y^ -lSx^-\-SOxij- 57/ The sum, l4:x'^-i-15xy-{-lSy^ is obtained by adding similar terms in the polynomials.
Frequently the work is arranged as follows:
(27a;2 -15xy-\- 18y^) + ( - ISx^+SOo:?/ - 5?/) = + Ux^-\- 15xy-^lSy'~
353. Degree of a number. The degree of a number
is denoted by the exponent of the number. The numbers x^, y^, z^ are of the second, third, and fourth degree respectively.
The monomial 5a¥x^y^ is of the first degree in a, of the second degree in b and y, and of the third degree in x.
354. Degree of a monomial. The sum of the expo- nents of the literal factors of a monomial is the degree of the monomial.
Thus, x^, 4:xy^, Sx^yz^ are respectively of second, third, and sixth degree.
355. Degree of a polynomial. When a polynomial has been reduced to the simplest form, the degree of the term having the highest degree is the degree of the poly- nomial. Thus, x'^-\-xy-\-2x-\-2y-\-4: is of the second degree; x^+2x^-]-2x^-Sx^-Qx-3x^-{-7 is of the third degree, althouglt in form it is of the fourth degree.
356. Descending power. A polynomial is said to be arranged in descecding powers of x when the term of the highest degree in x is placed first, the term of next lower degree second, etc., and the term not containing x placed last. Thus, the polynomial x^+2x^-Sx^-\-4:X^+7-\-x, when arranged according to descending powers of x, takes ihcionn2x^+5x^-Sx^+x+7.
ADDITION AND SUBTRACTION 205
357. Ascending powers. When arranged in the order 7-\-x—Sx'^-\-5x^-\-2x'^, the polynomial is arranged according to ascending powers of x.
Find the sum of the following polynomials:
-ix''-\-2x^-Sx+Q, 7-x\ 6x3-4a;+3, -2^:3-2:^2-7
Arranging according to descending powers of .t:
+2.r3-4a:2-3.T+ 6
- x' +7
Qx^ -4x+ 3
-2a;3-2a;2 + 7
Adding Qx^-7x^-7x+2S
Check: Let x = l. Then
2x3 -4^2 -3a: +6= i
- x2 +7= 6
Qx^ -4x+3= 5
-2a;3-2a:2 +7= 3
6 -7 -7+23 = 15
358. The problems in §§352 and 357 illustrate the following law for adding polynomials:
Addition of poljniomials. Polynomials are added hy adding similar terms.
EXERCISES
Add the following polynomials and reduce the sum to the simplest form. Test by substituting values for the letters.
1. 5a+^b+9c,2a-l0b-2c
2. Sx-\-iy-\-2z, 5x-j-6y-5z
3. 2x-^5ij-\-7z,x-8y-\-Sz
4. 6a+15c-176-8d, -7a+21(i+156-12c
5. 7m — n-{-Sp, 5w — 4p, 2m+6n — Sp
6. -3a-76+14c, -lla+206-34c
7. 3m-fl6n+4p, m — 4n+6p, 2m — 2n — lOp
8. Uk-ni-\-l2m, -Sk+l2l-6m, -12k+l-27n
206 FIRST-YEAR MATHEMATICS
9. 25x-6y-\-Uz, -22x+ny-^2z, llx-lOy+Sz
10. 7a2-15c2+2362, -Qa^-\-12c''-2W
11. x'^-\-2xy-\-y^, x'^ — 2xy-\-if, x'^—^y, ixy+y^
12. 29xy-7y^-\-24:X^, -Uxy-3Qx^+3Qy^
13. Q.2x^-12.5xy-2.5y^, -4.1x2+6a:?/+ .03?/
14. h'-ixy-iy^ -x^-hy-^'^y^ h^-xy-^i/
15. 8a'-7a^-na, -6a''-\-2a+10, ^a'-\-Qa-5
16. 5x3-10a:2-2a:-12, -9x^-^x^-7x-\-U, Sx^-x-h5
17. 9x3-3x2+4a:-7, -4a;3+3:r24-2a:+8, 4a;3-2a;2+8a:-4
18. p3+3p2_|_4p_6^ _p2_2p-fi^p2_i^3p3_|_2p-|-2
19. -18a262_^i2a4-8a36, 4|a36-rt252_|_3|a4
20. -23a26+41a2c+56c26- 1562c, -6a26+26a2c+59c26-2662c, 25a2c+1962c- ISc^/;
21. 5(a+6)-7(a2+62)+8(a3+63), -4(a3+63)+5(a2+62)-5(a+6)
22. 5a3(a+6)-6a26(a2+62)+3a62(a3+63)^
23. f(a+6+c)-|(a-6+c)+|(a+6-c) + li(-a+6+c), -|(a+6+c)+|-(a-6+c)-|(a+6-c)-|(-a+6+c)
24. 5(a+6)3-7(a+6)2+4(a+6), 3(a+6)3-2(a+6)2-5(a+6)
25. 2(x+2/)2-6(x+?/) + l, -5(a;+?/)2+3(a:+?/)-6, (x+yy-(x-{-y)+2
26. 6(/r+0+7(Z-n)+/2, 5^2-8(Zr4-0-5(/-n), 3(/r+/.)-(Z-n)-4<2;
Subtraction of Monomials
359. It has been shown (§334) that subtraction of algebraic numbers may always be changed into addition by the following law :
ADDITION AND SUBTRACTION 207
To find the difference of two numbers the sign of the sub- trahend is changed and the residt added to the minuend.
Thus, -\-5x ] , 1 1 u [ +5a:
_ may be replaced by _
S A
S means subtract, A means add.
EXERCISES
Subtract the lower monomials from the upper :
1. IQx - 4:a -10s -7ia +15a:2 -15 ab -5x +17a - 2s -\-Sia -17a;2 -18fa6
2. +5m2pa: - Si(p-\-q) -7|(x-?/) +nm^{a-2b')
360. Instead of writing the subtrahend under the minu- end, it is often written on the same Hne with the minuend, connected by a minus sign, — . Thus, (-^5x) — { — 7x) is equivalent to {-\-5x)-\-{-\-7x) or -\-5x-\-7x or +12a:.
Omitting the second step the work may be written as follows :
{+5x) - {-7x) =5x+7x= +12x
EXERCISES
Reduce the following to the simplest form, doing all you can orally:
1. (+5afe)-(+12a6); {5x^y)-{-Sxhj); {-l2a¥)-{-\-Sab')
2. {-2xhj)-{i-2x^y); (-75aVZ2)-(-54aV/2)
3. (-18py3)-(63py3); (+25sW-(-75sW
4. {-S^hkH)-{-2AhkH); (+S.7pYs')-{-4:ipYs*)
5. \-5{a+b)\-\-\-7{a+b)\-\-\2(a-\-b)\
208 FIRST-YEAR MATHEMATICS
6. ■;+18f(a+6+c)5-S25|(a+6+c)(-
7. -) +4(^2 _ p) \-\-2{f -p) (■ + \ 10(^2 _j3) 5
8. j_3.4(2;2-/i2)[ + j_4.5(y2_/^2);_j2.1(2j2_/,2)j
Subtraction of Polynomials
361. When the subtrahend consists of more than one term the subtraction may be performed by subtracting; each term of the subtrahend from the minuend.
For example, when we wish to subtract 7 dollars, 4 quarters, and 10 dimes from 15 dollars, 8 quarters, and 30 dimes, we subtract 7 dollars from 15 dollars, leaving 8 dollars; 4 quarters from 8 quarters, leaving 4 quarters; and 10 dimes from 30 dimes, leaving 20 dimes.
The subtraction of algebraic polynomials is then not different from the subtraction of monomials and may again be reduced to addition.
Example :
T 9 , r 1 r 9 n^ay be replaced by , _ „ c- , i k o
S A
The result is: IQx^ - Idxy+Sy""
EXERCISES 1. Subtract the lower from the upper polynomial:
1. a2+2a6+62 3. a''b''-\-a'-b'-Sa%-^4:ab'
2. Ax''-2xy +8?y2 4. x'^-\-3x''y-\-3xy^-\- if
4a;2 -ixhj- y^ -Zxhj^^xif-Zy'^
2. From c3-2a2c-d3-r2 subtract -a^c-M^-r'^-c\ Without rewriting the following polynomial is obtained:
c3 _ 2a2c-d3_r2+a2c +3^3 4.^2 +c3 Combining similar terms, we have 20^ — 0^0 -\-2d^.
ADDITION AND SUBTRACTION 209
3. From -17x^+mi/-^4:xhj-2Qx7f-35xY subtract
+ 15y'^—4ilxhf+Sx^y — 18x'^—25xy^, without rewriting the poly- nomials.
4. From 12ab — Scd-\-12xy subtract Sab-\-2cd—llxy.
5. From Sxy — Sx-\-4.y subtract —4:xy-\-3x — 4y.
6. From lQm^—8mn^-{-'in^ subtract 7m^—4:7n^n-{-l4:7i^.
The preceding problems have illustrated the following law:
Subtraction of polynomials. Polynomials may he sub- tracted by changing mentally the sign of each term of the subtrahend and then adding the resulting polynomial to the minuend.
EXERCISES
Subtract as indicated, doing as much of the work as you can mentally:
1. {-{-17x^-Uxy-15y^)-{-mx^-\-12xy-9y^)
2. (45xV-27a:4^81?/4)-(73x4+45i/+65xy)
3. {27x^-Qx'^y-\-8tf) -i-15x^-\-8xy^-iy^)
4. (-56g'-\-27fh-Ug¥) -{-lZg%-^lSg%^-6gh^-\-25h')
5. i-da^x-{-10bxy-{-24bhj-18axy)
-(-Qbhj-\-12axy-4:a^x-2bxy)
6. ( - Qm^pq - bm^p - Um^q'^) - ( - 6m^p - lOm'^pq - 13mY)
7. (2r3+r2s+rs2+2s3)-(r3+3r2s+3rs2+s3)
8. (-\-^x^-Sxhj-\-l2y^-7xy^) - {2y^-3xt/-{-ixhj+7x^)
9. (iZ3-3|/m2+4im3-3/2m)-(f/m2+5|Z2^-2/3-3m3)
10. (5 |a6c - 7 la% - Sfft^c - 6 1 c^a)
-(ila%-5lc'a+Si-abc-\-7ib'c)
11. {3Avh^-5.7v^s^-\-9.8vh^)-{-1.7v*s^-S.2v^s^-Avh*)
12. {U'-7y')-{shY-7.Qt'+5itf'-\-¥')+i^h'f-^W')
210 FIRST-YEAR MATHEMATICS
13. {-2ik^ml+7 .5k^mH-S.24kmH)
- {SikmH - 3 . QBmH - 5 . 4khnl+Qkml)
14. (Sax^-4syz^) -{-2ax^-\-4:if-oz^-\-3syz^) - {27f-\-Sz^)
15. ( — 5mv^ — Smvu-\-4:mu^) — {-{-3mv^ — Qmuv — 4:nuv-\-9mu^)
16. Compare the signs of the terms of the subtrahend in the separate parts of exercises 1-15 before and after the parentheses are removed.
17. State a rule as to the effect of a minus sign in front of a polynomial in parenthesis.
18. State a similar rule as to the effect of a plus sign.
Removal of Parentheses
362. Parentheses. Numbers are grouped by inclosing them in a parenthesis, thus, ( ) . Other symbols used for this purpose are the brackets [ ], the brace \ \, and the vinculum . Thus, the expressions (a+6)-i-(a — 6),
[a-\-b]^la—b], \a-\-h\-i- \a—h\, and ——r all indicate
that a-\-b is to be divided by a—h. Similarly, to indi- cate that c+d is to be subtracted from a-\-h, any of the following symbols may be used: {a-\-h) — {c-\-d), la-\-h]-[c+d],\a+b\-\c+d\, or a+b-c+d.*
EXERCISES
1. Give the meaning of the following:
1. 7-\5+{S-2)\
2. -3-[5-j-4+6(]
* The Italian BombeUi in an algebra of 1572 used a symbol which amounted to a parenthesis. Vieta (1540-1603) was the first to employ the parenthesis ( ), the brackets [ ], and brace j | .
Girard (died 1632) perfected the parenthesis to the modern usage.
Descartes in 1637 introduced the vinculum , in the way we
use it.
ADDITION AND SUBTRACTION 211
3. 2{a-\-b)-A\a-2a-Sb\ 2. Perform the following operations and simplify results: 1. 4-S5-(^2_4)f
3. if-\t^-St^+W^^\
4. 2/-[6/-3^-4/-(2sr-4/)]
5. 9x-{5y-iQy+7z)-{7y-^z)i;
6. -3a4+[4a3- {Sa^-5a^) - (4a2+3a)]
7. 16e2-j-42e2-(3e2+2)J-(50e2+3)
8. 7.5p3-[3.4p3_4.2p3+i.6^2_3 4p2_4 5p]
9. 3A;3-j2p2-|-A:3-^2:hrS-b2-r-(/b3+r)] 10. 105s2-l4sH^2_j_j45^_2f2_(552_2s^)J
11. -[3rZ-(3r2-3Z2)]-j4r2-5r/-2^2j_^(2r2+2;2) Summary
363. The meaning of the following terms was taught in this chapter: degree of a number, of a monomial, of a polynomial; ascending and descending powers; brackets, brace, vinculum.
The chief uses of the vinculum in modern algebra are :
1. The dividing-line of a fraction is a vinculum over the denominator.
2. The bar of a radical sign is a vinculum over the number underneath it.
364. The following laws have been extended to apply to positive and negative numbers:
1. Commutative law. The value of a sum remains unchanged by changing the order of the addends.
2. Associative law. In adding several numbers the sum is the same in whatever way two or more of the numbers are combined into a sum before adding in the rest.
212 FIRST-YEAR MATHEMATICS
3. To add similar monomials prefix to the common factor the sum of the coefficients.
4. To add polynomials add the similar terms.
5. To find the difference of two numbers, change the sign of the subtrahend and add.
6. Polynomials may be subtracted by changing the sign of each term of the subtrahend and then adding the resulting polynomial to the minuend.
7. A parenthesis preceded by a -\- sign may be removed without making any other changes.
8. A parenthesis preceded by a minus sign may be removed if the sign of every term within the parenthesis is changed.
LEONARDO OF PISA
LEONARDO OF PISA
LEONARDO FIBONACCI, the greatest mathe- matician of Europe of the period from the fourth to the sixteenth century, was born at Pisa, Italy, about 1175 A.D., and died about 1250 a.d. While still a boy he went with his father to live at Bugia on the coast of Algiers, at which place he was educated. Here he learned the Arabic system of notation and numeration and became acquainted with the algebra of the great Arabian scholar, Alkarismi. He traveled later in Egypt, Syria, and other Mediterranean coun- tries, and acquired all the mathematical learning that he could. Returning to Pisa about 1200, he began writing a book on arithmetic and algebra, which he published in 1202 under the title Algebra et almuchabala, which he took from Alkarismi's great work. This book is commonly referred to as the Liber abaci (Book of the Abacus).
In Liber abaci Leonardo explains the Arabic system of numeration and emphasizes its advantages over the Roman system then in vogue. He gave an account of Arabic algebra and pointed out the convenience of using geometry to get demonstrations of algebraic formulas, just as geometry is used in chapter XV of this text. He also solved simple equations, showed how to solve a few quadratics, and illustrated his rules by prob- lems on numbers, much as is done here. The Liber abaci contained much material of historical value about Arabian mathematics, which was then more highly developed than European mathematics and was also unknown in Europe. This book was the source and authority for arithmetics and algebras in Europe for centuries.
Leonardo wrote a book of Geometrical Practise, another on Squares, a tract on determinate algebraic problems, and a few other minor works, none of which ever acquired the celebrity of his Liber abaci. For fuller details about Leonardo's work, see Ball's History of Mathematics, pp. 167-70.
CHAPTER XIV
MULTIPLICATION AND DIVISION
Multiplication of Monomials
365. In finding the product of two or more monomials it is assumed that the commutative law, § 152, holds for positive and negative numbers.
Find the product : {2a%) ( - Sab^) ( - 4¥)
The sign is determined by the sign law, § 336, and is found to be+. Why?
The factors are then rearranged as follows : — 2 • 3 • 4a2 • a • b • b^ • b^, i.e., the product of the arithmetical factors is formed first and then the product of the literal factors.
Multiplying, we have
(2a%) ( -3a62) {-A¥) = -\-2Aa%<'
This problem illustrates the following method of multi- plying monomials:
Multiplication of monomials. To find the product of two or more monomials:
1. Determine the sign of the product according to the law of signs in multiplication.
2. Find the product of the arithmetical factors.
3. Multiply this product hij the product of the literal factors.
Find the following product and test the result for a = 1, 6= 1,
{2ay){-^a'^hx){-Za¥xhf) = +2'^'M'a^'a'h'¥'X'xhj'y'^ = +2Aa^¥xHf Test: (6) (-8) (-216) = 10,368 24- 1.1. 16 -27 = 10,368 213
214 FIRST-YEAR MATHEMATICS
EXERCISES
1. Simplify the following products, doing as much of the work as you can mentally :
1. (+17)(+25)(-8)
2. {2mt^)i-SmH){-5m^)
3. (6a26c2) (4a262c) {3a%^c^) (5ab^c'^)
4. ( + 8 . 2a2m6) ( - 3 . 5amn%^) (4 . Smhf-b)
5. {-3xpqh)(-2lxp^qr)(-^zpqr^)
7. (-2ix2?/)(-5|xV2)(+16a:3|/24)
8. (- 5ip'g3) (+5igV3) ( - l\r^p^)
9. (4i< VW^) ( - 21^2^5^12) ( _ -Jg ^10^7^8)
10. (ia-'x) ( - lOaby) ( - 6a7/) ( ia:z/) (lab)
11. (iW(i«^6a;2^)(-V«^'?/)
12. (-2)2; (-2)3; (-2)^; (-2)^
13. (-3)1; (-3)2; (-3)=^; (-3)^
14. (-2)2(-3)2(-4)(2)3
15. (-a)2(-o2)(-a3)(-a)3
' 16. (-a2)(+a4)(_^3)(+«)2 .•
17. (-a)2(-a)4(-a)3(+a4)
18. (a5)2(-a3)4(_a2)5(c^3)
19. (3p2)2(-p3)5(_5p4)2
20. (2ix3)2(- 1x^)2(3x7)2
21. (a-j-by{a-{-b)'(a-\-by '
22. 3(x+?/)5(x-7/)^-4(rc+7/)3(.T-?/)«
2. Find the value of the following:
1. (-3)(-2)+(4)(-5)-(-3)(2)-(-2)3
2. a:2-6.T+9 for.T = 2; forx=-3
3. x*-3x3+2x2-f 4.T+6 for x= -2
MULTIPLICATION AND DIVISION 215
3. Find two factors of each of the following products:
SQxhjh; -72a^b^', 51??V'; -l^a^bx^
4. Solve the following equations :
x_ J_ x^_ ^ x_
6~~12 ' ~a2~~" ' a~~^
Multiplication of Polynomials by Monomials
366. Distributive Law of Multiplication. It was found in chapter V, § 158, that a polynomial may be multiplied by a monomial as follows:
To multiply a polynomial by a monomial multiply every term of the polynomial by the monomial and add the resulting products.
This is called the Distributive Law of Multiplication.
EXERCISES
Multiply as indicated and test by substituting values for the letters:
1. 5x(x^-Sxy-\-52f)
x^—'Sxy-\-5y^ 5x
Product: 5x^ - 1 6x^y -\-25xy^
Check: Let x = l, y = 2.
Then 5x(x2-3x?/+52/2) =5(1 -6+20) =5 • 15 = 75
and 5^3 -15x22/+25a:?/2= 5-30 + 100 = 75
2. 6|a6(3a2-6a6+1262)
3. 3a(4a-56)+56(6a-36)
4. - 5 . 7x\{Ubx - It Qaby - 2t9xyz+T\byz)
5. {3a%^+Aa¥-\-2¥)2ab^
6. (x3?/4-4a;^5_^6xV-9a;5i/4)3.5a^V
7. 2(3a-2b)-5{a+Sb)-\-2(Sa-b)
216
FIRST-YEAR MATHEMATICS
8. 2a+6(a-26) + (4a-6)3+fe
9. {x^-5x-\-Q)2x-{x+l)Sx^-x{x^-7x-\-i)
10. 5a2(2ia2-44a6-3.562)-462(3ia2+2a6-2i62)
11. 4:x''{3x-27j)-2xy(Sx-2y)-\-iSx-2y)6if
12. 2(3a2-4ja6+762)-(4a2+5f 06-8^62)
13. x^-[Sx{x^-2)-2x^{x-\-l)]
14. 4a[(l-a)2a2-f-(3a+l)3a2]
15. 5aJ4a-2(3a-46)+5(4a-36)f .16. 2aJ5(4a-76-3c)-6(5a+46-8c)S
17. -4a;[2x2+3a;i4(a;-l)-5(x-2)}]
18. 52/3-[32/3-2i/H2/(2/+3) -52/(22/4-6) n
Multiplication of Polynomials by Polynomials
367. When the dimensions of a rectangle are poly- nomials the area can be found by separating the rectangle into other rectangles whose dimensions are simpler numbers.
EXERCISES
1. Using the heavy dividing line, express the area of the rectangle of Fig. 244 as the sum of two rectangles. Express the area also as the sum of four rectangles.
c |
d |
|
a |
||
b |
(• d
a b
Fig. 244
Fig. 245
2. Express the area of the rectangle of Fig. 245, (1) as the sum of two rectangles; (2) as the sum of four rectangles.
MULTIPLICATION AND DIVISION 217
3. Write the areas of the rectangles whose dimensions are given below as the sum of two or more rectangles. Simphfy the polynomials thus obtained.
Length |
Width |
|
1. |
3+4 |
6+2 |
2. |
a+6 |
a-2 |
3. |
a2+62+2a6 |
a-\-b |
4. |
x'+2xy+7f |
x-y |
5. |
a-b |
a^-2ab+b^ |
6. |
5-\-x |
25+10a;+r |
The rectangles of exercises 1, 2, and 3 are first expressed as the sum of two rectangles. Thus, for the rectangle in exercise 1, (a+6)(c+rf) =a{c-[-d) +6(c+d). These separate products are called partial products of the given polynomials a+6 and c-{-d. To obtain the final product, the partial products are simplified.
This gives (a+6)(c+d) =ac+ad+bc-\-bd.
This illustrates the following method of multiplying two polynomials:
Multiplication of polynomials. Multiply every term of one polynomial by each term of the other and add the partial products.
EXERCISES
Multiply as indicated:
1. (a2-fa6+62)(a+6)
The work may be arranged as follows:
g+b
First partial product : a^-\-a%+ab^ =a (a^ + a6 + b'^)
Second " " a%-{-ab^-\-¥ =b(a^+ab+b^)
Hence, a^+2a^b+2a¥+b^ ={a+b)(a^-\-ab+b-')
2. (ax -\- by) (ax— by)
3. (a^+ab^+¥)(a-hb)
4. (p^^2p+l)(p-S)
218 FIRST-YEAR MATHEMATICS
5. {2ia^-Si¥)il2a^-Q¥)
6. (:ipt-hs)(Tpt-its)
7. (a+6+c+rf)2; {a+b-2c)^
8. (-2a-36+c)2; {a^-\-b''-^c')^
9. (2a2+fe2+3c2)(2a2+62-3c2)
10. i.5x- Ay- .Sz){l0x-20y-\-^0z)
11. (32/2-45a;?/+6a;2)(3t/2-5a:2)
12. (ax-\-by—cz)(ax+by-\-cz)
13. {ax-\-by-^czY', {ax — by — czY
14. (W-is^-3M(6m-12sH-180
15. (-1.4Z2-2.5m/+.9m2)(. 2^2-^7^-. Im2)
16. (a2+a6+62)(^-6)(a+6)
17. (4x2+2a;?/+2/2)(2x+?/)(2a:-?/)
18. (a+6)3; (a-&)3; {2x-yY', {x+2yY
19. (4.6a&c+1.2a62)(4a2c-5a26)
20. (p24-pr-r2)(p2-}-pr+r2)
21. (9A:2-3)b^+i2)(9A:2+5A;^-^2)
22. (9a;2-6a;2/+4|/2)(9a;2+6a:?/+?/)
23. (5a+26)3-(5a-26)3
24. 4jmw(4m2-6mn+97i2)-3mn2(13in-9m)
25. (27;2 + g2)2_(2p2_^2)2_4p2(p_2^)2
26. (2/-3/i)2- (2/+3/^)2+(2/-3/i)(2/+3/i)
27. (.5a;-.6?/)2--(.5x+.6?/)2-(.5x+.6|/)(.5x-.67/)
Multiplication of Arithmetical Numbers
368. Multiplication of arithmetical numbers is a special case of multiplication of polynomials.
For, in the polynomial a • lO^+b • lO^+c • 10+rf let a = 2, 6=5, c = l, d = 3.
Then the polynomial takes the form 2 • lO^+S • 10^ + 1 • 10+3 =2,000+500+10+3=2,513.
MULTIPLICATION AND DIVISION 219
EXERCISES
1. Multiply 482 by 347.
The number 482, written as a polynomial, takes the form 400+80+2, or 4 • 10^+8 • 10+2. The number 347 takes the form 3 . 102+4 • 10+7. Thus,
(482)(347) = (4 • 10^+8 • 10+2)(3 •.102+4-10+7)
Multiplying, (4 • 10^+8 • 10 +2) (3 • 10^+4 • 10+7)
12. 10^+24. 103+ 6- 102
16 . 10^+32 . 102+ 8 . 10
+28 . 102+56 .10+14
12 • 10* +40 . 103+66 . 10^+64 .10 + 14 This is equal to
120,000+40,000+6,600+640+14, or 167,254
2. Write 32,569 as a polynomial arranged in descending powers of 10.
3. Using the method of exercise 1, find the product 3,462 by 3.
4. Using the method of exercise 1, multiply 287 by 453.
369. Decimal system. Because the arithmetical num- bers are expressible as polynomials in 10, they have been called decimal numbers.
Thus, a decimal number is obtained by substituting 0^ = 10 in such a polynomial as 2x^+5x^+7x+8.
370. Other systems of numbers. Substitution of other values for x gives a system of numbers different from the one with which we are familiar; e.g., a: = 12 gives the number 2- 123+5- 122+7- 12+8. Such a system of numbers would even have certain advantages over the decimal system.* It is called the duodecimal system. We know that the Hindoos used 5 as a base for their sys- tem of numerals.
* See Cajori, History of Elementary Mathematics, p. 2.
220 FIRST-YEAR MATHEMATICS
Division of Monomials
371. In § 367 the area of a rectangle was found when both dimensions are known. It will now be shown how to find one of the dimensions if the area and the other dimension are given.
SQ. ft.
EXERCISES
1. If the area of a rectangle is 144 sq. ft. and the base 9 ft., Fig. 246, what is the altitude?
2. Find the altitude of a rectangle of area 144 sq. ft., if the base is 16 ft.; 12 ft.; 8 ft.;
9'
Fig. 246 72 feet.
372. Division. To divide 8 by 2 means to find a number which multiplied by 2 gives 8. (See § 343.) In general, to divide one number by another is to find a third number which, multiplied by the second number, gives the first.
EXERCISES
1. Show that a number divided by itself gives 1.
Since n-l=n, it follows by §372 that - = 1.
n
2. Show that a number divided by 1 gives the number.
3. Show that 0 divided by a number, not 0, gives 0.
4. Show that a product of two numbers divided by one of the factors gives the other factor.
5. The following sometimes purports to be a proof demon- strating that 4 = 7. Find the fallacy:
Two numbers a and b are given equal, a = b.
Then a-b = 0. Hence, 4(a-6) = 0 and 7{a-b) = 0.
It follows that 4(a-6) = 7(a-6). Why?
Dividing both sides of the equation by a—b gives 4 = 7.
MULTIPLICATION AND DIVISION 221
6. Divide 25a% by 5a.
First method: Since bah • 5a=25a'-^6, 25a-6-T-5a = 5a6 (by § 372).
Second method:
Since 250^6 = 5a . 5a6, — z — = — ^- — =5a& (by exercise 4)
5a P(^
7. Find the following quotients:
a*^
a"' a^' a2 '
8. Divide as indicated:
-25a%^^5a%^ {x-^yy-i-{x-hyy
SOa'¥^-Qa¥ i-ayb^{-cy-h{-ayb{-c)
Reduction of Quotients
373. Reduction of quotients. In algebra, as in arith- metic, the quotient is not altered if dividend and divisor are both divided by the same factor. Dividing dividend and divisor by the highest common factor is to reduce the quotient to the simplest form.
EXERCISES
Reduce to the simplest form: 1. QSa%^c'
The sign of the quotient is — . Why ? The numerical factors can be divided by 7.
The literal factors a' and a^ are divisible by a^; ¥, and ¥ are divisible by ¥; and c* and c^ by c^
The work of reducing the quotient may be arranged as follows:
9 . a • 1 . 1
6Sa^¥(^ ^^ / / / ^ 9q ^ 9^
-1 • 1 • 6 • c2
222
2 2M
xz
FIRST-YEAR MATHEMATICS
3.
4.
6.
56a^ a*
— Qxh/z* Sxifz^
2xy -25a'¥c'
- 1 . 69xYz^w^a^ -l.SaHv^xhfz*
10.
11.
12.
13.
14.
15.
16.
17.
i2{-ayb'{-cy
S(-ay¥{-cy
14(a+6)3 ' -7(a+6)2
-94(a;2-i/2)5 -2(x2-?/2)3
-3.43(a'^+2a6+&'^)' 49(a2+2a6+&2)4
2UxY(a''-\-h''y
— 4:0'm'hi^w^z
— blx^z'^
1 .lax^z^
Monomial Factors
374. In § 367 we have learned to find the product when the factors are given. We will now see how the factors may be found when the product is given.
7 10-7 |
2 L0.2 |
5 10-5 |
3 10-3 |
Fig. 247
EXERCISES
10 10-7 ho. 2 10-5 10-3 1. Show that the total area
of four adjacent flower-beds, Fig. 247, may be expressed in either of the following ways:
10 . 7+10 • 2+10 • 5+10 • 3 or 10(7+2+5+3)
Hence, 10 • 7+10 • 2+10 • 5+10 • 3 = 10(7+2+5+3)
Which way is the more advantageous? Give reasons for your answer.
MULTIPLICATION AND DIVISION 223
2. Represent in two different ways the combined area of four rectangles of length a and of bases a, Sab, 6,, and c, respectively. Express the equality of the two representations by an equation.
3. The total area of three adjacent lots is x'^-j-Sxhj -{-4:X^y^ sq. m., each term representing the area of a lot. The lots all have the same length. What may their dimensions be ?
4. Sketch a rectangle whose area is
4x2 — Sxy 3m^ — 1 2m^n + 6mn*
5x^- lOx^y-h 15x?/ 15x4- 10x^-\-5x^
Ua%^c^-21a%^c^-\-S5a%^c^ 10x^-^x^-\-6x+2
375. Exercises 1-4, § 374, illustrate the following method of finding the monomial factor of a polynomial:
Determine by inspection the highest monomial factor con- tained in each term of the polynomial. This is one factor of the polynomial.
Divide the polynomial by this factor. The result is the other factor of the polynomial.
EXERCISES
1. F'dctor i\ie poiynomiaX Hxhfz — 7 xhfz^-\-28xy^z^.
By inspection the highest common factor is found to be 7xy''z. Dividing each term of the polynomial by 7xy^z, the quotient is
2x—x'^yz-\-4:Z.
Hence, Ux^y^z - 7xYz^+2Sxy^z'' = 7xy^zi2x ^x^yz +4^)
2. Factor the polynomials in exercise 4, § 374.
3. Factor the following:
5a— 106 ax+ay — az
17x2-289x3 4a2x3- I2a^x^-20a'^x^
16x2 - 2a6x 5mx+ 10m2x2 - 40m3x3
224 FIRST-YEAR MATHEMATICS
Reduction of Quotients
EXERCISES
376. Reduce the following quotients to lowest terms 27a'b'>c^- 18a^6V-54a667c8
The factors of the numerator are 9a*b^c^ and iS—2ahc—Qa%'^c^). The fraction may now be written :
9a^6^c6 (3 - 2abc - 6a^6 V)
Dividing dividend and divisor by 9a^¥c^, gives _3j-2a6c-6aW
39a6+9a=^ _ 20a"b-lm''b^+^0a¥
2. n t> O.
Za? "• bob
36a^a;+8ax^ l{)xHj—\bx^y'^-\-bxhj
^' -12aV ^' -bxy
36x^2/^-42x^^2 16a=^x'^?/-44a2x%
-12x3?/5 '^^ 4a2x2
^ Zbab^c^ - ^2a%'c' - 49a864c34- 2 la^fo-^c^
o.
9.
7ab^c^ 12aw2n3-166m3n2+40a6m2w2-28a2m2n2
— 4m?i
- Ibp^y- 12p2V^+9a^P^V^-27a6cpV" ^^* 3pV
4a6c(3a6c - 5a^6V - 7a&^c+ 6a6^c) -2a26c2
2x(4?/2-6xV) -81/(5x2^- Ux^)
4x2/2 ISa^fe^c^- 9a^fe'^c'^ - i5a%'c'+21a%'c^-Sa*b'c^
2labdpqs — S5abcpqt — A2acdpts ^^' -lav
.. 15(x+i/)^-25(x+y)«-35(x+y)« -5(x+?y)^
MULTIPLICATION AND DIVISION 225
Division of Polynomials
377. In the division of polynomials, all the preceding operations find application. It is therefore a subject by means of which addition, subtraction, multiplication, and division by binomials are reviewed.
Divide 86,932 by 412.
The process in full is as follows:
86,932 1412
824 I Quotient =211
453
412
412
412
Arranging the numbers 86,932 and 412 in the form of polynomials, they may be written 80,000+6,000+900+30+2 and 400 + 10+2.
The process of dividing 86,932 by 412 may now be arranged as follows :
80,000+6,000+900+30+2 1400 + 10+2 . 80,000+2,000+400 [Quotient =200+10 + 1
4,000+500+30 4,000+100+20
400+10+2 400 + 10+2
Writing the polynomials arranged according to descending powers of 10, the division takes the form:
8-10^+6.103+9.102+3.10+2 14.102 + 1.10+2 8.10^+2-103+4.102 I Quotient = 2. 102 + 1. 10 + 1
4.103+5.102+3-10 4-103 + 1.102+2.10
4.102+1.10+2 4.102 + 1.10+2
226 FIRST-YEAR MATHEMATICS
378. Inspection of the process of dividing one poly- nomial by another brings out the following facts:
1. The first term of the dividend divided by the first term of the divisor gives the first term of the quotient.
2. The divisor is multiplied by the first term of the quotient and the product subtracted from the dividend.
3. The first term of the remainder divided by the first term of the divisor gives the second term of the quotient.
4. The divisor is multiplied by the second term of the quotient and the product subtracted from the remainder.
5. To get the other terms of the quotient proceed in the same way as for the first and second terms.
6. Throughout the process all polynomials are to be ar- ranged according to powers of the same letter.
EXERCISES
1. Divide 8x'+Qx^i-9x''-\-Sx-^2 by ^x''-\-x-\-2.
8a^+6x3+9a;2+3xH-2 8x* +40^3+4x2
Quotient ^2x'^4-a: + l
2x3+5x2+3x
4x2+2x4-2 4x2+2x+2
2. Divide a^-\-Sa%-\-3a¥-\-b^ by a+6.
a^+ a% I Quotient = a^ -\-2ab +^2
2a%+Sa¥ 2a26+2a62
a¥+b' 062+53
MULTIPLICATION AND DIVISION 227
3. Divide x^-\-7j^ by x-\-y.
x^-\-y^ \x-\-y
X? -\-xHj I Quotient =x'^—xy-{-y^
-x^y+y^
+xy^-\-y^ +xy^-\-y^
379. Checking division. Whenever the division is exact, there is no final remainder. The equation: quotient times divisor equals dividend, or, in symbols, QXd = D, may then be used as a check.
EXERCISES
Divide and test by multiplying:
1. (a3-a2-4a+4) by {a^-3a+2)
2. (a3-5a2+10a-12) by (a-3)
3. (l+5xH-6a;2) by (H-2a;)
4. (9<2+24s«-|-16s2) by (3«+4s)
5. (1 . 2^2+ .5xy-2. Sy^) by (^x+Jy)
6. {Qk'-31kH+4:7kt^-4:2t^) by (2k-7t)
7. (l0x^-5xy^-5^^-^'^y) by ix-2y)
8. (27a3-54a26+36a62-863) by (3a-26)
9. (27a3-54a26+36a62-863) by (9a2-12a6+462)
10. (9««-12/3s3+4s«)by (3i3_2s3)
11. (4a:6+12xV+9?/«) by (2a;3H-32/3)
12. {Sy'-\-l5x^y-9xy^-25x^+10xY) by (5a:2-3?/)
13. (7a;5+10a:4-26x3+17a:2-lla;+3) by (7a;3- 4^2 +3^-1)
14. (15+8a-32a2-|-32a-^-15a4) by (3+4a-5a2)
228 FIRST-YEAR MATHEMATICS
15. (64w«+14.4wV4-1.08?<V+.027tJ«) by {4:u'+ .'Sv)
16. (x'-if) by (x-y); {27f'-}-Ms') by (3^-4s)
17. (^c^-l) by (a;-l); (a;«-l) by (x-1)
18. {x^-y^)hy (x-y); (x^-y^) by (x-y)
19. (64a«-66) by (8a3-63); (16a:4-25r4) by (4.T2-5r2)
20. (8x3-2/3) by ('^x'^+2xy-\-y^)
21. (.008s3^3_^t'3) by ( 2s«+?')
22. (64a6-66) by (16a4+4a262+64)
23. (.008s3^3_}_^3) by {.04:sH''-.2stv+v^)
24. (27m6-8n6) by {9m'^+Qm^ri'-\-4n')
25. (27w64-8n«) by (3m2+2n2)
26. (a3+a2&+a62_j_ttc2+6c2+63) by (a+6)
27. (a3+a26+a62+ac2+6c2+63) by (a2+624-c2)
Summary
380. The chapter has taught the meaning of the following terms: partial products; decimal system of numbers; factoring; division.
381. The chapter has taught the processes for the following operations:
Multiplication of monomials, of polynomials by monomials, of polynomials by polynomials.
Division of monomials; reduction of quotients to simplest form ; division of polynomials.
382. Distributive law of multiplication. Polynomials are multiplied by monomials according to the following law:
Multiply every term of the polynomial by the monomial and add the resulting products.
MULTIPLICATION AND DIVISION 229
383. It was seen:
That a product is divided by a number if one of the factors is divided by that number.
That a number divided by itself gives 1.
That a number divided by 1 gives the number.
That 0 divided by a number, not 0, gives 0.
384. Arithmetical numbers may be arranged in the form of polynomials according to powers of 10.
385. Monomial factors of a polynomial are found by inspection. The other factor is then found by dividing the polynomial by the monomial factor.
386. The process of dividing one polynomial by another is essentially the same as the process of dividing arith- metical numbers.
CHAPTER XV
SPECIAL PRODUCTS. FACTORING. EQUATIONS
QUADRATIC
The Square of a Binomial
387. It is of advantage to be able to carry out rapidly certain multiplications that occur frequently in algebra.
388. Quadratic trinomial square. Let A BCD, Fig. 248, represent a square whose side is 25 ft. long. Show that the area of A BCD is equal to the sum of the areas of the four parts, 202, 20 -5, 20 . 5, and S^. This may be expressed by the equation (20+5)2 = 202+2 • 20 . 5+52.
Similarly show that the
) 20 |
5 |
c |
20-5 |
5' |
5 |
20' |
20-5 |
20 |
Fig. 248 (20+5)2 = 202+2.20-5+52
area of the square in Fig. 249 gives
the equation
ia-\-by = a'-{-2ab+b\ The trinomial a'^-\-2ah-{-¥ is of
the second degree and is called a
quadratic trinomial. Since this trinomial represents the area of a
E |
^-b- |
a-b ^C |
|
a |
D A |
{a-h)' |
B |
b' |
a |
b |
ab |
h' |
a'^ |
ab |
Fig. 249
(a+6)2=rt2+2a?>+/>2
K G
Fig. 250
(o-?))2 = a2-2a/)+fe2
square, it is called a quadratic H trinomial square.
389. Let ABCD, Fig. 250, represent a square (a — h) ft. long. 230
SPECIAL PRODUCTS 231
Show that the area of A BCD equals
EFGC+GHIB-FKDE-KHIA
Therefore {a-by = a^-\-h^-ah-ab. Why? This may be
written {a — by = a^ — 2ab-\-P.
The trinomial a^ — 2ab+b^ is also a quadratic trinomial
square.
Express by means of a drawing the following squares of binomials as quadratic trinomial squares:
(p+qy; (p-qY; i^+yY; i^-yY; (m+ny-, {m-ny 390. Squaring binomials. The square of binomials
can be obtained without a figure, by multiplication. Express by multiplication the following squares in the form
of trinomials :
1. {m-xY 4. {v+kY (m — x)2 = (m — x) (m —x) = m^ —2mx-\-x^
2. {a^hY 5. {x-yY
3. {v-kY 6. {m+xY Notice that the squares of these binomials may be
found by inspection from the terms of the binomials by squaring the first term, adding the square of the second term, and then adding or subtracting 2 times the product of the two terms, according as the sign of the second term is + or — .
EXERCISES Find by inspection the trinomials equal to the following squares :
1. {x^vY 9. {k-rY 17. (5a+2fe)2
2. {v-^-aY 10. {r+yY 18. {Ixy+zY
3. {x-vY 11. {k+nY 19. (3a-46)2
4. {x^-bY 12. {r-yY 20. [(a+6)+3p
5. {y-7Y 13. {h-vY 21. {{a-b)-^zf
6. (4a- 1)2 14. {g-fY' 22. [{k+b)-sY
7. ih^hY 15. {^px-^yY 23. [{a-b)-2cf
8. (.4a-. 30' 16. {.%xyz+lY 24. [{ax+y)-zY
232 FIRST-YEAR MATHEMATICS
391. Squaring arithmetical numbers. Squares of arith- metical numbers are easily found by writing the numbers in the form of binomials.
EXERCISES Square the following numbers :
1. 53
532 = (50+3)2 = 502+2.50.3+32 = 2,500+300+9 = 2,809
2. 68
682 = (70-2)2 = 4,900-280+4 = 4,624
3. 13 6. 21 9. 43 12. 91
4. 14 7. 31 10. 72 13. 89
5. 15 8. 38 11. 81 14. 103
Factoring Trinomial Squares
392. It was found in § 390 that the square of a binomial is a trinomial consisting of the square of the first term of the binomial, plus or minus (according as the binomial is a sum or a difference) twice the product of the first and second terms of the binomial, plus the square of the second term. It follows that a trinomial, in which two terms are squares (and positive) and the other term is plus or minus twice the product of the square roots of those two terms, is the square of the sum or difference of those two square roots according as the remaining term is plus or minus.
This enables us to find by inspection the two equal factors of a trinomial square.
EXERCISES
Factor the following trinomials by inspection:
1. k^-{-6kl-\-9l^ fc2 and 9^2 are the squares of k and 31 respectively, and Qkl is twice the product of their square roots.
Hence, k^+m+9l^ = {k-\-3l) {k+Sl) = {k-\-'Sl)\
FACTORING
233
2. x^+2xy-\-if
3. a2-2a&+62
4. m?-\-'iinn-\-^n^
5. 62-66+9
6. A;2+8A;+16
7. a2+10a6+256'-^
8. 49-140n2+100n*
9. a''h''c^+d>abc-\-l^
10. 16x2+24x^+91/2
11. 25A;2+70A:s+49s2
12. 16p2+72pg+81g2
13. 121a2-242a6+12162
14. 81x2-126x?/+49?/2
15. 169r2-260rs+100s2
16. 256m2+96mn+9n2
17. 49x6- 154x3?/+ 121?/
18. 625m4+50?^2y2_^^4
19. 169r2-78r^2_|_9^4
20. 25x2?/2+40x?/2+16s2
21. 36r2s2-84rsfw;+49^2^2
22. 9a262+48a6mn+64w2n2
23. 49a2A;2+42aA;x?/+9x2?/2
24. 25p2g2r2_60pgfs+36s2
25. 81a2+180a6cd+10062c2d2
26. 64m2n2+320mnp+400p2
27. 225A;2-120A;Z+16?2
28. 100x2+280x?/+196?/2
29. 64x2?/2+80x?/2+2522
30. 16x2-56x?/2;+49?/222
31. 4p2_36pgr+81g2r2
Product of the Sum of Two Numbers by Their Difference
393. The product of the sum of two numbers by their difference can be found from a rectangle whose dimensions are the sum of two segments and the difference of the same two segments.
Let ABCD, Fig. 251, represent a rectangle having the dimensions (a+6)(a-6).
ABCD = AEFD+EBCF. Why? = AEFD+IFGK. Why? = AEGH-DIKH-IFGK+IFGK. Why? = AEGH-DIKH.
Hence, (a +6) (a - 6) = a2 - 62
t'' |
6 K a-b 1 |
-f |
1 6 Id |
b" [ {a-b)b 1 |
\f |
\ \ |
/ |
\a-b)b |
L |
||
A |
a |
E b |
Fig. 251 |
||
• ( |
a+b)(a-b) = |
= a2-62 |
234 FIRST-YEAR MATHEMATICS
From a drawing express the following products as the differ- ence of two squares:
{a-^x){a-x); (m-\-n){m-n); (?/-f-4)(?/-4)
394. The product of the sum of two numbers by their difference can be readily found by multiplying.
EXERCISES
Express as the difference of two squares by multiplying:
1. {a-\-x){a—x)
{a-\-x){a~-x) = a^-{-ax—ax—x^ = a^—x^
2. (r4-4)(r-4) 4. {p+s){p-s)
3. {t-\-q){t-q) 5. {y-{-b)iy-b)
The results obtained in these problems suggest the following way of obtaining the product of the sum of two numbers by their difference by inspection: Square each number and subtract the second square from the first.
EXERCISES
Express by inspection the following products as the difference of two squares:
1. (2a+36)(2a-36) 6. (4a6c-3)(4a6c+3)
2. {SX-^Z){SX + 4Z) 7. (l+5p2g2)(l_5p2^2)
3. {5p+Ss)(5p-Ss) 8. {l-a){l-\-a)(l-\-a')
4. ihr-^hXir-h) 9. {u-v)(u+v){u^+v')
5. (ixy-ixz){lx'ij-\-Uz) 10. (10a-h76c)(10«-76c)
Factoring the Difference of Two Squares
395. The equation {a-\-b){a—b)=a^ — b^ suggests a method of obtaining the factors of a^ — b^. In words, the factors of the difference of two squares are the sum of their square roots and the difference of their square roots.
FACTORING 235
EXERCISES
Factor the following binomials and test the results by multi- plication:
1. 256a:2-2/2
The square root of 25Qx^ is 16a: and the square root of y^ is y. Hence, 256.t2-?/ = (16x-2/)(16x+?/)
2. |
x'-y' |
13. |
r^-s^ |
3. |
16A:2-2562 |
14. |
81m4-16nV |
4. |
49a2-962 |
15. |
X^-2/ |
5. |
m2/i2-144r2 |
16. |
256a4-625c4d8 |
6. |
2892/2-811)2 |
17. |
m'^—n^ |
7. |
16-257/ |
18. |
0^10 _ 510 |
8. |
169^2/^2^2 _ 225^4 |
19. |
64x6-9 |
9. |
9m4-121n2^4 |
20. |
(r+3s)2-16^2 |
10. |
49a2- 1006 V |
21. |
(2a+62)2-9c4 |
11. |
196-361a4A;«:r2 |
22. |
idx^-Sy^y-lQz' |
12. |
225b'^-fYh'' |
23. |
{x'-yy-x^ |
396. Factoring the difference of two square numbers enables us to find the value of the difference of two squares of arithmetical numbers by inspection.
Find the value of the following:
1. 812- 192 4, 1262-262
2. 1372-372 5. 1,0172-172
3. 1372-632 6. 5112-4892
397. To find the square of a trinomial. Draw a square whose side is a-{-b-\-c and show that
ia+b-\-cy = a'+P+c'+2ab+2bc+2ac
How may a trinomial be squared by inspection ? Find by inspection: (a+26+3c)2; (x-Sy+z)^
236 FIRST-YEAR MATHEMATICS
The Product of Two Binomials of the Form
{ax+h){cx-\-d)
398. A quick way of multiplying two binomials of the form ax-\-h and cx-\-d can be seen from a study of the result obtained by multiplication.
Multiply {ax-\-h){cx+d)
ax-\rh cx-\-d acx^-\-hcx -\-adx-\-hd
acx^ + {hc-\-ad)x-\-hd
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 two cross- products.
EXERCISES
Multiply by inspection, doing all you can mentally:
1. (3x+5)(2a;+3)
Writing the second binomial under the first.
3a: +5
2x+3, we see that the product of the first terms is ^x-, the product of the last terms is 15, the sum of the cross-products is ^x + lQx = \^x.
Hence, (3x+5)(2x+3) =6a:2 + 19a:+15
2. (5a:+4)(3a;+2) 5. (2a:-7)(4:c+3)
3. (3x-5)(2x+3) 6. (3a: -5) (2a: -3)
4. (5a;+4)(3a:-2) 7. (5a: -4) (3a: -2)
Factoring Trinomials of the Form ax^-\-hx-\-c
399. The trinomials found in exercises 1-7, § 398, are all of the same form: ax'^-\-hx-\-c. The method of
FACTORING
237
factoring a trinomial of this form can be seen from the following illustration: Factor 3^2+ 17a: + 10. The various possibilities are
+3x^+10
+ x^+ 1
+3a: + 1 + a:^+10
+ x^+2
+3rc +2 + 0:^+5
The last case being the one with the sum of the cross-products equal to 17a; gives the required factors.
Hence, 3a:2+17a;+10 = (3a;+2)(a:+5)
EXERCISES
In the same way, factor the following trinomials:
1. 2a;2+llx+12
2. 8c2+46c-12
3. 3a;2-17j+10
4. 832-31^+21
5. 5x2-38ar+21
6. Ila2-23a6+2fe2
7. 7A:2+123A:-54
8. 12<2+31s«-15s2
9. 5w2-29mn+367i2 10. 10r2-23r-5
11. 662-296+35
12. 6/2-/-77
13. 102-lla-a2
14. 15+373-8^2
15. l-6a:?/+5a:2i/2
16. 2a;2+llx+12
17. 14a:2+53a:?/+14?/2
18. 5a:2+13a;-6
19. 17a2+6a-ll
20. 8p2-14p-39
Miscellaneous Exercises for Practice
400. Factor the following:
1. C2-C-56
2. 48a;2-3?/2
3. a262+a26+a62
4. 72r2+41r-45
5. a;27/2— 5a;i/+4
6. a2-64
8. a}hc'-2ahc^-^hc^
9. 4aa:2-9a?/
10. 18a:2+37x+19
11. 76a:2+426a;?/+636?/2
12. 6a26+12a62+18a262
13. 2?*- 16
7. xhjz-\-x}fz-\-xyz^
14. 762+416c-6c2
238
FIRST-YEAR MATHEMATICS
15. |
SlOa'c^-lOb'c' |
28. |
562+106-15 |
16. |
piq^r — r |
29. |
5a2-562 |
17. |
38a: V+ 57a: V- 19:^3 |
30. |
6a;2-56?/2+41a:2/ |
18. |
Qx^-^xy-lSy'^ |
31. |
a%^-2abk^^-¥k^ |
19. |
a%^c'^-\-a%^c'-{-a%'c' |
32. |
3a2-9a6- 21062 |
20. |
450 -2a2 |
33. |
ax^-\-ax'^-\-ax-\-a |
21. |
c2-16c+64 |
34. |
30c2-31c+8 |
22. |
6+1562-16 |
35. |
ax^- 100a |
23. |
a'b''-Qa%-^9a^ |
36. |
9m2-24mn+16n2 |
24. |
12a^xy-{-12axy-\-Sxy |
37. |
d'+a'^a |
25. |
x^-\-ix-5 |
38. |
36a2+27a6+262 |
26. |
15+19a2-34a |
39. |
a262+18a6+80 |
27. |
81a2-l |
40. |
4.'r2+32a:?y+39?/2 |
\y^ |
||||
\v |
||||
\> |
||||
\ |
||||
B |
||||
- |
||||
The Theorem of Pythagoras
401. There exists a simple relation between the sides of a right triangle which is used in solving some geometric problems.
1. Construct a right triangle,
making the sides including the right
angle 3 and
4 units long
respectively,
as /\ABC,
Fig. 252. Using the same unit find the length of AC. On each side draw a square and divide each into unit squares. Counting these squares, find how the square on the hypotenuse compares in size with the sum of the squares on the other two sides. Express the result by means of an equation.
2. How does the square on the hypotenuse. Fig. 253, compare with the sum of the squares on the other sides ?
Fig. 252
X |
X |
X |
X |
|
X |
X |
K |
X |
|
X |
yi |
% |
X |
|
X |
X |
X |
X |
|
X |
X |
X |
X |
X |
Fig. 253
FACTORING
239
402. The last two problems illustrate the following theorem :
Theorem of Pythagoras: In a right triangle the sum of the squares on the sides including the right angle is equal to the square on the hypotenuse.
This theorem is one of the most famous theorems of geometry. It was named after the Greek mathema- tician, Pythagoras (569-500 B.C.). He is believed to have been the first to give a general proof of it.
EXERCISES
1. Show by counting the small squares, Fig. 254, that the square on the hypot- enuse equals the sum of the squares of the other two sides.
2. Let c, Figs. 255 and 256, denote the length of the hypot- enuse and let a and b denote the lengths of the sides including the right angle.
Draw squares on line-segments equal to (a+b) . Divide these squares as indicated.
1 |
|
it -^-i |
|
i^^ T\ |
|
^^'^ w _r. |
|
_<^^^- i |
|
'"^^ \ |
|
L- i J V |
|
^L^M ^o\ |
|
\ ^ IP Jl^v |
|
\ 1 1 ^''"^ |
|
^r\ 26-^^" -" |
|
-- |
\t -^ |
--_--=^ " it |
|
r |
|
1 |
Fig. 254
^"-^ |
^ |
U=«iB=^ |
III/ / |
a b
Fig. 255
240 FIRST-YEAR MATHEMATICS
Show that A I, II VIII are congruent, (s.a.s.)
Show that the square on (a-\-b), Fig. 255, equals
a2+62+(H-IH-III+IV) Show that the square on {a-\-b), Fig. 256, equals
c2+(V4-VI+VII+VIII) Hence, if from each square on (a +6) four of the congruent triangles are taken, the remainders a^+b^ and c^ are equal. Therefore a^-\-b'^ = c''
3. The two short sides of a right triangle are 6 and 8. What is the length of the hypotenuse ?
Denote the hypotenuse by x, Fig. 257.
Then x2 = 82+62 = 64+36
0:2 = 100 X =10
^^* Hence, the hypotenuse is equal to 10.
4. A wall is 16 ft. long and 12 ft. high. How long is a string stretched from a lower corner to the opposite corner ?
5. The hypotenuse of a right triangle is 30 ft. and one of the other sides is 18 ft. long. How long is the third side ?
6. The sides of a right triangle are as 3 is to 4 and the hypot- enuse is 35 ft. long. How long are the sides ?
Call the sides Sx and 4a;.
7. A ladder 20 ft. long just reaches a window 16 ft. above the ground. How far is the bottom of the ladder from the foot of the wall if the ground is level ?
8. The diagonal of a square is 12. What is the side ? The perimeter ?
The square root of a number, as 80, or a,_or x+y, that is not a perfect square, is indicated thus: V^80, or Va, or Vx-\-y.
9. The diagonal of a square is a. Find the perimeter.
10. Find the hypotenuse, /i, of a right triangle whose area is 54 sq. ft. and whose base is 12 feet.
11. Find the hypotenuse, h, of a right triangle whose area is r sq. rd. and whose base is b rods.
FACTORING 241
12. Find the perimeter of a right triangle whose area is 216 sq. rd. and whose base is 48 rods.
13. Find the perimeter of a right triangle whose area is s sq. rd. and whose base is b rods.
14. A tree, standing on level ground, was broken 36 ft. from the ground and the top struck the ground 27 ft. from the stump, the broken end remaining on the stump. How tall was the tree before breaking ?
15. Under conditions of exercise 14 suppose the top piece t ft. long and that the top struck the ground /ft. from the stump. How tall was the tree before it broke ?
16. The sides of a rectangle are a and b and the diagonal is d. Show that d^ = a'^-hb^ and d=Va'^-\-¥.
17. Show that the diagonal, d, of a square of side a is given by the formula d = a^2.
18. The hypotenuse and one side of a right triangle are c and b. Show that the other side a equals ^c^ — 6^ = i/ (c-f 6) (c—b).
19. Using the result of exercise 17 as a formula, find the diagonals of squares whose sides are 4 in.; 1.2 cm.; 3.42 centimeters.
20. Using the result of exercise 18 as a formula, find the side a when c = 40 in., 6 = 24 in.; c= .625 cm.; 6= .375 centimeters.
In the case c = 4, & = 3, show that a = i/7. The length of the side cannot be expressed exactly without the radical sign, but can be approximated by extracting the square root of 7.
Square Root of Arithmetical Numbers
403. The square of a number is found easily by multi- plying the number by itself. Thus, the square of 346 is 346-346 = 119,716. However, the process of finding the number which multiplied by itself gives 119,716 is much more complicated.
242
FIRST-YEAR MATHEMATICS
404. The square of a binomial is obtained from the
formula
(a+«2 = a2_^2ai+i2 I
This formula is represented geometrically in Fig. 258.
-2a+b-
ab
b b' b
ab
ab
ab
Fig. 258
Fig. 259
Show from Fig. 259 that by placing one of the rectangles ab in a different position, formula I can be changed to the form II:
ia+by = a'+i2a+b)b II
Formula II will be used in finding the square root of a given number.
405. The number of digits in the square root. The
following exercises show how to determine the number of digits in the square root of a given number.
EXERCISES
1. Show that the square of a number of 1 digit contains 1 or 2 digits.
Give the squares of the integral (whole) numbers from 1 to 9.
2. Show that the square root of a number of 1 or 2 digits contains 1 digit in its integral part.
3. Find the integral part of the square roots of the following numbers: 3, 5, 7, 18, 27, 39, 50, 65, 89.
4. Show that the square of a number containing 2 digits in the integral part has 3 or 4 digits in its integral part.
FACTORING
243
5. Show that the square root of a number of 3 or 4 digits has 2 digits in the integral part.
6. Find the integral parts of the square roots of the following numbers: 110, 150, 209, 630, 1,625, 8,274.
The following table is a summary and extension of exercises 1 to 6.
Number of digits in a given number |
1 |
2 |
3 |
4 |
5 |
6 |
Number of digits in the square |
lor 2 |
3 or 4 |
5 or 6 |
7 or 8 |
9 or 10 |
11 or 12 |
Thus, 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 in the integral part of the square root of a number. Beginning at the decimal point mark off toward the left periods of 2 digits. Then the number of digits in the square roots will be the same as the number of periods. Thus, since 54,783 is divided into three periods: 5'47'83, the period farthest left containing only 1 digit, the square root of 54,783 contains 3 digits in its integral part.
Find the square root of 729.
Beginning at the decimal point mark off to the left periods of 2 digits: 7'29.
Hence, the square root contains 2 digits. Why ?
The tens digit of the square root is the largest integer whose square is less than 7, i.e., 2.
Denoting the unit digit by x, 1^729 = 20+3:
and 729 = (20 +xy = 400 + (2 • 20 -\-x)x, by formula II
Hence, 729 -400 = (2 • 20+a:)a;, or 329 = (2 • 20+x)a;
The value of x is found by trial : Since 2 • 20 = 40 and since 8 • 40 = 320, 8 might be tried. But (2-20+8)8 = 384 and is larger than 329. Trying the next sjnaller number, 7, we find (2 • 20+7)7 =329. Therefore a: = 7 and i/729=20+7=27.
244 FIRST-YEAR MATHEMATICS
406. The process of finding the square root of 729 may be given in the following condensed forms:
_w ^
l/7'29 = 20+7 Omitting zeros: l/7'29 = 27 202=400 22=4
329 329
(2-20+7)7 = 329 47-7 = 329
407. The process of finding the square root. The
process of finding the square root, omitting zeros, consists of the following steps:
(1) Point off periods of 2 digits, beginning at the decimal point 7'29
(2) Find the largest integer whose square is less than 7, i.e., 2.
This is the tens digit of the square root.
(3) Subtract the square of the digit just found
from 7 4
and bring down the next period of digits 329
(4) Neglecting the units digit in the remainder, divide the number so formed by 2 times the tens digit of the root, i.e., by 4.
(5) The result found in (4) is used tentatively as unit digit of the roots. It is also adjoined to 2 times the tens digit of the square root and the result obtained is then multiplied by the unit digit of the root 329
(6) The product in (5) is subtracted from the remainder obtained in (3). If the product is larger than the minuend the unit digit of the root is decreased by 1 and step (5) repeated with it. If there is a remainder the given number is not a perfect square.
EXERCISES
Extract the square roots of the following numbers: 4,096, 1,444, 676, 2,116, 784, 4,761
QUADRATIC EQUATIONS
245
Quadratic Equations
408. Quadratic equations. Geometric problems some- times lead to equations in which the highest power of the unknown is the second power. Such equations are of the second degree and are called quadratic equations.
The diagonal of a rectangle is 8 units longer than one side and 9 units longer than the other. Find the length of the diagonal.
Denoting the length of the diagonal by x, Fig. 260, the sides of the rectangle are a;— 8 and x — 9.
Therefore a:2 = (a; -8)2 + (a: -9)2. Why?
This equation reduces to the normal form a;2-34a:+145 = 0
It is our next aim to learn how to solve the equation. ^^^' ^^^
409. Quadratic function. The expression x^ — 34a; + 145 is called a quadratic function of x, or a function of the second degree, and x-— 34a; +145 = 0 is a second degree, or quadratic, equation.
410. Graphical solution. The equation x^—Z^x + 145 = 0 may be solved graphically.
Denoting the function a;2 — 34a: + 145 by y, we find the following table of corresponding values of x and y (Fig. 261):
The graph shows that the func- tion becomes 0 at two places, for
X |
y |
0 |
145 |
2 |
81 |
4 |
25 |
6 |
23 |
10 |
- 95 |
15 |
-140 |
20 |
-135 |
25 |
- 80 |
30 |
25 |
35 |
175 |
\ 150\ 100 ' 50 ' 0 |
1 |
t |
||||||
\ |
||||||||
\ |
1 |
|||||||
\ |
||||||||
V |
J / |
) |
0 i |
5 1 S |
0 a |
s |
||
-100 i^J-SO |
\ |
J |
||||||
V |
^ |
/ |
Fig. 261
Fig. 262
246 FIRST-YEAR MATHEMATICS
X = 5 and for x = 29. Thus, the equation x^ — 34.r + 145 = 0 is satisfied by a: = 5, and by .t = 29. Test by substituting these values in the equation.
PROBLEMS AND EXERCISES
1. A stone, falling from rest, goes in a given time 16 ft. multiplied by the square of the number of seconds it has fallen; i.e., s = 16«2
Find s, if (1) ^ = 4 seconds
(2) ^=11.5 seconds Findi, if (1) s = 64 feet
(2) s= 1,600 feet Make a graph of the function 16^^
2. A stone, thrown downward, goes in a given time 16 ft. 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; i.e., s = vt-\-\%f.
Find s, when v = Z and ^ = 3, 7, 12. Make the graph of vt+l^f^ for v = Z.
3. Solve the following equations graphically:
1. x'-^x+l2 = 0 5. a;2-10a;+24 = 0
2. a:2-6a;-f 5 = 0 6. x2-10a;4-25 = 0
3. x^-^x- 6 = 0 7. 4a:2-12a;+ 5 = 0
4. a:2-3x'-10 = 0 8. 4a:2+ 8x- 5 = 0
411. Quadratic equations solved by factoring.
Solve the equation x'^ — ^x-\-\2 = Q
a:2-8x+12 = 0 Factoring,
(a;-2)(a:-6)=0 (1)
This equation is satisfied if x— 2 = 0 (2)
and if x-6=0 (3)
because:
The product of two or more numbers is zero only if nt least one of the numbers is zero.
From equation (2), x = 2
Check: 22-8-2 + 12 = 4-16 + 12=0
Likewise from (3),
Check as above. a: = 6.
Consequently, both 2 and 6 are roots of tlie equation.
QUADRATIC EQUATIONS 247
PKOBLEMS AND EXERCISES
Solve the following quadratic equations by the method of factoring, and test the results:
1. x2-3a;H-2 = 0
2. ?/-4?/+3 = 0
3. m2 = m+2
4. n2+5?z = 6
5. a24-7a+6 = 0
6. m2 = 4m+12
7. k^+k = 5Q
8. r2+51 = 20r
9. 62 = 46+77 10. c2+112 = 23c
11- 15 = 5+3
x^ X ,2 3
^"^^ 10 ^~ 10
x^ |
4a; |
4 |
|||
13. |
21" |
-y= |
"3 |
||
14. |
x^ 6" |
= 1+9 |
|||
15. |
4x 5 ' |
_x^ 7 15^3 |
|||
16. |
i+ |
15 7 ~ |
x^ 14 |
||
17. |
13^13 |
= 7 |
|||
18. |
4.7 a: — |
'i+- |
^-10 _ |
4 |
|
19. |
52/- |
P^+ |
32/- 1 5 |
102/ 9 |
4 "92/ |
20. |
a+7 9-4a2- |
1-a ~2a+3 |
+^- |
^ |
Solve the following problems:
21. The base of a triangle exceeds the altitude by 4 in., and the area is 30 square inches. Find the base and altitude.
22. A rectangular field is twice as long as wide. If it were 20 rd. longer and 24 rd. wider, the area would be doubled. What are the dimensions ?
23. The perimeter of a rectangular field is 60 rods. The area is 200 square rods. Find the dimensions.
24. A tree standing on level ground was broken over so that the top touched the ground 50 ft. from the stump. The stump was 20 ft. more than two-fifths of the height of the tree. What was the height of the tree ?
248 FIRST-YEAR MATHEMATICS
412. Quadratic equations solved by completing the square.
EXERCISES
Solve the following problems:
1. The area of a rectangle 10 units long is 21 square units greater than the area of a square whose side equals the unknown dimension of the rectangle. Find the unknown dimension.
Show that x2 — 10a;+21=0 is the algebraic statement of the problem. *
Write the equation thus:
a;2-10x = -21 Add 25 to both sides to make the first side a trinomial square:
or, (x-5)2 = 4
Take the square root of both sides, remembering that 4 has two square roots, +2 and —2; thus:
x-5=±2*
I) sing the + sign
x — 5= 2, whence a: = +7
Using the — sign,
x — 5= —2, whence x= +3
Check both values of x by substituting in the equation, x2-10a:+21=0
The algebraic method just given is called the method of solving the equation by completing the square.
2. One dimension of a rectangle is 5 units and the other is equal to the side of a square. The sum of the areas of the rec- tangle and of the square is 36 square units. Find the unknown dimension of the rectangle.
Show that 1/2 +5i/— 36 = 0 is the algebraic statement of the problem.
* Square-root axiom: Any number has two square roots of the same absolute value, but of contrary sign.
QUADRATIC EQUATIONS 249
Solve the equation, y^-\-5y—SQ=0 by the method of completing the square:
„ , - ,25 _ , 25 169 2/^+%+^ = 36+- = -^
= 4, or -9 Hence, the unknown dimension is 4 units.
413. Though a quadratic equation generally has two solutions, this does not mean that every problem that leads to a quadratic equation has two solutions. The nature of the conditions of the problem may be such as to make one, or even both, of the solutions of the quadratic impossible, or inappropriate, 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 are contradictory, or that the problem is erroneously stated. To decide which solution, if either, meets the conditions stated in a prob- lem, it is necessary to substitute the solutions in the conditions of the problem, and to reject solutions of the equation which do not meet the conditions.
The graphical method of solving quadratic equations may furnish only approximate values of x for the corre- sponding quadratic equations. The algebraic method furnishes exact solutions for quadratic equations.
PROBLEMS AND EXERCISES
Solve the following quadratic equations by the method of completing the square, and check: 1. 4x2-12a;+5 = 0 Dividing both sides of the given equation by the coefficient of x\
This may be written x^—3x= — -:
250
FIRST-YEAR MATHEMATICS
Adding to both sides of the equation the square of one-half of the coefficient of x,
4
3\2
H)
1
Extracting the square root of both sides of this equation,
3
Whence,
x = |±l
and
5 1
Check by substituting in the given equation :
17. w^-9l = Qiv
18. r2+3r+2 = 0
19. m2-h5m+6 = 0
2. 6x^-17x-U = 0
3. 6a;2-f-7x-20 = 0
4. 9x2+30a:-24 = 0
5. 4s2+45s-36 = 0
6. .10s2-21s-10 = 0
7. 12s2-71sH-42 = 0
8. 4x''+Sx-5 = 0
9. a;2+2x-3 = 0
10. x''i-4:X+S = 0
11. a;2-|-4a;-5 = 0
12. a2+8a-20 = 0
13. y^+Uy +4:5 = 0
14. ^2+14^-51 = 0
15. A;2-85 = 12A;
16. 22=102+24
20. 10x2+21a:-10 = 0
21. /i2+40 = 13/i
22. a;2+x = 42
23. x2+6a;+5 = 0 x^ X _1 24~3~~2
24
25. - =
5 2—1 z
26.
27.
Z Z ' 2—1
2^+3 2^+9
= 0
^+8 3^+4 k±l_k±S_S k-h2 A:+4"3
Solve the following problems:
28. The length of a rectangular field is 4 yd. more than the width, and the area is 60 square yards. Find the dimensions.
QUADRATIC EQUATIONS 251
29. The hypotenuse of a right triangle is 10 ft. and one of the sides is 2 ft. longer than the other. Find the length of the sides.
30. The sum of the area of two square fields is 61 sq. rd., and a side of one is 1 rd. longer than a side of the other. Find the sides of both squares.
31. What must be the dimensions of a coal-bin to hold 6 tons of coal, if the depth is 6 ft. and the length is equal to the sum of the width and depth, allowing 40 cu. ft. of space per ton of coal ?
32. Telegraph poles are placed at equal distances along a railway. In order that there be two less per mile it would be necessary to increase by 24 ft. the distance between every two consecutive poles. Find the number of poles to the mile.
Summary The chapter has taught the following:
414. Certain multiplications occurring frequently in algebra may be performed mentally. They are: the square of a binomial, the square of a trinomial, the prod- uct of the sum of two numbers by their difference, and the product of two factors of the form ax-\-h and cx-\-d.
415. Factors of the following polynomials may be found by inspection: The quadratic trinomial square, the difference of two squares, and the quadratic trinomial of the form ax^+hx+c.
416. Arithmetical numbers may be squared according to the formula a'^-{-2ab-\-¥.
417. Several illustrations of the theorem of Pythagoras.
418. The extraction of the square root of arithmetical numbers.
419. Quadratic equations may be solved by graph, by factoring, and by completing the square.
CHAPTER XVI
PROBLEMS LEADING TO EQUATIONS OF THE FIRST DEGREE IN ONE UNKNOWN
Solution of Problems and Equations
420. Arithmetic and algebraic solution of problems.
The problems of the foregoing chapters have shown the advantage of using letters to represent numbers when stating and solving problems. Many problems in arith- metic are simplified when solved by algebra, as may be seen by comparing the following arithmetic and algebraic solutions.
A man after traveling 9 mi. finds that he has yet yV of his trip to make. How long is his trip ?
Arithmetic Solution
\ 0 of his trip equals the whole distance
-\o " " " " " distance still to be traveled
J^_^\=i'\ " " " already traveled
9 mi.
Hence, 1% of his trip equal 9 mi.
iV " " " " 3 mi.
1 § '' " " " 30 mi. Therefore his trip is 30 mi. long.
Algebraic Solution
Let X be the number of miles in the whole trip.
7x Then 9+t^ =the whole distance
7x Hence a: = -T:+9
10a: = 7a;+90 3a: = 90 a: = 30 Therefore the trip is 30 mi. long.
252
JOSEPH LOUIS LAGRANGE
JOSEPH LOUIS LAGRANGE
JOSEPH LOUIS LAGRANGE, the greatest mathe- matician of the eighteenth century, was born at Turin, Italy, January 25, 1736, and died at Paris, April 10, 1813. It was by his own industry and ability that he rose to the front rank of mathematicians. He showed little taste for mathematical studies before he was seventeen. Happening upon a memoir by Halley that interested him, he threw himseK into the study of mathe- matics, and after a year of hard work became so accom- plished a mathematician that he was made a lecturer in the artillery school. He at once took up an isoperimetrical problem of long-standing difficulty among mathematicians, solved it by methods of his own devising, winning the admiration of Euler, and, by nineteen years of age, a place in the very front rank of living mathematicians.
In 1758 he established the Turin Academy, and most of his early writings are found in the five volumes of its transactions. In 1761 he stood without a rival as the foremost living mathematician. In 1766 Euler left Berlin for St. Petersburg, and Frederick the Great wrote to Lagrange saying "the greatest king in Europe" wished "the greatest mathematician in Europe" to reside at his court. Lagrange accepted the invitation, and during the 22 years of his stay here did a prodigious amount of work. In 1787 Frederick the Great died, and in 1788 Lagrange accepted an offer of Louis XVI of France to move to Paris. Here he remained, and wrote and taught mathematics during those troublous times in France, until his death. He was befriended by three rulers — Frederick the Great, Louis XVI, and Napoleon. For a fuller account of the labors of Lagrange at Berlin and at Paris read Ball's History of Mathematics, pp. 404-11 (5th ed.).
Ball says of his personal appearance: "he was of medium height, and slightly formed, with pale blue eyes and a colorless complexion." Of his character Ball remarks: "he was nervous and timid, he detested con- troversy, and to avoid it willingly allowed others to take the credit for what he had himself done."
EQUATIONS WITH ONE UNKNOWN 253
421. Stating and solving problems. Stating a prob- lem in algebra usually means expressing in algebraic, symbols certain number relations that are given in words in the problem.
Skill in stating and solving problems can be acquired best through much practice. The statement of a prob- lem in most cases takes the form of an equation. To obtain this equation the following rules may be useful:
I. Denote the unknown numhei^'hy a letter, then trans- late the verbal statement of the number relations into a symbolic statement in equation form.
■ The fir^t letters of words are convenient letters to denote numbers while they are yet unknown, as n for number, t for time, a for age, w for weight, etc.
• • . EXERCISES
Give statements of the following problems, then solve and check.
1. What number increased by 6 gives 13 ? .
Statement:
n + 6 = 13 •
Solve the equation and check by substitution.
2. Two-thirds of a number diminished by 12 equals 4. Find the number.
Statement:
Two-thirds of a number diminished by 12 equals 4 I X n - 12 = 4
Solve the equation and check by substitution.
3. Translate the following equations into words and find the value of x:
1. a:-17 = 19 4. 2x-S = Sx-7
2. 66 = x+48 5. lS-x = 4:X-S
3. 2.8 = 2x+1.9 6. 8x-m = 7x-Q
254 FIRST-YEAR MATHEMATICS
Sometimes the equation may be obtained as fol- lows:
II. By the aid of literal and of arithmetical numbers obtain two different number expressions for some number of the problem, and write the two expressions equal to one another.
Thus, an equation states in symbols that two different number expressions stand for the same number.
EXERCISES
1. A rectangle is 3 times as long as wide, and the perimeter is 48 inches. Find the width and length.
Equate two different expressions of the perimeter.
2. The angles of a triangle are x, 2x, and Sx degrees. Find the numerical values.
3. The acute angles of a right triangle are 4x and 5x. Find the acute angles. •
422. Notation for unknown numbers. While any letter may be used to denote the primary unknown it is customary to use x, y, or z or some one of the later letters of the alphabet for this purpose.
The earlier letters of the alphabet, as a, b, I, m, k, etc., are commonly used to denote known numbers.
Solve the following equations for a:; i.e., find the value of X in terms of the other letters in the equation:
1. 2x-2Sa = x-na
2. 4:x+b = 5x-l%
423. Solving equations. The solution of an equation consists in getting another equation in which the unknown
EQUATIONS WITH ONE UNKNOWN 255
is alone on one side. This equation is obtained by means of the axioms stated in § 89.*
One of the important form changes is the removal of parentheses, as in the solution of the following problem:
The difference of double a boy's age and 3 times his age 10 years ago is his present age. How old is the boy ?
Statement: 2x-3(x-10)= x (1)
Since - 3 (re -10) = -3a: +30, hence 2x-3rc +30= re (2)
Since 2rc—3rc = — re , hence — x +30= x (3)
Adding x to both sides, 30 = 2rc (4)
Dividing by 2, 15 = re (5) The boy's age is 15 years.
Check by testing in the conditions of the problem. State the axioms used to get equations (4) and (5).
EXERCISES
Solve the following equations, in each case checking the result :
1. 5rc+9 = 3x+17
5re+9 = 3re+17 Subtracting 3rc from both sides, 3rc =3re
Subtracting 9,
2rr+9 = 9 = |
17 (State axiom used.) 9 |
2x = 2rc 2 X = |
8 8 2 4 (State axiom used.) |
Dividing both sides by 2, or.
Check: 5rc+ 9=5-4+ 9=29 3rc + 17 = 3 -4+17 = 29 Hence, 4 is a root of the equation 5rc+9=3x+l7
* Mohammed ben Musa Al Hovarezmi, who lived during the reign of Caliph Al Mamun (813-33), was the first notable Arabian author of mathematical books. The title of a book in which he explains the solution of equations is Aldshehr walmukahala. By "aldshebr" is meant the transposing of terms from one side of an equation to the other. This is the origin of the word "algebra." See Ball, pp. 156-57.
256 FIRST-YEAR MATHEMATICS
The preceding solution may be shortened by doing mentally some of the work written out in full.
Thus, oa;+9 = 3.c + 17
Subtracting 3x+9 from both sides, 2a; = 8 Dividing by 2, x= 4
2. 5a:-hl7 = 8x-2 Sx ^^ ^ ^^
16. -r — 91= — 5 — lOo;
3. 3a;+5 = 5a;-15 4
4. 6x-7 = 10x+l 8+2a;+^-l^+?^
5. 5x-17+3:.-5 = x-l ^^' 8+2a;+4-l^+3
6. 6x-7-8a:+115 = 0 2x+3_^-2_7
7. 5(a:-l) = 3(a:+l) ^^' 5 "3~~5
8. 8(3-2a;) = 2(5-a;) = 28
9. -3a;-24 = 33(2-a:) 19- ^^V^-"^— = %T
10. 5(1 -13a;) = 35 -105a;
11. ll-3(x-2)=a;-8 20. ~~7-^+^^ = 2x
3a;-2 |
l-4a; |
7 |
3 |
3(5-a; |
) ,6-a; |
4 |
' 5 |
21. i(52/-3)-i(57/-2) = 5
12. 3a;-2(a;+5) = 6a;-20
13. 3(a;-2)4-15 = 5a;-3 3x _ 22. f(7n+4)+f(7n-l) = 10M+2
* 4+^-"^+^ 23. (t/-7)(2/-8) = (2/-5)(y-9)
Multiply both sides by 4
24. (4m-5)(3m+l) = 12m(m+l)
X
15. ^H-2x-8 = 3a;-5 ^^ x{x-l)-x{x-2)=4:{x-S)
Geometric Problems
424. The following problems contain geometric rela- tions. Besides expressing these relations in algebraic symbols, a figure representing them should be drawn in each case before obtaining the equation.
1. In an isosceles triangle the exterior angle at the base is twice as large as the exterior angle at the vertex.
Find the interior angles of the ^ ,„ triangle.
Denoting the vertex angle, Fig. 263, Fig. 263 by x, the exterior angle at the vertex
EQUATIONS WITH ONE UNKNOWN 257
is 180 — ..T. The exterior angle at the base is therefore 2(180— x-). Hence, the interior angle at the base* is 180 — 2(180— x). Why? What is the third interior angle of the triangle ?
Show that a;+2[180 -2(180 -.r)] = 180 and solve the equation.
2. One of two supplementary adjacent angles is 4.5 as large as the other. How large is each?
3. The difference of the acute angles of a right triangle is 18°12'. How large is each ?
4. A rectangle is 2 units longer than 3 times the width, and the perimeter is 60. Find the length and width.
5. The perimeter of a rectangle is 20 inches. If the base is de- creased by 4 in. and the altitude increased by 2 in., the area remains unchanged. What are the dimensions of the rectangle ?
6. Find the sides of an isosceles triangle whose perimeter is 360 in. and whose base is 75 inches.
7. A classroom of a certain high school is | as long as it is wide. If the length were diminished 3 ft. and the width in- creased by the same amount the room would be square. Find the dimensions.
8. The area of a square is equal to that of a rectangle having its base 12 ft. greater and its altitude 4 ft. shorter than the side of the square. Find the dimensions of both figures.
9. The area of a triangle, whose altitude is 5 units less than the base, is equal to 10 square units less than half the area of a square on the base of the triangle. Find the base and the altitude of the triangle.
10. The United States Treasury building in Washington, D.C., is 222 ft. longer than it is wide. The entire distance around the building is 1,500 feet. Find its length and width.
11. The length of the Pennsylvania Station in New York is 80 ft. less than twice the width and J as long as twice the distance around diminished by 700 feet. How long and how wide is this station ?
258 FIRST-YEAR MATHEMATICS
12. The perimeter of a rectangular steel plate is 240 in. and the length is 85 . 24 inches.* Find its width and area.
What is the weight of this plate, if it is f in. thick, and if 1 sq. ft. of 8 -in. plate weighs 5 . 1 pounds ?
13. The sides of a triangle are in the ratio 3:4:5. Find the three sides, if the perimeter is 108 inches.
14. One of the acute angles of a right triangle is 8 times as large as the other. Find the two acute angles.
15. Find the angles of an isosceles triangle if one of the base angles is 50°.
16. One angle of a triangle is 5° greater than 5 times another. Find the third angle, if the smaller of the first two angles is 19°. Make a scale drawing of the triangle, if the side opposite the greatest angle is 60 feet.
17. The interior angles on the same side formed by two parallel lines cut by a transversal are ^x and 88+~5X. Find X and the angles.
18. The acute angles of a right triangle are denoted by 3 a: +7 and 41— Jx. Find x and the angles. Draw the triangle.
19. Two supplementary adjacent angles are denoted by 4^^+29 and 97— fa;. Find x and the angles.
20. The interior angles of a triangle are 69°, (t+17) , and ( ^+80y . Find x and the two unknown angles of the triangle.
Problems Involving Number Relations
425. The following problems give training in express- ing number relations in algebraic form.
1. Express that one-eighth of the double of n, increased by 25, equals one-half of n.
2. Write in symbols an expression for double a number, increased by 3 times the sum of the number and 4.
EQUATIONS WITH ONE UNKNOWN 259
3. Write an expression for double a number, decreased by 3 times the difference between the number and 4.
In the phrases "the difference between 8 and 4," ''the difference of 8 and 4," and the Hke, it is understood that the first-mentioned number (the 8) is the minuend; thus 8—4, not 4—8, is the difference between 8 and 4.
4. Write the double of a number, decreased by 3 times the difference of the number and 2.
5. Write in symbols 4 times the difference of a number and 3, decreased by 3 times the difference of the number and 1.
6. Show, in symbols, that the double of a number, increased by 3 times the difference of the number and 4, equals 13.
7. One-fourth of the difference of 3 times a number and 8 is 10. Find the number.
8. Three times a number is 56 greater than one-third of the number. Find the number.
9. What number multiplied by 2 . 5 gives 40 ?
10. Three times a number, increased by 5 times the number, gives 72. Find the number.
11. If 1 is added to 4 times a number and subtracted from the number, the ratio of the results is 5. Find the number.
12. Five times a number is increased by 3 and the sum is divided by the sum of the number and 4. The result is then equal to 4. Find the number.
13. I have in mind a certain number. You can determine it from the following data: If you multiply the number in- creased by 3 by the number decreased by 5 and divide the product by the number decreased by 7, the result is the same as the number increased by 1.
14. A father is 40 years old and his son is 7. In how many years will the father be twice as old as the son ?
15. The combined age of a father, mother, and son is 75 years. The mother is 3 times as old as the son and the father H times as old as the mother. How old is each ?
260 FIRST-YEAR MATHEMATICS
16. Double the number of years in a boy's age is 16 more than his age 2 years ago. How old is the boy ?
17. The difference of double a boy's age and 3 times his age 10 years ago is his present age. How old is the boy ?
18. Express in symbols that the sum of three consecutive integers, differing by 3, equals 27. Find the three integers.
19. The difference of the squares of two consecutive numbers is 19. Find the numbers.
20. The difference of the squares of two consecutive numbers is 273. Find the numbers.
21. The difference of the squares of two consecutive numbers is a. Find the numbers.
22. The difference of the squares of two consecutive even numbers is 28. Find the numbers.
23. The difference of the squares of two consecutive even numbers is 100. Find the numbers.
24. The difference of the squares of two consecutive even numbers is a. Find the numbers.
25. The difference of the squares of two consecutive odd numbers is 48. Find the numbers.
26. The difference of the squares of two consecutive odd num- bers is s. Find the numbers.
Motion Problems
426. Rate. Distance. Time. If a train travels 30 mi. an hour, it is said to move at a rate of 30 mi. an hour. If a man walks 35 yd. a minute, he is said to walk at the rate of 35 yd. a minute.
The distance traveled by a body depends upon the time it travels and upon the rate at which it travels. The following problems will show the relation between the distance, rate, and time.
EQUATIONS WITH ONE UNKNOWN 261
EXERCISES
1. The rate of a train is 40 mi. an hour. If it leaves the station at 1:00 p.m. how far is it at 2:00 p.m., 3:00 p.m., 4:00 P.M., 5:00 p.m., etc.? How far is it at 3:20, 4:30, 6:45?
2. Denoting the distance traveled by d, find d when the rate is 60 mi. an hour and the number of hours is 5.
3. Let d be the distance traveled in t hr. at the rate of r mi. an hour. Show that d = rt. Translate this equation into words.
4. Show how to obtain from the equation d = rf, the equations r= - and t = -. Translate these equations into words.
427. Relation between distance, rate, and time. The
equations of problems 3 and 4 above show that if two of the three numbers d, r, and t are known or expressed in algebraic symbols, the third can always be expressed in terms of them.
EXERCISES
1. A bird flies a distance of 80 mi. in 2 hr. 30 minutes. Find the rate, supposing it to be uniform.
2. Sound travels 1,080 ft. a second. If the sound of a stroke of lightning is heard 3 . 5 sec. after the flash, how far away is the stroke ?
3. If sound travels/ ft. a second, how far away is a Ughtning stroke if the sound is heard s seconds after the flash ?
4. A tree 2,376 ft. distant was struck by lightning. It took 2i sec. for the sound to reach the ear. Find the rate at which the sound traveled.
428. Graphical solution of motion problems. Motion problems may be solved graphically.
262
FIRST-YEAR MATHEMATICS
1. A man walks along a straight road, AB, Fig. 264, in the direction of the arrows (->), at the rate of 3 mi. an hour. How far from the house, H, is he :
4 hr. before reaching it ? 3 hr. before reaching it ? 2 hr. before reaching it ? 1 hr. before reaching it ?
1 hr. after reaching it ?
2 hr. after reaching it ?
3 hr. after reaching it ?
4 hr. after reaching it ?
Denoting the number of hours before reaching the house by — and the number of hours after by +, lay off these numbers on the time-line to the right or left of 0 according as they are + or — . At each of the points —4 to +4 on the time-line lay off vertically the corresponding distances from the house, letting the vertical side
of a square represent 3 mi. In this way the points P,Q, R, S
X are obtained. If these points are joined by straight lines, it will be found that they all lie in the same straight line, PX.
Time |
Distance |
-4 |
-12 |
-3 |
- 9 |
-2 |
- 6 |
-1 |
- 3 |
1 |
3 |
2 |
6 |
3 |
9 |
4 |
12 |
B ''12 9 |
1 |
n |
\A |
|||||||
/ |
||||||||||
— 1 |
/ |
|||||||||
J'' |
/ |
|||||||||
/ |
Time- Line |
|||||||||
-- |
h- |
3- |
2- |
i/o |
1 |
2 |
3 |
u |
5 |
|
v: |
/ |
s |
||||||||
/ |
R |
|||||||||
-12 A |
/ |
Q |
||||||||
[/ |
h |
_ |
|
|
Fig. 264
The straight line PX represents geometrically the uniform motion of the man walking from A to B.
2. Show from the graph how far from the house the man was 3| hr. after reaching it. Show from the graph when the man was 4^ mi. from the house; 3^ mi. from the house.
3. At 10:00 A.M. a freight train leaves a station 0 going north at the rate of 30 mi. an hour. At 1:00 p.m. an express passes the station going in the opposite direction at the rate of 60 mi. per hour. When and where did the trains meet?
EQUATIONS WITH ONE UNKNOWN
263
Graphic Solution
On the time-line (Fig. 265) lay off the time, letting one side of a square represent one hour. At each hour-point lay off vertically the distance from the station, letting a vertical side of a square represent 30 mi. The points A , B,
C thus obtained are joined
by the straight line PQ. PQ is the graph for the freight train.
Similarly, the graph LM for the express train is obtained. The point of intersection C of PQ and LM gives the distance of each train from the station at 12:00 o'clock. Why ? Thus, the trains meet at 12:00 o'clock 60 mi. from the station.
120 90 60 30 0 30 -60 -90 |
1 |
n |
V |
Q |
|||||||
/ |
E |
||||||||||
/ |
D |
||||||||||
^irm |
-in |
e |
to |
^ |
^n |
i\ |
2 |
s |
|||
/ |
A |
\ |
|||||||||
/ |
\ |
||||||||||
/ |
\ |
||||||||||
/ |
\ |
M |
|||||||||
150 |
\A |
\ |
Fig. 265
Algebraic Solution
(1) Let X be the number of hours from 10:00 a.m. until the trains meet.
(2) Then the time t) rate r; and distance, d, for each train are as follows :
f I =x For the freight train <^ r = 30 [ d = 30.c
f t =3— X, since it passed the station 3 hr. For the express train <j r=60 after the freight train left
[ d = 60(3-x)
(3) The fact that the distance from the station to the meeting- point is the same for both trains gives the equation
30x = 60(3-a-)
(4) Solve the equation in (3) .
4. Solve graphically the following problem:
A, starting from point P, moves downstream at the rate of 8 mi. an hour. B starts 3 minutes later and moves at the rate of 10 mi. an hour. When and how far from P will B overtake A f
264 FIRST-YEAR MATHEMATICS
EXERCISES
Solve the following problems algebraically:
1. A local train goes at the rate of 30 mi. an hour. An express starts two hours later and goes at the rate of 50 mi. an hour. In how many hours, and how far from the starting- point, will the second train overtake the first ?
(1) Let X be the number of hours it takes the second train to overtake the first train.
(2) Then, according to the data of the problem,
for the express train for the local train t = x ■t = x+2
r = 30 r = 50
Hence d = SOx Hencec^ = 50(x+2)
(3) Since both trains travel the same distance, the following equation holds:
50(a:-h2)=30a;
(4) Solve this equation to determine the value of x, and then solve for d.
2. An express train whose rate is 40 mi. per hour starts 1 hr. 4 min. after a freight train, and overtakes it in 1 hr. 36 min. How many miles per hour does the freight train run ?
3. Two friends, A and B, five at the distance of 33 mi. from each other. In order to meet A, B leaves home an hour earlier than A. li A travels at the rate of 5 mi. and B at the rate of 4^ mi. an hour, when and where will they meet ?
4. Two towns, A and B, are 288 mi. apart. Two men travel toward each other, starting at the same time from A and B respectively, and going at the rates that are to each other as 3:5. They meet in three days. How many miles a day does each travel ?
5. ^ is 160 yd. east and B 112 yd. west of a gate. Both start at the same time to walk toward the gate, A going 3 yd. and
EQUATIONS WITH ONE UNKNOWN 265
B 2 yd. a second. When will they be at equal distances from the gate?
The drawing, Fig. 266, will be helpful in obtaining the equation.
-1^0—
y
B' G A'-
Fig. 266
6. Two trains start at the same time from Chicago and Kansas City respectively, 498 mi. apart. The train from Chicago travels at a rate of 40 mi. an hour, the other at a rate of 50 mi. an hour. When and where will they meet ?
Make a drawing representing the distances and the equation.
7. At 6:00 A.M. a train leaves New York for Buffalo on the New York Central. At 9:00 a.m. a second train follows. The first train travels 30 mi. an hour, the second 50 mi. an hour. When and how far from New York will the second overtake the first?
8. A man rows downstream at the rate of 6 mi. an hour and returns at the rate of 3 mi. an hour. If he has 9 hours at his disposal, how far downstream can he go and return ?
9. Two trains start at the same time from S, one going east at the rate of 35 mi. an hour, and the other going west at a rate 1 faster. How long after starting will they be 100 mi, apart ?
10. A, B, and C are three towns on a straight road. The distance from A to B is 20 miles, A man leaves B for C and travels at the rate of 5 mi, an hour. At the same time an automobile traveling at the rate of 30 mi, an hour leaves A to go to C When and where will the auto overtake the man ?
11. At 10: 00 A.M. a freight train leaves the station A , traveling at the rate of 32 mi. an hour. An express train going at the rate of 72 mi, an hour passes the station at 11:00 a,m. and follows the freight train. When and where will the trains pass ?
266 FIRST-YEAR MATHEMATICS
12. A train moves at a uniform rate. If the rate were 6 mi. an hour faster the distance it would go in 8 hr. is 50 mi. greater than the distance it would go in 11 hr. at a rate 7 mi. an hour less than the actual rate. Find the actual rate of the train.
429. Motion along a circular path. In the following problems the path of motion is assumed to be circular and the motion uniform.
EXERCISES
1. Two boys were running in the same direction along a cir- cular path at the rates of 110 yd. and 98 yd. a minute, respec- tively. The length of the circle is 36 yards. Supposing they start from the same point, when will they be together again ?
(1) Show that in x minutes they run 11 Ox yd. and 98.T yd., respectively.
(2) Then 110:c = 98a:+36 expresses the fact that the faster runner has made a gain of a complete circle, which is necessary to enable him to meet the slow runner.
(3) Solve the equation in (2).
2. Suppose the boys in exercise 1 are running in opposite directions. How long will it take them to meet ?
3. Two automobiles travel over a road which runs around a lake practically circular in shape. They make the circuit in 2 hr. 45 min. and 3 hr. 30 min., respectively. How long after passing each other will they meet ?
(1) Let X be the required number of hours
-^ is the part of the circuit made by one automobile ^'f in 1 hr.
(2)
(3)
— is the part of the circuit made by the other automobile ^^ in 1 hr.
X
— is the number of circuits made by the first automobile ^4 in X hr.
— is the number of circuits made by the second automobile ^2 in X hr.
EQUATIONS WITH ONE UNKNOWN
267
[Since the faster automobile makes the circuit once more
(4) <! X X
I than the slower, it follows that ^ =— j +1.
4, Venus makes its orbit in 224.7 days, or about 7| months, the earth starting as shown in Fig. 267. In how many days will Venus next be in line between the earth and the sun ?
The rate per month of Venus is 1^5^ of the orbit, that of the earth iV- Let x be the required number of months.
The equation may be obtained as in exercise 3.
5. Calling the time of revolution of Venus about the sun 7| months, and that of Mercury 3 months, how many months after Mercury is in the line between Venus and the sun will it next be in the same relative position ?
Earth
Fig. 26/
Clock Problems
430. The movement of the hands of a clock furnishes a mechanical illustration of circular motion.
1. At what time between 3 : 00 and 4 : 00 o'clock are the hands of the clock together ?
(1) Let a:, Fig. 268, be the number of minutes after 3 : 00 o'clock when the hands are together; i.e., x is the num- ber of minute-spaces over which the minute hand passes from 3:00 o'clock until it first overtakes the hour hand.
Fig. 268
(2) Then —^ is the number of
minute-spaces over which the hour hand passes in the same time. Why ?
(3) Since the number of minute-spaces from 12:00 to 3:00 is 15,
X
it follows that x = W-\-~. (The whole is equal to the sum of the parts.)
(4) The equation in (3) determines the value of x.
268 FIRST-YEAR MATHEMATICS
2. At what time between 2:00 and 3:00 o'clock are the hands of the clock together ?
Represent in a drawing the number of minute-spaces passed over by the hour hand and by the minute hand.
3. At what time between 3 : 00 and 4 : 00 o'clock are the hands of the clock at right angles ?
4. At what time between 3:00 and 4:00 o'clock are the hands of the clock 16 minute-spaces apart ?
5. At what time between 7 : 00 and 8 : 00 o'clock are the hands of the clock pointing in opposite directions ?
6. At 12 :00 o'clock the hands of a clock are together. When will they be together the next time ?
7. What angle is formed by the hands of a clock at 2 : 30 ?
8. In how many minutes does the minute hand of a clock gain 15 minute-spaces on the hour hand ?
9. At what time between 5:00 o'clock and 6:00 o'clock will the hands be together?
10. When between 8:00 o'clock and 9:00 o'clock is the minute hand 10 minutes behind the hour hand? When is it over the hour hand ? When is it 10 minutes ahead of the hour hand?
EXERCISES
Solve the following equations:
1. 4(2x+9)+3(a:-9) = i2^y=^
2. 2(a:-l)+3(a:-2) = 4(.r+3)
3. 5x-2k = 2k-\-Sx
4. ax-\-ad = bd—bx
5. a^x+b = b'^x+a
6. {21J-5) (42/-7) =81/2+52
7. (a+x)2-a;2 = 62
8. (n+4) (n+3)-(n+2) (n+l)=42
EQUATIONS WITH ONE UNKNOWN 269
St 2x
9. ^+37+3- + 109 = 180
10. ^+93+|-180 = 18
11.2(|+|)-180 = ^(309-|)
12. 2(^+0;) -90+^(x+210) = 90
13. 100-4(x+^)=^(261-5a:)-80
14. 1(12^-1) -80 = 4+f (147/+ J)
15. i(5n-l)-i(4n-2) = 8
16. i(l-s) = i(2-s) + J-(3+s) '
17. «-15-T(9«-2)-|<-i = 0
18. 5.8a;+3.69 = 3.96+2.8.T
19. .374a:-. 53+1. 2a;+. 06=. 8+1. 32.x-
20. .3(1. 5a:-. 8)=. 6(5.1+. 2a:)
21. .05(20^-3. 2)=. 8(4a:+.12)-ll. 256
22. 1.4.-1.61-^^1^+^^^ 3^
l-2a: 2a:-.5 2x-\ 6. 35-. 5a:
^'^- OK 10 C I
24
3 ^
.25 12.5 ' 5 3
4a:+.39 .2a:-. 66 .08x+.38
7 .9 .2
Problems on Percentage and Interest
431. Some problems on percentage and interest lead to equations of the first degree.
1. Find the percentage of $120 at 1%.
The expression 1 per cent means r-r^. Thus, 1 per cent of $120
means ^ of $120 or $120 multiplied by t^.- lUU lUU
Therefore the percentage of $120 at 1 per cent is zr^ $120 or
$120
100*
270 FIRST-YEAR MATHEMATICS
2. Find the percentage of $120 at 4%.
4 . $120
Since the percentage of $120 at 1% equals ^7^7; • $120, it follows
100
that the percentage at 4% equals 4 • — -r • $120 .
3. Find the percentage of $120 at 6%; at 7^%; at r%.
4. Find the percentage at 8% of $25; of $250; of $6.
5. Find the percentage at r% of $20; of $80; of $6.
6. Calling p the percentage, b the base, and r the rate, show that
p-bX^ (1)
and show that, by one of the laws of multiplication of fractions, this may be written
^ = 100 (2)
7. Read (2) of exercise 6 and translate it into words.
8. Find the interest on $175 at 4% for 2 years; for 5 years; for f of a year; for 2| years; for t years.
9. Find the interest on $600 for 5 years at 3%; at 5%; at 8%; at 61%; at r%.
10. Find the interest on $160 for t years at 6%; at 3|%; atr%.
11. Find the interest at 6% for 3 years on $200; on $360;
on $756; on %p.
12. Find the interest at 5% for t years on $/;.
13. Find the interest i, at r% for t years on $p.
14. State in words the meaning of
._ r pXrXt
^"^^100^^" 100
432. The result of exercise 14 above is a general formula for finding the interest when principal, rate, and time are given.
EQUATIONS WITH ONE UNKNOWN 271
EXERCISES
1. Find the interest on a principle of $225 for 5 years at 2|%.
Let p = 225, r = 2.5, ^ = 5, and substitute these values in the formula of exercise 14 above.
2. Find the income from an investment of $2,800 at 5% for 8 years.
3. Find the percentage of $450 at 5%; of 375 bushels at 20%; of 1,800 men at 15%.
4. On a certain day a grocer sells $315 worth of goods. His profit is 12% of the purchase price. How much did he gain ?
Show that x+'^j^=315.
5. A merchant lost $345 in a certain sale. His loss was 9% of the amount invested. How much was the investment ?
6. The number of pupils enrolled in the United States in 1910 was 17,813,852. During the following year there was an increase of 1.25%. How large was the enrolment in 1911?
7. Into what two parts can $1,000 be divided so that the income of one at 6% shall equal the income of the other at 4%?
8. How can a man divide $2,000 so that the income of part at 4% shall be the same as that of the rest at 5% ?
9. How many dollars must be invested at 4% to give the same income as that of $2,500 at 6% ?
10. A certain sum invested at 5|% gave the same interest in 4 years as $3,300 gave in 8 years at 3%. How large was the sum?
11. Show how to divide $1,400 into two parts so that one part at 4% shall produce twice as much income as the other at 3%.
12. A sum of $2,200 is divided and invested so that the simple interest on one part at 5% equals the interest on the other at 6%. Find how the money is divided.
272 FIRST-YEAR MATHEMATICS
Mixture Problems
433. The total amount of the mixture in the following problems may be expressed in two ways. Solve the equations thus obtained.
1. Bell-metal is by weight 5 parts tin and 16 parts copper. How many pounds of tin and copper are there in a bell weighing 4,800 pounds ?
First method: Let x be the number of pounds of tin, then show
16a: 16a:
that x-\ — — = 4,800, and solve for x; then find —^ . 5 5
Second method (without fractions): Let 5a: be the number of
pounds of tin, and solve: 5a: + 16a: = 4,800.
After finding x, calculate 5a: and 16a:.
2. Gunpowder contains, by weight, 6 parts saltpeter, 1 part sulphur, and 1 part charcoal. How many pounds of saltpeter, of sulphur, and of charcoal are there in 120 lb. of gunpowder?
3. If gunpowder were composed of 4 parts, by weight, of saltpeter, 2 parts sulphur, and 3 parts charcoal, how many pounds of each would there be in 200 lb. of gunpowder ?
4. With ingredients as in exercise 3, how much saltpeter is burned in the discharge of a cannon using 50 lb. of powder to the cartridge ?
5. Baking powder is composed of 4 parts, by weight, of cream of tartar, 1 part starch, and 1 part soda. How much of each ingredient is there in 18 lb. of baking powder ?
6. A certain mixture weighing 240 lb. contains 3 parts, by weight, of copper, 5 parts of iron, and 4 parts of carbon. How may pounds of each ingredient are there in the mixture ?
7. In a watchcase weighing 2 oz. the gold is 14 carats fine; i.e., there are 14 parts of gold in every 24 parts of the whole alloy. How many ounces of pure gold are there in the case ?
8. A certain compound contains, by weight, 5 parts carbon to every 3 parts of iron, and 7 parts of iron to every 2 parts of copper. In 124 lb. of the compound how many pounds are there of carbon, of iron, and of copper?
EQUATIONS WITH ONE UNKNOWN 273
Lever Problems
434. Law of leverages. In order to solve by means of the equation certain problems arising out of the common uses of forces it is necessary to know a law of these forces.
A bar of lever (Fig. 269) has _ ^
loadings as follows : A force of "^ '"i^
(+3) on an arm of ( — 6) and . , , , ^ ^ , ^ ^ _ ^ a force of (—3) on an arm of I i <^
^ ^ +3 -3 Ti
( — 2). Adding the turning- / V
tendencies, the total turning- ^^^==^
^ , . Fig. 269
tendency is
(+3)(-6) + (-3)(-2) = (-18) + (+6) = -12
Thiss ays in mathematical language that the bar turns _5 in negative direction, i.e., clock-
^-£^ wise.
^ ' ' ' ' ' ' If a force of (+2) on an arm of (—3) and a force of jTjQ 270 (~3) on an arm of (—2) are
on the bar, Fig. 270, the total turning-tendency is
(+2)(-3) + (-3)(-2) = (-6)+(+6)=0
which means that the bar does not turn.
If a force of ^..f.i.j
(+3) on an arm of I- rJ. ^..tt.
(+2), a force of | '''''''' A
1t
+ 2-5
J I I I f
(—4) on an arm of !^
(—9), and a force Fig. 271
of ( — 12) on an arm
of (+3) are on the bar, Fig. 271, the total turning-tendency is
(+3) (+2)+(-4) (-9)+(-12) (4-3) = +6+36-36= +6
which shows that the bar turns in the positive direction.
Experiments like the one above show that: If two
or more forces are acting on the bar at the same time
274 FIRST-YEAR MATHEMATICS
the total turning-tendency is found by adding algebraically the separate turning-tendencies. If the algebraic sum is zero, the bar balances. If the sum is not zero, the bar turns in the direction indicated by the sign of the sum.
Thus, the following law will furnish an equation in lever problems:
Law of turning-tendencies or leverages: For balance, the algebraic sum of all the turning-tendencies must equal zero.
EXERCISES
Using the law of leverages solve the following problems :
l-__rJL_^ 1. If a force of —4 on an arm of
H"-^H -9, Fig. 272, and a force of +/ on
I I an arm of —2, are on the bar, what
-^ -"-^ must / be for balance?
Fig. 272
If the bar is balanced, the sum of the
turning-tendencies must be zero. We may then write
(/)(-2) + (-4)(-9) =0 (1)
Multiplying, -2/+36 = 0 (2)
Subtracting 36, -2/ = -36. What axiom? (3)
Dividing by -2, /= 18. What axiom? (4)
Check: From equation (1),
( + 18)(-2) + (-4)(-9) = (-36) + (+36)=0
2. A bar, Fig. 273, is balanced _6 +5
by a force of +10 on an arm of m
— 6, and a force of s+3 on an arm L ^ ^ ^
t t
of +5. Find the values of s and
' -no s+S
s+3. Fig. 273
<----*— ^--^—> 3. A force of Sw - 15, Fig. 274, on
— ^ j^ :^ an arm of —4 is balanced by a force
l_ of +12 on an arm of +3. Find the
Sw-15
Fig. 274 values of w and 3?y— 15.
EQUATIONS WITH ONE UNKNOWN
275
e%
..&'/,—.
I^
i
■1800
Fig. 275
4. AB^ Fig. 275, is a cr()wbar,'6-2 ft. long, supported at F, \ ft. from A. A stone presses down at A with a force of 1,800 pounds. How many pounds of force must be exerted by a man pressing down at B to raise the stone ?
The weight of the crowbar is to be disregarded.
5. With other conditions as in exercise 4, what would be the pressure at B if the fulcrum F (point of support) were 3 in. from A f
6. With the fulcrum § ft. from A (Fig. 275) what weight would be held in balance by a pressure of 200 lb. at B f
7. A suction pump, Fig. 276, is a device for raising water from wells. The handle, OB, works
==> against a pin at ^, so that the hand pushing downward at B raises a mass of water by the aid of a piston C connected with the handle OB by the rod OC. If 0A = 2 in. and 05 = 3 ft., what load at 0 will be raised by a force of 20 lb. pushing downward at Bf
8. With other conditions as in exercise 7 what force will be exerted at 0 by a downward force of 681b. at 5 f
/^i
O A
f-
Fig. 276
9. A stone slab S, Fig. 277, weighing 2,400 lb., rests with its edge on a point B, 6 in. from the fulcrum /^ of a crowbar FA 6 ft. long. How many pounds of force must be exerted at A to raise the slab?
¥
I
-2m Fig. 277
+ 'Af
276 FIRST-YEAR MATHEMATICS
10. Solve the following equations:
1. 3a;+15 = x-+25 ,. -x 2 ^5
6. X — ^— ^=/Y
2. 9x-8 = 25-2.x ^ '
3. lQ+Sx = Q+x 7 7H-5_!±5_0
11 ^ ~
4. 20-4x = 8-10a:
^ a:-3^x+2 _ 8. ^ = 'xil
9. x-^ = 2-{x-5)
a\x-S) 3a2 a-c
11. s(s-l)-s(s-2)=3
12. (s+3)=s+3(s+8)
s-1 s-2 ^
13. -T— ^ = 2
14. (8+1) (s+3) = (s+2) (s+5)-13
15. 4(s+6)-2(s-3) = 38
1 1
^^- (s+3)(s+6) (s+2)(s+8)
1^ g-4 , s+4 s+20 s-5 ^^- 3 "^ 7 ~ 10 "^ 5
_ 2s-3 s+5 s-1 ^3 18- -4 ~2 2"="^^
19. (57/+6) (22/+3) = ll?/+26+10i/2
20. 0=(9?/+4) (8i/+9)-72i/2+39i/-75
21. (1202/2+752/-165)^15 = 27/(4?/-3)
22. (78?/+37i/-63)^(6?/+7) = 64?/-5(ll?/-15)
y^2a y—Sb
23. ay-\-by-\-cy = d-^e 25. -35- = -2fr
V c y , c oA ^ I ^ ^_i_^
04 •^4-- = "-4-- 26. — — = — hr
EQUATIONS WITH ONE UNKNOWN 277
Summary
435. The chapter has taught the following suggestions * for obtaining the equation of the problem :
I. Denote the unknown number by a letter and trans- late the verbal statement of the number relations of the problem into a symbolic statement.
II. Obtain two different number expressions for some number of the problem and write them equal to one another.
436. The letters x, y, z or some of the other later letters of the alphabet are generally used to denote the unknown numbers. The earlier letters, as a, h, k, I, m, etc., are commonly used to denote known numbers.
437. An equation is solved by obtaining another equation in which the unknown is alone on one side. This is done by means of the addition, subtraction, multi- plication, and division axioms.
438. Special devices have been taught, as being help- ful in the solution of problems of certain types. Thus, in problems containing geometric relations a figure should be drawn. Often the equation is found by express- ing algebraically some geometric theorem. Motion problems are solved by expressing algebraically two of the three elements — distance, rate, and time. The third is then found from these two by means of the equation: distance equals rate times time, or, d^r-t.
439. Clock problems, mixture problems, percentage and interest problems, and lever problems all give practice in expressing algebraically the verbal statements of problems.
440. Law of leverages. For balance, the algebraic sum of all the turning-tendencies must equal zero.
CHAPTER XVII
LINEAR EQUATIONS CONTAINING TWO OR MORE UNKNOWN NUMBERS
A System of Two Linear Equations
441. Some problems lead to two linear equations in two unknowns.
Find two numbers such that 3 times the first diminished by 2 times the second is equal to 9, and 3 times the second is 4 greater than 2 times the first.
Let X and y denote the required numbers; then, by the first condition of the problem
3x-22/ = 9 (1)
By the second condition of the problem
3?/ = 4+2a; ,.(2)
Solving both equations for y
.=V-.-.- (!')
and yJ-^p '. (2')
Equations (1') and (2') express y in terms of x. To every value assigned to x there is a /corresponding value of y; e.g., from (!') it follows that i/ = 6, when x = 7.
442. Solution of a system. A pair of values of x and y satisfying either equation is a solution of that equation. A pair of values of x and y satisfying both equations is a common solution of the two equations. Two equations having a common solution are called simultaneous equa- tions and the pair of equations is referred to as a system of equations.
278
EQUATIONS WITH TWO OR MORE UNKNOWNS 279
To solve the problem in § 441 means to find the solution of the system of equations (1) and (2), or of the system (10 and (20.
In the following paragraphs, two methods of obtaining the solution will be presented.
Graphical Method of Solving a System of Equations
443. Graphs of equations in two unknowns.
A quick way of obtaining the graph of the equation Sx-9 .
y=—2~ ''
9
(1) to let a: = 0. This gives !/= —2= —4.5
(2) to let y = 0. Then ^^^=0, 3x=9, x=S
Thus, we have two solutions of the equation: x=0, 2/=— 4.5, and x = 3, y = 0.
This is sufficient for drawing the graph. For the pur- pose of greater accuracy a third value of x, as :z; = — 5, may be taken. The corresponding value oi y is —12. Tabu- late these three solutions of the equation, as in Fig. 278 (table 1).
Similarly obtain three solutions of the equation
2/=?^. (See table 2 to the left in Fig. 278.) The
. . 4
value ?/ = 1 . 3 + in this table is an approximation for ^.
o
Each solution in the tables locates one point on the graph; e.g., to graph the solution ( — 5, —12) pass from 0 5 units to the left and from there 12 units downward,
280
FIRST-YEAR MATHEMATICS
locating point A. In the same way points B, C, D, E, and F are determined. Then draw lines ABC and DEF.
y" |
A- J^ ^ |
t ^^ |
y ^ |
T ^ At |
V-^-^ .'JZ^' |
SO.UTON \y =6 ^/y^> |
'^' i ^sr |
^&- |
^h^hiy J |
^^Z |
_,^,_.v.^^ / |
CorV:J) ^ '^ |
4: _ ^^S:cz ._ _ ±: |
(^.PJ^'F ' >!!(•>! 0 1 |
^ ■ ^ i |
„ IT Z |
^^ n. / |
-h ^1^ -BA-&u^sr- |
~ y 1 |
^^ 1 |
TABLE 2 ^' / TABLE 1 |
-. ^^ z it |
_^.J^ • J. - . *I^_ |
^-^ -^ A/i-5, m -^ ^ |
^1 |
.9 0/ -5 12 |
Fig. 278
444. Co-ordinates. The pairs of values of x and y
locating points A, B, C are the co-ordinates of
A, B, C , etc.
EXERCISES
1. If the drawing, Fig. 278, is made accurately, any solution
Sx — 9 of the equation y = — ^ — ^^^^ locate a point on line ABC. To
show this, find various solutions of the equation, not found in the tables, and graph them.
2. Any point on ABC, Fig. 278, will furnish a solution of the equation 3x — 2y = 9. E.g., the co-ordinates of point G are x= —7,y= —15. Show that they satisfy the equation. Deter- mine other solutions of the equation from points on ABC.
EQUATIONS WITH TWO OR MORE UNKNOWNS 281
445. Indeterminate equations. Exercises 1 and 2 above show that there are as many solutions of the equa- tion as there are points on the graph, i.e., an indefinitely large number. For this reason equations in two unknowns (variables) are called indeterminate equations.* How- ever, a system of two equations may be determinate.
For, the co-ordinates of any point on the graph ABC
3a;— 9 determine a solution of y = — ^ — and the co-ordinates of
any point on DEF determine a solution of y = — - — .
o
Hence, the point of intersection, H, oi ABC and DEF
determines the common solution of these equations,
i.e., x = 7, y = Q.
446. Summary. The following is the process of solving a system of equations graphically:
1 . Find at least two solutions of each equation, preferably three.
This is done quickly by substituting for x the value 0 and finding the corresponding value of y and then substituting y = 0 and finding the corresponding value of x. Any third value of x may be used to obtain the third solution. If any values come out fractional, it is best to use approximations to one decimal place.
2. Tabulate the solutions.
3. Graph the solutions.
* Diophantus of Alexandria (fourth century a.d.) made a study of indeterminate equations. Indeterminate equations are therefore known as Diophantine equations.
Rene Descartes (1596-1650) made the discovery that a point in a plane could be completely determined if its distances, say x and y, from two perpendicular lines were known. He saw that though an equation in two unknowns was indeterminate, the values of x and y satisfying the equation determined the points on a line representing the equation. (See Ball, pp. 272, 273.)
282 FIRST-YEAR MATHEMATICS
4. Determine the point of intersection of the graphs and the co-ordinates of that point.
5. The co-ordinates of the point of intersection are the solution of the given system of equations.
EXERCISES
Solve the following systems by the graphical method :
1. |
x-\-2y=17 |
Sx-y =2 |
|
2. |
2.-1 = 3 |
x-h2y = 7 |
|
3. |
7x-\-3y=-SQ |
5x+2y = 7 |
|
4. |
9x-6i/ = 36 |
lSx-\-5y = n |
|
5. |
2a:-3?/ = 4 |
4a:+5i/ = 30 |
x-{-2y=n |
.-1=1 |
8x-\-5y = U |
2x- y = 2 |
llx+7?/ = 40 |
3:^-5^ = 4 |
x-2y = 0 |
a: = 4 |
a:-2 = 0 |
2/= -3 |
10.
447. Normal form. When a linear equation in two or more unknowns has all terms containing the unknowns on one side and all other terms on the other side, and when all similar terms have been combined, as in 2x—Sy = 7, the equation is said to be in the normal form.
EXERCISES
Put the following equations into normal form:
1. 7x = H-9y 2. ~w+5 = ^
m+3p 3m-p x-\-y-S x-y-\-4: ^
'^^ 5 2 ~* ^- 2 ~ 3 ~^^
5. (a:+2)(?/-4)-(a:+3)0y-6) = 14
448. Disadvantages of the graphical solution. One of the disadvantages of the graphical method of solving a system of equations is the difficulty of finding accurate
EQUATIONS WITH TWO OR MORE UNKNOWNS 283
solutions of some systems. In the following pages an algebraic method will be shown which always gives accurate results. The algebraic method also has the ad- vantage that it is easily extended to solve equations in three or more unknowns. For practical purposes, as in the solution of problems in science, the approximate solutions obtained by the graphical method are generally sufficient.
Algebraic Solution of Equations in Two Unknowns
449. Elimination by addition or subtraction.
(a) A boy sets out on a walking trip and travels at a uniform rate for 5 hours, when he meets with an accident. He continues, however, at a slower pace, and 3 hours later reaches a point 26 miles from home. (6) If he had turned back at the time of the accident he would have reached in 3 hours a point 14 miles from home. What was his rate of speed both before and after the accident ?
Letting x and y denote the rate before and after the accident respectively.
Condition (a) |
5x-\-dy = 2Q |
Condition (b) |
5x-Sy = U |
lOx =40 (Add. Ax.) |
|
, x =4 (Div. Ax.) |
|
6t/ = 12 (Sub. Ax.) |
|
y= 2 (Div. Ax.) |
|
I the solution x -- |
= 4, y = 2 by substituting in the problem. |
The equation 10a: =40 was obtained from equations (1) and (2) by what operation? What terms were elimi- nated from equations (1) and (2) by this operation?
What is true of the coefficients of the y-terms which makes possible the elimination of these terms by addition f
The equation 6?/ = 12 was obtained from equations (1) and (2) by what operation? What terms were elimi- nated from equations (1) and (2) by this operation? What
284 FIRST-YEAR MATHEMATICS
is true of the coefficients of the ?/- terms which makes possible the ehmination of these terms hy subtraction f
This method of ehmination is called elimination by- addition or subtraction.
Solve by this method the following systems:
1. Z+s = 24.5 2. m+4n= 4
l-s= 8.5 m-2n=16
450. In the preceding systems the coefficients of the unknown numbers were numerically equal. We must learn what to do in case the coefficients are not equal, as in the following example:
Solve this system :
5x-{-Sy = 2Q (1)
^x-7y= 2 (2)
If equation (1) is multiplied by the coefficient of x in equation (2), and equation (2) by the coefficient of x in equation (1), what is the coefficient of x in each of the resulting equations?
What operation will then eliminate the a:-terms from the resulting equations ?
How may the y-ierms be eliminated from equations (l)and(2)?
The following gives the complete solution:
7X(1) 35x+21i/ = 182 4X(1) 20a: + 12i/ = 104
3X(2) 12a:-21i/= 6 5X(2) 20x-35i/= 10
Hence, 47x =188 Hence, 47y= 94
X = 4 y= 2
Having found the value of x as shown above, the value of y might have been found by substituting the value of x in one of the given equations, thus,
5a:+3z/ = 26 Check in (1) 5-4+3.2=26
5-44-3^ = 26
31/ = 6 Check in (2) 4-4-7.2=2
y= 2
EQUATIONS WITH TWO OR MORE UNKNOWNS 285
451. Summary. To eliminate by addition or sub- traction proceed as follows:
1. Make the coefficients of one of the unknown numbers numerically the same in both equations.
This is done by multiplying one or both equations by the proper number.
2. Then eliminate one of the unknowns either by adding or by subtracting the equations according as the coefficients of the unknown have unlike signs, or like signs.
EXERCISES
Solve the following systems of equations:
l2x-\-l5y = m lQx-25y=-2
Sh-21y = 33 Qh+S5y= 177
7x+5y=17 nx-3y = 5
6a-46 = 2 5a+76 = 43
1.
2.
5.
10.
Ux-^Sy=lS \Sx+2y = 9
j5x-\-4y = 22 \Sx-7y=-l5
Sx-5y = bl
2x-^7y = S
f 2t-7s = 5S
ilSx+10y = 59 [Ux- 9?/= 15
|-5s+9/; = 4 ] 7s+6p=-80
25ie+14r = 385 -15R-\- 9r = 30
21u+692J=lll 14w-26z^ = 2
(SSA-2SB = SS i22A+355 = 79
f66m+55n = 308 i77m-15n = 201
11.
12.
13.
14.
15.
16.
17.
18.
19.
-lla;+97/=16
4a;+8?/ = 28
Su- 2z? = 4 -7u-\-lSv=-]
9R-2r = U QR- r = 31
lSu-Qv = 22 4ii+9y = 61
3^_2^_
286
FIRST-YEAR MATHEMATICS
20.
21.
22.
\3x-2ij=-l
Sy
5
j7x-2y = 8 \3x+4i/ = 18
23.
24.
25.
2/w+lU = 50 llm+ 2A; = 41
s+6/> = 4s-f 5p (s+4)-(5p-s)-6- = 0
4ig 1 ,
T-4^=^
5^_2r^ 2 3"
Geometric Problems
452. The following geometric problems lead to equa- tions in two unknowns:
i(x -f V)}-yy
Fig. 279
1. With parallels, transversal, and angles as shown in Fig. 279 find x, ?/, and all the 8 angles.
2. With parallels, transversal, and angles as shown in Fig. 280 find x, ?/, and all the 8 angles.
3. With two parallels and a transversal, a pair of corresponding angles are {x-\-2y)° and 2{x — y)°, the angle adjacent to the latter being 120°. Find x, y, and the un- known angles.
3Z-2V,
c.-^
X.
4. With two parallels and a transversal a pair of alternate exterior angles are Fig. 280 {by — 2x)° and (9x4-2/)°. The angle adjacent
to the latter is 86°. Find x, y, and the unknown angles.
5. With two parallels, the interior angles on the same side of a transversal are {l4^ — y)° and (Qx-\-y)° and their difference is 14°. Find x, y, and all the 8 angles.
6. With two parallels, the interior angles on the same side of the transversal are {ix — y)° and 5(2i/+x)°. Two alternate
EQUATIONS WITH TWO OR MORE UNKNOWNS 287
Fig. 281
interior angles are 5{2y-^x)° and 125°. Find x, y, and the unknown angles.
7. One angle of a triangle is 64° less than the sum of the other two and 16° less than the difference. How large is each?
8. The angles made by two pairs of parallels intersecting as in Figs. 281 and 282 are designated as shown. Find x, y, and all the angles.
9. In the trapezoid ABCD, Fig. 283, angle C is 3 times as large as angle A and angle D is 5 times as large as angle B. How large is each angle ?
10. The difference of the acute angles of a right triangle is 36°. Find the number of degrees in each acute angle.
11. One dimension of a rectangle is 5 and one dimension of another is 3. The sum of the areas is 65 and the difference 35. Find the dimensions of the rectangles.
12. The sum of the areas of two rectangles of dimensions 5 and X, and 3 and y, is 49, and the area of a rectangle of dimensions 3 and x exceeds the area of a rectangle of dimensions 5 and y by 9. Find x and y.
13. The areas of two triangles having
equal bases are 72 sq. in. and 60 sq. in. / \
Twice the altitude of the first plus 3 a^ \b
times the altitude of the second is equal Fig. 283
to 54 in. Find the altitudes.
14. The altitude of a trapezoid is 8 and the area is 56. If the lower base is increased by a length equal to the upper base, the area is 72. Find the bases of the trapezoid.
15. The lower base of a trapezoid is 24 and the area is 150. If a length equal to e of the lower base is added to the upper base the area is 170. Find the altitude and upper base of the trape- zoid.
Fig. 282
FIRST-YEAR MATHEMATICS
16. The triangles in Fig. 284 have equal corresponding parts as indicated. The letter x has the same value throughout.
The same is true of y. Find x and ?/.
17. Theratioof the lengths of two circles is 2. Six times the radius of the first minus 4 Find the radii and
10x-6y
Fig. 284
times the radius of the second is equal to 14.
lengths of the circles.
22 The length of a circle is 2'jrr, where T = -=r approximately.
18. The areas of two circles are to each other as 4 : 36. One- half the radius of the first plus \ of the radius of the second is equal to 6|^. Find the radii and areas.
4
The area of a circle is irr^. Hence — -„ =
root of both sides of the equation.
7r?/2 36
Extract the square
Motion Problems
453. Absolute and relative velocity. While a freight train is moving at the rate of x mi. an hour a brakeman walks along the top of the cars of the train from rear to front at the rate of y mi. an hour. Show that the velocity with which the brakeman moves over the ground, his abso- lute velocity, is equal to his own velocity on top of the cars, his relative velocity, increased by the velocity of the train. Express his absolute velocity in terms of x and y.
Determine the absolute velocity of a brakeman walking along the top of the cars of a train in the direction opposite to that of the train if his relative velocity is y and the velocity of the train x.
EXERCISES
1. Two trains A and B pass each other going in the same direction at rates of 30 mi. and 40 mi. an hour
EQUATIONS WITH TWO OR MORE UNKNOWNS 289
respectively. What is the speed of A relative to B ? What would it be if the trains were moving in opposite directions?
2. Two trains pass each other going in the same direc- tion with a relative speed of 10 mi. an hour. Going in opposite directions they would pass with a relative speed of 70 mi. an hour. Find the speed of each train.
3. A boat crew rows 4 mi. downstream in 20 minutes and the same distance upstream in 35 minutes. Find the rate of the boat crew in still water and the rate of the current.
4. A steamboat can run 7| mi. per | hour upstream and 4| mi. per J hour downstream. What is its rate in still water and what is the rate of the current?
5. A and B run a race of 450 yards. In the first trial B begins with a start of 60 yd. ahead of A and A wins by 18 seconds. In the second trial B begins with a start of 30 seconds and wins by 10 yards. Find the rates of A and B.
Miscellaneous Problems
454. Problems involving number relations. The fol- lowing exercises give practice in expressing nurdber rela- tions in the form of equations :
1. The difference of two numbers is 17 and the sum is 167. Find the numbers.
2. If 13 times one number be subtracted from 3 times another the difference obtained is 41. The sum of 1 1 times the first num- ber and 8 times the second is 18. What are the numbers ?
3. Three tons of hard coal and two tons of soft coal cost $32. The price remaining the same, 2 tons of hard coal and 6 tons of soft coal cost $43 . 50. What were the prices per ton of the two kinds of coal?
4. A street railway company receives a certain sum for each cash fare and a different sum for each transfer. On one trip 13 cash fares and 18 transfers were taken, amounting to $1 . 10 for
290 FIRST-YEAR MATHEMATICS
the company. On the return trip there were 7 cash fares and 24 transfers and the amount for the company was $0.95. What does the company receive for a cash fare ? For a transfer ?
455. Digit problems.
1. The tens digit of a number is 4 and the units digit is 3. Indicate in symbols the numbers of units in the tens. Indicate the number of units in the number.
2. The tens digit of a number is t and the units digit is u. Indicate the number of units in the number.
3. A number is denoted by 10t-\-u; what will denote the number formed by reversing the order of the digits ?
4. Show, in symbols, a number whose digits are x and 4; X and y; y and x.
5. Show, in symbols, a three-digit number whose digits are a, b, and c; c, a^ and b; x, z, and y; y, z, and x.
6. Express, in symbols, that a number whose units digit is 3 less than the tens digit equals 27 more than the number obtained by writing the digits in the reverse order.
7. A number having 2 digits is 4 times as large as the sum of the digits. If the order of the digits is reversed the resulting number is 27 larger than the original number. What is the number?
8. The difference of a number of 2 digits and a number having the same digits but in reverse order is 18. The sum of the digits is 8. What are the numbers ?
9. A number of 2 digits is equal to 9 less than 7 times the sum of the digits. If the order of the digits is reversed the number obtained is 18 less than the original number. What is the original number ?
456. Income problems.
1. A man gained 8 per cent on one investment and lost 3 per cent on another. If the money invested amounted to $22,000 and the net gain was $440, what was the amount of each investment ?
2. Two investments, one at 3^ per cent, and the other at 5| per cent, yield annually $150. If the first had been at 8f per
EQUATIONS WITH TWO OR MORE UNKNOWNS 291
cent, and the other at 31 per cent, the annual income would have been $175. What was the amount of each investment?
3. A man invested two sums at 4 per cent and 5 per cent respectively, receiving annually $158. He reinvested the money at 5 per cent and 6 per cent respectively, receiving $36 per year more than before. Find the sums invested.
4. A part of $2,500 was invested at 3.5 per cent and the other at 4 per cent. The second investment brings $25 less per year than the first. How much was each investment ?
5. A part of $4,000 is invested at 3 . 5 per cent and the remain- der at 4.5 per cent. The first investment yields in four years $15 less than the second investment in two years. Find the sums invested.
Fractional Equations
457. Solve the following equations:
fx+y_x-y
2 3
x-\-y
x-y
3 ^ 6-47?
3 372-4
7 6 5 4 4
3H-8
2 8/7-2
4.
5.
[7u'Av = 77'A()
fr-2 v-\-4: r-S~v-\-5
r±l_v±S (r-h2~v-4: 2m 5n 1 5 "6"'"~2 m 5n 5 6"^"9'"2 2.2^+.8 ■5;^-.4
1.5 ~ .3 .5^-. 3
1.5/1-
5a:- 18+
2x 1 , 7-5y 9.
•27 =
3 17 -2c
2"^ 2
y-gC^ + l)
2.5
1
0^-1)-
16
„, 2c-d ■2d=-^--
5d-8
2 55+c
5
10.
i{w+l)-u = ^{w-l)- 2a-b a+7 4 ~ 5
36-5+
3— 2+5a-
18 = ^^
292 FIRST-YEAR MATHEMATICS
^j U+^(3a-6-l)=i+l(6-l) ' [TV(76+24) = H4a-f36)
[6-^ R±5_ ^^ 12. 2-5 --^^
li(12+i/) -1(72-4) =4
20-2X . 5ii: H AK-IQ
13.
3 ' 6 6"^ 3
2K±21__R±5K 3 ~ 6
458. Systems of three or more linear equations.
Solve the following problem: " The sum of 3 times the first, 5 times the second, and 3 times the third of three numbers is equal to 22. The sum of 5 times the first and 3 times the second, diminished by 4 times the third, is equal to — 1. If from the sum of 4 times the first and twice the second, 5 times the third is subtracted, the remainder is —7. What are the numbers ?
Letting x, y, and z denote the first, second, and third numbers respectively, the equations are:
3x+5^+32= 22 (1)
5x+3z/-42=- 1 (2)
^x+2y-bz=- 7 (3)
Eliminating z from (1) and (2), and again from (1) and (3) or (2) and (3), what two unknowns will the resulting equations contain ? Solve these equations for a: and 2/ : Eliminating z from (1) and (2) :
4X(1) l2x-\-2Qy+l2z= 88 (4)
3X(2) \bx+ ^-\2z=-^ (5)
(Add. Ax.) 27.T+29?/ = 85 (6)
Next eliminate z from (1) and (3), thus:
5X(1) 1.5x+25i/ + 15z= 110 (7)
3 X (3) \2x+ 6y-152=-21 (8)
27a: +311/ = 89 (9)
Solving (6) and (9), a: = l and i/ = 2 Substituting in (2), 5.1+3-2-42 = -1
z= 3
EQUATIONS WITH TWO OR MORE UNKNOWNS 293
(x = l Thus, the sohition of the given system is the set of numbers :l y=2
[z = 3 The solution is checked by substituting these values of x, y, and z into equations (1), (2), and (3):
Check in (1);
3-l-f5. 2+3-3= 22; i.e., 22=22
Check in (2) Check in (3)
5.1+3.2-4.3 = -l; i.e., -1 = -1
4.1+2-2-5.3= -7; i.e., -7= -7
The method used in solving the system of equations of the foregoing problem consists of three principal steps:
1. Make two different pairs of equations out of the three equations, and eliminate the same unknown from both of these pairs.
In the problem the unknown was eliminated by the method of addition and subtraction.
2. This gives two equations in two unknowns. Solve the two resulting equations as a system of two linear equations.
3. Substitute the values of the two unknowns just found in any one of the given equations containing the third unknown, and solve the resulting equation.
EXERCISES
1 . The sum of two angles of a triangle exceeds the third angle by 26°. Five times the difference of the first two is 8° more than the third angle. Find the number of degrees in each angle.
2. The sum of two numbers is 2 greater than 3 times a third
number, the difference is equal to the third number, and twice
the third number increased by the second is 4 greater than the
first. Find the numbers.
1 R
3. The angles of a triangle are A, B, and C; jA-\--^ = C,
and 2A+T7jB = 2^+30. Find the values oi A, B, and C.
294
FIRST-YEAR MATHEMATICS
4. Solve the following systems: [ Sx-h y-2z=-\
1. -4x+2y/+30 = 9 I 5x+3?/-22; = 5 (2u-{-2v-{- w = 9
2. ] u-\-Sv-\-2w=l^ [Su-Sv-\-^w = 9
{1„ I l.,i 1_ 2 3
9.
10.
11.
13
14
f a-f36+5c = 21 ^3a-26-4c = 22 t4a-36-6c = 28
(2p-\-2s+3v = 4: \3p-\-As-\-6v = 7 [ p-\-2s+Qv = 4: [4a:-4|/+82 = 3 I x-\- y-\- 2 = 0 I X— y — 4:Z = 2
7. 3x+4i/+6<; = 2x+6?/+52 = 4x+2?/+9s = 68
F+K = 2 K-\-R = ^
2^4 5
B . C
A_
3
ABC
2 + 3
12
= -13
4 4 2
2a;+3i/+ z = i dx-2y-{- z = i} Sx-\- 2/+22 = |- 7x+52/-2z = V^ 5x-27/+32 = 5 2x+9!/-52; = -V- a:+ 2/+2 = 3 X— y-\-z = Q [-y-{-Sz-^x=12 {^u- v= 0 12. { bv -2w = 40 [Gw— w=15 [5a:+3?/ = 45 7y-2z = 27 Sz-^=-l2 3v-2u = l 7u-^w = i
15.
16.
17.
18.
19. ^
ie+i7=6
x-h y-\- z=l 2x+3i/+42=-3 3a:— 4?/— 52 = 14 x-{-2y + 32=14
2.C+ ■?/ + 2 = 7 .3a;+.22/+.22 = 1.3
a:+l =2(?/+l)
y+2 =2+1 1(2+3) =x+l
2 +2"'^
10
20.
1-^ = 4 ^ 2
32-2.c = t
^-2 = '
^+ 2=14 X+?/ + 2 + lA=10
a; — ?/+2+it = 6 x+t/— 2+M = 4 x-\-y+z—u = 2
. EQUATIONS WITH TWO OR MORE UNKNOWNS 295
Problems
459. Solve the following problems:
1. The sum of the 3 digits of a number is 16. If the order of the digits is reversed, the new number is 396 less than the original number. If the middle digit be placed first, the resulting num- ber is 90 less than the original number. What is the number ?
2. The sum of the 3 digits of a number is 15. The second digit is 2 times the first less the third. If the order of the digits is reversed the resulting number is 198 less than the given number. What is the original number?
3. A sum of $18,000 is invested as follows: One part at 3.5 per cent, a second part at 5 per cent, and the rest at 4 per cent. The total interest is $730. If the first part had been invested at 4 per cent, the second at 3 per cent, and the third at 6 per cent, the total annual interest would have been $840. How much was each part ?
4. A man invested three sums of money in the following ways : The first sum at 5 per cent and the second at 4 per cent, giving an income of $490. The second at 5 per cent and the third at 3 per cent, giving an income of $540. The three sums together amounted to $19,000. How much was each sum ?
Summary
460. The meaning of the following terms was taught in this chapter: system of equations in two or more un- knowns; solution of a system of equations; method of elimination by addition and subtraction; co-ordinates of a point; indeterminate equations; normal form of an equation in more than one unknown; absolute and rela- tive velocity.
461. A system of equations in two or more unknowns may be solved by the method of elimination by addition
296 FIRST-YEAR MATHEMATICS
and subtraction. A system of equations in two unknowns may be solved by the graphic method.
462. Some of the following typical problems lead to equations in several unknowns: geometric problems, motion problems, digit problems, and income problems.
463. To solve a system of equations graphically, find two, or three, solutions of each equation, graph these solutions, draw the lines determined by them, determine their point of intersection, and find the co-ordinates of that point.
464. To solve a system of equations by elimination, make the coefficients of one of the unknown numbers numerically the same, then by adding or by subtracting one equation from the other, one of the unknowns may be eliminated.
465. A system of equations in three unknowns is solved by eliminating the same unknown from both of two pairs of equations made from the three given equations. This gives a system of two equations in two unknowns. By solving this system the values of two of the unknowns are found. The value of the third unknown is obtained by substituting the values of the other unknowns (now known) in any of the given equations. Systems in more than three unknowns are treated similarly.
l.EONHARD ElILER
LEON HARD EULER
LEONHARD EULER was born at Basle on April 15, 1707, and died at St. Petersburg on September 7, 1783. He was educated under the direction of the great mathematical teacher, John Bernouilli, at Basle in Switzerland. In 1725 he went to St. Petersburg as pro- fessor of mathematics, taught there until 1741, when he moved to Berlin, Germany, at the request of Frederick the Great. He lived in Berlin until 1766, when he returned to St. Petersburg. Within two or three years of his return to Russia he became totally blind, but in spite of this he worked on until his death.
He wrote an immense number of papers dealing with all sorts of mathematical subjects. He reformed almost all the branches of pure mathematics that were known, added numerous details and proofs, created much of what is now called analysis, and arranged the entire material of mathematics in the consistent form that we have today. He was probably the most prolific mathematical writer the world has known, and most of his work was of a very high order of merit, though his style was diffuse and prolix.
In 1748 he wrote his Introductio in analysin infinitorum, the first part of which contains about what our modern texts contain on algebra, theory of equations, and trigo- nometry. He separated trigonometry from astronomy and made it a branch of mathematics. He here introduced the use of e for the Napierian base and tt for the ratio of a circumference to the diameter of a circle. The second part was on analytic geometry.
He wrote a book on differential calculus, one on integral calculus, and much too many other books and memoirs to permit of even mention here. See an account of Euler in any good encyclopedia, or in Ball's or Cajori's histories of mathematics, and note what is said about his ways of working in mathematics; for he is one of the very best mathematical authors to read and study to learn how to become a really independent worker in mathematics.
CHAPTER XVIII
THE FORMULA
The Formula as a General Rule
466. Formula. The ideas of mathematical statements can often be expressed in algebraic symbols with gain in clearness and conciseness. Thus, the statement, ''The area of a rectangle equals the length by the height," takes the form A=lh, I denoting the length, h the height, and A the area; the statement ''distance = rate X time" may be expressed by the equation d = rt; etc. The symbolic form of such statements not only is more comprehensible than the verbal form, but may easily be applied to a par- ticular case by assigning definite values to the letters; e.g., the area of a rectangle 2 in. long and 3 in. high is found by letting 1 = 2 and h=3, in the equation A =lh.
A general statement, or rule, in which letters and sym- bols are used in place of numbers or words is a formula. A formula is an algebraic equation. If the values of all but one of the letters are known, the value of the unknown is then found by solving the equation. Thus, the volume
4 of a sphere is expressed by the formula F=^7rr\ where V
o
22 is the volume, '^ = -tj- approximately, and r is the radius.
To find the radius of a sphere whose volume is 14y cubic
inches, let V = U\. Then 14 J = ^ X ^r^ or ^ = | X ^H.
o 7 7 o 7
9 4 2 Dividing both sides by 11, ;^=5X;^r^, from which it
7 o 7
follows that r = l\ inches.
297
298 FIRST-YEAR MATHEMATICS
467. Motion-problem formula. The formula is used in solving motion problems.
EXERCISES
1. An express train whose rate is 45 mi. per hour starts 1^ hr. after a freight train and overtakes it in 2| hr. Find the rate per hour of the freight train.
The formula for uniform motion is d = r -1,4 denoting the distance, r the rate, and t the time.
Let r be the rate of the freight train
Then 4:j =time the freight train travels
Hence, 4jr = distance the freight train travels
45 = rate of the express train
2f = time of the express train Hence, 123f = distance the express train travels Therefore, 4jr = 123f r = 29T\
2. An express train whose rate is R mi. per hour starts h hr. after a freight train and overtakes it in t hours. Find the rate per hour of the freight train.
Let . r= rate of the freight train
Then h-\-t = tim.e the freight train travels
(^-f-0^ = distance the freight train travels R = rate of the express train t = time the express train travels Rt = distance the express train travels
Hence, {h-\-t)r = Rt ^ Rt ^ h+t
Rt
3. Using the equation r= t—t^. as a formula, solve exercise 1.
Lets = 45; <=2f; h = l\
4. An express train whose rate is 40 mi. per hour starts 1 hr. 4 min. after a freight train and overtakes it in 1 hr. 36 min. How many miles per hour does the freight train run ?
Use the formula in exercise 2.
THE FORMULA 299
5. Find h in terms of R, t, and r.
Clearing the equation r = j— of fractions, we have hr-\-tr = Rt.
Hence, hr = Rt — tr and h =
' r
468. Work-problem formula. The work-problem is a type of problem for which a formula is easily obtained.
EXERCISES
1. A can do a piece of work in 5 days and B in 7 days. How long will it take them to do it together ?
Let n be the number of days it will take them together
Then - = the amount of work they can do in 1 day n
- = the amount A can do in 1 day 5
= = the amount B can do in 1 day
Hence, h\ = \
Multiplying by 35n, 7n+5n = 35 12n=35 n = 2H It is clear that numbers other than 5 and 7 would be used just as the 5 and 7 are used here.
2. A can do a piece of work in a days and B in 6 days. How long will it take them to do it together ?
Let n be the number of days it will take them together
Then - = the amount of work they can do in 1 day n
- = the amount A can do in 1 day
the amount B can do in 1 day
Hence, " + l=~
' a 0 n
Multiplying by afen, hn-\-an = ab {b-\-a)n=ab
ab
'"'= — TT
300 FIRST-YEAR MATHEMATICS
Any problem of the type of exercise 1 may be solved by using the equation n = —jj^ as a formula. Thus, to solve exercise 1,
leta = 5, 6 = 7. Then n = -^=j^ = 2{i
3. A can build a wall in 10 days and B in 14 days. How long will it take them to do it together ?
Use the formula of exercise 2.
4. A and B can build a fence in 7 days. B can do it alone in 12 days. How long will it take A to do it alone ?
Use the formula.
5. Solve the equation n = ^rT for a; for 6.
Clearing of fractions, na-\-nh = ab na—ab = nb a(n — b) = —nb nb n — b
EXERCISES
1. Express in symbols the following laws of arithmetic:
(1) The product equals the multiplicand times the multi- plier.
(2) The dividend equals the divisor times the quotient plus the remainder.
(3) The product of a fraction by a whole number is the product of the whole number by the numerator, divided by the denominator.
(4) The sum of two fractions having the same denominator equals the sum of the numerators divided by the common denominator.
(5) The quotient of a fraction divided by a whole number is equal to the numerator divided by the product of the whole number by the denominator.
(6) The quotient of two fractions equals the dividend multi- plied by the inverted divisor.
THE FORMULA 301
(7) The square root of a fraction equals the square root of the numerator divided by the square root of the denominator.
(8) The square of a fraction is the square of the numerator divided by the square of the denominator.
2. Translate into words the following formulas expressing laws of arithmetic:
(1) P = Tqq (percentage law)
lovt
(2) ^ = Tqq (interest law)
(3) -r-\-:r = — j-i (addition of fractions)
(4) -i — -T = — -ri (subtraction oi tractions)
tti fl2 UlCl2
.^x ni W2 nin2 , , . ,. . » .
(^) w" ^ T ~ 'TT (multiplication of fractions)
(6) :T--j--r = ^X:f (division of fractions)
Cli Cf2 Cfi tti
3. Express the following statements in symbols:
(1) The distance passed over by a body is equal to the rate times the time.
Denote the distance by d, the rate by r, and the time by I.
(2) The power (or rate of doing work) is equal to the amount of work divided by the time in which it is done.
Denote the pDwer by P, the amount of work done by W ^ and the time by t.
(3) The time of vibration, ^, of a pendulum is equal to ir times the square root of the quotient of the length, /, divided by the acceleration, gr, due to gravity.
(4) The velocity is equal to the space passed over divided by the time.
(5) The momentum of ^ mass is equal to its velocity multi- plied by the mass.
302 FIRST-YEAR MATHEMATICS
(6) The time it takes a body projected upward to reach the greatest height is equal to the velocity of projection divided by the retardation due to gravity. The greatest height is equal to the square of the velocity of projection divided by 2 times the retardation due to gravity.
4. Translate the following formulas into words:
(1) If the velocity obtained by a falling body in t seconds is denoted by Vt and the acceleration due to gravity by g, then i)t = gt.
(2) The distance, s, passed over by a falling body in t seconds is s = igf. The velocity obtained in the same time is v = V'2gs.'
(3) If a fluid of density d moves with a velocity v, the diminu- tion of pressure due to the motion is p = 2dv'^.
(4) A body of volume V immersed in a liquid of density D is buoyed up by a force F=DgV (Archimedes' principle).
(5) For a perfect gas, changing from pressure p and volume V to pressure p' and volume v' without change of temperature, pv = p'v' (Boyle's law).
Evaluation of Formulas
469. Find the value of the letter called for in the following formulas:
1. The distance traversed by a moving body is equal to the rate multiplied by the time; that is, D = rt.
Find Z), if (1) r = 30 ft. per second, and t = 5 seconds. (2) r = 5 mi. per hour, and t=l7 hours. .
2. The area of a rectangle is equal to the product of the base and the altitude; that is, A = ba.
Find A, if (1) 6 = 13 ft., and a = 24 feet.
(2) 6=10.2 in., and a = 3. 5 inches.
3. The area of a triangle is equal to i the product of the base by the altitude; that is, A = 2ba.
Find A, if (1) 5 = 12 ft., and a= 16 feet.
(2) 6 = 8.2 rd., and a = 7. 78 rods.
THE FORMULA 303
4. The area of a i)arallelogram is equal to the product of the base by the altitude; that is, A = ba.
Find A, if (1) 6 = 28, and a=19.
(2) 6 = 16.3, anda=14.6.
5. Two weights, Wi and W2, will balance on a beam that lies across a stick when the distances, di and ^2, of the weights from the stick are in the inverse ratio of their weights; i.e., when
dl_W2
di ~ Wx'
Find dx, if (1) da =18 ft., ^2 = 60 lb., w;i = 50 pounds.
(2) c?2 = 27 in., w^^^M lb., Wi = 24 pounds.
Find ^2, if (1) c?i = 40 in., wi=\^ lb., iyi = 18 pounds.
(2) rfi = 25 in., w;2 = 3.8 lb., Wi = 2.85 pounds.
6. The weight, w, of any mass is equal to the volume, v, multiplied by the density, d; i.e., w = vd.
Find w, if 2; = 64 cu. in., and d=16.2 pounds. Find v,i{ w = 648 lb., and rf = 12 . 2 pounds. Find d,ifw = 800, and v = 160.
7. The weight, w, that a force, p, pulling up a smooth slope, h ft. high and I ft. long, will just move is given by w = tP.
Find h if w;=120 lb., / = 20 rd., p = 60 pounds.
8. A stone falling from rest goes in a given time 16 ft. mul- tiplied by the square of the number of seconds it has fallen; i.e., s = 16<2
Find s, if ^ = 4 sec; 11.5 seconds. Find <, if s = 64 ft.; 1,600 feet.
9. A stone, thrown downward, goes in a given time 16 ft. 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; i.e., s=l6t'^-^vt.
Find s, if ^= 12 sec, and v = 3 ft. per second, if i= 8 sec, and 2^ = 7 ft. per second. Find V, ii t= 5 sec, and s = 500 feet.
304 FIRST-YEAR MATHEMATICS
10. The time, t, taken for a pendulum to make a single vibration equals tt ^i- , where I is the length of the pendulum.
We take g approximately equal to 32, and v equal to V .
Find t,iil=^8 ft.; if / = f|f feet. Find I,' if ^ = 1 sec. ; if ^ = 4 seconds.
11. The length of a circle is approximately equal to \^ of the diameter; i.e.,C = 7rd.
Find C, if d = 21 ft.; 7 ft.; i foot. Find d, if C = 88 ft.; 66 ft.; 16 feet.
12. The volume of a sphere equals ^tt times the cube of the radius; i.e., V^iirr^.
Find V, if r = | ft.; 11 ft.; 2 feet.
13. The area of a circle is tt times the square of the radius; i.e., A = 7rr^.
Find A,ii r = 3 ft.; 7 ft.; 21 feet.
Find r, if A = 154 sq. ft.; 220 square feet.
14. Find the length of a belt connecting two pulleys whose diameters D and d are 22 in. and 24 in. respectively, the distance, a, between the centers being 9 ft. The length is given by the formula
15. The diagonal, d, of a rectangular parallelopiped of dimensions a, b, and c is given by the formula d = ^a^-\-¥-\-c'^ . Find d, if a = 5, 6 = 4, c = 3.
16. Find the area A, of a. circular ring, formed by two concentric circles of radii R and r, if R = 12, r=10, and
A = 7r(/22-r2).
17. li a,b, and c denote the lengths of the sides of a triangle and s one-half of the perimeter, the area is ^ s{s—a){s — b){s — c).
THE FORMULA 305
Find the area of a triangle whose sides are 10, 6, and 8; 10, 17, and 21; 3, 4, and 5; 5, 12, and 13.
1}'
18. h = ^ (law of falling bodies). Find h,\i v==ll,g = Z2.2;
if ?; = 2, ^ = 32.2
Wl shh^
19. x = -g-- Find h, if TF = 6,748, / = 5.5, s = 3,500, 6 = 6
iflF = 4,954, Z = 3, s = 2,500, 6 = 5.5
Er
20. X = 7--^. FindX, if ^ = 0.056, r = 2, 7=1.4
rv'
21. R = ——,. Find/?, if r=11.5, r' = 6.5
r-\-r ' '
if r=13, r' = 15
22. E = -^. Y'm^E, ifM = 12, F = 5
ifM = ll, F = 9 FindM, if £;= 8, F = 4 if £^ = 50, F = 5 F h ^. , ^ ._ _ .. , 3
23.
p = 2^. Find P, if F = 25, r=lS,h = i
24. h = j77^ TT. Find r, if s = 425, h = S
P dr '^^' W^2M' Find TF, if 72 = 20, /=10, r = 9, P = 75, rf = i
470. Expressing one of the letters of a formula in terms of the others.
Solve the following equations :
1. A = - 2~ • ^, for h; for 6; or 6' (area of a trapezoid)
7)vt
2. ^ = Tqq for p; for r; for t; for pr; for rt (interest
formula)
306 FIRST-YEAR MATHEMATICS
3. V = -lbh for b; for h (volume of a pyramid)
4. C=| (F— 32) for F (centigrade in terms of Fahrenheit)
5. pd = PD, for p; for (i; for P; for D (law of levers)
E
6. (7= D , for E; for R; for ?- (current in a circuit includ-
ing external and internal resistance)
7. v, = 7r+T- for /^; for/i; for/2. (F = principal focal dis-
tance of lens, /i and /2 = conjugate focal distances)
8. p?; = po^o(l+27q/ for po^o; for t (expansion of gases)
9. R =
273
]^ . d
+^ for I (/^ = radius of curvature from spherom- eter readings)
10. C = ^+^ forCi; for Co
1 7Sf 7
11. ^17L = — for 1^; for .S; for - o c c
12. ax — bx = Qa—Qb, for a:
13. cx+ax = a^-{-c'^-\-2ac, for a;
14. aa; = a2 for .t
c '
15. — hr = -r,fora a b ab^
a^ix-S) 3a2 a-c ,
16. ca; = , for a:
c c c '
^^ XX b'^-a? .
17. -—T = —T—,10VX
a b ab '
X X 6a-86 -
THE FORMULA 307
Summary
471. The chapter has taught the following:
1 . The formula expresses in algebraic symbols the ideas of a mathematical statement with gain in clearness and conciseness.
2. The formula may be used to solve any particular problem of a type by assigning definite values to the letters.
3. The laws of arithmetic may be expressed con- veniently by formulae.
4. Any letter of a formula may be expressed in terms of the others, by solving the equation for that letter.
CHAPTER XIX
REVIEW AND SUPPLEMENTARY QUESTIONS AND PROBLEMS
CHAPTER I
472. Measurement of line-segments.
1. Give the meaning of the following terms: straight line, geometric line, geometric point, line-segment, and unit of length.
2. Draw a line-segment and measure the length approxi- mately to two decimal places.
3. Draw a triangle and measure the sides approximately to two decimal places.
4. A train travels at the rate of 24 mi. an hour. Represent graphically the distances passed over in each of the first 12 hours.
5. In 1913 and 1914 a certain magazine contained approxi- mately as many lines of advertisements as given in the table below. Make the graph and tell what it shows.
Months |
Jan. |
Feb. |
Mar. |
Apr. |
May |
June |
[1913 Number of lines< U914 |
12,000 |
11,700 |
13,600 |
15,900 |
18,400 |
16,600 |
13,600 |
16,000 |
19,000 |
20,200 |
21,400 |
20,100 |
Months |
July |
Aug. |
Sept. |
Oct. |
Nov. |
Dec. |
fl913 Number of lines< U914 |
14,800 |
12,700 |
13,300 |
16,000 |
18,700 |
25,000 |
15,800 |
15,000 |
15,700 |
16,900 |
20,000 |
26,100 |
308
REVIEW AND SUPPLEMENTARY
309
6. The following table gives the temperature readings for a certain day:
2:00 A.M. |
4:00 |
6:00 |
8:00 |
10:00 |
12:00 |
2:00 P.M. |
4:00 |
6:00 |
8:00 |
10:00 |
12:00 |
70° |
66° |
65° |
66° |
68° |
77° |
81° |
81.5° |
83° |
82° |
80° |
77° |
D] |
raw t |
he t€ |
mpe |
rature curve and tell what the g CHAPTER II |
raph shows. |
473. Graphical addition and subtraction.
1. Give the reasons for the following conclusions and illustrate them graphically:
(1) If a = S, 6 = 5, then a+6 = 8.
(2) Ifm = 8, n = 2, thenm-n = 6.
(3) li a = x, b = y, then a-\-b = x-^y.
(4) If a = s, b = t, then a—b = s — t.
2. Construct 3a+46-2c.
3. If m = 2.4, construct 3m-h4//i.
474. Perimeters.
1. Find the perimeter of a triangle whose sides are 21, 0/, and 5/, respectively. What is the length of the perimeter if /=1.2?
2. Give the meaning of the following terms: triangle, perim- eter, polygon, quadrilateral, pentagon, hexagon, n-gon.
3. Give the meaning of the following: coefficient, term, monomial, binomial, trinomial, polynomial.
4. Show by a sketch figures whose perimeters, p, are given by the equation: p = 4:x; p = Qx-{-lSy; p = 3x-\-4x-\-5x.
5. Find the value of p in problem 4 for a: = 4, ?/ = 2; a; = 1 .3, 1/-2.7.
310 FIRST-YEAR MATHEMATICS
475. Algebraic addition and subtraction.
1. State how to add similar monomials.
2. Add as indicated:
4a4-7a+10a; 2a+56+6a+36; 12Ax-\-S.Sx
3. State how to find the difference of similar monomials.
4. Reduce to the simplest form: 12x — 8a;; 5.7y—2.8y; (8a -a) -(5a -2a).
5. Illustrate by an example the commutative law of addition.
6. Illustrate the associative law of addition.
7. Simplify:
a+{10a+(8a-6a) + (4a-a)} 106-6+[{ 166-26- (36+6)+56| -6]
CHAPTER III
476. Use of axioms in solving equations.
1. Give the meaning of the following terms : equation, solving an equation, checking, root of an equation, satisfying an equa- tion.
2. Show how the value of p in the equation 4/?-|-9 = 17 may be found with the balance.
3. State the axioms used in solving equations.
4. Solve the following equations, giving in each step the axiom used, and check:
10a:-9 = 65; ^=14; a+|a=12; ^ = 20; x+5.37x-8.73 = 61.34
Solve the following problems:
5. Three times a certain number exceeds 40 by as much as 40 exceeds the number. What is the number ?
6. Six times a number less 5 is the same as 4 times the number increased by 1. Find the number.
REVIEW AND SUPPLEMENTARY 311
7. A line 20 in. long is to be divided into 2 parts, one of which increased by 14 in. shall be equal to the other increased by 10 inches. Find the length of each part.
8. Find two consecutive numbers whose sum is equal to 247.
9. The length of a rectangular field is 3 times the width, and the length of a fence surrounding the field is 744 feet. How wide is the field ?
CHAPTER IV
477. Classification of angles.
1. How many degrees are there contained in the angle formed by the hands of a clock at 2:00 o'clock? at 6:00 o'clock? at 8:00 o'clock?
2. Draw a right angle, a straight angle, a perigon.
3. Draw an acute angle, an obtuse angle, an oblique angle.
478. Measurement of angles.
1. Using only ruler and pencil, draw angles containing 40°, 75°, 135°, 270°. Check the accuracy of your angles by measuring them with a protractor.
2. What part of a right angle is 1 degree ?
3. How many minutes are there in a degree ? How many seconds in a minute?
479. The use of the protractor.
1. Give the meaning of the following: circle, arc, radius, semicircle, quadrant. Illustrate each by means of a figure.
2. Draw an obtuse triangle. Measure each angle and find the sum of the three angles.
480. The sum of the angles of a triangle.
,1. li X, y, and z denote the number of degrees in the angles of a triangle, what equation expresses a relation between x, y, and z ?
312 FIRST-YEAR MATHEMATICS
2. Show that the sum of the interior angles of a triangle is 180°.
3. One angle of a triangle is 30° greater than another. The third angle is 10° greater than twice the first. How large is each angle ?
4. Show that the sum of the exterior angles of a triangle is 360°.
5. Find the exterior angles of a triangle if the first is 30° greater than the second, and the third is | of the first, plus 15°.
6. Give the meaning of the following: obtuse triangle, acute triangle, right triangle, equiangular triangle, equilateral triangle. Illustrate each by means of a figure.
7. Show that in the same, or equal, circles equal central angles intercept equal arcs.
8. Show that in the same, or equal, circles equal arcs are intercepted by equal central angles.
481. Problems of construction.
1. Draw an angle equal to a given angle, using the protractor.
2. Using ruler and compass only, construct at a given point on a given line an angle equal to a given angle.
3. Draw a right angle, using the protractor.
4. Construct a right angle, using ruler and compass.
5. Construct an angle equal to the sum of two given angles, using ruler and compass.
6. Construct an angle equal to the difference of two given angles, using ruler and compass.
7. Bisect a given angle.
8. At a given point in a line construct a perpendicular to the line.
SIR ISAAC NEWTON
SIR ISAAC N E A\ T O N
SIR ISAAC NEWTON, the world's greatest mathema- tician and physicist, was born at Woolsthorpe, Lincoln- shire, England, December 25, 1642, and died at London, March 20, 1727. At Trinity College, Cambridge, where he was educated, his genius for mathematics showed itself even before graduation. He was prepared for his life-work by reading Oughtred's Clavis, Descartes' Geometrie, Kepler's Optics, Vieta's works. Van Schooten's Miscellanies, and Wallis' Arithmetica.
He took his B.A. degree in 1665, and within a year had discovered the binomial theorem and invented the fluxional calculus. He discovered numerous laws of physics, the law of gravitation, and wrote numerous works, which he usually published only at the urgent insistence of his friends, and generally much later than he finished them. He wrote a work which he entitled Universal Arithmetic, which was really an algebra. In this work we find the first use of positive and negative fractions and irrational numbers used as exponents. His masterpiece, and one of the greatest works of all times, is the Principia, in three books. It was published in 1687.
Newton at one time represented his university in Parlia- ment, in 1705 he was knighted by Queen Anne, and for twenty-five years he was president of the Royal Society. He held the Lucasian chair of geometry at Cambridge from 1669 to 1701.
Ball says of his appearance: "Newton was short, and towards the close of his life rather stout, but well set, with a square lower jaw, brown eyes, a broad forehead, and rather sharp features. His hair turned gray before he was thirty, and remained thick and white as silver till his death."
Bishop Burnet called him "the whitest soul" he ever knew.
Ball says, "Newton was always perfectly straightforward and honest; but in his controversies .... though scrupu- lously just, he was not generous."
Newton said of himself: "I do not know what I may appear to the world: but to myself I seem to have been only like 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 undis- covered before me."
Ball's History of Mathematics gives an excellent account of Newton's life and works from p. 319 to p. 352 (5th ed.). Read it for the sake of the interest and inspiration it will give you.
REVIEW AND SUPPLEMENTARY 313
CHAPTER V
482. Area of a square.
1. Draw figures representing the following quadrilaterals: parallelogram, rectangle, square, trapezoid, rhombus.
2. State the formula for the area of a square.
3. Find the area of squares of sides 2.4 cm.; 5f cm.; 13 cm.
4. The perimeter of a square is 16a. Find the area.
483. Area of a rectangle.
1. State the formula for the area of a rectangle.
2. The frame of a picture of rectangular form of dimensions 18 in. by 12 in. is 2j in. wide. Find the area of the frame.
484. Cube and parallelopiped.
1. State the formula for the volume of a cube.
2. State the formula for the volume of a rectangular paral- lelopiped.
3. The edge of a cube is 12 inches. Find the area of the total surface.
4. The length of a parallelopiped is 6 in., the width is 8 in., and the height is 10 inches. Find the area of the total surface.
5. The length of an edge of a cube is /. Find the surface and volume.
6. The three dimensions of a rectangular parallelopiped are a, b, and c. Find the area of the surface and the volume.
485. Graphing equations.
1. Graph the equation y = 10x—2.
2. Graph the equation y = 2x'^-\-l.
486. Multiplication of monomials.
1. Give the meaning of the following: exponent, base, power. Illustrate each by giving an example.
314 FIRST-YEAR MATHEMATICS
2. Find the values of the following for x = 2\ 4a;, x^, 4x^,
{2xY, ^^, X^+X' + X-1, X^'X', X'X'
3. Illustrate by an example the commutative law of nmltiphcation.
4. Simplify the following products: 3a'^bc'2ab'^C' 4:abc'^; 10a'' ' 5a' ' 2a; {3ab)(2a''b)il0a%^)', {Qa'^vxif)'
487. Addition of monomials.
1. What is meant by the coefficient of a factor in a term ?
2. In 2a'^bx, what is the coefficient of a%x ? Of a; ? Of a'^x ?
3. Reduce the following polynomials to the simplest form:
4a+7a+15a mx-\-4:X-\-nx -\-3x 4{x+y)+2{x-\-y) + {x+y)
488. Multiplication of a polynomial by a monomial.
1. Find by means of a rectangle the product of x4- 4 by a.
2. Multiply as indicated : x{a-\-5b); x-\-y - m;
a4+3a2. 262+64. 26(a+6-3c); 5+a;(2a;+3); 2m{Sm^-{-2m-i-7)
3. Factor the following: ab-\-ac', x^—xy; 5xy — 15x^] a62c+a26c2; 60^27/+ 1 2a;?/ +18x-y; SSaY-\-o7aY-l9a'
4. Solve the following equations:
H-3_^-2 4 3 1 ,1_.
7 6 ' n'^'^'n"^'*' 3/"^2/~^
489. Multiplication of polynomials by polynomials.
1. By means of figures express as polynomials the following products: (a+6)(a+6); {m-{-n){2x-\-'Sy)
2. Draw rectangles whose areas are expressed by the following trinomials: ab-{-ac; 2xy-^4:x''
REVIEW AND SUPPLEMENTARY 315
3. Draw squares whose areas are expressed by m?-\-2mn-\-n^]
4. Multiply as indicated:
{x''^-2x+\){x+l)] (a'-+2a6+62)(a2+2a6+6"')
490. Area of parallelogram and triangle.
1. Show that the area of a parallelogram is equal to the 'product of the base and altitude.
2. Show that the area of a triangle is equal to one-half of the product of the base by the altitude.
3. Show that the area of a trapezoid is equal to one-half of the product of the altitude by the sum of the bases.
CHAPTER VI
491. Adjacent angles.
1. What are adjacent angles? Make a drawing of two ad- jacent angles.
2. What are perpendicular lines?
3. Show that at a given point in a given line only one per- pendicular can be drawn to the line.
4. What is the sum of the adjacent angles about a point, on one side of a straight line ?
5. The angular space about a point, on one side of a straight hne, is divided into angles denoted by 3^;^ 2{x-\-9), x, and 42— a:. Find X and each angle in degrees. Draw the figure.
6. What is the sum of the angles at a point just covering the angular space about the point ?
7. All the angular space about a point is divided into angles represented by x, 36+5a:, and 3a; — 9. Find x and each angle in degrees. Draw the figures.
492. Supplementary angles.
1. What are supplementary angles? Illustrate your answer with a drawing.
316 FIRST-YEAR MATHEMATICS
2. The diiference of two supplementary angles is 110. Find them.
3. If an angle is increased by 12°, and the supplement is divided by 5, the sum of the angles obtained is 80°. Find the supplementary angles.
493. Complementary angles.
1. What are complementary angles ? Illustrate your answer with a drawing.
2. The difference between an angle and its complement is 27°. Find the angle.
3. If an angle is increased by 15° and the complement is divided by 3 the sum of the angles obtained is 75°. Find the complementary angles.
494. Opposite angles.
1. What are opposite angles? Make a drawing of opposite angles.
2. Show that if two lines intersect, the opposite angles are equal.
Sx 5x
3. Two opposite angles are denoted by 5x+-j- and ^+130.
Find X and the angles. Draw a figure representing these angles.
495. The acute angles of a right triangle.
1. Show that the acute angles of a right triangle are com- plementary.
X
2. The acute angles of a right triangle are denoted by ^+2x
87 5x and -w — -n- Find x and the angles.
3. If in a right triangle the acute angles are 30° and 60° respectively, how does the hypotenuse compare with the shortest side?
REVIEW AND SUPPLEMENTARY 317
496. Equations.
Sa-2 1 3
1. Solve
14 2 14
2. Solve ^+30 -§ = 57
3. Solve 2(a-2)+~^-4 = 4
497. Angle pairs formed by two lines intersected by a third.
1. If the corresponding angles are equal prove that the alternate interior angles are equal.
2. If the corresponding angles are equal prove that the interior angles on the same side are supplementary.
3. If the alternate interior angles are equal show that the corresponding angles are equal.
CHAPTER VII
498. Parallel lines.
1. When are two lines said to be parallel?
2. If two parallel lines are cut by a transversal how do the corresponding angles compare ? If two given lines are cut by a transversal making the corresponding angles equal, what is known about the given lines ?
3. Show that two lines perpendicular to the same line are parallel.
4. Show that two lines are parallel if two alternate interior angles formed with a transversal are equal. . (Use problem 3, § 497.)
5. Show that two lines are parallel if the interior angles formed with a transversal are supplementary.
6. Show that two lines parallel to the same line are parallel to each other.
318 FIRST-YEAR MATHEMATICS
7. Show that if two parallels are cut by a transversal the alter- nate interior angles are equal and the interior angles on the same side are supplementary. How may this be used to draw a line parallel to a given line ?
8. Prove that if two angles have their sides parallel they are either equal or supplementary.
9. The corresponding angles formed by two parallels and a transversal are denoted by 2(4a: — 3)° and (79+3a:)°. Find x and the angles.
10. Show that the sum of the interior angles of a triangle is 180°.
11. Show that an exterior angle of a triangle is equal to the sum of the two remote interior angles.
12. What is meant by a parallelogram ?
13. Prove that the opposite angles of a parallelogram are equal.
14. Prove that the consecidive angles of a parallelogram are supplementary.
15. The consecutive angles of a parallelogram are so related that 3 times one angle diminished by the other is equal to 30°. Find the number of degrees in each angle.
16. Make drawings representing the following: cube, parallelopiped, prism, pyramid, cylinder, cone, sphere.
17. Name on the figures of problem 16 the following: parallel lines, parallel planes, lines parallel to planes, parallel lines cut by a transversal, non-parallel lines cut by a transversal, lines perpendicular to each other, lines perpendicular to a plane.
CHAPTER Vin
499. Drawing to scale.
1. What is meant by indirect measurement?
2. To measure the width, AC, of a stream an engineer lays off a line, BC, on one side of the river, and measures the angles ACB&ndABC. If B(7 = 45rd., ZACB=nO°,am\ ZABC = 40°, find AC hy means of a scale drawing.
REVIEW AND SUPPLEMENTARY 319
3. The angle of elevation of the top of a pole is 38°, the ob- server standing 20 yd. from the pole. How high is the pole ?
4. From the top of a cliff 150 ft. high the angle of depression of a boat is 25°. How far is the boat from the top of the cliff ?
5. The view from a battery at B to the enemy's fort at F is obstructed. A point P is located from which F is observed to bear 4 mi. northeast. P is 6.24 mi. northwest of B. Find the distance and bearing of F from B.
500. Ratio.
1. What is meant by the ratio of two numbers? Illustrate.
2. What is the ratio of two line-segments ?
3. What number added to 12 and subtracted from 30 gives results that are in the ratio t^ ?
4. Divide 81 into three parts that are in the ratio 2:3:4.
501. Similar figures.
1. What are similar figures?
2. State the relation existing between the sides of similar triangles. How do the angles compare ?
3. The sides of a triangle are 8, 10, and 13. The shortest side of a similar triangle is 11. Find the other sides.
4. A tree casts a shadow 90' long. At the same time a verti- cal stick 4' long casts a shadow 50' long. How high is the tree ?
5. In the adjoining figure CD represents the distance across a swamp. AB is par- aZ—^-^h aUel to CD. If AB = 100', AE = SO\ and C^=150', find CD.
CHAPTER IX
502. Trigonometric ratios.
The angle of elevation of the top of a tower is 27° and the distance from the foot of the tower is 259 ft. Find the height of the tower: first, by a scale drawing; second, by trigonometry.
320 FIRST-YEAR MATHEMATICS
503. Ratio.
1. In the adjoining figure, DE is parallel to AB. Find EB, if DC = 4, AD =.27, and CE=l^DC. State the principle used in obtaining the equation.
2. In the adjoining figure CD bisects angle C. If AC = 3", BC = Q", and AD is 2" less than DB, find AD. State principle used in obtaining the equation. c
3. Bisect a line-segment.
4. Divide a line-segment into parts
3 having the ratio j.
5. Using ruler and compass only, find the ratio of two given line-segments.
6. Find the greatest common divisor of 3,542 and 5,016.
7. Reduce to simplest lorm „ ^.^ ^
8. How is a ratio reduced to the simplest form ?
504. Variation.
1. Give the meaning of the following terms: variable, con- stant, function. Illustrate by an example the meaning of each of these terms.
2. What is meant by the statement : x varies directly as ?/ ?
3. The simple interest on an investment varies directly as the time. If the interest for 6 years on a sum of money is $200, what will be the interest for 8 years ?
4. The distance, d, through which a body falls from rest varies directly as the square of the time, t, in which it falls. State the equation for d and t for a body observed to fall 400 ft. in 5 seconds. Graph the equation.
5. What is the meaning of the statement : x varies inversely as ?/?
REVIEW AND SUPPLEMENTARY 321
6. The apparent size, s, of an object varies inversely as the distance, d. Express the statement by means of an equation.
505. Proportion.
1. What is a proportion? Illustrate your statement by giving several examples.
2. Prove that in a proportion the product of the means equals the product of the extremes.
5
3. Divide $2,400 into two parts having the ratio ^.
4. bolve the equation ~r5="— rr
x~\~o x~\~ i
5. If 80 lb. of sea- water contain 4 lb. of salt, how much fresh water must be added to make a new solution of -which 45 lb. contain f lb. of salt ?
6. If x = ~jfinddiif 6^2 = 27 ft., t<;2 = 36 lb., and w;i = 24 pounds.
7. What per cent of evaporation must take place from a 90 per cent solution to produce a 95 per cent solution ?
8. Prove that the areas of two rectangles are to each other as the products of the dimensions.
9. Prove that if two rectangles have equal bases they are to each other as the altitudes.
10. Prove that the areas of two triangles are to each other as the product of the bases and altitudes.
11. Prove that if two triangles have equal bases they are to each other as the altitudes.
CHAPTER X
506. Congruence.
1. What are congruent figures ?
2. Prove that two triangles are congruent if two sides and the included angle of one are equal respectively to two sides and the included angle of the other.
322 FIRST-YEAR MATHEMATICS
3. How may the theorem in problem 2 be used to find inac- cessible distances ?
4. Prove ihsittwo triangles are congruent if two angles and the side included between their vertices in one triangle are equal respec- tively to the corresponding parts in the other.
5. Prove that the base angles of an isosceles triangle are equal.
6. Prove that an equilateral triangle is equiangular.
7. Prove that the bisector of the vertex angle of an isosceles triangle bisects the base and is perpendicular to it.
8. Prove that all points on the perpendicular bisector of a line-segment are equidistant from the end-points of the segment.
9. Prove that if a line bisects an angle of a triangle and is perpendicular to the opposite side the triangle is isosceles.
10. Prove that if two angles of a triangle are equal the triangle is isosceles.
11. Prove that an equiangular triangle is equilateral.
12. Prove that if the perpendicular bisector of one side of a triangle passes through the opposite vertex the triangle is isosceles.
13. Prove that if two sides of a triangle are unequal the angles opposite them are unequal.
14. Prove that if two angles of a triangle are unequal the sides opposite them are unequal.
15. Prove that if three sides of one triangle are equal respec- tively to three sides of another the triangles are congruent.
16. Prove that if each of two points of one line is equally distant from two points of another line the line joining the first two points is the perpendicular bisector of the segment joining the other two.
17. What is the locus of all points equidistant from the end- points of a line-segment ?
18. Prove that two right triangles are congruent if the hypot- enuse and a side of one are equal respectively to the hypotenuse and a side of the other.
REVIEW AND SUPPLEMENTARY 323
19. Prove that the shortest distance from a point to a line is the perpendicular from the point to the line.
20. Prove that obUque Unes drawn from a point on the per- pendicular to the hne and making equal angles with the perpen- dicular are equal.
CHAPTER XI
507. Fundamental constructions.
1. To bisect an angle. What method of proof is used to prove this construction? Make the construction and give the proof.
2. At a point mi a given line to construct a perpendicular to the line. Make the construction. State the principal theorem upon which the proof of this construction is based. Give the proof.
3. To bisect a given line-segment. Make the construction. What is the principal theorem used to prove the construction? Give proof.
4. To construct the perpendicular bisector of a line-segment. Make the construction and give the proof. Compare this construction with that in problem 3. Compare the proofs of problems 3 and 4.
5. From a point outside of a line to construct a perpendicular to the line. Make the construction. Upon what principal proposition is the proof based ? Give proof.
Q. At a given point on a given line to draw a line making an angle with the given line equal to the given angle. Make the con- struction. What method of proof is used? Give the proof.
508. Application of the fundamental constructions.
1. Construct a triangle having given the following parts: two sides and the included angle; two angles and the side between their vertices; three sides; two sides and the angle opposite one of them.
324 FIRST-YEAR MATHEMATICS
2. Construct a right triangle having given the following parts : the hypotenuse and one of the other sides; the hypotenuse and one of the acute angles.
3. Construct an isosceles triangle having given one of the base angles and the altitude.
4. Construct an equilateral triangle having given the alti- tude.
5. Construct a right triangle whose acute angles are 30° and 60°.
6. To trisect a right angle.
7. Construct angles of 60°, 30°, 15°, 120°, 90°, 45°, 22°30', 135°, 75°, 165°.
8. Through a point outside of a given line to draw a line parallel to the given line.
509. Theorems.
1. Only one perpendicular can be drawn to a line at a given point on the line. Give reason.
2. Only one perpendicular can be drawn from a point to a line. Give reason.
510. Symmetry.
1. Give several illustrations of symmetric bodies.
2. Name several single bodies symmetric with respect to a plane.
3. Draw several figures symmetric with respect to a line.
4. What is an axis of symmetry ?
5. Establish the following theorems by means of the sym- metry of figures :
a) A point on the perpendicular bisector of a line-segment is equidistant from the end-points.
b) A point not on the perpendicular bisector of a line-segment is not equidistant from the end-points.
REVIEW AND SUPPLEMENTARY 325
c) A point on the bisector of an angle is equidistant from the of the angle.
d) A point not on the bisector of an angle is not equidistant from the sides.
6. Show that the locus of points, within an angle, equidistant from the sides is the bisector of the angle.
511. The circle.
1. Give the meaning of the following and draw a figure for each to illustrate your statement: tangent, point of contact, regular polygon, inscribed polygon, circumscribed polygon.
2. Prove the following theorems:
a) The radius drawn to the point of contact of a tangent is perpendicular to the tangent.
b) A line perpendicular to a radius at the outer end-point is tangent to the circle.
3. Make the following constructions:
a) At a point on a circle construct the tangent.
b) Find the center of a given circle.
c) Draw a circle passing through two given points.
d) Circumscribe a circle about a triangle. Give proof.
e) Inscribe a circle in a triangle. Give proof. /) Inscribe a square in a circle. Give proof.
g) Circumscribe a square about a circle. Give proof.
h) Inscribe a regular hexagon in a circle. Give proof.
i) Prove that perpendicular bisectors of the sides of a triangle pass through the center of the circumscribed circle.
k) Prove that the bisectors of the angles of a triangle pass through the center of the inscribed circle.
CHAPTER XII
512. Uses of positive and negative numbers.
1. What are positive numbers? Negative numbers?
2. Give the meaning of absolute, or numerical, value.
326
FIRST-YEAR MATHEMATICS
3. State some of the uses of positive and negative numbers. Illustrate each case.
4. Graph the following hourly temperature readings:
8:00 A.M. |
9:00 |
10:00 |
11:00 |
12:00 |
1:00 P.M. |
2:00 |
3:00 |
4:00 |
5:00 |
6:00 |
-6° |
-4° |
-2*' |
+2° |
+4° |
+3° |
+3° |
+2° |
+0° |
-2° |
-3° |
513. Addition of positive and negative numbers.
1. Find the following sums graphically:
(+5) + (-h3); (+5) + (-3); (-5) + (+3); (-5) + (-3)
2. State a rule for adding positive and negative numbers.
3. Find the sums in problem 1 algebraically.
4. What is the most advantageous way of adding three or more algebraic numbers?
5. Find the sums of the following -236; 25a:, -38a:, -20a:, 30a:, -6a:.
6. Add the following: +25a +3ia: -3.14/c
+49, +35, -45, +75,
7a
2ta:
+4.26A;
-2Aa% + 18a26
— 6 . ^x^yz^
514. Subtraction of positive and negative numbers.
1. State the rule for subtracting algebraic numbers.
2. In the following subtract the lower number from the upper : +1 +18c -6.3a2 -4.2(a+6)
7 — ^
+ 3
-12c
-7.2a2
+2.7(a+6)
515. Multiplication of positive and negative numbers.
1. State the law of signs in multiplication.
2. Find the value of the following products: (+2)(+8); (-f)(+|); (+3)(-a:); (-7ia)(-3j6)
REVIEW AND SUPPLEMENTARY 327
3. Express the turning-tendency, or leverage, in terms of the force and arm.
4. In the following the first factor is the force, the second is the arm. Find the leverage.
(+2)(-3); (-2)(-3); (-2)(+3); (+2)(+3)
5. Give the value of the following products :
0X&; 6X0; (4-3)(Q)(+4); (1.4)(-3.2)(0)
6. Findthevaluesof 3a;3-2x2+7x-4fora; = 0, 3, 1, -2, -4.
516. Division of positive and negative numbers.
1. State the law of signs for division.
2. Show that — = "~r~ = — 1~
-y -\-y +y
3. Find the quotients of the following:
(-l)-(-l); (-2a;)-^(+6a:); (+4a6)^(+6); (-3.6x2)^(+4a;?/)
CHAPTER XIII
517. The laws of addition.
1. Illustrate the commutative law of addition.
2. Illustrate the associative law of addition.
518. Addition of monomials.
1. Add the following monomials:
+ 12a:, -3a:, -f 14a; {-bak), {+2bk), {-{-ck)
2(m+n), -6(m+7i), (m-\-n) a(x-\-y), b{x+y), -c{x-{-y) U{a^b-c), -8ia%-c) 3a% -5xy, -\-7a% -\-2xy
519. Addition of polynomials.
1. State how to add polynomials.
2. Arrange according to ascending powers of x] according to descending powers of x:
x^+x^ - 2xhf - 77/ - SxY-^^^y^
328 FIRST-YEAR MATHEMATICS
3. Add the following polynomials:
-4ia26+16a62-763+a3; -4:a^J^8ab''-2¥-2ia''b
4. Add 7{x-y)-4:{x+y)+4:'7; 9(x+y)+3(x-y)-9 - 7; Q{x — y)-\-2' 7 — Z{x-\-y). Verify the results by letting x = 2, y = l. (Chicago)*
520. Subtraction of monomials.
1. State how to find the difference of two monomials.
2. Subtract the lower from the upper:
-12a -7.5x -{-16a%c -{-17 .2{x-3y^)
+ 3a -2.3a; -20a%c -\-22. 6ix-3y^)
3. Subtract as indicated: {-SAa'^b^c^)-{-6.Sa%''c^); \-A.5{x'-if)]-{-7.Q{x'-y')\
521. Subtraction of polynomials.
1. State how to find the differences of polynomials.
2. Subtract as indicated:
{2Ax''-7.Sxy-{-lQy^)-{-S.7x''-\-2Axy-10i/)
3. From 16 - 15-30 +14 (a; - 5?/2;) - 13 {5y-x) subtract 32 — 16-30+8(5?/ — 2) without performing any of the indicated multiplications. (Chicago)
522. Removal of parentheses.
Add and subtract as indicated and simplify:
1. Sx-[Ay-{Sy-\-7x)]
2. 5p3- {3p2- (2p2 + 77?3) _ (3p2-|-4p3) j _ (p3_^i0p2)
3. Remove the parentheses: a'^—{2ab — [b'^ — {c^—2cd—d^)]\ (Chicago)
* (Chicago) means : taken from an entrance examination given by the University of Chicago.
REVIEW AND SUPPLEMENTARY 329
CHAPTER XIV
523. Multiplication of monomials.
1. State how to multiply monomials.
2. Find the following products and test the result by sub- stituting values for the letters:
(+22)(-14)(-3); {-2iax'y)i-^Wxy); {-^x'n-2xy{x'y
3. Find the valueof :c2-12a;+4fora: = 2; fora;=-3.
524. Multiplication of polynomials by monomials.
1. State how to multiply a polynomial by a monomial.
2. Muhiply as indicated and test by substituting values for the letters:
4:X-\-Sx{x-2y) + (2x-^y)x; 2a|26-3(4a-6)+4(3a-o6)}
525. Multiplication of polynomials by polynomials.
1. State how to multiply polynomials by polynomials.
2. Multiply pa'^-\-qb^-rc^ by a-\-b-c. (Chicago)
3. Multiply x^-{-2xY-^y^ by x^-2xhf-\-i/.
4. Multiply lSx-12y-x7j-\-3 by 5x-3y-\-xy-\-5.
526. Reduction of quotients.
1. State how to reduce quotients to the simplest form.
2. Reduce the following to the simplest form:
-25a'^bd mx^y-4:5xy^+90xy
12a%c^ 15xy
{-x)hj{-zy 20{a-^by- 15{a-{-bY-\-S0{a+hy
'5{-x)Y{-z) -5(rt+fe)3
330 FIRST-YEAR MATHEMATICS
527. Division of polynomials by polynomials.
1. Divide x^-\-x'^y-\-xy~-\-xz'^-\-yz^-{-z^ by x-\-y.
2. Divide 10xY-^^^^+^'^x^-xhf-^^y^-\-'^y^
by 2y^ — 3xy^—^xhj-\-Qx^, and check the result by substituting x = y = l. (Chicago)
CHAPTER XV
528. Special products.
1. Draw a figure to show that ic-\-dy = c'^+2cd-{-d\
2. State how to find the square of a binomial without a figure, i.e., by inspection.
3. Find the trinomials equal to the following products:
(p-rY; {Axy-.3zy; [{m-\-n)-tY
4. Give by inspection the following products as the difference of two squares:
(36-2c)(36+2c); {l+U')(l-lx'); (x-^y){x-y){x'+y'^)
5. Give by inspection the following squares of trinomials:
{2m-is-\-ty; (a-. 36+2)2
529. Factoring.
1. Factor the following trinomials:
49 - 140n2+ lOOn^; ^9m''n^+^2mnxy-\-9xhf
2. Find the factors of the following binomials:
l-25?/2; lQQ-225d%Y; x^-y^; {a'-by-c'
3. Find the factors of
Qx^-29xy-\-35i/; 2a2+lla+12; 102-llm-m2
530. The theorem of Pythagoras.
1. State the theorem of Pythagoras.
2. Express the diagonal of a square in terms of the side.
REVIEW AND SUPPLEMENTARY 331
531. Square root.
Find the square roots of 6,241 ; 643,204.
532. Quadratic equations.
1. Solve the following equations graphically:
2 o ^- „ 3m 27
2. Solve by factoring :
n2+28 = lln; 3c2+c-2 = 0
3. Solve by completing the square:
a;24-10a; = 24; 4?/H-20?/=-9
4. The hypotenuse of a right triangle is 5 ft. and one of the sides is 1 ft. longer than the other. Find the length of the sides.
CHAPTER XVI
533. Solution of equations.
1. State how to solve an equation of the first degree in one unknown.
2. Solve the following equations:
(1) 4a;+16 = 9a:-hll
(2) 5+3a:+f =1|+|
.^. 3^1 x-S 7 ^'^^ 4 ~ 2 "4
(4) {2x-l){x-3) = {x-5){2x+4:)
(5) i{l-x)=ii2-x)+U^+x)
(6) .8(10a:-2.3)=.05(5.x+.4)-2.45
^ ^ -^ a a a
332 FIRST-YEAR MATHEMATICS
3. Solve the following problems:
(1) The difference of the acute angles of a right triangle is 22°14'. How large is each ?
(2) The perimeter of a rectangle is 24 inches. If the altitude is increased by 4 in. and the base decreased by 2 in. the area remains unchanged. What are the dimensions of the rectangle ?
(3) The sides of a triangle are in the ratio 5:6: 13. Find the three sides, if the perimeter is 144 inches.
(4) The angles of a triangle are denoted by (a:-|-8)°, 2(x+8)°, and (2a:— 4)°. Find x and the angles.
(5) The interior angles on the same side formed by two parallel lines cut by a transversal are ^x and fa;— 20. Find x and the angles.
(6) Three times a number is increased by 21 and the sum is divided by the sum of the number and 7. The result is then equal to 3. Find the number.
(7) A father is 46 years old and his son is 8. In how many years will the father be 3 times as old as the son ?
(8) The difference of the squares of two consecutive even numbers is 108. Find the numbers.
(9) Two men are 25 mi. apart and walk toward each other at the rate of 3| and 4 mi. an hour respectively. After how long do they meet ? (Yale)
(10) A man travels 50 mi. in an automobile in 3j hours. If he runs at the rate of 20 mi. an hour in the country and at the rate of 8 mi. an hour when within city limits, find how many miles of his journey is in the country. (Yale)
(11) The length of a room is 8 ft. greater than its width. If each be increased by 2 ft., the area of the room will be increased by 60 sq. ft. What is the actual area of the room ? (Yale)
(12") Show how to divide $2,000 into two investments, one at 4% and the other at 3%, so that the former shall produce twice as much income as the latter. (Chicago)
(13) Find two consecutive odd numbers such that the differ- ence of their squares shall be 152. (Chicago)
REVIEW AND SUPPLEMENTARY 333
(14) The tickets of admission to a game are 25 cents each for adults and 10 cents each for children. It is found that $18.75 is taken in and the turnstile shows that 117 persons attend. How many children are there at the game ? (Yale)
(15) A sum of $1,050 is divided into two parts and invested; the simple interest on the one part at 4% for 6 years is the same as the simple interest on the other at 5% for 12 years; find how the money is divided. (Princeton)
(16) At what time between 8:00 and 9:00 o'clock are the hands of the clock together? At right angles? In opposite directions ?
(17) A beam 18 ft. long and weighing 40 lb. is supported at a point 3 ft. from the center. What force must be exerted at the end farthest from the fulcrum for balance ?
(18) A, standing 5 ft. from the fulcrum, balances B, who stands 7 ft. from it. A weighs 126 pounds. Find the weight of B.
CHAPTER XVII
534. System of two linear equations in two unknowns.
1. What is a simultaneous system of equations?
2. State the graphical method of solving simultaneous equa^ tions.
3. Solve the following systems graphically:
f2x-3i/ = 4 /4a;+77/=-27
^^^ Ux+2ij=l ^"^^ \ x-2y=12
4. When is a linear equation in several unknowns said to be in normal form ?
5. State how to solve simultaneous equations by elimination.
6. Solve the following systems by elimination by adding or subtracting:
(1)
X ?y_37 5"^3~15
3 3
^ 1
.+1 = 4
334 FIRST-YEAR MATHEMATICS
7. With two parallels the interior angles on the same side of a transversal are {8y—10)° and (5x-\-y-{-5)°. Two alternate interior angles are (x+4?/)° and 70°. Find x, y, and the unknown angles,
8. A man has two sons, one six years older than the other; after two years the father's age will be twice the combined ages of his sons; and six years ago his age was four times their com- bined ages. How old is each ? (Princeton)
9. The rates of two trains differ by 5 mi. an hour.' The faster requires one hour less time to run 280 miles. Find the rate of each. (Yale)
10. The sum of the two digits of a 2-digit number is 10. If 54 be subtracted from the number the result will be equal to the number obtained by reversing the digits of the original number. Find the number.
11. A man invested two sums, one at 4%, the other at 5%, and received annually an income of $500. He then reinvests these sums at 5% and 6%, respectively, receiving an income of $600 annually. Find the two sums invested.
12. Two angles of a triangle are equal and the third angle is equal to their sum. How many degrees are in each angle ?
13. |
Solve for m and n: |
|
7m+8 7n-l 5 4 |
= —2 |
|
2w-4 n-1 2*3- |
1 3 |
|
Make a graph for the equations and then illustrate and verify i solution. (Chicago) |
||
14. |
Solve for a, b, and c: |
|
3a-2c+ 6=- |
1 |
|
2a- 6+3c = 9 |
||
6-f3a- c = 2 |
(Chicago) |
REVIEW AND SUPPLEMENTARY 335
15. Solve the following simultaneous equations: 2a;+ 32/+5 = 0 Qy+ 52 =7 3x+102-l = 0
Indicate your answers clearly, and verify by substituting in the given equations. (Harvard)
CHAPTER XVIII
535. The formula as a general rule.
1. What is a formula ?
2. A man rides a distance of p mi. and walks back at a rate of q mi. an hour. The entire trip took t hours. Find his rate of riding. (Yale)
3. An express train whose rate is 45 mi. an hour leaves a station a hr. after a freight train. The express overtakes the freight train in b hours. Find the rate of the freight train.
536. Evaluation of formulas.
1. If s = lQt^+vt, find s for ^=10, 2^ = 5.
4 d
2. If v = ^Trr^ and r = ^ , find v for d = 7.
3. If h = ^ , find V for h = 2b and g = Z2.
4. If /2 = ^,, find r for 72 = 50, / = 15.
r-\-r ' '
537. Express any one of the letters of each of the following formulas in terms of the others:
1. Solve C = |(F-32) for F.
2. Solve 8 = 2(^+0 for rt.
336 FIRST-YEAR MATHEMATICS
3. Solve A = 2(61+6.) for 62.
4. Solve s = r for I.
r — l
^ „ - X a — 2x , ^ „
5. Solve — I ^ |-4 = 3aforx.
a a
7. Solve jj-j> = l for E; for P; for ,^.
INDEX
[References are to sections, not to pages]
Absolute value
Acute angle
of a right triangle
Acute triangle
Addition; algebraic 52,
of angles
associative law of 63,
commutative law of
61, 332,
graphical 32,47,
of monomials 58,
of polynomials 352,
of positive and negative numbers
Adjacent angles
Algebra, origin of word. . . .
Algebraic expression, evalu- ation of
Alloy problems
Alternate exterior angles. . .
Alternate interior angles. . .
Altitude: of parallelogram .
of trapezoid
of triangle
Angle
acute
bisection of
central 124,
construction of. .121, 123,
exterior
generated
interior angle of a triangle measurement of
324 notation for 96
100 oblique 122
184 obtuse 100
119 of depression 223
328 °^ elevation 222
JQ3 of triangle 112
348 right 99
sides of 97
348 straight 99
328 trisection of 298
35J vertex of 98
358 Angles: about a point. . 179, 180
adjacent 174
33Q addition and subtraction
174 423
469 259
of 103
alternate exterior 188
alternate interior 188
comparison of 101
complementary 182
consecutive 201
188 corresponding 182
188 equal 101
163 notation for 96
166 opposite in quadrilateral. 201
164 opposite or vertical 183
94 positive and negative .... 325
100 supplementary 181
127 Arc 109
125 degree 107
125 intercepted 124, 125
115 Archimedes 245
94 Area 137
280 of parallelogram 163
111 of rectangle 141
337
338
FIRST-YEAR MATHEMATICS
[References are to sections, not to pages]
of square 137
of trapezoid 166
of triangle 164
unit of 139, 140
Areas, proportionality of
260-266
Ascending powers 357
Associative law of addition
63, 348
Axis of symmetry 302
Axiom 23
Axioms. . 30, 37, 66, 89
;: of power 150
of trapezoid 166
Bearing: of line 224
of point 225
Binomial 57
square of 390
Bisection of angle 127
Bombelli 362
Braces 362
Brackets 362
Center of circle 105
Central angle 124, 125
Chain, surveyor's 221
Checking 76,162,379
Circle 105,305
arc of 109
center of 105
circumscribed 311
inscribed 312
radius 103
tangent 306
Circumscribed circle 311
Clearing of fractions .... 87, 92
Coefficient 54,153
Coinciding lines 20
Commutative law
61, 152, 332, 348 Compass 8, 9, 247
mariner's 117
Complement of an angle . . 182
Cone 203,213
Congruence 273
symbol of 278
Consecutive: angles 201
numbers 85
Constant 252
Constructions, fundamental 296
Contact, point of 307
Co-ordinates 444
Corresponding angles 182
Cube 142, 203, 209
volume of 145
Cylinder 203
Data 8,327
Decimal system 369
Degree : of angle. 104
of arc 107
of monomial 354
of polynomial 355
Depression, angle of ..... . 223
Descartes 327, 446, 362
Descending powers 356
Difference 38
Diophantus 45, 446
Direct variation 254
graph of 256
Directed line-segments. ... 321
Direction of turning 94, 338
Dissimilar terms 155
Distance: between parallel
lines 192
between two points 22
from a point to a line . . . 285
INDEX
339
[References are to sections, not to pages]
Distributive law 366
Dividend 344
Division, checking of 379
law of signs in 345
of monomials 372
of polynomials 377
Divisor 343
greatest common 248
Drawing to scale 220
Elevation, angle of 222
Elimination by addition or
subtraction 449
Equal: angles 101
segments 101
symbol of equality 28
Equation 40
fractional 87,92
graph of 147
indeterminate 445
linear 148
members of 70
normal form 447
of two unknowns 441
of three or more un- knowns 458
quadratic 408
roots of 74
simultaneous 441
solving 75
system of 441
use of 69
Equiangular triangle 119
Equidistant points 283
Equilateral polygon 51
Euclid 45
Eudoxus 244
Euler 325
Evaluation of formulas . . . 469
Exponent 150
Exterior angle 1 15
Extremes 259
Factors : common mono- mial 375
of difference of two squares 395
of trinomial 392, 399
Formulas 138, 466, 467
evaluation of 469
Fractions: clearing of . . . .87, 92
reduction of 250, 373
Fulcrum 434
Functions 251
quadratic 409
Fundamental constructions 296
Gauss 325
Generation of angles 94
Geometry, origin of 135
Girard 362
Graph of data 18, 327
linear equations 147, 443
quadratic equations . 147, 410
variation 256, 257
Graphical: addition 47,328
multiplication 335
representation 16
solution of equations. 443, 410 subtraction 16
Hexagon 50
Hypotenuse 185
Hypsicles 104
Indeterminate equations. . . 445
Indirect measurement 219
Inequality, symbols of . , . . 28
340
FIRST-YEAR MATHEMATICS
[References are to sections, not to pages]
Inscribed polygon 311
Instruments 27
surveyor's 221
Intercepted arc 124, 125
Interior angles: of a tri- angle 280
alternate 188
Intersection, point of 24
Inverse variation 255
graph of 257
Isosceles triangle 186, 280
Latitude 326
Laws 67, 152, 336, 345, 441
Lever: problems 259
arm 339
Line 1
bearing of 224
Linear equation 148
Line-segment 5, 13
ratio of 227
directed 321
Lines: parallel 191
perpendicular 175
perpendicular to a plane. 208
Literal number 14
Locus 284
Mariner's compass 117
Means, of proportion 259
Measurement: indirect. ... 219
of angles Ill, 120
of areas 139, 140
of line-segments 6
Members of an equation ... 70
Methods of proof 276, 281
Mixture problems 259
Minutes, of angle 104
Models of solids 209-213
Monomial factors 375
Monomials 53
addition of 351
degree of 354
division of 372
multiplication of 365
Motion, formula 427, 467
Multiple 46
Multiplication: by zero.. . . 341
commutative law of 152
distributive law of 366
graphical. . . 156, 159, 160, 335
law of signs in 336
of arithmetical numbers. 368
of monomials 342, 365
of a polynomial by a
monomial 156, 158, 366
of polynomials 161, 367
of positive and negative
numbers 340
symbols for 149
Negative numbers 323
Normal form 447
Notation : for angle ....... 96
corresponding parts 277
line-segments 13, 321
point 4
ratio 228
unknown numbers 422
Number: absolute value of . 324
degree of 353
literal 14
negative 325
prime 249
ratio of 226
representation of 15
scale 31,322
systems 370
INDEX
341
[References are to sections, not to pages]
Oblique angle 122
Oblique lines 122
Obtuse angle 100
Oenopides 126,296
Opposite angles 183
in quadrilateral 201
Pacioli 45
Parallel lines 191, 194, 195
axiom of 193
construction of planes . . . 206
Parallelism, symbol of 193
Parallelogram 136, 200
area of 163
Parallelopiped 143, 203, 210
volume of 145
Parentheses 62
removal of 362
Partial products 157
Pentagon 50
Perfect trinomial square . . . 388
Perigon 99
Perimeter 49, 50
Perpendicular 122, 175
Perpendicular bisector, con- struction of 128
Pi (t) 466
Plane 204
Plato 297
Plutarch 235
Point 3
locus of 284
notation for 4
of intersection 24
of tangency 307
Polygon 50
circumscribed 312
inscribed 311
regular 310
Polynomial 56
addition of 356, 357
degree of 355
division of 377
multiplication of 361
subtraction of 361
Positive numbers 323
addition of 328,330
angles 325
Power 150
Powers: ascending 357
descending 357
product of 151
Prime number 249
Prism 203,211
Problems: alloy 259
clock 430
consecutive number 85
digit 455
expressing number rela- tions 84,425,454
income 456
lever 434
geometric 83, 424, 542
mixture 433
motion 426, 429
of similar triangles 235
percentage and interest. . 431
solution of 82, 114
star 429
Producing line-segments. . . 21
Product: of powers 151
of two binomials 398
of sum and difference of
two binomials 393
of polynomial by mo- nomial 158, 159
partial 157
Proof, method of 276
342
FIRST-YEAR MATHEMATICS
[References are to sections, not to pages]
Proportion 258
Proportionality of areas
260-266
Protractor 106
Pyramid 203,211
Pythagoras 190, 402
theorem of 401
Quadrant 110
Quadratic equation 408
Quadratic function 409
Quadrilateral 50, 136
Quotient 165,373
Radius. 103
Ratio 226, 244
found by use of compass . 247
of angles 230
of line-segments 227
of numbers 226
reduction of 250
trigonometric 242
Rectangle 136
area of 141
Regular polygon 310
Removal of parenthesis. . . . 362 Representation of numbers 15
Review 472
Rhombus. 136
Right angle 99
Right triangle 119, 184
congruence of 285
Roots of equation 74
of square 403
Ruler, test 19
Satisfying equations 73
Scale 31,322
drawing to 220
Seconds of angle 104
Segments : line 5
equal 11
unequal 12
notation for 13, 321
Semicircle 110
Side: of angle : . . 97
of triangle 48
Signs, law of 336,345
Similar: triangles 232
terms 55,154,349
Similarity, symbol of 234
Simultaneous equations. ... 441 Simultaneous equations. . . . 442
Solids : geometric 203
models of 209-213
Solving: equations . . 75, 423, 442
problems 82,114,420
Sphere 203
Square : of binomial 390
of trinomial 397
of arithmetical numbers. 391
quadratic trinomial 388
Square root 403
Squared paper 10
Straight angle 99
Straight line 1
Substitution 72
Subtraction : graphical
38, 47, 333
algebraic 334
Sum 32
of angles 103
of angles of a triangle. . . 117 of angles about a point
179, 180
Superposition 276
Supplement 181
Surface 203,204
Surveying 221
INDEX
343
[References are to sections, not to pages]
Symbols, table of 344
Symmetry 299-301
axis of 302
System: of equations 442
of numbers 370
Table of tangents 243
Tangency, point of 307
Tangent 306
table of 243
Tape 221
Term 53
Terms: similar 154, 349
dissimilar 155, 350
Testing ruler 19
Thales 113,235,280
Theorem 113
of Pythagoras 401
Transit 120
Transversal 187
Trapezoid 136, 202
area of 166
Triangle 48
acute 119
area of 164
equilateral 50
equiangular 119
exterior angle of ....... . 115
interior angle of 280
isosceles 186,280
right 119,184
sides of 48
similar 232
vertex of 48
Trigonometric ratio 242
Trinomial 57
square 388
factoring 392
Trisection of angle. ....... 298
Turning-tendency 337
Unequal angles. 101
Units of measure. . . .7, 139, 144
Value 52
absolute 324
Variable 253
Variation: direct 254
inverse 255
Velocity 17,426
Vertex : of angle 98
of triangle 48
Vertical angles 183
Vidta 54,327
Vinculum 362
Volume 144
of parallelopiped 145
of cube 146
Wallis 193
Widman 327
Work problems 468
Zero, operations with . .341, 372
344 FIRST-YEAR MATHEMATICS
[References are to sections, not to pages] SYMBOLS
= equals, is equal to 28 !| parallel 193
> is greater than 28 J- perpendicular 208
< is less than 28 Z angle 95
4= is not equal to 28 A angles 95
+ addition . 39 A triangle 234
— subtraction 39 A triangles 234
X multiplied by 28 ^ similar 234
-j- divided by 343 ^ is congruent to 278
( ) parenthesis . 62 (°) degree 104
[ ] bracket 62 (') minute 104
I [ brace 62 (") second 104
FORMULAS
Right triangle a^-{-b'-=c^-
Circle c = 27rr, ird
Area of rectangle A=bh
Area of square A=s^
Area of parallelogram A=bh
Area of triangle A= \hh
Area of trapezoid A = \h{h-\-h')
Area of circle A =Trr^
Motion d = rt