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iiqilizBd by G00gle
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3,g,1 EE d by GoOgle
i BV Google
JUNIOR HIGH SCHOOL MATHEMATICS
THIRD COURSE
i BV Google
A SERIES OF MATHEMATICAL TEXTS
BAKLE BAYKOHD HEDRICK
THE CALCULUS
By Ellkky Williams Davis and William Charles
BRENKE.
PLANE AND SPHERICAL TRIGONOMETRY WITH COM-
PLETE TABLES
By Alfred Monroe Kbhton and Louis Ingold.
PLANE AND SPHERICAL TRIGONOMETRY WITH BRIEF
TABLES
By Alfbrd Monrok KbhtON and Louis Inoold.
MATHEMATICS FOR AGRICULTURE AND GENERAL
SCIENCE
By Altrkd Mohroe Kenton and William Vernon Lovtet.
CONSTRUCTIVE GEOMETRY
Prepared under the direction of Eable Raymond Hrdbiok.
JUNIOR HIGH SCHOOL MATHEMATICS
By William Lrdlbt Vosbvroh and Frederick William
Gentleman.
^Google
JUNIOR HIGH SCHOOL
MATHEMATICS
THIRD COURSE
WILLIAM LEDLET VOSBURGH
AND
FREDERICK WILLIAM GENTLEMAN
Stto gnk
THE MACMILLAN COMPANY
1919
JSrtgUtr—nid
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£A59
o
Oorraanr, lfilB,
Bi THE MACMILLAtr COMPART
Published July, 1915.,
SgilzedBy GoOgk
PREFACE
This book has been planned to meet the needs of the
first year mathematics in the ordinary high school, as
well as to serve as a Third Course in Junior High School
Mathematics.
Comparison with the traditional freshman course in
ordinary high schools, will show that certain geometric
matter of admitted value has been inserted, and some
relatively useless topics have been omitted from the
algebraic portions. This renders the book particularly
suitable for use as a freshman text in mathematics in
ordinary high schools, and it does not detract from its
value as a Third Course in Junior High School Mathe-
matics.
The review of arithmetic and of elementary geometric
and algebraic notions, with which the book begins, is
very desirable for any course of this type ; and it makes
the work usable either with or without the preceding
books in the series.
The authors have been guided in their work by the
following principles :
1. That there should be a high degree of continuity in
the subject matter of mathematics and in the methods of
presenting it during the three years of the Junior High
School.
2. That by the completion of Courses I and II, the
pupil has acquired the following ideas and habits :
459972
vi PREFACE
(a) By checking, the habit of assuming responsibility
for the correctness of Mb results.
(b) By estimating bis results in advance of the computa-
tion, a rational idea of number values.
(c) By the systematic and concise methods of handling
the equation, an appreciation of its value as a mathe-
matical tool in the solution of problems.
(d) By the use of the compasses, protractor, and ruler
in geometric constructions and scale drawings, a familiar-
ity with the properties and relations of the more common
geometric figures and solids.
(e) By the graphic interpretation and representation of
number data and of equations, an appreciation of the
value of the graph in science and industry.
3. That the course in mathematics should bring the
pupil who leaves school during, or at the end of, his ninth
school year, in contact with adult activities that lend
themselves to mathematical interpretation ; and it should
afford him an opportunity for the exercise of his mathe-
matical powers through the handling of a variety of
mathematical tools used in the solution of problems of
everyday life.
4. That the course should aid the pupil who continues
in school, in deciding whether or not he is capable of
continuing his work in mathematics with profit; and it
should aid him in acquiring a keener interest in the
further study of mathematics. He should get from the
course a clear idea of the meaning of mathematics and a
vision of its manifold applications to the world's impor-
tant work.
The attention of teachers is directed to the following
features of this course :
i BV Google
PREFACE vu
Part I, Algebra.
1. The emphasis upon the idea that per cent is a ratio
per hundred; the introduction of the term per cent error
with its interesting applications ; the use of the percentage
formula as an application of the simple equation.
2. The continued emphasis placed upon rational methods
of locating the decimal point in multiplication, division,
and square root.
3. The determination of the reliability of numerical
results when computed from data obtained by measure-
ment and the number of significant figures to be retained
in such results.
4. The introduction of such metric measures as have
come to be generally used and whose equivalent values
the pupil, as a result of his reading, should want to be
able to determine.
5. The construction of the formula as a shorthand
mathematical sentence and its appreciation as such when
its subject is changed.
6. The simple and relatively late introduction of the
negative number.
7. The simple direct presentation of the four funda-
mental processes with algebraic expressions.
S. The special emphasis on those forms in factoring
which are applicable in the solution of equations of the
second degree.
9. The practical use made of tables of square roots in
evaluating irrational roots of quadratic equations.
10. The presentation of the idea of a function as de-
veloping from the idea of a rabio, which has been kept in
the foreground throughout the course, and variation as
commonly known in the quantities of industry and science.
viii PREFACE
11. The use of the graph in interpreting functions of
the first and second degree.
12. The number and variety of the problems presented
in the several chapters. The problems have been selected
with the idea that by means of them the pupil shall de-
velop the ability to apply general principles to new situa-
tions, shall become proficient in the use of a variety of
mathematical tools, and shall acquire an appreciation of
the quantitative phases of his environment.
Part II, Geometry.
1. The organized presentation and summary of geo-
metric facts and constructions previously studied.
2. The reliance upon drawing and measurement of
figures in the first geometric proofs.
3. The inductive presentation of all proofs.
4. The introduction and use of the tangent ratio with
the use of the table of natural values to three figures.
5. The natural coordination of arithmetic, algebra, and
geometry in the numerous applications presented.
6. The emphasis placed upon the "shop" methods in
geometric constructions.
7. The considerable amount of formal geometry pre-
sented through the wise selection of major theorems.
William Ledley Vosburqh.
Frederick William Gentleman.
is, Google
XI.
• XII.
XIII.
l/ XIV.
CONTENTS
PART I. ALGEBRA
Percentages
Approximate Computation
Construction or Formulas
Metric Measures
Linear Equations
Positive and Negative Numbers
Addition and Subtraction op Algebraic
Expressions
Multiplication and Division or Alge-
braic Expressions
Pairb of Linear Equations
Factors and Equations .
Radicals and Roots ....
Quadratic Equations
Ratio, Proportion, and Variation .
Fractions and Equations
1-12
13-26
27-12
43-49
60-62
70-80
81-104
105-112
113-130
131-137
138-151
152-168
169-190
PART n. GEOMETRY
Lines and Angles ....
Congruent Triangleb
Parallel Lines and Parallelograms
Circles 234-243
Similar Triangles 244-261
Mensuration . . . . . . 262r-273
Formulas of Mensuration . . . 274-284
191-199
200-220
221-233
XV.
XVI.
XVII.
XVIII.
XIX.
XX.
XXI.
Appendix 285-292
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JUNIOR HIGH SCHOOL
MATHEMATICS
THIRD COURSE "
PART I. ALGEBRA
CHAPTER I
PERCENTAGES
§ 1. Ratios. A ratio is the expression of the quotient
of two quantities. The quantities must be alike in kind
and their quotients may be expressed as integers, or as
fractions (common or decimal).
A. Express the ratio of each of the following pairs of
quantities, (a) in the order given, (6) in the reverse order.
For example, (a) the ratio of 3 in. to 1 ft. is fy, or J,
(6) the ratio of 1 ft. to 3 in. is J£, or 4.
1. * in. to 1 ft. 11. 500 lb. to 1 T.
2. 18 in. to 1 yd. 12. 750 lb. to 1 T.
3. 27 in. to 1 yd. IS. 144 cu. in. to 1 cu. ft.
4. 2.4 in. to 1 ft. 14. 1.5 sq. ft. to 1 sq. yd.
5. 440 yd. to 1 mi. IB. 5.4 cu. ft. to 1 cu. yd.
6. 3.6 in. to 1 yd. 16. 2 lb. 8 oz. to 15 lb.
7. 220 yd. to 1 mi. 17. 24.5 ft. to 100 ft.
8. 100 yd. to 1000 ft. 18. 9$ sec. to 1 min.
9. 4 oz. to 2 lb. 19. 15J sec. to 1 min.
10. 12 oz. to 1 lb. 20. 39.37 in. to 1 yd.
b 1
L .^.Google
.2.. i .'.JUftfOR.^IGH SCHOOL MATHEMATICS [1, 1 1
B. Express the ratio of each of the following pairs of
numbers, (a) as common fractions in lowest terms, (b) as
decimals.
For example, (a) the ratio of 1-J to 2 is t = $, (b) this
ratio may also be written decimally, as 0.625.
1. 2$ to 4 11. 25.5 to 375
2. l}to5 12. 12$ to 5$
3. 2| to 7J 18. 75.75 to 187.5
4. lito2J 14. 84.15 to 78.75
5. 10} to If IS. 0.125 to 2.25
6. 4^to2£ 16. 3.375 to 5.625
7. 4A to 8| 17. 0.075 to 1.2
8. 15|to60 18. 0.025 to 0.75
9. 12.5 to 9$ 19. 0.005 to 0.0002
10. 16£$ to 12$ 20. 0.00375 to 0.00625
§ 2. Ratios as Per Cents. For purposes of comparison,
ratios are often expressed as per cents, that is, with the
denominator 100.
Express the ratio of each of the following 'pairs of
quantities as per cents.
For example, the ratio of 3 in. to 1 ft. may be expressed
as -^, or £. This may be expressed also with a denominator
100, that is, as ^fo. Using the term per cent, we may say
that 3 in. is 25% of 1 ft. Since the ratio of 1 ft. to 3 in.
is £, we may also state that 1 ft. is 400% of 3 in.
1. 8 in. to 1 ft. 8. 7.2 in. to 1 yd.
2. 9 in. to 1 yd. 4. 880 yd. to 1 mi.
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I, { 3] PERCENTAGES 3
6. 800 lb. to 1 T. 11. 576 cu. in. to 1 cu. ft.
6. 1200 lb. to 1 T. 12. 16.2 cu. ft. to 1 cu. yd.
7. 36 sq. in. to 1 sq. ft. IS. 4500 lb. to 1 T.
8. 108 sq. in. to 1 sq. ft. 14. 300 sq. in. to 1 sq. ft.
9. 2J sq. ft. to 1 sq. yd. 15. 120 in. to 2 yd.
10. 6| sq. ft. to 1 sq. yd. IS. 18 qt. to 3 gal.
§ 3. Per Cent Error. In the arithmetic of science
and industry we reason and compute with the estimated
or measured values of various quantities. As neither
our estimates nor our measurements of such quantities
can ever be exact, the numbers used to express them can-
not be exact. For the purpose of comparison, a common
way of expressing the error in such numbers is by means
of what is known as the per cent error.
The per cent error in any quantity is the ratio of the
amount of error in it to the true value of the quantity,
expressed as a per cent. This will be made clear in the ex-
amples which follow.
Example 1. It was estimated in advance that 45
desks would be needed in a certain classroom. It was
' later found that 40 desks were sufficient. What was the
amount of error in the estimate? Express this error as a
common fraction ; as a per cent.
Solution. 45 (Estimated number)
40 (Actual number, or true value)
5 (Amount of error)
-fa -•- $ (Error expressed as a common fraction)
i-12J% (Per cent error)
Check. 12$% of 40=5
40+5=45
" : Ans. 5 desks; i; 12+%.
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4 JUNIOR HIGH SCHOOL MATHEMATICS |I, I 3
Example 2. The estimated cost of laying a sidewalk
was $62.50, whereas the actual cost was 175. What was
the amount of error? Express the error as a common
fraction ; as a -per cent.
Solution. $75.00 (Actual coat, or true value)
62.50 (Estimated cost)
$12.50 (Amount of error)
12.50 1 _ ' m . ,
7 - oq = 5 (Error expressed as a common fraction)
*-16f% (Per cent error)
Check. 16$% of 75= 12.50
75-12.50 = 62.50 Arts. $12.50; J; 16%%.
Example 3. A pupil measures the length, of a line and
reports it as 4.47 in. By a more careful measurement, it
is found to be 4.53 in. Express the error as a common
fraction ; as a per cent.
Solution. 4.53 in. (Considered as the true valve)
4.47 in. (Pupil's measurement)
.06 in. (Amount of error)
-06 6 2 __ . ,
4^3 "453" 151 (Fractional part)
y'rr
= 1.3%
(Per cent error)
.013*
151)23500
151
490
453
37
Check.
1.3% of 4.53-0.059
4.53-
-0.06-4.47
Ana. ^j 1.3%.
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PERCENTAGES
1. The advance estimate of the registration in a cer-
tain school was 525 pupils. The actual registration was
500 pupils. Express the amount of error in the estimate
as a common fraction ; as a per cent.
2. The estimated number of books needed for a cer-
tain class was 85. The actual number needed was 80.
Express the amount of error in the estimate as a common
fraction ; as a per cent.
3. The length of a rug is reported as 7 yd. By a more
careful measurement the length is found to be 6.6 yd.
What is the per cent error to within a tenth of one per cent ?
4. The length of the rug in Prob. 3 is reported to be 20
ft. By a more careful measurement it is found to be 19.8
ft. Express the error in per cent, to within a tenth of one
per cent,
5. The estimated cost of food for a club of 10 boys
for a week was $28.50. The actual cost of the food was
$32. Express the error of this estimate as a per cent, to
within a tenth of one per cent.
6. In a certain retail store, a weight known to be one
pound is registered on one of the scales as 16.2 oz. What
is the per cent error in all weights registered on this scale?
Who is the loser, the retailer or the customer?
7. A yardstick is found to be 35.8 in. long. What
is the per cent error in its length, to within a tenth of one
per cent?
8. In checking up a record made in a race of 100 yd.,
it was found that the course was too short by 9 in. Ex-
press this error in per cent.
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6 JUNIOR HIGH SCHOOL MATHEMATICS [I, S 3
9. When the length of a quarter-mile running track
was. carefully measured, it was found to be too long by
2 yd. Express the error in its length in per cent, to
within a tenth of one per cent.
10. The length of a meter in inches, expressed to four
figures, is 39.37. When the value is taken as 39.4, what iB
the per cent error, to within a hundredth of one per cent?
11. A bushel is commonly regarded as containing 1-J
cu. ft. ; a more accurate value is 1.244 cu. ft. When the
former value is used, (a) what is the per cent error?
(b) what is the amount of error in 400 bu. ?
12. A cubic foot is commonly said to contain 7.5 gal. ;
a more accurate statement is 7.48 gal. When the former
value iB used, what is the per cent error, to within a hun-
dredth of one per cent ?
13. The value of r (pi) that you have been using is
3.14 ; a more accurate value is 3.1416. When the former
value is used, what is the per cent error, to within a hun-
dredth of one per cent?
14. Several pupils attempt to measure the length of
a certain line correct to .01 in. A reports 4.50 in., while
B reports 4.58 in. The average of all lengths reported is
found to be 4.56 in. If we conaidcc the average length
as the true value, determine the per cent error in A's meas-
urement ; in B's measurement.
15. A reports the length of a desk as 2 ft. 3.8 in. ; B
reports it as 2 ft. 3.5 in. The average of all the separate
measurements made by individuals in this class was
27.6 in. Assuming this to be the true value, determine
the per cent error in A's measurement ; in B's measure-
ment.
3,g,! EE d by GoOgk
I, § 4] PERCENTAGES 7
§ 1. The Percentage Formula. All computations of
percentage may be expressed by the statement,
percentage = rate X base.
The formula for this statement is :
p = rXb,
where
b-- the bane (the number on which the percentage is
found),
r = the Tate (the number of hundredths to be taken),
p =* the percentage (the number found by taking a cer-
tain per cent of the base).
When the base is not known, the percentage formula
gives us a direct way of finding it. This is shown in the
solution of the examples that follow.
Example 1. 240 is 75% of what number?
Solution. Let 6 (base) = the number, then,
® 240 = .756
® 240 = J6 ®^*
® 960 = 36 ©X4
® 320-6 ®-=-3
Check. Substitute, in the statement of the given
example, 320 for the required number. Ans. 320.
Note. The symbol, © =, means that the members of equa-
tions (T) and @ have identical values.
The symbol, @ X4, means that each member of the equation.
® is multiplied by 4.
The symbol, ®-:-3, means that each member of equation ®
is divided by 3. .,
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8 JUNIOR HIGH SCHOOL MATHEMATICS [1, { 4
Example 2. The total registration in a certain college
in October, 1917, was 374. This was 32% below the regis-
tration in October, 1916. What was the registration in
October, 1916?
Solution. Let 6 — the number registered in 1916, then
1.006- .325= .686 (Number registered in 1917)
then,
® .685=374
(5) 6=550 ®+.68
Check. Substitute 550 for the registration in 1916.
Ana. 550.
Example 3. Suits which cost $15 are to be marked
bo as to gain 25% on the selling price. Determine the
selling price.
Solution. Let e (cost) represent the percentage, and
s (selling price) represent the base in the
formula,
p=rX6, which then becomes
c=rXs.
As 25% of a ia to be the profit,
75% of s must be the cost.
Hence the equation is :
® 15 = . 75s
CD 20=e ©+.75
Check. Substitute $20 for the selling price.
Ana. $20.
;, S ,:z K i:vC00gIe
PERCENTAGES
[In checking, substitute in the statement of the given
problem the result obtained.]
1. A has been able to save, during the past year, 15%
of his income. If he saved $420, what was his' income ?
2. B has discovered that his living expenses in 1916
were $1567.80. This amount was 67% of his income.
What was his income?
3. The receipts of an athletic association, from the
sale of tickets for the year, amounted to $774.90. This
amount was 82% of the total expenses for the year. What
were the total expenses for the year ?
4. In winning the pennant in the National League, in
1917, the Giants won 98 games, or 63.6% of the total num-
ber they played. What was the total number of games
played by the Giants?
5. In winning the pennant in the American League,
in 1917, the Whit* Sox won 100 games, or 64.9% of the
total number they played. What was the total number
of games played by the White Sox?
6. The Pittsburgh team won 33.1% of the games it
played in 1917. It won 51 games. How many did it
play?
7. The Philadelphia team (American League) won
36% of its games. It won 55 games. How many did
it play?
8. Baseball suite, on which a discount of 16|% was
allowed, cost 912.50, net. What was the original price
of these suits?
3,g,1 EE d By GoOgle
10 JUNIOR HIGH SCHOOL MATHEMATICS [I, j 4
0. An automobile was sold at the end of the season
for $750. This was 14$% below what it cost. What
did it cost?
10. The production of oats in Algeria for 1917 was re-
ported as 18,946,000 bu., or 144.2% of the 1916 crop.
What was the number of bushels produced in 1916 (to the
nearest thousand) ?
11. New York State produced 877,000 bu. of beans in
1917. This was 121.8% of the number of bushels it pro-
duced in 1916. What was the production in 1916 (to the
nearest thousand) ?
12. The country price of a bushel of wheat on October 1,
1917, was published as $2,066, or 151.6% of the price on
October 1, 1916. What was the price on October 1, 1910?
IS. The country price of a bushel of corn on the same
date, October 1, 1917, was $1,751. This was 212.8%
of the price on October 1, 1916. What was the price on
October 1, 1916?
14. At the end of the season an automobile was sold
for $700. This was 24.3% below what was paid for it.
What was the amount paid for it (to the nearest dollar) ?
15. A dealer sold two cows at $95 each. On one he
gained 25% of the cost while on the other he lost 20%
of the cost. Did he gain or lose on the transaction?
How much?
16. A retailer sold two suits at $30 each. On one he
made a profit of 50% of the cost while on the other he
lost 10% of the selling price. What was hia net profit?
;, S ,:z K i:vC00gIe
I, M]
PERCENTAGES
11
17. The following articles are to be marked to sell at
the per cent of profit indicated on the selling price. Find
the selling price in each instance.
**_
CO.T
Pbofit ok Snj^
mo Pricb
Simuo Puoa
(•)
(»
(a)
Chair . . .
Table . . .
Davenport .
Rug . . .
$2.25
$22.50
$75.00
$84.00
25%
37i%
62)%
75%
?
?
r
?
18. A merchant marks his goods 50% above cost and
gives his customers a discount of 10% of the marked
price. Determine his per cent of profit : (a) on the cost,
(6) on the selling price. (Suggestion. Let c - the cost.)
19. A merchant marks his goods 37£% above cost and
allows a discount of 8% of marked price. Determine his
per cent of profit : (a) on the cost, (fc) on the selling price.
20. If water expands. 10% when it freezes, how much
does ice contract when it turns into water ?
21. In settling with his creditors, A was able to pay
92f( on $1. One of the creditors received $529. What
was the original amount of his bill ?
22. After receiving an increase of 40% in wages, car-
penters are getting $7.14 per day. What was their wage
per day before the increase?
23. A coat which cost $12 was marked to sell at a profit
of 50% on the selling price. When the coat was sold
the price was reduced 25%. What was the per cent of
profit realized on the price at which it was sold?
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12 JUNIOR HIGH SCHOOL MATHEMATICS [I, * 4
24. A merchant marks his goods to sell at wholesale at
a profit of 20% on the selling price. His retail price is
20% above the wholesale price. What is his per cent of
profit on the selling price when he sells at retail?
26. A table was sold on a special sale for $22.50. This
was at a discount of 25% from the marked price. What
was the marked price?
26. Two successive discounts of 10% and 2% on a bill
of goods amount to $17.70. What was the original amount
of this bill?
27. At what price must an article costing $1.20 be
marked so that, after deducting 20% from the marked
price, a profit of 25% may be made on the cost?
28. At what price must an article costing $1.20 be
marked ao that, after deducting 20% from the marked
price, a profit of 25% may be made on the selling price ?
29. How much is gained on one dozen articles that cost
$2.50 each when they are sold so as to gain 20% on the
selling price?
SO. The publisher's price of a certain book is 75ji.
The publisher sells the book to retailers at a discount of
20% from this price. If the retailer sells this book for
75 £, what per cent does he gain, (a) on the cost ? (6) on
the selling price ?
3,g,1 EE d by GoOgle
APPROXIMATE COMPUTATION
S 6. Approximate Products. When numbers that are
derived by measuring are met in computation, the degree
of accuracy obtainable in results depends upon the kind
of quantity measured and upon the precision with which
the measurements are made, — not upon the number of
figures actually used in performing the computation. A
result can be no more accurate than the least accurate
datum.
Note. Commercial problems dealing with dollars and cents
should not be confused with problems dealing with numbers
obtained by measurement. In a commercial problem, when
the number of units is known and the price per unit is given, the
result should be computed to the nearest cent.
When we measure the length of a room and record
it as 37.4', we generally mean that the true length is
nearer to 37.4' than it is to 37.3' or to 37.5'. The length
might be nearly 37.45' or slightly more than 37.35'.
Similarly, we ought record the width as 23.8', meaning
that it represented the width to the nearest tenth of a foot.
We shall find that the area, computed from these measure-
ments, will not be dependable beyond the nearest square
foot. (See Ex. 1, which follows.)
Example 1. The length of a room is 37.4' and its width
is 23.8'. Find the area of the floor in square feet.
Formula. A=bh
A «=37.4X23.8
13
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14 JUNIOR HIGH SCHOOL MATHEMATICS [II, § 5
Specimen Solutions.
A. The work appears as follows, if the customary
method of multiplication is used :
37.4
23.8
1122 20X40=800
748
890.12
The area of the floor appears to be 890 sq. ft. to the
nearest square foot.
B. If the order of the multiplying is reversed ; that is,
if 37.4 is multiplied by the left-hand digit of the multiplier
first, the work will appear as follows :
37.4
23.8
748. (20X37.4) Estimate
112.2 (3X37.4) 20X40-800
29.92 (.8X37.4)
890.12
In this method the decimal point is located in the first
partial product to correspond with the estimate.
Note. A more complete explanation of this method of mul-
tiplying can be found in Chapter I, First Course.
C. The measurements of the floor, of which we are
finding the area, were made only to the nearest tenth of
a foot. It will now be shown that this area cannot be
accurate beyond the third figure, and that even that
figure is in doubt.
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II, i 5] APPROXIMATE COMPUTATION 15
37.4
23.8
748.** Estimate
112.2* 20X40=800
29.92
890.12
The area of the floor is computed as 890.12 sq. ft. In
the firBt partial product, however, the places marked with
stars (*) are not necessarily zeros. Had the length been
measured to the nearest hundredth of a foot, some figure
would have appeared after 748 in the column following
the decimal point. The stars show that the figures in
these columns are doubtful. In fact the first partial
product, 748, might he either 747 or 749 ; for 20 times
37.35 is 747, and 20 times 37.44 is 749 to the nearest third
figure. (Any measurement between 37.35' and 37.45'
would be called 37.4', to the nearest tenth.)
In the final product, therefore, the figures that appear
in the doubtful columns (which happen, in this example,
to be at the right of the decimal point) are of little use, as
their retention does not make the answer any more accu-
rate than it would have been without them. This final
product might be as small as 887.1 (37.35x23.75), or as
large as 892.6 (37.44X23.84), depending upon the data.
Hence the third figure of this answer is in doubt and we
waste labor by keeping figures in the partial products
that do not appear in our final product.
D. If we cut the partial products, by not writing any
figures which would come in the doubtful columns, we
shall obtain the figures necessary for the final product.
The work is shown on page 16.
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16 JUNIOR HIGH SCHOOL MATHEMATICS III, § 6
37.4
23.8
748. Estimate
112. (3X37+1. See Note 1.) 20X40 = 800
30. (.8X30+6. See Note 2.)
890. Ans. A =890 sq.ft.
Note 1. (3X37.4-112.2-) As the .2, if written, would
oome in a doubtful column (see Solution C), we do not write it;
but in multiplying 37 by 3, we notice that we have 1 to carry
(since 3X.4-1.2, which ia nearer 1 than 2). We now place a
mark (') over the 4 in the multiplicand. This is to show that
we have not written a figure in the partial product that would
oome in a doubtful column, but that we have noted, in multiply-
ing, its effect upon the next column to the left.
Note 2. (.8 X37.4 -29.92.) As the 9 and the 2, if written,
would both come in doubtful columns (see Solution C), we do
not write them; but in multiplying 30 by .8, we notice that we
had 6 to carry (since .8X7 = 5.6, which is nearer 6 than 5).
We now place a mark (') over the 7 in the multiplicand to indi-
cate again that we have not written a figure that would oome in
a doubtful column; but that we have noted, in multiplying, its
effect on the next column to the left.
Note 3. Measurements, like the above, made to the nearest
third figure, give a result not reliable beyond three figures, hence
figures beyond the third may be neglected. Even the third
figure may not be reliable. If the first three figures of a result
are reliable, such a result is said to be of three-figure accuracy.
Example 2. A table top is 8.28' long and 3.26' wide.
How many square feet does it contain? (Find your re-
sult to tenths of a square foot.)
Formula. A=bk
A = 8.28X3.26
A =27.0
(The work follows.)
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II, i 6] APPROXIMATE COMPUTATION 17
Specimen Solution.
8.28
3.26
24.84 (3X8.28) Estimate
1.66 (.2X8.28. See Note 1.) 3X9-27
.49 (.06X8.2. See Note 2.)
27.0 Am. A -27.0 sq.ft.
Note 1. (.2x8.28.) Place mark (') over 8 ; multiply 8.2
by .2 and add .02 (.2 X .08 - .016, which ia nearer .02 than .01).
Note 2. (.06X8.2.) Place mark (') over 2; multiply 8
by .06 and add .01 (.06 X-2 -.012, which ia nearer .01 than .02).
Note 3. Since the reault was required to tenths, it was
necessary to write after the decimal point in the answer.
■XEROSES
A. By the above method, in each of the following exer-
cises, find the product to three figures.
1. 3.76X25.2 7. 4.08X67.2
2. 8.34X3.61 8. 8.05X9.03
3. 32.5X4.24 9. 775X4.06
1. 8.46X3.14 10. 8.70X2.06
0. 42.5X16.7 11. 68.7X0.S04
«. 8.39X6.06 12. 4.04X20.4
B. In each of the following exercises, find the product
to four figures.
1. 39.37X18.75 6. 7.854X24.34
2. 25.72X62.44 7. 65.75X0.7854
S. 523.8X12.56 8. 24.75X3.142
4. 45.36X61.16 ». 40.08X3.142
5. 4.837X12.18 10. 25.08X20.05
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18 JUNIOR HIGH SCHOOL MATHEMATICS [II, 1 6
§ 6. Approximate Quotients. The ratio of the circum-
ference of a circle to its diameter is «■. This is expressed
by the equation
c
*"■
When the circumference of a circle is known, the above
equation may be transformed into a useful formula for
finding the diameter. The steps in the transformation
are as follows :
® |=r (Given)
© c=*d ®Xd
® Z =d ® + .
This formula is a brief way of saying that, when the
circumference of a circle is known, the diameter may be
found by dividing the circumference by x.
In performing the process of division, since the values
of both r and of the circumference are known only approxi-
mately, we may contract our work as shown in the model
examples which follow.
Example 1. The circumference of a circle is 30.6 in.
Find its diameter to three figures.
Solution. j = £ (Formula)
. 30,6
a = o~7| Estimate
d=9.75 5^=10
3
The circumference is given as 30.6 in. to the nearest
third figure, hence it might be nearly 30.65 in. or slightly
more than 30.55 in.
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n, i 6] APPROXIMATE COMPUTATION 19
Dividing 30.64 by 3.1416 in the usual way, the quotient
is 9.753 (9.75 to the nearest third figure).
Dividing 30.55 by 3.1416, the quotient is 9.724 (9.72
to the nearest third figure). •
Since the third figure of the quotient is in doubt, the
computation will be sufficiently accurate if done as follows.
Determine the location of the decimal point in the quotient
by means of the estimate.
, , , 9.74 6
3.14)303*
28 26 (9X3.14 to four figures. See Note 1.)
234
220 (7X3.14 to three figures. See Note 2.)
T4
12 (4X31 to two figures. See Note 3.)
~2
2 (6X3 to one figure. See Note 4.)
Note 1. In the first partial product (2826), although the 6
comes in a doubtful column, it should be retained. It is advis-
able to keep all figures of the first partial product.
Note 2. (7x314-2198.) The 8 in the partial product, if
written, would come in a column not to be retained. Do not
write it, but instead carry 3 (as 7x4-28, which is near 30) to
7 in the next column. Place the (') over the 4 of the divisor to
show that we have noted, in the partial product, its carrying effect.
Note 3. (4x31-124.) Write 12 in the partial product
and place the mark (') over the 1 in the divisor for the reason
given in Note 2.
Note 4. (6x3-18.) As 18 is near 20, place the mark
(') over the 3 in the divisor and write 2 to carry in the partial
product.
Check. In checking, multiply the quotient by the
divisor, not the divisor by the quotient. In multiplying,
reject figures that would come in doubtful columns as
shown in the work in multiplication. (See page 20.)
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20 JUNIOR HIGH SCHOOL MATHEMATICS [H, 5 6
9.75
3,14
29.26
.98
.39
30.62
(30.6 to three figures)
Aras. The diameter is 9.75".
The value of r is 3.14159 ---. In all problems in which
t is the divisor, use at least as many figures for the value
of n- as are given in the data of the problem. It is cus-
tomary to use not less than three figures in any case.
Example 2. The circumference of a circle is 51.26 in.
Find its diameter.
_ 51.26
3.142
-16.32
, ,',, 16.316
3.142)5136"
3142
19 84
18 85
99
94
5
3
2
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II, 1 6] APPROXIMATE COMPUTATION 21
Check. 16.32
3.142
48.96
1.63
.65
.03
51.27 Ana. d- 16.32 in.
Example 3. The surface of a rectangular plate is 42.3
sq. in. The length of the plate is 8.27 in. Find its width.
Solution. Transforming the formula,
® A-bh
® ^-h ®+b
Substituting the values of A and b in ©.
® 43;2 =ft Estimate
8.27 40
n-5.11 8
,, 5.114
8.27)42.3'
4135
5.11
8.27
40.88
1.02
.36
42.26
(42.3 to three figures) Ans. A -S.ll in.
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22 JUNIOR HIGH SCHOOL MATHEMATICS [II, J 6
EXERCISES
A. In each of the following exercises with circles whose
circumferences are known, find the required values.
Formula. d=-.
-
d
CH.C
1.
56.4"
j
J
2.
69.7"
•>
?
3.
8.58'
?
?
4.
9.45'
?
?
8.
27.6"
?
?
6.
30.8"
?
?
7.
1.67'
7
?
8.
1.23'
7
7
9.
43,26"
7
?
10.
82.78"
7
7
11.
31.29"
?
?
12.
21.17"
7
?
IS.
31.2'
7
14.
29.8'
7
?
16.
31.29'
?
?
B. In each of the following exercises with rectangles
whose areas are known, find the required values.
a
b
ft
ct*.
1
45.3 aq. ft.
18.4'
,
9
49.1 eq. ft.
2.82'
?
3
82.7 aq. ft.
7
24.6'
4
64.2 sq. ft.
7
4.67'
ft
308 aq. in.
73.6"
?
6
350 aq. in.
7
65.4"
7
115.6 sq. in.
23.4"
7
e
201.5 sq. in.
7
56.4"
i
35.48 sq. ft.
7
76.2'
10
41.89 sq. ft.
83.7'
7
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n, 8?I APPROXIMATE COMPUTATION 23
§ 7. Square Root. The square root of a number is one
of the two equal factors of that number. Hence, when a
number is divided by its square root, the quotient equals
the divisor. It necessarily follows that when the divisor
of a number is smaller than its square root, the quotient
is larger than the square root J when the divisor is larger
than the square root, the quotient is smaller. In these
cases, the true square root will generally be very nearly
halfway between the divisor and the quotient.
The application of this principle gives us a method of
finding, by trial, the square root of any number. A study
of the examples which follow will make this clear.
Example 1.
figures.
Find the square root of 34.3 to three
Solution. Since 5* = 25 and 6
6, say 5.8.
!6, try a divisor n
, , 5.92
5.8)34.3
290
53
52
Since the quotient is 5.92 and the divisor
is 5.8, try a divisor halfway between ; that
is, 5.86. Repeat the process of division.
,, 5.85
5.86)34.3* Evidently the square root is very nearly
29 30 5.855. Retaining three figures (the last one,
5 00 as an even number), we have 5.86 for the
4 69 square root.
31 This should now be checked by multipli-
29 cation. (See page 24.)
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24 JUNIOR HIGH SCHOOL MATHEMATICS [II, | 7
Check. 5.86
5.86
29.30
4.69
.35
34.34 Am. V3T3-5.86.
Example 2. Find the square root of 657.4 to four
figures.
Solution. Since 25* -625 and 26* -676, try a divisor
near 26, say 25.8.
, r 25.48
25.85B573
516
1414
129
12 4
We shall now try a divisor halfway be-
tween 25.48 and 25.8; that is, 25.64.
10 3
21
2J.
, ,, 25.64
25.64JB573
512 8
144 6
128 2
ThiB answer may now be checked by
iel
multiplication, as before.
15 4
10
10
Am. V65T4- 25.64.
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II, j 7] APPROXIMATE COMPUTATION 25
Note. To find the square root of a number to three figures,
you will have to repeat the process of division if, in the first,
division, the divisor differs from the quotient by more than one
in the second figure. To find the square root of a number to
four figures, you will have to repeat the process of division if
the first two figures of the divisor and quotient do not agree in
the first division. It is advisable to check the answer in all
cases by contracted multiplication.
Example 3. Find the square root of 7 to four figures.
Solution. Consider that 7 is an exact number and
place three zeros after the decimal point. Since 2.6*=
6.25, try 2.6 as a divisor.
1st Division 2d Division
, 2.692 , , , 2.646
2.6J7150S 2.646)71500
52 5 292
180 1708
156 1588
240 120
234 106
6 14
5 16
This answer (2.646) may be checked by multiplication,
or by consulting a Table of Square Roots.
Ans. V7 = 2.646.
EXERCISES
A. Find the square root of each of the following num-
bers to three figures.
1. 237. 6. 52.4 11. 2.43
2. 538. 7. 87.6 12. 5.24
3. 910. 8. 0.910 IS. 8.76
4. 876. 9. 0.876 «■ 0.0538
5. 24.3 10. 0.243 15- 0.0524
;, S ,:z K i:vC00gIe
26 JUNIOR HIGH SCHOOL MATHEMATICS [II, { 7
B. Find the square root of each of the following num-
bers to jour figures.
1. 24.36 6. 8985. ML 2.
2. 28.54 7. 9234. 12. 3.
5. 57.96 8. 7.892 IS. 5.
4. 45.84 9. 0.7892 14. 8.
6. 2.854 10. 0.5642 ID. 12.
'Note. Consider the numbers in Exs. 11-15 as exact numbers,
and place as many zeros after the decimal point as are needed
to find the required number of figures in the root.
PROBLEMS
1. The area of a square lot of land is 458 sq. ft. Find
the length of one side.
2. Find the side of a square that has an area of 35.2
sq. in.
3. Find the radius of a circle whose area is 145 sq. in.
(Formula. r=-y— •)
4. Find the radius of a circular metal plate if its area
is 41.5 sq. in.
6. Find the diameter of a water pipe if its cross sec-
tional area is 5.30 sq. in. (Suggestion. Find the radius
first, using the formula in Prob. 3.)
6. A rectangle is 13.4" long and 12.3" wide. Find
the side of the square which has the same area as the
rectangle.
7. The sides of a right triangle are 12£" and 18^-".
Find its hypotenuse. (Formula: c = vV+5 ! , where c is
the hypotenuse and a and b are the two sides.)
8. The sides of a right triangle are each 4.25". Find
its hypotenuse.
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CHAPTER III
CONSTRUCTION OF FORMULAS
EXERCISES
§ 8. Introduction. The construction of formulas is
illustrated by the mensuration formulas in Course II,
by the percentage formula (§ 4), and other rules already
given. The following exercises give practice in actually
making new formulas.
Solve each of the following exercises and explain your
solution.
1. A boy walks at the average rate of 3 i miles an hour.
How far does he go in 3 hours? In x hours?
2. A train travels at an average rate of 40 miles per
hour. How far does it go in 2J hours? In 3x/4 hours?
3. A steamship moves at a uniform rate of 22£ miles
an hour. How far does it go in 4 hours ? In y hours ?
4. A boy runs at an average rate of 6 miles an hour.
How far does he go in 2 hours and 15 minutes? In x/4
hours?
5. An airplane is traveling at a uniform rate of 48 miles
an hour. How far does it go in 45 minutes? In y min-
utes?
6. In each of the exercises 1-5 there are three quantities
that have a certain relation to each other. The quantities
are distance, rate and time. Using the letters d, r, and (
write a formula to show their relation. Explain what
each letter stands for.
7. Find the average rate of a train that goes 150 miles
in 5 hours. (Use the formula obtained in Ex. 6.)
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28 JUNIOR HIGH SCHOOL MATHEMATICS [III, S 8
8. Find the average rate of an automobile that goes
198 miles in 9 hours.
9. Find the average rate of a bicyclist who goes 32
miles in 2 hours and 30 minutes.
10. Find the average rate of a runner (in feet per
second), who goes a mile in 5 minutes and 12 seconds.
Why is the average rate required rather than the uniform
rate?
11. An express train travels at an average rate of 44
miles an hour, and passes two towns 22 miles apart. How
long does it take the train to go from one town to the
other?
IS. An express train travels at an average rate of 40
miles an hour, and passes two towns 12 miles apart. How
long does it take the train to go from one town to the
other?
13. A freight train is traveling at an average rate of
25 miles an hour. How long will it take to go from one
city to another, if the cities are 70 miles apart?
14. In going from Boston to New York, 233 miles, an
express train takes 6 hours. What is its average rate per
hour?
15. A train leaves Boston at 9.00 A.M. and reaches
Portland, Me., at 12.10 p.m., a distance of 108 miles.
What is the average rate per hour to the nearest mile ?
16. A special train of cars is to take a party of men
a distance of 550 miles in 10J hours. What must be
the average rate per hour to the nearest mile?
17. Which is the greater speed, a runner going 100
yards in 9.5 seconds, or a train going 25 miles an hour?
J, S ,:z K i:vC00gIe
Ill, 5 8) CONSTRUCTION OF FORMULAS 29
18. I Bee a flash of a gun and hear the sound 5 seconds
later. How far away is the gunner? (Sound travels
at the rate of 1080 feet per second. Since light travels
at a speed of approximately 186,000 miles per second,
do you need to consider the time that it took the flash
to reach you T)
19. The echo of a sound returns from a cliff in 3 seconds.
How far away is the cliff?
20. A flash of lightning is seen, and the thunder peal is
heard 5 seconds later. How far away did the lightning
strike?
SI. A flash of lightning is seen, and the thunder peal is
heard 9 seconds later. How many feet away did the
lightning strike? Is your result probably the correct
distance to the nearest foot? Explain your answer.
22. A flash of lightning is seen, and the thunder peal
is heard 4 seconds later. Find the distance to the nearest
tenth of a mile. Is the accuracy required here reasonable ?
Why?
23. A tramping party is walking at the average rate of
3£ miles an hour. They leave a certain town at 8 a.m.
Taking out one hour for lunch and rest, they reach another
town at 5 p.m. How many miles apart are the towns?
Does this result show the exact distance or the approxi-
mate distance ? Why ?
24. What is a pedometer? ■ (Consult a dictionary.)
Explain how it works. The leader of a tramping party,
after traveling for 5 hours, finds that his pedometer regis-
tered l'li miles. What has been the average rate of the
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30 JUNIOR HIGH SCHOOL MATHEMATICS [III, £ S
party per hour? What can you tell about the accuracy
of the number of miles registered by the pedometer ?
25. If it is necessary for you to walk to a place exactly
one mile distant and to be there at a definite time, how
accurately can you tell when you must start?
26. Could you find the distance between two places by
walking from one to the other with a watch in hand?
Explain. If the distance is over a mile how accurately
do you think you could measure it in this way?
27. A street rises 20 ft. in a horizontal distance of 400
ft. What is the slope?
The grade or slope of a street tells how steep the street is.
It is often expressed in per cent ; that is, as a rise of a cer-
tain number of feet to the hundred.
In this problem, since the street rises 20 ft. in a horizon-
tal distance of 400 ft., it is evident that it rises 5 ft. to the
hundred. Why ?
This is called a 5% grade.
28. The formula for the relation expressed in Ex. 27 is :
£=£• (Fig. 1.) Explain.
29. What is the grade of a street that rises 32 ft. in 256
ft. ? Express it as a decimal and as a per cent.
SO. A bridge is to be 10 ft. above the level of the road.
If the grade must not be more than 20%, how long must
the approach to the bridge be ?
is, Google
Ill, ft 8] CONSTRUCTION OF FORMULAS 31
31. A street rises 43 ft. in a horizontal distance of
137 ft. Find its slope or grade, within a tenth of one
per cent.
82. Balance a yardstick at its center. Place a 3-lb.
weight on one part 12 in. from the center and a 4-lb.
weight on the other part, so as to balance the 3-lb. weight.
How far is the 4-lb. weight from the center?
r^n
Figure 2 represents the lever described in Ex. 32. The
balancing point is called the fulcrum of the lever.
The product of one weight by its distance from the fulcrum
equals the product of the other weight by its distance from the
fulcrum; that is, 4X9 = 3X12.
33. Two boys are balanced on a teeter board. One
of the boys weighs 80 lb. and is 3 ft. from the fulcrum.
If the other boy weighs 60 lb., how far will he be from the
fulcrum? (The equation is : 80X3 = 60d.)
34. One boy weighing 100 lb. is 4 ft. from the fulcrum
of a teeter board. Another boy just balances him at a dis-
tance of 5 ft. from the fulcrum. What is the weight of
the second boy ?
36. A and B are 5 ft. and 7 ft. from the fulcrum of a
balanced teeter board. A weighs 84 lb. What is the
weight of B?
36. In each of the Exs. '32-35 there are four quantities
that have a certain relation to each other. The quantities
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32 JUNIOR HIGH SCHOOL MATHEMATICS [III, § 8
are the two weights, wi and w t ; and their respective
distances from the fulcrum, di and dt. Write a formula
to show the relation, and explain what each letter stands
for.
37. What is the weight of an object 10 in. from the
fulcrum of a lever, if it balances a weight of 6 lb. at
14.5 in. from the fulcrum? (Use the formula obtained
in Ex. 36.)
36. What is the weight of an object 7£ in. from the
fulcrum of a lever, if it balances a weight of 4£ lb. at 10 in.
from the fulcrum?
39. To raise a stone of 250 lb., a man places a crowbar
so that the fulcrum is 6 in. from the point at which the
crowbar touches the stone (Fig. 3). What force must
be applied to the crowbar 24 in. from the fulcrum to
raise the stone?
40. In Ex. 39, suppose the force is applied 3 ft. from
the fulcrum, what force is necessary ?
41. A pole 10 ft. long is used to lift the wheel of an
automobile. One end of the pole is placed under the hub
of the wheel 1} ft. from the fulcrum. If the wheel iB
carrying a load of 800 lb., what force will need to be ap-
plied at the other end of the pole to lift the wheel?
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HI, (91 CONSTRUCTION OF FORMULAS 33
§ 9. Mathematical Symbols. Formulas are used to
express in abbreviated form the rules of mensuration and
mechanical laws. Thus long rules and tedious explana-
tions are avoided, and the solution of problems is made
simple and direct.
In order to construct a formula from a rule, the ability
to do the two things that follow, is essential :
(1) To interpret correctly from the language of the rule
what mathematical operations are required and in what
order they are to be performed.
(2) To express accurately these operations and rela-
tions by the use of necessary mathematical symbols.
EXERCISES
Express mathematical symbols for each of the follow-
ing phrases.
1. The sum of two numbers, a and 6. Ant. a+b.
2. The difference of two numbers, a and b.
3. The product of two numbers, a and b.
(The product of two numbers, such as a and 6, may be
written aXb, or a-b, or db. The dot (-) is often used
instead of the sign X to avoid confusion with the letter
»•>
4. The quotient of two numbers, a and b.
6. The ratio of a to 6.
0. The sum of three times x and twice y.
T. The difference between five times x and three
D
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34 JUNIOR HIGH SCHOOL MATHEMATICS [111. f 9
8. The product of three times x and four times y.
9. The quotient of x and four times y.
10. The sum of one half a and two thirds 6.
11. The ratio of 3 times xtoy.
12. The square of c.
13. The sum of the squares of o and b.
14. The difference of the squares of z and y.
15. The quotient of the square of a and three times b.
16. The ratio of the square of b t to the square of b±.
IT. Three times the difference of c and d.
IS. One half the sum of m and p.
19. The square of the sum of z and y.
20. The square of the difference of p and q.
21. The sum of the squares of a and b plus twice their
product.
22. The sum of the squares of a and b minus twice their
product.
23. The cube of a.
24. The sum of the cubes of a and 6. ^ -
26. The ratio of the cubes of x and y.
§ 10. Formulas of Mensuration.
Example. Write a formula for the following rule and
draw a figure. The area of a rectangle equals the product
of its base and height. (A, b, A.)
Solution. The formula is
A-bh,
;, S ,:z K i:vC00gIe
Ill, J 10] CONSTRUCTION OF FORMULAS
where
A=the area of a rectangle (in
square units),
b = thc base (in linear units),
A = the height (in linear units). F" 3 - *■
Note. In explaining the letters in a formula, it is necessary
to name the kind at unil used, when possible.
EXERCISES
Write a formula for each of the following rules, ex-
plaining each letter used, and draw a figure.
(The letters in the parentheses suggest the letters that
may well be used in each exercise. For definitions of
geometric terms see Part II.)
1. The area of a triangle equals one half the product
of its base and height. (A, b, h.)
2. The area of a trapezoid equals one half the product
of the sum of its bases by its height. (A, b,, bi, h.)
3. The area of a square equals the square of one side.
(A, 8.)
4. The perimeter of a rectangle equals twice its base
plus twice its height, (p, 6, A.)
6. The perimeter of a square equals four times one side.
(p. ..)
6. Circumference of a circle.
(a) The circumference of a circle equals r times its
diameter, (c, d.)
(ft) The circumference of a circle equals 2r times its
radius, (c, r.)
(Formulas (a), in Exs. 6-7, are commonly used in
science.)
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36 JUNIOR HIGH SCHOOL MATHEMATICS [III, g 10
7. Area of a circle.
(a) The area of a circle equals \x times the square of
its diameter. (A, d.) (See note at foot of page 35.)
(b) The area of a circle equals *- times the square of its
radius. (A, r.)
8. The three sides of a right triangle.
(a) The square of the hypotenuse of a right triangle
equals the sum of the squares of its two sides, (c, a, b.)
(6) The square of one side of a right triangle equals
the square of its hypotenuse minus the square of the
other side.
9. The sum of the three angles of a triangle equals
180°. (A, B, C.)
10. The volume of a rectangular block equals the prod-
uct of its length, width, and height. (V, I, w, h.)
11. The volume of a cube equals the cube of one edge.
{V, e.)
12. The volume of a regular pyramid equals one third
of the area of its base times its height. (V, B, A.)
15. The area of a curved surface of a right circular
cylinder equals 2t times its radius times its height.
OS, r, A.)
14. The volume of a right circular cylinder equals *
times the square of its radius times its height. (F, r, A.)
16. The area of the curved surface of a right circular
cone equals t times its radius times its slant height.
OS, r, I.)
16. The volume of a right circular cone equals £r times
the square of its radius times its height. (V, t, A.)
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Ill, ! Ill CONSTRUCTION OF FORMULAS 37
17. The area of the surface of a sphere equals 4r times
the square of its radius. (S, r.)
18. The volume of a sphere equals |r times the cube
of its radius. (V, r.)
Id. The area of a circular ring equals the difference
between the areas of the two circles inclosing the ring.
(A, ri, rj.)
20. Write a formula for the volume of a hollow cylinder.
{V, r It r a , ft.)
§ 11. Formulas of Science and Industry.
Write a formula for each of the following laws, or rules,
explaining each letter used.
1. The distance passed over by a body in uniform
motion equals the rate times the time, (d, r, (.)
2. The space passed over by a falling body equals
16.08 times the square of the time, (a, (.)
Note. In order to write the formula Tor Ex. 2, and the other
scientific facta that follow, it is not essential that you understand
the subject matter connected with the facts.
3. The law of the lever is : The product of one force
by its distance from the fulcrum equals the product of the
other force by its distance from the fulcrum. (J u & u / 2 , d t .)
4. The density of a substance equals the ratio of its
weight to its volume. (D, W, V.)
6. The horse-power of a gasoline engine is found by
multiplying the square of the diameter (in inches) of one
cylinder by the number of cylinders, and dividing that
product by 2.5. (A.p., d, n.)
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38 JUNIOR HIGH SCHOOL MATHEMATICS [III, 5 11
8. Ohm's Law for electric current is : The intensity
of the current (in amperes) equals the quotient of the elec-
tromotive force (in volts) divided c F
by the total resistance (in ohms). ■
(I, E, R.)
90
7. The law for changing the ■
reading on , the centigrade ther-
mometer to the reading on the
Fahrenheit thermometer is: The
Fahrenheit reading equals nine **_
fifths ($) the centigrade reading *'
plus 32°. {F, C.) (See Fig. 5.) »'
8. The law for changing from t0
Fahrenheit reading to centigrade 10'
reading is ; The centigrade reading
equals five ninths of the remainder _ M *
obtained by taking 32° from the
Fahrenheit reading.
9. The surface speed of a wheel
(in feet per minute) equals * times
the diameter (in inches) times the number of revolutions
per minute divided by 12. (F , d, R.)
10. The horse-power of a steam engine is found in the
following way. Find the product of the four quantities,
mean effective steam pressure (in pounds per square inch),
length of piston stroke (in feet), area of the piston (in
square inches), and number of strokes per minute ; divide
this product by 33,000. (H. P., P, L, A, N.)
11. The law of the jackBcrew is: The ratio of the
weight to be lifted to the force required equals the ratio
of the circumference of the circle traced by the end of the
Fig. 5.
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Ill, i 12] CONSTRUCTION OF FORMULAS 39
handle to the pitch of the screw (the distance between
two successive threads of the screw. See Fig. 6.). (W,
F, c, p.)
§ 12. Applications. The problems in this section are
solved by the aid of the formulas constructed in sections
10 and 11. All formulas of mensuration will be found
in Part II, Chapter XXI.
When solving problems by the use of formulas, take the
following steps :
(1) Write the formula.
(2) Substitute the known values in the formula.
(3) Estimate the result and record it with its proper
label.
(4) Compute the result and record it with its proper
label.
(5) Note how nearly your estimated aad computed
results agree.
Example. A cylindrical can is 8.0" high and 5.0" in di-
ameter. Find its contents.
Solution. V=rr t h Estimate
7=3.14X(2J)*X8 V=150cu. in.
F-157 Ana. 157 cu. in.
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40 JUNIOR HIGH SCHOOL MATHEMATICS [III, 8 12
1. Find the perimeter and the area of a rectangle, ii
its base is 7.5" and its height is 5.2".
2. Find the perimeter and area of a square, if its side
is 3*".
3. Find the area of a triangle, if its base is 8£" and its
height is 5.4".
4. Find the area of a trapezoid, if its bases are 7\"
and 8-}" and its height is 9.40".
6. Find the circumference and area of a circle whose
diameter is 9.0". (Use 3.14 for the value of *■.)
6. In a certain right triangle the two sides that include
the right angle are 60" and 11". Find its hypotenuse.
7. The area of a rectangle is 48.3 sq. in. and its length
is 11.5". Find its width.
8. A square lot of land has an area of 576 sq. ft. How
many feet of fence will be required to inclose it ?
9. A square lot of land has an area of 420 sq. ft. How
many feet of fence will be required to inclose it?
10. The perimeter of a square and the circumference
of a circle are of the same length, 42" each. Which
incloses the larger area? How much larger? What per
cent larger?
11. In a certain city there are 80 miles of street rail-
ways. The wire carrying the electric current has a f"
diameter. How many cubic feet of wire are being used
to carry the current ?
(Suggestion. The wire is a cylinder. See that the
units are of the same kind before using the formula.)
12. A certain steel ball is 3" in diameter.
(a) What is its volume?
J, S ,:z K i:vC00gIe
Ill, 1 12] CONSTRUCTION OF FORMULAS 41
(b) What is its weight, if a cubic inch of the steel weighs
.28 1b.?
13. Find the equal sides of an
isosceles triangle, if its base is 20"
and its height 12". (Fig. 7.)
Note. The altitude of an isosceles tri-
angle bisects the base.
14. Find the altitude and area
of an equilateral triangle, if each
side is 12". (Fig. 8.)
15. A baseball diamond is a
square 90' on a side. What is
the distance from first base to
third base? From second base
to the " home plate " ?
16. A circle has a 4" diameter. How long is an arc of
40°? (Suggestion. An arc of 40°=^yfe or £ of the cir-
cumference.)
17. Find the length of a 60° arc on a 12" circle.
Note. A 12" circle means a circle having a 12" diameter.
18. A belt is in
contact with a 36"
pulley for 200° of the
circumference. What
is the length of the
arc of contact? (Fig.
n \ i' us. a.
19. The arc of contact of a belt on a 15" pulley is 170°.
Find the length of the belt in contact with the pulley.
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42 JUNIOR HIGH SCHOOL MATHEMATICS (III, S 12
SO. A belt connects two IS" pulleys and the centers
of the pulleys are 10' 6" apart. How long is the belt?
(Fig. 10.)
21. What is the horse-power of the engine of an auto-
mobile having 6 cylinders, each 4" in diameter? (See
Ex. 5, page 37.)
22. Find the horse-power of a 4-cyiinder engine, each
cylinder being 4J" in diameter.
23. Make the following changes in thermometer read-
ings:
(a) Fahrenheit readings for 45° C. ; 100° C. ; 27° C. ;
10° C. ; 0° C. (See Ex. 7, page 38.)
(6) Centigrade readings for 77° F. ; 212° F. ; 67° F. ;
40° F. ; 32° F. (See Ex. 8, page 38.)
2*. Find the surface speed of a grindstone if its diameter
is 30" and if it is running at 90 revolutions per minute.
(See Ex. 9, page 38.)
;, S ,:z K i:vC00gIe
CHAPTER IV
METRIC MEASURES
§ 13. Metric Measures. The metric measures are used
by nearly all the countries of continental Europe and
South America. They were first used in France shortly
after the French Revolution. With the increasing busi-
ness relations with these countries it is necessary that we
become familiar with the measures used by them. The
metric measures are also used in scientific laboratories.
When you become familiar with them, you will see that
they are much more convenient to use than our measures.
§ 14. Linear Measure. The unit of linear measure is
the meter. The meter = 39.37", a little more than a
yard.
The meter (m.), is divided into 100 equal parts, called
centimeters (cm.). Each centimeter is again divided
into 10 equal parts, called millimeters (mm.). For long
distances the kilometer (Km.) is used. The kilometer-^
1000 meters and is about .6 of a mile. The four measures
named above are the ones commonly used.
Tablb
10 millimeters (mm.) '1 oentimeter (cm.) -.3937 in.
100 centimeters -1 meter (m.) -39.37 in.
1000 meters -1 kilometer (km.) - .621 mi.
The tables for metric measures are on page 292.
43
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44 JUNIOR HIGH SCHOOL MATHEMATICS [IV, { 14
EXERCISES
1. Express a meter in feet to the nearest tenth of a
foot. Express a kilometer in feet.
2. What is the diameter, in inches, of a 42-centimeter
gun? A 310-millimeter gun? A 75-millimeter gun?
A 36-centimeter gun?
S. An army captures 700 meters of trenches. Express
this distance in yards.
4. An army advances 4| kilometers. How many
miles does it advance?
5. The wing-spread of a certain 80-horse-power French
biplane is 28 meters. Express this in feet.
6. The wing-spread of a certain 220-horse-power
French biplane is 13 meters. Express this in feet.
7. To become a pilot of a scouting airplane "at the
front," one of the requirements was that the aviator must
make an altitude of 7000 meters. Express this height in
kilometers ; in miles.
8. The unit of area is the square meter.
How many square centimeters does it con-
tain?
9. The unit of volume is the cubic
meter. How many cubic centimeters does
it contain?
10. Measure your desk cover in centi-
meters and find its area.
11. Measure the length and width of the floor of the
classroom in meters. Find its area.
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IV, S 14]
METRIC MEASURES
45
12. How much longer is a 100-meter dash than a 100-
yard dash?
13. Suppose you walk at the average rate of 3£ miles
per hour. What is your average rate in kilometers per hour ?
14. What is your height in centimeters? In meters?
15. A group of French boys wish to lay out a baseball
diamond of the same size as the American diamond.
(a) How long, in meters, will they make each side of
the square? (See Ex. 15, page 41.)
(b) The pitcher's box is 60.5' from the home plate.
How far, in meters, will they locate the pitcher's box from
the home plate ?
(c) In making a home run, what part of a kilometer
will be traveled?
(d) What is the area of the baseball diamond in square
meters?
16. Figure 12 is the plan of a double tennis court,
drawn to the scale of 40 feet to the inch,
t
J.
71
\
Fiq. 12.
(a) How many meters would a French boy make each
of these distances, if he were laying out a court ?
(b) Draw a plan of this tennis court to the scale of 20
centimeters to one centimeter.
(c) How many square meters are there in the tennis
court?
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46 JUNIOR HIGH SCHOOL MATHEMATICS [IV, f 15
§ 16. Weight. The metric unit of weight is the gram.
The gram is the weight of one cubic centimeter of water.
It is a little more than .03 of an ounce, so you see that it
is a very small weight. The weight that is in most com-
mon use, corresponding to our pound, is the kilogram
(1000 grams) ; it is equivalent to very nearly 2.2 pounds.
Table
lgram(g.) -.03527 oz.
1000 grams -1 kilogram (Kg.) -2.20462 lb.
1000 kilograms -1 metric ton (T.) =2204.62 lb.
= 1.102 U.S. tons.
1. An order for 5 kilograms of sugar would be for
how many pounds?
5. A grocer receives the following order :
3* Kg. beef
12 Kg. potatoes
$ Kg. pork
200 g. pepper
How would he fill it using pounds and ounces ?
3. What is the weight in pounds of a 700-kilogram
Bhell?
4. How many kilograms of flour are there in a barrel?
(lbbl. flour=1961b.)
6. How much does a ton of coal weigh in kilograms?
6. A certain British war tank weighs 75 metric tons.
What is its weight in U. S. tons?
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IV, i 16) METRIC MEASURES 47
7. A certain type of steel merchant vessel weighs 8000
metric tons. What is its weight in U . S. tons ?
8. What is your weight in kilograms ?
9. Express in kilograms the weight of the United
States' standard bushel of each of the following grains :
(a) wheat, 60 lb., (6) corn, shelled, 56 lb., (c) barley,
48 lb., (d) oats, 32 lb.
10. What price per kilogram is equivalent to each of
the following prices per bushel : (a) oats, $1.00, (6) barley,
»1.75, (c) corn, shelled, $2.00, (d) wheat, $2.20?
§ 16. Capacity. The metric unit of capacity is the
liter. The liter is equivalent to very nearly 1.06 quarts
(liquid measure) .
1000 cubic centimeters -1 liter (1.)
Since one cubic centimeter of water weighs 1 gram, a liter
of water weighs 1 kilogram.
1. What is the weight of a liter of water in pounds?
2. A gallon jug holds how many liters?
3. If milk is selling at 12 cents a quart, what will 5
liters of milk cost ?
4. A liter of milk is sold for 15 cents. How much will
eight quarts cost ?
5. The capacity of the gasoline tank on a certain auto-
mobile is 10 gallons. What is the capacity of the tank
in liters?
6. What price per liter is equivalent to 28 cents per
gallon for gasoline ?
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48 JUNIOR HIGH SCHOOL MATHEMATICS [IV, 5 17
§ 17. Specific Gravity. One cubic centimeter of water
weighs one gram. One cubic centimeter of mercury
weighs 13.6 grams ; that is, mercury is 13.6 times as heavy
as water. This result, 13.6, is known as the specific
gravity of mercury.
The specific gravity of a substance is the ratio of the
weight of a certain volume of the substance to the weight
of the same volume of water. For example, the weight
of one cubic foot of a certain kind of brick is 125 pounds
and the weight of one cubic foot of water is 62.5 pounds,
hence the specific gravity of brick is 125-5-62.5, or 2.
1. Explain the statement: The specific gravity of a
certain kind of steel is 7.8.
2. A block of steel (Ex. 1.) contains 200 cu. cm. What
is its weight in grams? In kilograms?
3. The specific gravity of a certain kind of maple
wood is 0.8. What is the weight in kilograms of a beam
5 m. long, 10 cm. wide, and 8 cm. thick?
4. A certain rectangular iron plate is 1 cm. thick, 18
cm. long, and 3 cm. wide. How many grams does it weigh ?
(The specific gravity of the iron is 7.2.)
5. A circular iron plate is £ cm. thick and 30 cm. in
diameter. Find its weight in grams, if the specific gravity
is 7.2.
6. A block of iron is 28 cm. long, 15 cm. wide, and 10
cm. thick. How many kilograms does it weigh if its
specific gravity is 7.1? How many pounds?
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IV, { 17) METRIC MEASURES , 49
7. The specific gravity of a certain kind of copper ia
8.9. Find the weight of a solid copper cylinder 15 cm.
high and 10 cm. in diameter.
8. Find the weight of a copper block 12 mm. high,
14 mm. wide, and 20 mm. long, if its specific gravity is 8.9.
8. The specific gravity of mercury is 13.6. Find the
weight of the mercury that fills a circular tube 5 mm.
inside diameter and 20 cm. long.
10. The specific gravity of ice is .92. What is the
weight of a block of ice 24 cm. by 18 cm. by 10.5 cm.?
11. In the metric system the density of a certain kind
of iron is 7.3 g. per cubic centimeter. The specific gravity
of the iron is 7.3. Is the density of the iron 7.3 lb. per
cubic foot?
12. The density of water is 62.5 lb. per cubic foot. What
ia the density of the iron in Ex. 11, in lb. per cubic foot?
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CHAPTER V
LINEAR EQUATIONS
§ 18. Equations. An equation is a statement of the
equality of two number expressions. The two number
expressions are called the members of the equation.
There are two kinds of equations, the equation of condi-
tion and the identity.
The equation of condition is an equation in which the
members are equal only when the letters have particular
values. For example, 5x = 10 is an equation, the particu-
lar value of x being 2. Equations of condition are the
equations used in the solution of problems.
The identity is an equation in which the members merely
represent different ways of writing the same number.
For example, 5x+3x»8x is an identity.
In expressing identities involving letters, the symbol «
is often used instead of the symbol = . The symbol ■ is
read "is identical to."
In Chapter VII, Second Course, equations were solved
by using one or more of the following axioms :
(1) // the same number is added to equal numbers, the
sums are equal.
(2) If the same number is subtracted from equal numbers,
the remainders are equal.
(3) // equal numbers are multiplied by the same number,
the •products are equal.
(4) If equal numbers are divided by equal numbers, the
quotients are equal.
»
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V, 1 18] LINEAR EQUATIONS 51
The root of an equation is the value of the unknown
quantity which satisfies the equation. The check shows
whether or not the value obtained from the solution is the
root.
In checking the root of an equation, the following axiom
is used:
(5) A number may be substituted for its equal in an equa-
tion.
answer.
Solution.
®
3o 9
2 "4
©
6o-9
®X4
®
0-1.5
®+6
Check.
3X1.5 .9
2 4
4.5 .9
2 4
2.25-2.25
Ans
. 0-1.
In checking the root of an equation, the following rules
should be observed :
(1) Substitute the value *of the root obtained in the
original equation, — not in any subsequent step.
(2) Simplify each member of J.he equation by itself.
(3) Retain the question mark over the equality sign in
the check until the two members of the equation are shown
to be equal. In case the value of the root is an approxi-
mation, the two members can be shown to be only ap-
proximately equal ; hence, retain the sign A throughout.
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52 JUNIOR HIGH SCHOOL MATHEMATICS [V, g 18
Example 2. Solve the equation 5n— 6=2n+12, and
check the answer.
Solution.
® 5n-6=-2n+12
® 5n-2n+18 ®+6
® 3n=18 ®-2n
® n = 6 ®-^3
The symbol, ®+6, means that 6 is added to each mem-
ber of equation ®.
The symbol, ®— 2n, means that 2n is subtracted from
each member of equation ©.
Check. (5X6)-6=L(2X6) + 12
30-6il2+12
24 = 24 Ana. n = 6.
Is 6 the root of the equation 5n-6 = 2n+12? How
do you know?
Solve each of the following equations and check each
answer.
1. 3m=12 10. x+3 = 7
2. 5a+3a=16 g 11. 14 = wi+ll
3. 3j/+y=10 12. 28 = 0+15
4. 7y-5y=S IS. 2y+3 = 7
6. 4m+3m=21 14. 26+12 = 17
6. 3a+2a+a=24 16. 36+4 = 16
7. 76+56-86=16 16. 15 = 106+10
8. 5z+2x-x=36 17. 38-2c+9
9. a+8=ll 18. y-3 = 8
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LINEAR EQUATIONS
19. 4«m-5
42.
?K_3
20. 3o-4-8
3 5
21. 56-3 = 7
43.
10 _ 2 (Multiplying
x by*, 10 = 2s.)
22.~4ifc-6 = 16
23. 44 = 8c-4
44.
!?-4
24. 5a~2 = a+26
45.
«-4
25. Zn+2 = n+S
to
26. 5d+7=9d+3
46.
i-3
27. 4fc+13=6*:-ll
5-12
a
28. 5u>+3 = 9m>-17
47.
29. 5o+6 = 7a-8
- 12
48.
4.8=—
30. 7«)-2=4to+19
r
31. 3*+2=27-2x
49.
9 . . (Multiplying
32. 7j/-4=16-3j/
2i by2x,9-3».)
33. 15-2y = 3y+10
50.
f"' 3
4j|
5 "°
"2.5J
2-1?
34. 5w-6=40-3«>
35. 2=21
6 3
51.
52.
36. 2*2 = 12
5»
63
4_ 1 (Multiply by
37. f -12
w 2 2«J.)
38. ^-10
54
5 10
2 y
3 *' T 5
56
3_ 5
l"l2
54
15 = 45
40. — = .8
56
4
V 1
,« 5 2«j
_2_=A (Multiply
"■ ris
57
5ro 15 by 15m.)
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54 JUNIOR HIGH SCHOOL MATHEMATICS IV, i 18
68. -2-.J5
62. 4(m-2)-20
10* 100
66. 3(<J+7)-45
... 1-4
64. 15-3(1-1)
5 51
66. 12-3(jl+l)
eo ".L§
66. 2(3fc+4)-26
12 4x
67. 5(6+2)-2(6+14)
61. 3(26+3)-
21
68. 4(m-2)-3(i»+l)
(Removing parentheses,
68. 8(o+5)-2(3o+26)
66+9-21.)
70. 5(2s+l)-3(4-i,)
PROBLEMS
1. The average rate per hour that a certain boy travels
on his bicycle is 15 miles. This is one more mile than
four times his average rate when walking. What is his
average rate when walking ?
Solution. Let r=the boy's average rate in miles
per hour when walking, then 4r+l--- his average rate on
his bicycle.
The equation is :
® 4r+l = 15
® 4r=14 ®-l
® r-3i ®+4
Check. Substitute, in the statement of the given
problem, 3J miles per hour for the average rate of the boy
when walking. Am. 3£ miles per hour.
I 2. By vote of the United States Senate on the Food
Control Bill, August 8, 1917, the bill was carried by a
majority of 59. The total number of senators voting was
73. (a) How many voted for the bill? (&) What per
cent voted for the bill ?
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V, S 181 LINEAR EQUATIONS 55
3. On a certain Y. M. C. A. track a boy has to go 10
laps and 100 yd. besides, for a mile run. (a) What is the
distance around the track ? (b) What per cent of a mile
is one lap ?
4. The bill for the Suffrage Federal Amendment was
carried by a majority of 138 in the House of Representa-
tives on January 10, 1918. The total number of repre-
sentatives voting was 410. (a) How many voted for the
bill ? (b) What per cent ?
5. A freight train running 22 miles an hour is 154
miles ahead of an express train running 50 miles an hour.
In how many hours will the express overtake the freight ?
6. A transport averaging 21 knots an hour leaves a
certain port when a merchant ship, averaging 17 knots
an hour, is already 160 knots out. In how many hours
will the transport overtake the merchant ship?
7. A fleet averaging 15 knots an hour is 630 knots from
a certain port when a destroyer, averaging 22 knots an
hour, starts out to overtake it. In how many hours will
the destroyer overtake the fleet ?
8. A, traveling in an airplane at 95 miles an hour,
sets out to overtake B who is traveling in an airplane at
80 miles an hour. ' If B had a start of 135 miles, in how
many hours will A overtake him?
9. A certain- 12" I-beam used in building
construction weighs 525 lb. The density of the
iron is 450 lb. per cubic foot. How many cubic
feet of iron are there in the beam?
(Suggestion. Use the formula D = — .)
" Fio. 13.
3,g,1 EE d by GoOgle
56 JUNIOR HIGH SCHOOL MATHEMATICS [7, 5 18
10. A certain piece of steel weighs 16.8 lb. The density
of the steel is .28 lb. per cubic inch. How many cubic
inches are there in the piece ?
11. A certain ball of cork weighs 35 lb. The density
of cork is 16 lb. per cubic foot. How many cubic feet are
there in the ball?
12. A part of the trunk of a certain chestnut tree weighs
148 lb. The density of the chestnut is 33 lb. per cubic
foot. Find its volume in cubic feet.
13. The sum of two consecutive integers is 47. Find
them.
n=the smaller integer, then
the larger integer, and
their sum.
Solution. Let
n-
= the
n+1.
= the
2n+l-
= thE
The equation is
®
2n+l.
■47
®
2n.
■46
®
n =
-23
©
n+1-
-24
CD-l
®-r2
® + l
Check. Substitute 23 and 24 for the two integers.
The work is left to the student. Ana. 23 and 24.
11. In a certain number of two digits, the units' digit is
3 more than the tens' digit. When the number is divided
by 4, the quotient is 9. Find the number.
[Thus, 47 is an illustration of a number of two digits ;
4 is the tens' digit and 7 is the units' digit. To form the
number, you multiply the tens' digit by 10 and add the
units'digit; forexample, 4X10+7=47.]
isy Google
V, ! 18] LINEAR EQUATIONS 57
Solution. Let (--the tens' digit, and
(+3 = the units' digit, then
10f+(f+3) = the number.
The equation is :
q 10H-<+3 _
© lOH-f+3-86 ©X4
® ll(+3 = 36 ®=
® 1M-33 ®-3
® ( = 3 ®-s-ll
® f+8-6 ®+3
The number is: 3X10+6, or 36.
Check. Substitute 36 for the number. The work is
left to the student. Arts. 36.
IB. There are two consecutive integers whose sum is
75. Find them.
16. There are three consecutive integers whose sum is
246. Find them.
17. There are two consecutive integers whose sum in-
creased by 5 is equal to 42. Find them.
18. There are two consecutive integers whose sum de-
creased by 4 is equal to 45. Find them-
19. There are two consecutive integers whose sum
divided by 7 gives a quotient of 9. Find them,
20. If you add 3 to a certain number and then divide
this sum by 7, the quotient will be 13. Find the number.
21. If you subtract 3 from a certain number and then
divide this result by 3, the quotient will be 27. Find the
number.
lilted By GoOgle
58 JUNIOR HIGH SCHOOL MATHEMATICS (V, S 18
22. If a certain number is divided by 5, the quotient
is 8 and the remainder is 3. Find the number.
(The equation is : ?-8+|, or 5^=8. Why?)
5 5 5
23. If a certain number is divided by 7, the quotient
is 20 and the remainder is 4. Find the number.
21. There are two consecutive integers whose sum
divided by 6 gives a quotient of 9 and a remainder of 3.
Find the integers.
25. If 9 is added to a certain number and this sum is
divided by 7, the quotient is 8. Find the number.
26. There are two consecutive even numbers whose sum
is 42. Find them.
27. There are two consecutive odd numbers whose sum
diminished by 2 is 30. Find them.
28. If 4 is added to twice a certain number and the re-
sult is divided by 5, the quotient is 3 and the remainder is 1 .
Find the number. ,
29. In a certain number of two digits, the units' digit
exceeds the tens' digit by 2. The number is 4 times the
sum of its digits. Find the number.
30. In a certain number of two digits, the tens' digit
is twice the units' digit. The number is 12 more than 5
times the sum of its digits. Find the number.
31. In a certain number of two digits, the units' digit
exceeds the tens' digit by 5. When the number is divided
by 3, the quotient is equal to the sum of the digits. Find
the number.
J,S,:z K i:vC00gIe
V, { 19) LINEAR EQUATIONS 59
32. In a certain number of two digits, the tens' digit
exceeds the units' digit by 5. If 6 is subtracted from the
number and the result is divided by 6, the quotient is 2
more than the sum of the digits. Find the number.
§ 19. Transformation of Formulas. The formula for
the area of the circle is : A =^-- A is called the subject
of the formula, just as in the rule for the area of the circle,
the word "area" is called the subject of the sentence.
To transform a formula means to change the subject of the
formula. This requires the solution of the given formula
for the new subject.
When it is necessary to find the diameters of circles which
shall have given areas, time and labor can be saved by
changing the subject of the formula for the area of the
circle from A tod.
Example. Change the subject of the formula A=~
f rom A to d.
Solution.
®
*-?
®
®
M.J.
®X4
®+i
©
vf=«
®v^
®
©-
An.. <i=\/B
;, S ,:z K i:vC00gIe
60 JUNIOR HIGH SCHOOL MATHEMATICS IV, { 19
BXKRCISES
Change the subject
aa required.
of each of the following formulas
1. A=bk, 6=?
16.
S-*rl, !-?
2. A = ibh, ft-?
17.
7-4Bft, B-?
S. A-*, s=?
18.
V-j^ft, ft-?
i. c=-rd, d= ?
18.
K-Jn'ft, r-?
5. c = 2w, r=?
20.
S = 4rH, r-?
6. d=rt, r= ?
91
F-4n-,r-?
7. Wiiii = Wfdt, uii =
? 22.
A-l(oi+6.)ft, ft-?
8. uiidi = wtdi, rf s =
? 23.
^A+ZB-180, ^A-?
..,-£r=?
24.
c"-o'+()', o-?
10. >-prt, j>-?
23.
c'-o'+d', 6-?
11. A=*r*, r=?
26.
H-H,,.?
12. V=lwh,u>="[
27.
I-fj-t
13. S = 2rTfc, r=?
28.
14. F-n*», ft-?
29.
S-i »l',l= ?
H. F-«T'ft, r-?
30.
F-JC+32, C-1
§ 20. Graphs of Linear Equations. The equation
x+j/=4 contains two unknown numbers. One solution
of this equation is x = 1, y = Z; a second solution is x— 2,
j/=2; a third solution is x = 3, 9=1; etc. The value
obtained for x depends upon the value given to y ; that is,
x changes with y or fanes with y.
U, S ,:z K i:;G00gIC
V, S 20) LINEAR EQUATIONS 61
In Chapter VIII, Second Course, you learned how to
locate these pairs of values on squared paper, and you
found that, when you joined the points thus located, you
had a straight line, as in Fig. 14. This is the graph of the
equation x+y = 4.
■XEROSES
1. From the equation x+y=4, when x—3, y= ?
Which point does this
pair of values of x and y
locate on the graph?
2. From the equation
when x=4, y= ? Which
point does this pair of
values locate ?
3. What value of y
is paired with x = + 5 ?
Which point is it?
An*. y= - 1, Point h. Pw " "'
4. What value of y is paired with i-+7? Which
point is it?
8. What value of x is paired with y=+57 Which
point is it?
6. What value of x is paired with y=+6? Which
point is it ?
7. What value of x is paired with y=+7? Which
point is it?
8. What value of * is paired with y=+8? Which
point is it?
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62 JUNIOR HIGH SCHOOL MATHEMATICS [V, §20
Note. From exercises 1-8 it appears that :
(a) All values of x to the right of the axis YiYt arc plus,
so all values of x to the left of YiYi are minus.
(b) AH values of y above the axis X S X 2 are plus, so all
values of y below XiX 2 are minus.
9. Show that all pairs of values obtained in Exs. 3-8
satisfy the equation x+y = i.
10. What are the pairs of values for the following
points in Fig. 14 ?
(a) Point Ml
(6) Point O?
(c) Points?
(d) Point A?
(«) Point G?
11. Find four pairs of values for the equation x+2y= 9.
Let y = l, 2, 3, and 4, aud tabulate these values.
v 1 2 3 4
12. Locate on squared paper the four points for the
pairs of values obtained in Ex. 11. Draw the graph.
13. From the graph fill in the following table.
»
-1
-3
-5
?
?
?
y
?
?
'
+7
+8
14. Find out whether or not the pairs of values in the
table in Ex. 13 satisfy the equation x+2y9.
Jigilized by GoOgk
CHAPTER VI
POSITIVE AND NEGATIVE NUMBERS
5 21. Signed Numbers. In checking the pairs of values
in the equations in the last section you were dealing with
numbers having + and — signs in front of them. Such
numbers are signed numbers. Those having plus signs
in front of them are positive numbers ; those having minus
signs in front of them are negative numbers.
On the squared paper all values of x to the right of the
YiYt axis (Fig. 14) and all values of y above the XiXi
axis are positive; all values of a; to the left of the Y\Yt
axis and all values of y below the XiXi axis are negative.
Hence, when numbers are represented by points on lines,
negative numbers are measured in the opposite direction
from positive numbers. In general, the negative number
is opposite in quality to the positive number.
Concrete Illtjbtbationb of Ioeao Opposite in Qualiti
p^™
Nehativk
Cash on hand
Bills to be paid
Gain
Assets
Liabilities
Rise in temperature
Fall in temperature
Direction East
Direction West
Direction North
Direction South
3,g,1 EE d B y G00gle
64 JUNIOR HIGH SCHOOL MATHEMATICS (VI, $ 22
§ 22. Addition of Signed Numbers of the Same Quality.
Example 1. (a) A man gains S300 and later gains
$200 more. What is the result?
A gain is expressed by a positive number, hence com-
bining +300 and +200, we get +500.
Ana. $500 gain.
Example 2. A man loses $300 and later loses $200
more. What is the result?
A loss is expressed by a negative number, hence com-
bining -300 and -200, we get -500. Ans. $500 loss.
Example 3. (a) The temperature riseB 12° and later
rises 15° more. What is the total change?
A rise in temperature is expressed by a positive number,
hence combining + 12 and + 15, we get + 27.
Ans. 21° rise.
Example 4. The temperature falls 12° and later falls
15° more. What is the total change?
A fall in temperature is expressed by a negative number,
hence combining — 12 and — 15, we get —27.
Ans. 27° fall.
Combine the following pairs of numbers, placing the
necessary sign in front of the answer.
1. +7 and +4 6. -7 and -7
2. -3 and -4 7. -15 and -1
3. -9 and -15 8. +3 and +7
4. +13 and +3 9. -2and-l
5. +4and+l 10. -40 and -3
i BV Google
71,122] POSITIVE AND NEGATIVE NUMBERS 65
11. +7 and +9 16. +2J and +5J
12. -7 and -8 17. -J and -J
IS. +15 and +8 18. +i and +$
14. -12 and -18 19. -3*and-5|
IB. -land-1 20. +12$ and +6J
Note. In each of the exercises 1-20 the signs of the num-
bers to be combined are the same. In each exercise the sum of
the two numbers is obtained and the sign of the answer is the
same as the sign of the two numbers.
21. A ship travels north 80 miles the first day, 72 miles
the second and 68 miles the third. Find the average
distance per day and express the result as a signed
number.
22. A man owes bills of $25, $32.50, $17.20, and $8.90.
Find his total debts and express the result as a signed
number.
23. A provision dealer has five unfilled orders valued at
$4.50, $8.75, $18.25, $6.80, and $11.40. Find the total
amount of unfilled orders and express the result as a
signed number.
24. The temperature readings taken every three hours
for a certain day were : +36°, +42°, +50°, +68°, +72°,
+65°, +57°, +48°. Find the average temperature for
the day.
25. The temperature readings taken every three hours
for a certain day were: —12°, —10°, —6°, —4°, —1°,
-5°, —6°, —4". Find the average temperature for the
day.
J, S ,:z K i:vC00gIe
«6 JUNIOR HIGH SCHOOL MATHEMATICS [VI, J 23
§ 23. Addition of Signed Numbers of Opposite Quality.
Example 1. During a certain day the temperature
rises 10° and then falls 25°, What is the direct change in
the temperature for the whole day ?
A fall in temperature is opposite in quality to a rise.
A rise of 10°, written +10, and a fall of 25°, written -25,
when combined, give a fall of 15°, written — 15.
Ans. 15° fall.
Example 2. A man gains $500 and then loses $700.
What is the net result?
A loss of money is opposite in quality to a gain. A gain
of $500, written +500, and a loss of $700, written -700,
when combined, give a loss of $200, written —200.
Ans. $200 loss.
Example 3. A man loses $350 and gains $750. What
is the net result ?
Combining -350 and +750, we get +400, the plus
number being the larger. Ans. $400 gain.
Example 4. A motorcyclist travels 40 miles east
and then 65 miles west. How far is he from his starting
point and in which direction ?
Direction east is expressed by a positive number, hence
combining +40 and —65, we get —25, the minus number
being the larger. Ans. 25 miles west.
In combining a positive number with a negative num-
ber:
(1) Note which is the larger, the positive or the negative
number.
;, S ,:z K i:vC00gIe
VI, §23) POSITIVE AND NEGATIVE NUMBEHS 67
(2) Note how much larger it is.
(3) Note which sign is to be placed in front of the
answer.
Combine the following pairs of numbers, placing the
necessary sign in front of the answer.
1. +12 and -7 11. -4 and +4
2. +10 and -16 12. -12 and +15
3. -4and+8 13. -9and+12
4. -3 and +7 14. -12 and +14
5. +8 and -3 IB. -10 and +5
6. +15 and -11 16. +14 and -2
7. +9 and -12 17. +11 and -18
8. +8 and -17 18. +32 and -33
9. -10 and +3 19. +12 and -12
10. -15 and +1 20. -land +9
Note. In each of the exercises 1-20 the signs of the num-
bers to be combined are oppotite. In each exercise the differ-
ence between the two numbers is obtained and the sign of the
answer is the sign of the larger number.
21. The temperature at 1.00 p.m. on a certain day was
+ 12". During the next six hours it fell 15°. What was
the temperature at 7.00 p.m.?
22. A man loses $35 on one transaction and gains $47
on another. What is the net result? Express the result
as a signed number.
;, S ,:z K i:vC00gIe
68 JUNIOR HIGH SCHOOL MATHEMATICS {VI, 1 23
23. An automobilist travels 45 miles east and then
120 miles west. How far is he from his starting point?
Express the result as a signed number.
24. Emperor Augustus Csesar was born in 63 B.C. and
died when he was 77 years old. What was the date of
his death? Express the result as a signed number.
20. The water in a reservoir rises 5 in., then falls 7 in.,
rises again 12 in., then falls again 9 in. Find the net rise
or fall. Express the result as a signed number.'
26. The midnight temperatures for a certain week
were as follows : +8°, -4°, -9°, +1°, +12°, +8°, +6°.
Find the average of these temperatures.
27. The midnight temperatures for the next week were
as follows: +4°, 0°, -3°, -8°, -12°, -1°, +2°. Find
the average of these temperatures.
28. The latitude of New Orleans, La., is +30°, and of
Boston, Mass., +45°. Find the latitude of Richmond,
Va., halfway between them.
29. The latitude of Chicago, 111., is 41° 50' and of Rio
Janeiro, Brazil, —22° 50'. Find the latitude of Panama,
halfway between them.
80. Pythagoras was born about 600 B.C. and died when
he was 65 years old. Using signed numbers, find the
probable year of his death.
31. Thales was born about 640 b.c, and died when
he was 92 years old. Using signed numbers, find the
probable year of his death.
32. Plato was born about 429 b.c. and died when he
was 81 yeare old. Find the probable year of his death.
is, Google
VI, S24] POSITIVE AND NEGATIVE NUMBERS 6
§ 24. Drill Table — Addition o* Signed Numbers
>
II
III
rv
V
VI
VII
VIII
A
+3
+3
-4
-5
3
-12
-3*
3
B
-7
-8
+7
-3
-5
4
-3
-4
C
+4
+fi
+3
-6
-3
3
-3
12
D
-3
-3
-6
-9
6
-6
-3
-8
Note. Any number having no sign in front of it is positive.
Suggested drills with this table :
(a) Add each number in line B to each number in line
A ; each number in line C to each number in line A, in
line B ; etc.
(b) Add each number in column I to each number in
column II ; in column III ; etc.
(c) Find the sum of all the numbers in column I ; in
column II ; etc.
(d) Find the sum of all the numbers in line A ; in line
B; etc.
Note. In adding more than two signed numbers, add the
first to the second, then that sum to the third, etc.
jigiiized by Google
CHAPTER VII
ADDITION AND SUBTRACTION OF ALGEBRAIC
EXPRESSIONS
§ 26. Algebraic Expressions. Algebra as well as arith-
metic deals with numbers. In algebra letters are used
to represent numbers.
The formula A =oft, the equation 5a= 10, and the sum
of two numbers a+b are algebraic expressions. In alge-
braic expressions, letters are used to represent numbers.
In the product bh, b and h may be made to represent any
numbers that we wish them to represent, b and A are the
factors of the product bh.
If an expression is the product of two or more numbers,
then these numbers are factors of the expression.
If an expression is the product of two factors, either of
these factors is the coefficient of the other.
In the product bk, b is the coefficient of A, and ft is the
coefficient of 6. In the product box, the coefficient of x
is 5a ; the coefficient of a is bx.
EXERCISES
1. In the following products what is the coefficient of
as: 5a:? 7ox? 3o5x? Axyl xzl xt fmxyt
2. The expression 2a+3y is the sum of %a and Sy.
What is the coefficient of o? The coefficient of y?
§ 26. Polynomials. A polynomial is an algebraic ex-
pression composed of parts connected by the signs + or — .
Each of these parts of a polynomial, together with the
sign in front of it, is a term. A polynomial of two terms
.70
i pr^dsyGoogle
VII, 1 271 ALGEBRAIC EXPRESSIONS 71
is a binomial ; of three terms, a trinomial. One term by
itself is a monomial.
For example, 3a+4ft — 5c is a trinomial, in which the
terms are 3a, +46, and — 5c.
EXERCISES
From the following group of polynomials, select and
write in separate columns the monomials, the binomials,
and the trinomials: bh, 2a+3y, 3a-2fe+c, 10a-8fc,
5c, 2a+4, &r+2y-4, 6a-2, Aaxy, 3, 7a-36-7, x,
a-fc-2c, a-26, 3a6-2ac, m+2, 2a-3o-H, 3x, a-5-
§ 87. Exponents and Powers. The formula for the
area of a square is: .4=8*. s ! means sXs. 8* is read
"s square" or "the second power of »" ; the 2 is called
an exponent.
The formula for the volume of a cube is: F~=e*. e*
means eX«Xe. e* is read "e cube" or "the third power
of «" ; the 3 is called an exponent.
The power of a number is the product obtained by using
the number as a factor one or more times. The number
is often called the root, or bate, of the power.
The exponent of a power is the number written at the
right of, and slightly above, the factor to show how many
times the factor is used.
Example 1. Write the power a* without an exponent.
Name the exponent and the root.
Solution. a'=aXaXaXa.
In the power a 4 , 4 is the exponent, a is the root.
Example 2. Write the power 5 1 without an exponent.
Name the exponent and the root.
Solution. 5*=5X5X5.
In the power 5", 3 is the exponent, 5 is the root.
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72 JUNIOR HIGH SCHOOL MATHEMATICS [TO, 1 27
Example 3. Write the product a*x*y without using
exponents. Name the exponents and the roots.
Solution. a t x*ysa -a -x ■ x -x -y.
In the power a*, 2 is the exponent, a is the root.
In the power x*, 3 is the exponent, x is the root.
In the power y, 1 (not written) is the exponent, y is the
root.
Note. In algebra any letter, or any number, not having
an exponent written, is considered as the first power of itself ;
that is, its exponent is 1.
EXERCISES
A. Write each of the following products without ex-
ponents. Name the exponent and the root of each power.
1. a* 4. 2» 7. aW 10. x*y*
2. 6* 5. m 6 8. x*y 11. a*6V
3. a: 6 6. 8 1 9. xy> 12. 3o*& 4 c l
B. In each of the following expressions, name the
numerical coefficient and the exponent of each letter.
1. 3a* 4. 4m 7. 5a&* 10. mnj^
2. 5i* 6. y» 8. y 11. 4a*
8. 3a*&» 6. 3a s 9. 2ax*y 12. 6a f b*c'd«
§ 28. Degree of a Polynomial. The degree of a term
is the number of letters that are factors of it. The degree
is found by adding the exponents of the literal factors.
For example, 2af>* is of the fourth degree.
The degree of a polynomial is that of the term of the
highest degree in the polynomial. For example, 3a6+
2a6*+6 2 is of the third degree, this being the degree of
the highest term (the second term).
;, S ,:z K i:vC00gIe
ALGEBRAIC EXPRESSIONS
1. Name the degree of each of the following mono-
mials : a*, ab, ty', Zab, bob*, 20*6*, 3ab s c, 4a.
2. Name the degree of each of the following polyno-
mials: a , +2ab+b 1 , x+y, a?-a, 2a*+3b*, a ! +2a+l,
a*b+ab+a, 4x+5, xhj+xy.
§ 39. Addition of Monomials. Terms that contain
the same factor are similar terms or like terms. For
example, +5x, —Zx, and +7x are similar, since each
contains the factor x. Their sum is +9*.
In adding like terms, the coefficient of the common factor
in the sum is found by adding its coefficients in the different
terms.
p,i. | v EXERCISES
In each of the following exercises find the sum.
1. -6a, +4a Ans. -2a. 6. 8m 1 , -3m*
2. -13n, +7n 7. -lfa, -2*
8. +8x», +3x» 8. D, — ID
4. -3a, -2a 9. -(, +t
5. -5a:, -2x 10. -17a, +a
11. — 5x, +6ar, —2x, +x, —x Ans. —x.
12. -Ax*, +3a?, -2x*, +Sx}, -Mte*
13. -Sab, -Zab, -4ob, +7ab, +ofc
14. 14t, -St, -St, -45(
15. 68fc, +34fc, -16fc, -3*
16. 4a, -12s, +6a, +6*, -9a
17. 2&to , ) -f-SOuJ 1 , -60w)», -10w»
18. 7b, +86, -20b, +Zb, -b
19. 5x*, -3a*, -2x», -as", +3z*
20. 3o6, -7o6, -2a6, +a&, +2a6, -10a6
3, g ,i EE d by Google
74 JUNIOR HIGH SCHOOL MATHEMATICS [VII, 5 30
§ 30. Addition of Polynomials.
Example 1. Add: 31— I, 2i+l, x-7.
Solution. Arrange the polynomials so that the tike
terms are in the same column.
3i-4
2z+l
x-7
6z-10 Ana. ftr-10.
Example 2. Add: 3x-2y+3, -2x+3y, x-y-2.
Solution. 3x— 2y+3
-2x+3s/
x- y-2
2x +1 Am. 2i+l.
Note. A method of checking the addition of polynomials
is by assigning number values to the letters.
EXERCISES
In each of the following exercises add the given poly-
1. 3as+2, 41-1,31-12, 2i+8
2. 5o-3, 2o+7, 3a-5, 4o+l
3. 2i/+7, j-4, 1/-2, 3J+1
i. 46+3, -26+4, -36-12,26+6
5. 3» ! -7, 2* ! +4, -2Ji ! +5, 4ft"-2
6. 3o-l, 4o, -7o+3, 2o-l
7. 0-2, 4+3o, 7-2o, 0-5
8. z+2j, 2x-3y, x-7y, 3x+y
9. 2i 1 -y>, 4»>-2x', »"+7j ! , Sx'-y'
10. o'-2a, 2a-3o*, a*+2o, 7a a -a
JigilizBd by GoOgle
Vn, j31] ALGEBRAIC EXPRESSIONS 75
11, c+d,c-2d, d-3c, c-d
12. 4m, 2m— 3p, p— 4m, 5p
18. 7a-26, a-o,&-2a, 26-7a
14. :e*-3x, s+2x*, x-3x*, ~3x
IB. 3a, a-56, 6+3a, a-76
§ 31. Parentheses. The symbols [ ], called brackets,
J J, called braces, and ( ), called parentheses, are used to
group terms together. Thus the expression (2a+5b) +
(3a— 2b) means that we are to add 2a+5b and 3a— 2b.
Example 1. Simplify (2a+56) + (3a-26) + (6-3a).
Solution. (2a+56)+(3a -26) +(6 -3a)
T + ft
2a+5& +3a-26 +6-3a
(Parentheses removed)
2o+4b (Like terms collected)
Ana. 2a+46.
Note. The plus signs between the parentheses indicate that
the quantities are to be added. Hence, in removing the paren-
theses, the signs of all the terms remain the same. This is shown
by the arrows.
I Example2. Simplify (3as-2)+(-4-7a;)+(3-x).
Solution. (3x-2)+(-4-7x) + (3 -x),
+ + T +
Zx-2 -4-7x +Z-x
(Parentheses removed)
— 5x— 3 (Like terms collected)
Ans. — 5x— 3.
Note. In Algebra it is customary to add terms horizontally,
u in examples 1 and 2, instead of arranging the expressions in
Jigiiized by Google
76 JUNIOR HIGH SCHOOL MATHEMATICS [VII, §31
EXERCISES
Simplify each of the following expressions.
1. <a-2)+(a+3) + (2a-7)
2. {x+5)+(2x-3) + (x-2)
8. (2x*-7)+®x*-l)+(2x*-S)
4. (3a+26)+(a-26) + (6-3a)
5. (c-2d)+(-d-2c) + (c+M)
6. (2x t -x) + (Zx i -ix)+{x-7x i )
7. (a-6+3) + (2a+6-l)
8. (3s-2j,+l)+(-2-2j,)+(3a:-7)
8. (2a*-3a+7)+{o-2o*)+(o+4)
10. (x+y+l) + (x-y-l) + (y-2x)
11. (-3z l -5)+(2s l -7)+(-5s'-3:r+12)
12. (-5+8m 1 +3m)+(-3m a +4m)+(-7m-5m 1 )
18. (lV-13)+(-7+6!/-8j^)+(-6i/+7y»)
14. {2;3i*-3.4i+1.7)+(-2.4a! 1 +4.5*-2.1)
16. (-5.6m s +3.&m-4.3)+(-3.4m*-4.1m+5.4)
§ 32. Subtraction of Monomials. Subtraction is the
process of taking one number (called the subtrahend)
from another number (called the minuend).
The process of subtraction is directly opposite to the
process of addition. Subtracting a positive number is
the same as adding a negative number. Subtracting a
negative number is the same as adding a positive number.
From these statements the rule for subtracting one
signed number from another is :
Change the sign of the subtrahend and combine as in the
addition of signed numbers.
is, Google
VII, S 32] ALGEBRAIC EXPRESSIONS 77
The following examples show all the different combina-
tions of signed numbers for subtraction.
SnaiHAHBND,
Process, Add
Answbb
Chinged
1.
Take +4 from +7
-4
-4, +7
+3
2.
' +7 " +4
-7
-7, +4
-3
3.
' -4 " -7
+4
+4. -7
-3
4.
• _7 '■ _4
+7
+7, -4
+3
5.
' -4 " +7
+4
+4, +7
+11
6.
' +7 " -4
-7
-7, -4
-11
7.
' +4 " -7
-4
-4, -7
-11
8.
" -7 " +4
+7
+7, +4
+11
In subtraction, to check your answer, add it to the
subtrahend ; the result should be the minuend.
BZEKCISES
A. In each of the following exercises, subtract the first
number from the second.
1. +2, +7
a. +7, +3
5. -9, -15
4 -4, +8
». +8, -3
6. +4, +1
7. +9, -12
8. -15, -1
». -10, +3 19. -1,9
10. -3, -12 JO. -2J, -5J
Note. For additional drill in the subtraction of signed
numbers, the Drill Table on page 69 may be used.
11.
-4, +4
12.
-9, 12
13.
+7, +7
14.
-12, -1
15.
-10,5
16.
-1, -1
17.
11, -18
16.
15,8
;, S ,:z K i:vC00gIe
+33, -2x
7.
4c, 7c.
-3a, -2a
8.
2<j», -3d>
+3y, +2y
9.
— 5x, — 52
-36, +26
10.
»*, -2j"
-Sy',-W
11.
7*, 7ifc
+5m, —2m
12.
15nt, — m
78 JUNIOR HIGH SCHOOL MATHEMATICS [VII, 1 32
B. In each of the following exercises, take the second
monomial from the first.
13. -2<xP, -2cd*
14. Ibr, -20r
IB. 3(, -3i
16. x*y,ihl
17. 36", -136»
18. -2^>,-\^j
% 33. Subtraction of Polynomials. To subtract one
polynomial front another, change the sign of each term of the
subtrahend, and then add.
Example 1. Take 8x— 3 from 2x+7, and check the
result.
Solution. Think the sign of each term of the subtra-
hend (8x — 3) changed, making it — 8x+3; then add
-81+3 to 2a;+7, giving the sum, -6x+10.
The work appears as follows, the changing of signs being
done mentally.
2x+7 (Minuend)
&c-3 (Subtrahend)
— 6x+10 (Remainder)
Check, by adding upward (Remainder+subtrahend =
minuend). Ans. — 6x+10,
Example 2. From2i— y+3takex— Zy— 4, and check,
the result.
Solution. Here the subtrahend is x— 3y— 4.
2x- y+3 (Minuend)
x — Zy — 4 (Subtrahend)
x+2j/+7 (Remainder)
Note. As the sign of each term of the subtrahend (x — 3y —4)
has been changed mentally, the change of signs is not shown in
the work. Check by adding upward. Ans. X+2t/+7.
;, S ,:z K i:vC00gIe
VII, S 33) ALGEBRAIC EXPRESSIONS 79
Example: 3. From 12a— 36 take 5a— 5c, and check the
result.
Solution. Here the subtrahend is 5o— 5c.
12a -36
5a —5c
7a-36+5c
Note. Although there is no term containing e in the minuend,
the mgn of — Be must be changed, because it is one of the terms
of the subtrahend. Cheek by adding upward.
Am. 7a -36+ 5c.
EXERCISES
In each of the following exercises find the remainder
and check.
1. Take 3x+2 from 4x-l.
2. Subtract 5a— 3 from 3a— 5.
3. From 2y+7 take y-4.
4. From 46+3 take -26+4.
6. From 3ft* -7ft take 4ft*- 2ft.
6. Take 3a from 4a- 1. (See Example 3.)
7. Take 2x+5 from 5x. (See Example 3.)
8. From x+2y-4 take x— 3y-5.
9. From 3a 1 +2a-3 take a ! +8.
10. From 5x- 8 take 15.
11. Subtract a+26 from -2a-76+4.
18. From 2a*-7 take a-8. .
13. Take 2c+d from d-7c.
14. Subtract a*-26* from 6*+2a*.
15. From x"-2x l +7x take 3x*-x*.
16. Take 3x-y from 7x-12.
17. From x — y—7 take x—y+4.
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80 JUNIOR HIGH SCHOOL MATHEMATICS [VII, 5 33
18. Take 3a -7 from 3a -7.
19. From 3x'-5 take y*-12.
20. From 2a-56 take 3a-56+4.
21. Simplify (4a-2)-(2a+4).
Solution. (4a-2) - (2a+4)
+ +
4a— 2 —2a— 4 (Parentheses removed)
2a— 6 (Like terms collected)
Ans. 2a -6.
Note. The minus sign between the parentheses indicates
that the quantity following (the subtrahend) is to be subtracted.
Hence, in removing the parentheses from the quantity to be
subtracted, the signs of all its terms are changed. This is shown
by the arrows.
22. Simplify (3x-8)-(7;z-2). Ana. -4r-6.
23. Simplify (46+l)-(76-4).
24. Simplify (a-2) + (a+3)-(2a-7).
28. Simplify (i*+3)-(2a:*-3) + (3x 1 -4).
26. Simplify (2x-7)+(-3-x)-(2ir-8).
27. Simplify (3a+26)-(a-26)-(-b+3a).
28. Simplify (^-2^)+ (-0^-2^) -(^-3^).
29. Simplify (2x-y)-(Zx-2y)-(y-7x).
30. Simplify (3a-b+2)-(2a+b-l).
31. Simplify (2x*-2x+l) + (2x-2)-(3x*-7).
32. Simplify (2a-6+3)-(6-3a) + (— 6+b).
33. Simplify (x i +xy+y t )-(-y*+x t -xy) + {y*-3x*).
34. Simplify 4-(a;-y) + (2x+v+l).
38. Simplify 2a-(a+&)-36+(a+56)+a.
J, S ,:z K i:vC00gIe
CHAPTER VIII
MULTIPLICATION AND DIVISION OF ALGEBRAIC
EXPRESSIONS
§ 31. Law of Signs in Multiplication
Example 1. Find the product of +4 multiplied by +3.
Solution. Consider +3 the multiplier, then
(+3)X(+4)- + 12, for this means
(+4) + (+4) + (+4) ; that is, +4 is to be added three
times. Ana. +12.
Example 2. Find the product of —4 multiplied by +3.
Solution. Consider +3 the multiplier, then
(+3) X (- 4) = - 12, for this means
(-4) + (-4)+(-4); that is, that -4 is to be added
three times. Ana. — 12.
Example 3. Find the product of +4 multiplied by —3.
Solution. Consider —3 the multiplier; as this multi-
plier is the negative of the multiplier in Example 1, this
product is the negative of that product ; that is,
(-3)X(+4) = -12.
(Compare this answer with that of Example 1.)
Ans. -12.
Example 4. Find the product of —4 multiplied by —3.
Solution. Consider — 3 the multiplier ; as this multi-
plier is the negative of the multiplier in Example 2, this
product is the negative of that product ; that is,
(~3)X(-4) = +12.
(Compare this answer with that of Example 2.) Ana. + 12.
i^v Google
82 JUNIOR HIGH SCHOOL MATHEMATICS [VIII, { 34
We may state these principles of signs as follows :
Plus times plus gives plus (Ex. 1).
Plus times minus gives minus (Ex. 2).
Minus times plus gives minus (Ex. 3).
Minus times minus gives plus (Ex. 4).
Summary.
These four principles may be further condensed into the
statement of the following Law of Signs in Multipli-
cation :
Two like signs produce plus.
Two unlike signs produce minus.
§3
:. Table — Combination of Signs
A + + + - + - + - + + - + +
B +- + + - + ++-+-+-+-+-
Suggestion. Use the signs in line A as the signs of
the multiplicand with those in line B as the signs of the
multiplier. Find the sign of each product.
§ 36. Multiplication of Signed Numbers.
EXERCISES
Find each of the following indicated products :
1. (+3)X(-5) 6. (-9)X(-9)
2. (-5)X(+4) 7. (+8)X(-9)
8. (-4)X(-6) 8. (-7)X(-9)
4. (-8)x(+9) «. (-4)X(+12)
B. (+7)X(-8) 10. (-6)X(-11)
;, S ,:z K i:vC00gIe
Vin, {37| ALGEBRAIC EXPRESSIONS 83
In Exs. 11-20, find the product of the first two num-
bers, then multiply by the third number.
11. (+2)X(-3)X(+4) 16. (-5)X(-4)X(-7)
U. (-3)X(-2)X(-6)
IS. (-4)X(+3)X(-4)
14. (-5)X(-4)X(+3)
18. (-2)X(+8)X(-3)
17. (+5)X(-6)X(-3)
1«. (+i)X(-6)X(-4)
">■ <-fiX(+8)X(-5)
ao. (-J)X(+12)X(-J)
§37.
Dull Table —
- Multiplication OF SlONSD Numbers
A
+1
-2
+3
-5
-4
+8
-9
+7
B
\ -1.
+2
+5
-2
-3
+4
+5
+6
C
' -6
+3
+5
+8
+9
-8
-7
-4
D
1*
+i
-i
-i
-1
+i
+«
Suggested drills with this table :
(a) Use line A as the multiplicand, and each number
in line B as a multiplier.
(6) Use line B as the multiplicand, and each number
in line C as a multiplier.
(c) Use each number in line D as a multiplier, and lines
A, B and C as multiplicands.
Note. When zero is one of the factors, the product is zero.
(d) Find the product of the first three numbers in line
A ; the first four, etc. ; similarly for the other lines.
(c) Find the product of all the numbers in each column.
Note. In finding the product of more than two signed num-
bers, the simplest way to determine the correct sign of the product
is as follows : Count the number of minus signs of all the factors.
If there is an odd number of minus signs the final product is
minus ; if there is an even number of minus signs, the finnl product
is pine. For example, the final product of all numbers in line 1
in plus ; in line 2, it U minus ; etc.
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84 JUNIOR HIGH SCHOOL MATHEMATICS [VIII, S 38
§ 38. Multiplication of Monomials. In finding the
product of two or more monomials, the factors may be
written in any order, as in arithmetic. For example,
just as . 2X3X4-4X2X3, so
3a6c=36oc.
Example 1. Find the product of 5a l 6 s and — 4oo',
Solution. The factors may be rearranged as follows :
5X(-4)Xo*XaXb a Xo'= -20a*b*.
Explanation. (+5)X(-4) 20 (Law of Signs.)
The total number of times that a appears as a factor is
3 ; 3 is obtained by adding the exponents of.a, (2+1).
The total number of times that b appears as a factor is
5 ; 5 is obtained by adding the exponents of b, (2+3).
Ana. -20a s f> 6 .
Example 2. Find the product of — Zxy, x*y*, and — 7x s .
Solution. The product of the numerical coefficients
(including signs) is +21. The product of the literal
factors is xy.
(-33j/)(xy)(-7x i ) = +21xY- A**- +2IiV-
From the preceding examples you have observed that
the exponent of any letter in the product is found by add-
ing the exponents of that letter in the factors. This is
the Law of Exponents in Multiplication.
Summary. In finding the ■product of two or more mono-
mials :
(1) The numerical coefficient is the product of the numeri-
cal coefficients in the given factors.
(2) The exponent of each letter in the product is the sum
of the exponents of that letter in all of the given factors.
(3) The combined product is a monomial.
;, S ,:z K i:vC00gIe
ALGEBRAIC EXPRESSIONS
EXERCISES
Find each of the following indicated products.
1. 3a«&X(-4o*>)
2. -4a ! b*X5o ! b s
J. -3&"x«X(-3!rV)
t. -7c 1 d , X(-2c , d)
6. -a*cd»X2a , cW
6. 2ctPX(-5a , aP)
7. a'6X2ao , X(-3<A')
8. -2a**X3a*:rj/ s X(-2
». i<?dXt.-$ap)
10. ItfyXiaci?
it. fiyxizy
II Sox??
13.
-12i=X
31/
»')
14. 7oX==
15. JaX46
§ 39. Multiplication of a Polynomial by a Monomial.
Example. Find the product of 2o*b-3a6 l multiplied
by3af.
Solution. Consider 3a'6> the multiplier, then
3o"6>(2o'6-3o6")«3a ! o>(2a ! o)+3o , o 1 (-3iii) , ).6o't"-9a , o'
Am. 6a<6'-9a"b'.
Figure 15 is
identity
i geometric illustration of the algebraic
o(b+c)=a*+oc.
Figure 16 is a geometrical illustration of the algebraic
identity
JigilizBd by GoOgle
86 JUNIOR HIGH SCHOOL MATHEMATICS [VIII, 539
EXEHC1SES
Find each of the following indicated products.
3(2o-5)
6.
— 2mn(m— 3n)
-2i(3i+1)
7.
-3oi>(2o ! -ot-36')
o'(3a'-2a-3)
8.
SaWtfa'-Sdb+W)
2at ! (3a-2tc)
9.
taHM-Gb)
Zxy(2*-ixy+y')
10.
|i , (6» 1 -3i!,+3f I )
§ 40. Multiplication of Polynomials.
Example 1. Find the product of 2x+3 and 3x— 5.
Solution. The work may be arranged as in arithmetic.
2x + 3 (Multiplicand)
3x - 5 (Multiplier)
6x ! + 9x [Zx(2x+3), lBt partial product]
-101-15 [-5(2i+3), 2d partial product]
6x*— x— 15 (Product^aum of partial products)
Arts. 6i*— x — 15.
Example 2. Find the product of 2a*— 3a— 1 and 5a— 2.
Solution.
2a*- 3a -1 (Multiplicand)
5a - 2 (Multiplier)
10a 8 — 15a ! — 5a (1st partial product)
- 4a*+6a+2 (2d partial product)
10a*— 19a*+ a+2 (Product^aum of partial products)
Am. 10a'-19a*+a+2.
Note. A method of checking the multiplication of poly nomiftls
is by assigning number values to the letters.
EXERCISES
Find each of the following indicated products.
1. {x+SKx+2) 4. (a:-5)(ar-3)
2. (x+5)(x-3) 5. (a-7)(2a-l)
3. (*-2)(*+4) 6. (2a+36)(3a-26)
;, S ,:z K i:vC00gIe
VIII, ! 40] ALGEBRAIC EXPRESSIONS
87
7. (3&-2a)(2o-3a)
8. (4a-l)(3a+5) .
8. (fiz-2|r)C3z+7v)
10. (7+3*)(5+2as)
11. (ll+3x)(7-l)
11 (2as'+3x+l)(3x-2)
18. (5a"-4a+3)(2a-l)
14. (3m"-5m+7)(2m-5)
15. iz?+xy+y')(z-y)
18. (o"-a&+f> ! )(a+b)
17. (4+2*+x , )(2-z)
18. (o+6)(a-6)
1». (a+6)(a+o)
20. (a-6)(a-&)
81. 7}X3|
7|X3i-(7+*)(3+»)
-3(7+})+i(7+l)
-21+2+51+1
-28}
27. 30*X27|
28. 75JX60|
28. 83&X571
SO. 951X1021
Ana. 28}.
22. 6}X9}
23. 8|X7(
24. 9iX5J
26. 74X29f
26. 13JX124
81. A room is reported to be 15 ft. by 13 ft. By a more
accurate measurement it is found to be 15 ft. 3 in. by 13 ft.
4 in. Find, in square feet, the amount of the error in the
area when it is computed from the first dimensions.
Solution. (15+l)(13+iJ-15X13-8J.
Ana. 81 sq. ft.
32. The dimensions of a room are reported as in Ex. 31.
A more accurate measurement gives the dimensions as
14 ft. .9 in. by 12 ft. 10 in. Find, in square feet, the
amount of the error in the area when it is computed from
the first dimensions.
Solution. 15X13-(15-})(13-1)-5H.
Ana. m sq. ft.
JigilizBd by GoOgk
88 JUNIOR HIGH SCHOOL MATHEMATICS [VIII, J 40
33. A room is reported to be 18 ft. by 12 ft. By a more
accurate measurement it is found to be 17 ft. 9 in. by 1 1 ft.
10 in. Find in square feet the amount of error in the
area when it is computed from the first dimensions.
34. The lighting area of a window pane is reported to
be 10 in. by 8 in. A more accurate measurement gives
it as 9$ in. by 7J in. There are 12 panes of glass in this
window. Find the total amount of error in the fighting
area when it is computed from the first dimensions.
36. The dimensions of the floor of a room are given as
18 ft. 2 in. by 15 ft. 1 in. A less accurate measurement
gives them as 18 ft. by 15 ft. Find in sq. ft. the difference
in the floor area of this room when computed from each of
these two sets of dimensions.
36. If the dimensions 18 ft. by 15 ft., in Ex. 35, are
used in finding the area of the floor, what is the per cent
error in the area, within a tenth of one per cent ?
37. If you neglect the product term ab in (l+o)(l+b),
what is the per cent error, when a=0.004 and b = 0.005,
within a thousandth of one per cent ?
38. The total lighting area of a room containing 5 win-
dows, each with 12 panes of glass, each pane 12 in. by 10
in., is to be found. What will be the total amount of error
in the fighting area as computed from these dimensions,
if each dimension as given is £ in. too long? What is
the per cent error, within a tenth of one per cent ?
39. The dimensions of a room are taken as 28 ft. by
24 ft. The error in each dimension is not over ^ ft. How
great can the error in the area be?
Suggestion. (28 ± J) {24 ± i) ■ -28 X 24 -?
3, g ,i EE d by Google
VIII, MM ALGEBRAIC EXPRESSIONS 89
§ 41. Special Products. Certain types of products
occur so frequently in algebra that, like the Multiplication
Table in arithmetic, they should be memorized.
The types of products which are most frequently met
have been shown in exercises 18, 19, and 20, page 87.
These are :
(1) (a+b)(a-b)m^~V.
(2) {a+b)(a+b), or (a+b) i =a i +2ab+b t .
(3) (a-b)(a-b), or (a-b)*m*-2ab+b:
These products should be memorized. Each of these iden-
tities may be translated into words as follows :
(1) The ■product of the sum and difference of two numbers
is the difference of the squares of the two numbers.
(2) The square of the sum of two numbers is the square
of the first, plus twice their product, plus the square of the
second.
(3) The square of the difference of two numbers is the
square of the first, minus twice their product, plus the square
of the second.
Figures 17-19 are the geometric illustrations of the
algebraic identities in (1), (2), and (3).
3ig.1iz.ed by GoOgk
90 JUNIOR HIGH SCHOOL MATHEMATICS [VIII, |41
EXERCISES
Write each of the following indicated products by in-
spection.
11. 31X29
(Write as: (30+l)(30-l).]
11. 51X49
13. 42X38
14. 59X61
18. 72X68
16. 51' (Write as: (59+l) ! J
17. 31*,- 41"; 91" '
18. 42=; 62"
1». 59" IWriteas: (60-1)'.]
JO. 49"; 69 1
1. (o+c)(o-c)
J. (x+4)(l-4)
S. (o+»)'
4. la-c)'
5. (1+9)'
8. (m+3)'
7. (c-2)'
8. (j+5)b-5)
». (»+5)(v+5)
10. (»-5)(j-5)
§ 42. Products of Two Binomials Having a Common
Term.
Example 1
Example 2
(z+5)(i+3)«?
(l-5)(:r-3)..?
»+5
»-5
z+3
x-3
j'+5x
i"-5i
+3i+15
-31+15
i'+Si+IS, or
i'-Sx+IS, or
«"+(5+3)l+15
»>+(-5-3)j:+15
;, S ,:z K i:vC00gIe
VIII, 1 42] Al
X5EBRAIC
EXPRESSIONS
Example 3
Example 4
(l+5)(z-3).
i?
(i-5)(i+3)»?
z+5
»-5
1-3
1+3
iH-oz
i"-5i
-3i-15
or
+3I-15
i 1 +2»-15,
x*-2x-15, or
x"+(5-3)*
-15
l"+(-5+3)l-15
From an examination of each product and a compari-
son of the four products (Examples 1-4), we observe
the following :
(1) That the first term of each product is x i , the square of
the common term,
(2) That the last term of each product is ±15, the product
of the other two terms,
(3) That the middle term of each product is the common
term x with a coefficient which is the algebraic sum of the
other two terms.
Such products as examples 1-4 may be expressed in the
general form,
This identity may be translated into words as follows :
The product of two binomials having a common term is
the square of the common term, plus the algebraic sum of the
other two terms for the coefficient of the common term, plus
the product of the other two terms. For example,
(x+7){x-S)mx*+4x-21; (*-7)(at-3)-*-10*+21;
JigilizBd by GoOgle
92 JUNIOR HIGH SCHOOL MATHEMATICS (VIII, 1 42
Figure 20 is a geometric illustra-
tion of the algebraic identity
(x+a)(x+b)^x*+ax+bx+db
BXSRCISES
Write each of the following indi-
cated products by inspection.
(x+4)(*+3)
(*-4)<x-3)
(x+4)0r-3)
(*-4)0r+3)
(»+6)<S/-l>
(V-6)(s/+l)
(y-e)(y-i)
(x+9)(x-5)
(l-9)(*-5)
(o-7)(o+2)
(o+7)(o+2)
(t+10)(6-3)
(o-10)(o-3)
(«-8)(»+7)
(x-7)(*+8)
(m-9)(m+10)
(c+2)(c-12)
(»+12)(s/+3)
(»+12)(!/-3)
(y-U)(s+3)
1
Fid. 20.
»■ (»+0)(»+7)
22. (»-9)(s,+7)
53. (x-8)(x-7)
24. (i-8)(i-9)
28. (2+l)(5+l)
26. (5+x)(2-x)
27. (4-x)(l+x)
28. (x+8)(x+8)
29. (x-7)(x-7)
SO. (10+3)(10+2)
31. (10+5)(10+8)
82. (20+1) (20+2)
38. (20+2)(20+2)
54. (20+l)(20+5)
So. (20+3)(20+2)
38. (30+2)(30-l)
ST. (100+2)(100+3)
88. (100-3X100+2)
39. (1000+3X1000-2)
40. (1O0O+5X10O0+2)
JigilizBd by GoOgk
VIII, j 431 ALGEBRAIC EXPRESSIONS 83
§ 43. Products of Two Binomials Having Correspond-
ing Terms Similar.
Example 2
Example 1
(2i+l)(3l+2).?
2i+l
3i+2
6V+3*
+4x+2
6i"+7z+2
ExamsmUJ,
(2i+1K3i-2)«?
2l+l
3l-2
ftr'+Si
-4z-2
(2l-l)(3»-2;
2x-l
6^-31
-4»+2
Example 4
(2x-l)(3»+2)«?
6»"-3i
+4*-2
at 1 - »-2
6V+ »-2
From an examination of each product and a compari-
son of the four products (Examples 1-4), we observe the
following :
(1) That the first and lout terms of each product are the
products of the corresponding similar terms of the binomials,
(2) That the middle term of each product is the algebraic
sum of the cross-products of the dissimilar terms of the bi-
nomiale.
Write each of the following indicated products by in-
spection.
1. (2i+3)(3i+2) 4. (3*-l)(2x+3)
1. (3l-H)(2*+3) 8. (5o-2)(2a-l)
3. (2i-3)(l+2) 6. (3x-4)(2l-l)
\
;, S ,:z K i:vC00gIe
94 JUNIOR HIGH SCHOOL MATHEMATICS [VIII, J43
7. @y-S)£fy+2) 10. (3x+7)(x-2)
8. (iy-l)(y~Q) 11. (26+5)(5fe-4)
9. (3x-7)(a;-2) 12. (2fc+3)(3fe-2)
§ 44. Law of Signs in Division.
Example 1. Find the quotient of +12 divided by +3.
Solution. (+12) + (+3) = +4, since (+3)X(+4) =
+ 12. Ans. +4.
Example 2. Find the quotient of — 12 divided by +3.
Solution. (-12) + (+3) = -4, since (+3)X(-4) =
-12. Am. -4.
Example 3. Find the quotient of — 12 divided by — 3.
Solution. (-12) + (-3) = +4, since (-3)X(+4) =
-12. Ans. +4.
Ex/mple 4. Find the quotient of +12 divided by —3.
Solution. (+12) + (-3)=-4, since (-3)X(-4) =
+ 12. Ana. -4.
We may condense these four principles of signs into the
following Law of Signs in Division (See page 82.) :
Two tike signs produce plus.
Two unlike signs produce minus.
§ 45. Division of Signed Numbers.
EXERCISES
Find each of the following indicated quotients.
1. +12-H-6) 4. -48+6
2. -32-H-8) B- -63-K-7)
3. +36+(-4) »■ -54+ (-9)
3ig.1iz.ed by GoOgk
VIII, 5 46) ALGEBRAIC EXPRESSIONS 95
7. -72 + (+8) 14. 144+(-6)
8. -96+8 15. -160+ (-10)
9. -84+(-12) 16. -625+25
10. 81 + (-9) 17. -225+(-15)
11. -100+ (+4) 18. +8+(-±)
12. -120+(-8) 19. -24+1
18. -132+6 20. -48+J
§ 46. Division of Monomials.
Example 1. Find the quotient of — 20a*6'+(— 4a6).
Solution. ( — 20a'b , ) + { — 4a&) = +5a i b, since
(+5o%) X (-4ao) = -20o'&*.
Ans. +5a«6.
Example 2. Find the quotient of (21xV) + (-3xj/).
Solution. (21x t y t )-i-(— 3xy)m—7x*y 1 , since
(-7xV)X(-3iy) = +21jV.
Ans. — 7i*y*.
From the preceding examples you have observed that
the exponent of any letter in the quotient is found by
subtracting its exponent in the divisor from its exponent
in the dividend. This is the Law op Exponents in
Division.
Summary. In finding the quotient of two monomials :
(1) The numerical coefficient is the quotient of the numeri-
cal coefficient of the dividend divided by that of the divisor.
(2) The exponent of each letter in the quotient is its ex-
ponent in the dividend minus its exponent in the divisor.
(3) The combined quotient is a monomial.
Note. Zero cannot be used as a divisor.
J, S ,:z K i:vC00gIe
96 JUNIOR HIGH SCHOOL MATHEMATICS [VIII, { 46
EXERCISES
Find each of the following indicated quotients.
12a*6 4 ^(-3fl6') 8.
-36zY-i-6xY
-24a*c J -5- (-4o*c)
-36o*e*+(-9<rtB)
-72m 4 i B -5-8mx !
836 V+ 76V
- 96a«m V + 12a*m*B
132n s :r. 4 -J-(-6n l )
-240o"b s c 4 +(-8a*6*c 1 )
-lOOOs'V'-^xV
§ 47. Division of Polynomials by Monomials.
Example 1. Find the quotient of (6x 4 — 2x'+4x 1 )-s-2;r , ,
and check.
Solution. The work may be arranged as in short divi-
sion in arithmetic, each term of the dividend being divided
by the divisor. The quotient is written above the divi-
dend.
3s*- x +2 (Quotient)
(Divisor) 2x t )6x i -^x t +4x i (Dividend)
Check. This work may be checked by multiplying the
quotient by the divisor. Am. 3* 1 — x+2.
EXERCISES
Find each of the following indicated quotients.
(24a*-18o , + 12o)-f-6o
(16W-12a , 6 s -8aoH4a&
(32z B -24x 4 +20r ! )-K-4x 3 )
(54o 6 fc» - 27a 4 b 4 + 18a'& b ) +9rt*
(63&V+496V - 356'c 8 ) * 76*c*
(7xV - 2 lxy - I4asy«) + 7ry
(6V-12a*-6a l K6V
(aV-3x , +kt s )-s-(-i«)
Jigilized by GoOgle
VIII, S 481 ALGEBRAIC EXPRESSIONS 97
§48. Division of Cue Polynomial by Another Poly-
nomial.
Example 1. Divide s^+5aH-fl by x+2, and check.
Solution. The work may be arranged as in long divi-
sion in arithmetic. The principal steps are :
(1) Divide, (2) Multiply, (3) Subtract, (4) Bring
Check the work by multiplying the quotient by the divi-
i+3
(Quotient)
(Divisor) x+2)x t +5x+6 (Dividend)
x*+2x [x(x+2)}
3x+6 (Subtract, bring down)
3s+6 [3(:f+2)]
(Bring down)
x+3
x+2
x*+3x
2s+6
x*+5x+Q
Explanation of each step :
qSotwht,™
a-
Id Tihii or Quotum*, 3
6
Divide
Multiply
Subtract
Bring down
3x+x
3<s+2>
(ar+6)-C3r+6)
Jigilized by GoOgle
98 JUNIOR HIGH SCHOOL MATHEMATICS (VIII,|48
Example 2. Divide 12i*-23xj/+5j/ ! by 3a=-5j/, and
■check.
Solution. 4x—y (Quotient)
(Divisor) 3x - 5y) 12x* - 23x3/+%= (Dividend)
12x 1 -20xy
-Sxy+Sy*
— Szy+Sy*
Check. Ax — y
3x -5y
12x»- 3ary
-20ij/+5y'
Xltf-IZxy+Sy* Ane - *x-y.
Example 3. Divide x l —5x i +8x — G by x-3, and
check.
Solution.
z>-2x +2
i-3)z"-5x ! +&r-6
st«— 3af
-2z"+8«
-2i"+6l
21-6
2»-6
Check.
i'-Sz+S
1-3
l'-2i"+2l
-3^+61-e
Arts. i I -2z+2.
;, S ,:z K i:vC00gIe
VIII, i 481 ALGEBRAIC EXPRESSIONS
Find each of the following indicated quotients*, and
check each answer.
1. (i*-2x-8)i>(i+2)
2. (o'+5a-14) + (o-2)
3. (m"+2»>-35) + (m+7)
«■ iy-3!/-54) + (!,-9)
5. (l=-13l+40)-i-(l-8)
6. (2o'+7a-30) + (o+6)
7. (6i=+7»-5) + (3a!+5)
8. (6l)'+176-14) + (3o-2)
9. (15a"-51(K>4-18f>") + (5a-2!>)
10. (4*"-27l9+44!fl-i-(i-4i/)
11. (»=-16) + (l+4)
12. (a'-4(V) + (<i-26)
13. (V-lfl+Os-If)
H. (16a ! -t ! ) + (4a-6)
IB. (64-9l") + (8+3z)
18. (x'-2l ! +4z-8)-!-l>-2)
17. W+ilf+y-6) + (y+3) .
18. (.a 1 -15a'+65a-m)+i"-T)
19. (3l"+28x ! +29»-140, + (3z-5)
20. (as*-8*'+10i ! +32*-35)-i-[>-5)
21. (I'+1) + (I+1)
22. (a'-J")-M>-&)
23. (i"+»") + (i+ ! ,)
2*. 6/-8) + (!/-2)
It. (27j*-8|/ , ) + (3»-2j<)
JigilizBd by GoOgle
100 JUNIOR HIGH SCHOOL MATHEMATICS (VIII, !48
„. <*.-,-«)«.(,-.!) Am 4
3!-2
27. (2^+5x-6)-^(2i-5)
28. (z a +5« i -3ar+8) + (x-3)
29. (a"+o*)-i-(a-&)
30. (xM-lR(s+l)
§ 49. Equations Involving Parentheses.
Example L Solve the equation 3(2i-2) = 2{i+3),
and check the answer.
SOLUTION.
® 3(2*-2)-2(l+3)
© ftr-6 = 2a:+6
®= (Parentheses removed)
® 6x-2l-6+6
®+6-2x
® it- 12
® =■ (Like terms collected)
® 1-3
®+4
Check.
3(2X3-
-2)
.2(3+3)
3(6-
-2)
A 2(6)
12
= 12 Ana. 1 = 3.
Note. In evaluating expressions which involve the four
fundamental operations, multiplication and division should be
performed before addition and subtraction. If parentheses
are involved, the expressions within the parentheses should be
evaluated first.
Example 2. Solve the equation 2(5z-7)-3(2s-9)
= 15, and check the answer.
Solution.
® 2(5x-7)-3(2x-9) = 15
® (10x-14)-(6x-27) = 15 ©=■ (Multiplying)
® 10x-14-6x+27=15 ©=■ (Parentheses
removed)
j, s ,:z K i V CoogIe
VIII, |4»] ALGEBRAIC EXPRESSIONS '101
® 4i+13 = 15 ® = (Like terms collected)
© 4i-2 ®-13
® x-i ®+4
Check.
2(5Xi-7)-3(2Xi-9)Al5
2(2i-7)-3(l-9)il5
2(-4J)-3(-8)il5
-9+24-15 Ant. x-f
Example 3. Solve the equation (x+3)(2x— 1)— 4
= (2x+5)(x— 7), and check the answer.
Solution.
® (x+3)(2«-l)-4-(2x+5)(x-7)
® (2x ! +5x-3)-4-(2x 1 -9x-35) ©-(Multiplying)
© 2x*+5x-7 = 2x*-9x-35 ©"(Parentheses
removed)
® +14X--28 ®-2x"+9x+7
© I- -2 ®+14
Check. (-2+3)[2(-2)-l]-4i[2(-2)+51(-2-7)
(+l)(-4-l)-4i(-4+5)(-9)
(+l)(-5)-4_(+l)(-9)
-9--9
Ana. x=-2.
IXERCISSS
Solve each of the following equations, and check each
answer.
1. 5(3x+l)-7x-3(x-7)+31
2. 3+2(1-3)- 3(1-3)
3. 5(x-2)-7x«3(2x-3)-7
4. 3+2(x-6)-3(2-x)
KigilizBd by G00gle
li)2'jUNldR-WGH-8CHOOL MATHEMATICS [VIII, (49
8. 14»+3(3+z)-2(3*-l)
6. 12z-2(4x-7)-18
7. 7*-12-2(l-5)=z+22
». 9s-3(2y-4)-8-(6-|f)
0. 4-3(l-*)-3-(3i-l)
10. 2(x-2)-3(6-i)-9(3+2i)+3
11. (l-2)(*-5)-(j!4-3)(l4-2)
IS. (x-5)(x-3)-(i-8)(x+2)
13. (2*+5)(4i+7)-8i(i+3)
14. 4j,(%-l)+27-8|f(3s<+2)+147
1>. (3z-5)(4x+3)-l«(4as-l)(3x+7)
1«. 12-3l(8i+5)-21-4i(te-3)
17. 13»-(i-5)(i+7)-37-(i-6)(i+2)
18. 2(3l-l)(2»+5)-(4i-7)(3i+2)-238-0
»• irD-irl)- 4 *
SO. 3(2.7x-.8)-I.2(5x-3)-11.7
.. 4»,2x_ 26
n . n n . n _
S3. fc+li-Iz
3, g ,i,z E d by Google
VIII, 4 491 ALGEBRAIC EXPRESSIONS
27.
7s-5 3s+l _ 6g+l
2 6 ™ 3
[3(7a:-5)-(3x+l) = 2(6x+l); x-3-1
9x+3 2x-5 _
3s+4
i+5 3x-4 3
4 5 "4
4j-1 2j+3 _j-1 3j+7
3 2 2 6
1. The length of a rectangle exceeds its width by 3 in.
and its area exceeds the area of a square constructed on its
width by 18 sq. in. Find the dimensions of the rectangle.
2. The length of a rectangle exceeds its width by 3 in.
and its area is 36 sq. in. less than the area of a square con-
structed on its length. Find the dimensions of the rec-
tangle.
3. Hazel is 25 years old and her brother Seth is 14
years old. How many years ago was Hazel twice as old
as Seth?
4. John's age is two thirds of his father's age. How
many years ago was John one fourth as old as his father,
if John is now 42 years old ?
5. My oldest brother is 7 years older than myself.
Thirty-three years ago he was just twice as old. Find
our ages now.
6. Three years ago John's age was twice that of his
brother Charles. Three years hence Charles' age will
be $ John's age. Find their ages now.
Jigilized by GoOgle
104 JUNIOR HIGH SCHOOL MATHEMATICS [TIH, |4S
7. The sum of the fourth, eighth, and sixteenth parts
of a certain number is 49. Find the number.
8. There is a certain number from | of which if you
take i of it, you get the number 7. Find the number.
9. Find that number which when it is divided by 15
will give a quotient of 13 and a remainder of 13.
10. Separate the number 120 into three parts such that
the second part shall be 2 more than the first part, and that
the third part shall be 3 times the sum of the first and sec-
ond parts.
11. A motorist rode 120 miles in 4£ hours.- Part of the
distance traveled was in the country at the average rate
of 30 miles per hour, and the rest within city limits at the
average rate of 10 miles per hour. For how many hours
was he riding in the country?
12. A steamship left port at the average rate of 15 knots
per hour. When it was at a certain distance from port,
it became disabled and returned at the average rate of
4 knots per hour. It left port at 11.30 a.m. and had re-
turned at 2.40 p.m. How far from port was the steamship
when the accident happened? What was its average
rate for the entire trip ?
3,g,1 EE d by GoOgk
CHAPTER IX
PAIRS OF LINEAR EQUATIONS
§ 60. Graphs of Linear Pairs. In Chapter VIII,
Second Course, pages 196-200, the graphs of linear equa-
tions were plotted on squared paper. To definitely
locate a straight line on squared paper two points were
necessary, and a third point was obtained to check the
location of the two points, hence it was necessary first to
tabulate three pairs of values. To get these values, it was
found convenient' to solve the equation for x or for y.
Example 1. Plot the graph of the equation x+2y = 8.
Using the same axes, plot the graph of the equation
2x— y = 6. From the graphs, find the pair of values that
satisfies both equations.
Solution. (I) x+2y = S Solving for x, x = 8 — 2y
Wheny=4,x=0. When y = 3, x = 2. When y=l,x=G.
x 2 6
1/431
(II) 2x-y =
Solving for x,
2
x 3 5 7
v 4 8
., S ,:z K i:vC00gIe
106 JUNIOR HIGH SCHOOL MATHEMATICS [IX, (50
The graphs of these equations intersect at (+4, +2);
that is, i=4, y= 2 is the pair of values that satisfies both
Check. (I) 4+4-8
(II) 8-2 = 6
Ana. i = 4, y=2.
These simultaneous equations are called independent
equations because they have distinct graphs.
XiXt and Y i _Y t are called coordinate axes. (+4, +2)
are coordinates of the point P. is called the origin.
Example 2. Plot the graphs of the -pair of equations
2x+y = 7 and 4x+2y = 8.
Solution. (I) 2x+y = 7
Solving for y,y = 7—2x
*
2
4
V
7
3
-1
(II) 4z+2y = 8
Solving for y, y = 4—2x
x 2 4
u 4 -4
These graphs are parallel, hence no pair of values
satisfies both equations. These simultaneous equations
are called inconsistent equation*.
jigiiized by Google
IX, ! 51] PAIRS OP LINEAR EQUATIONS
Plot the graphs for each of the following pairs of equa-
tions. In exercises where the two equations are independ-
ent, prove that the pair of values for the point of intersec-
tion of the two graphs satisfies both of the equations.
! [ x+y = 6 4 f x-2y=5
2 J2x+5j/=10 6 f x+iy = 9
\ x—y-4 [ 5x—y = 9
§ 51. Solution of Linear Pairs by Addition or Sub-
traction. When linear pairs are independent, they can be
solved by eliminating or getting rid of, one unknown,
thereby forming one equation. The process of elimination
that we shall use here is by addition or subtraction.
Example 1. Solve the pair of equations :
(I) ' 5x+Zy~8
(II) 4x+5y=-i
Solution. To eliminate y multiply (I) by 5 and (II)
by 3, then subtract the resulting equations, because the
signs of the terms containing y are alike.
®
25«+15j-40
(I)X5
©
12i+15j--12
(II)X3
®
131-52
®-©
®
1-4
®*13
is, Google
108 JUNIOR HIGH SCHOOL MATHEMATICS [IX, (51
Substituting 4 for x in (I),
®
20+3J/-8
(I>-
©
3s- -12
©-20
®
V--i
©+3
Check.
(I) 20-12-8
(II) 16-20- -4
Arcs
x =
4, V- -4.
Example 2. Solve the pair of equations
(I)
22^+ 9 _14
(ID
to il+8_,
3 2
Solution
First get the equations in
the form of those
in Ex. 1.
®
3x+5+2j-28
CDX2
®
3»+2s-23
®-S
®
12l-3»-24-24
(IDX6
®
12x-3»-48
©+24
©
12i+8»-92
®X4
©
llj-44
®-®
©
y-i
©+11
Substituting 4 for y in ®, x = 5
Check.
(I) 3x )j +5 +4-14
10+4-14
an »x5_4+8i 4
3 2
10-6-4
Arcs
x=5, j/=4.
Google
IX, S 51] PAIRS OF LINEAR EQUATIONS
109
EXERCISES
Solve the following pairs of equations and check the
answers.
|fa+2j-8
■ 13I+2J-4
f a+b-7
l2o+3i>-17
J5m+6n=16
(3m+4n-10
(5p+2»-22
' l3n-4io = 8
5.
(4I.+96-5
120+36-2
f6o+3b-12
l4o-56-8
J 3i+s 2
1 4I-J-12
J5m+4n=2
1 2m-»--7
Transform the equations in Exa. 9-16, so that they will
be in the form of those in Exs. 1-8, before eliminating.
(5I-1-4J
(4JI-15-3*
[ 7a- 15 t
1 5o-36-9
[q+6 ■ a— D
2^2
8 9
3 4 12
3 4
1
12
""I
= 1
i+3
2
9-5J
!/+9_
10
1-2
3
m-k2
3
3
2m-7
3
13-n,
6
x+5
6
fcti.O
i-3_
. 2
5-Ct2
4
J, S ,:z K i:vC00gIe
110 JUNIOR HIGH SCHOOL MATHEMATICS [IX, 5 51
1. A farmer paid 10 men and 8 boys $51 for a day's
work. Later be paid 4 men and 6 boys $26 for a day's
work. How much did be pay each man per day? How
much did he pay each boy ?
2. For an entertainment, tickets were sold for 35 cents
and 25 cents. The total proceeds were $100 for 320
tickets. How many of each kind were sold?
5. For the same entertainment the next night the total
proceeds were $140.10 for 450 tickets. How many of each
kind were sold ?
1. A grocer has two kinds of tea, one selling at 40 cents
per pound and the other at 50 cents. How many pounds
of each must be used to make a mixture of 10 pounds to
sell for $4.40?
6. Milk is sold at 12 cents per quart and heavy cream
at 60 cents per quart. How many quarts of each will be
needed to make 18 quarts of light cream to sell for $6?
6. A grocer has two kinds of coffee, one selling at 22
cents per pound and the other at 32 cents. How many
pounds of each kind must be used to make 12 pounds to
sell for $3.20?
7. A mixture of 7-cent rice and 11-cent rice is to be
sold at 3 pounds for a quarter. How many pounds of
each must be used to make up sixty 3-pound packages?
8. Nougatines selling at 40 cents per pound are to be
mixed with chocolate almonds selling at 60 cents per
pound to make a mixture to sell at 48 cents per pound.
If 15 pounds of the mixture are wanted, how many pounds
of each must be used?
Jigilized by GoOgk
IX, (51] PAIRS OF LINEAR EQUATIONS 111
9. Forty laborers were engaged to raze a building.
Some of them agreed to work for $2.25 per day and others
for $2.50. The total amount paid them per day was $92.
How many worked at each rate ?
10. A collection of nickels and dimes, containing 46
coins, amounted to $2.90. How many coins of each kind
were there?
11. A part of $1200 is invested' at 6% and the remainder
at 5%. The combined yearly income is $68.50. Find
the number of dollars in each investment.
12. A man invested $750 in Liberty Bonds, part of it in
Z\% and the rest of it in 4J% bonds. His annual income
is $29.25. How many dollars were invested in each kind ?
13. In playing teeter, two boys use a board 16 feet long.
One boy weighs 90 pounds and the other weighs 110
pounds. At what point must the board be supported to
balance ?
14. A man rows 12 miles down stream in 3 hours and
returns in 4 hours. Find the rate of the river and his rate
of rowing in still water.
Solution. Let m = tbe number of miles per hour that
the man rows in still water, and
r — the number of miles per hour that
the river flows, then
m+r = the number of miles per hour that
the man rows down stream, and
m— r = the number of miles per hour that
the man rows up stream.
The equations are :
(I) 3(m+r)«12
(n) 4(m-r) = 12
Solving, m-ii, r-J. Ana. m-3i, r-J.
3, g ,i EE d by Google
112 JUNIOR HIGH SCHOOL MATHEMATICS [IX, $51
15. A boat goes down stream 30 kilometers in 3 hours
and up stream 24 kilometers in 3 hours. Find its rate in
still water and the rate of the current.
16. A man rows in still water at the rate of 3 miles
per hour. It takes him 2 hours to go down stream to the
next village. He returns in 4 hours. How far away is
the village and what is the rate of the river?
17. A man rows in still water at the rate of 7 kilometers
per hour. It takes him 30 minutes to row down stream
to a certain island, and 1 hour and 15 minutes to return.
How far away is the island and what is the rate of the
current?
18. An airplane travels with the wind from one city
to another 270 miles away in 2| hours and returns in 4J
hours against a wind of the same velocity. Find the
velocity of the wind and the rate of the airplane, if no
wind is blowing.
19. A camping party sends a boy with mail to the near-
est post office at 6 a.m. At 6.45 a.m. another boy is sent
to overtake the first, which he does in 1£ hours. If the
second boy travels 1£ miles per hour faster than the first
boy, what is the rate of each?
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CHAPTER X
FACTORS AHD EQUATIONS
§ 52. Factors. If an expression is the product of two
or more numbers, then these numbers are factors of the
expression. Thus, pairs of factors of 30 are 2 and 15,
3 and 10, or 6 and 5. The prime factors of 30 are 2, 3, and
5, since each factor is a prime number. (A prime number is
a number that is exactly divisible only by itself and by one. )
The factors of 6ab — 15ac are 3o(2fc — 5c), since,
3a(2o-5c)=6ao-15ac
The factors of x , —Qy t are x+Zy and x — 3y, since
The factors of a 1 — 8a+16 are a— 4 and a— 4, since
(a-4) ! ~a s -8a+16
S 63. Monomial Factors in Polynomials.
Type I. am+bm — cm.
am+bm — cm=m(a+b+c)
In this polynomial m represents any monomial that is a
factor of each term. In factoring an expression of this
type, m should be the greatest common factor of all the terms ;
that is, it should contain every factor that appears in all
the terms of the polynomial.
Example 1. Factor 5i*j/-10xj/*.
5x*y-lQxy*m5xy(x~2y)
Check. Multiply x—2y by 5xy.
Why is the Hymbol a used? (See page 50.)
i 113
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114 JUNIOR HIGH SCHOOL MATHEMATICS [X, 5 53
Example 2. Factor 12a*-8a s +4a.
12a»-8a ! +4a*4a(3a l -2a+l)
Check. Multiply 3a ! — 2a+l by 4a.
Why do you take out the common factor 4a rather
than a ?
EXERCISES
Factor each of the following polynomials and check.
1. 3x+6 11. ftr'-a:
2. 8a*-3a 12. 16a*-12a»
3. ab+b* 13. 28ax-49&3
4. 5x+5 14. 2sy +8*^-12$*
5. e*-3c*-c 10. 10x*y-12xy*+4xy
6. a s +2a£>+a 16. &-X
7. 3y*-15y+9y» 17. 15a*y-10aY+3aY
8. S^+lOxy+lSj/ 1 18. 2m*n-2fltn*
0. 5i»-lftr s 19. a?-a
10. 4aV-6ay+4a 1 z* 20. 5a*-10a-5
§ 64. Binomials — The Difference of Two Squares.
Type II. *-V.
a*-b*=(o+b)(a-6)
In this type the expression consists of two squares con-
nected by a minus sign.
Since (a+frXa-ftJ^a^-ft'CSeepage 89.), it follows that
the factors of a 1 — b 1 are a+b and a— b.
Example 1. Factor m*— 9.
m* is the square of m, 9 is the square of 3, and the squares
are connected by a minus sign ; hence,
m*-9=(m+3)(m-3)
Check. Multiply m+3 by m— 3.
sy Google
X, J54] FACTORS AND EQUATIONS 115
Example 2. Factor 1— 26f»*.
l-25m*=(l+5m ! )(l-5m s )
Check. Multiply l+5m* by 1 — 5m*.
Could the factor 1— 5ro s be written first? Prove it.
EXERCISES
Factor each of the following binomials and check.
1. x'-y' 11. o J b»-4
2. m*—n* 12. 1-iV
3. 25 -s* 13. A*- 166'
4. a?-l 14. 9a*-25&'
B. lOO-o* 15. 25a t b i -x*
6. l&e»-9j/ ! 16. 144y 2 -l
7. m'-ftr* 17. 9a*b*-49<?
8. 49a i -9m i 18. df-dt*
9. D*-4d* 1». 100m*-l
10. 1-&* 20. fi I -4r,V 1 *
21. 9x*+4
Is 9x*+4 factorable under the type a 1 — b 2 ?
In what respect does it differ from the type?
22. 4s 4 -8
Is 4x*— 8 factorable under the type a 1 — b s ?
In what respect does it differ from the type ?
If it were 4i ! — 9, what would be its factors ?
In Exs. 23—32, some of the given binomials cannot be
factored as the difference of two squares, while others can
be. State in what respect each of the former differs from
the type.
23. 4o a -l 25. o*+4
24. d>~a* 28. 4x*-$y*
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118 JUNIOR HIGH SCHOOL MATHEMATICS [X, S 54
27.
9o"-206 !
30.
6 2 +i !
28.
49e*-16ci'
31.
lOO-zy
29
4i>-l
83. 2»"-
32.
-18v'
*y— 100
If the monomial factor 2 is
taken out, the other factor
is of the type a*—
k 3 , hence
2x<- 18»
'■2(x'-8lfl
-2(i+3!,)Ci-3»)
In Ejra. 34-43,
find the prime factors by removing
the
monomial factor first.
34.
aj? — ay*
39.
9x"-4x
35.
5o ! -20
40.
25i"-x
36.
aty-V*
41.
a*— a
37.
8o , -2oo I
42.
2^-2
38.
4l ! -36
43.
IV-4J/2 1
Solution.
In Exs. 45-52, find the prime factors by factoring any
factor that is not prime.
45. y*-lG 49. a*-16
46. x 4 -81 50. ax*-16a
47. y*~l 51. x»-y*
48. 1-asV 52. ar*-«
58. (a-fc) l -c*
Solution, (a— 6)*— <?=(<*— 6+c)(o— b— e)
64. (2a+3)*-16
Solution. (2a+3) ! -16=(2a+3+4)(2a+3-4)
«(2o+7)(2a-l)
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X, (55) FACTORS AND EQUATIONS 117
In Exs. 55-64, factor each expression as the difference of
two squares and collect terms in the factors when possible.
55. (m-n)'-& 60. (2a+3)"-4
66. (2o+6)"-c ! 61. (5z-2) ! -9
67. («-2j) I -9 62. (3o+7)"-25
58. (3o-o)"-25 63. (4o ! -5)"-16
5». (ro"-3m)'-36 64. (ftr'-10) ! -36
§66. Trinomials — The Squares of Binomials.
Type III. te±tab+l?.
o'+ioH-d'- (o+6)"
a'-JoiH- &"-(«-(>)■
A trinomial square contains two terms that are positive
squares and a third term that is twice the product of their
square roots. (See page 89.)
EXERCISES
In each of the following expressions, supply the missing
term that will make it a trinomial square.
1. o'+( )+&* 11. ( )-12i+36
2. m>-( )+l 12. 25o'+( )+9
3. «"-10«K+( ) 13. 9z'-6oxV+( )
4. 4o'+4oi+( ) 14. a'+»'+( )
5. ( )-6o+9 15. 4i"+V+( )
6. ( )-30iy+25jf" 16. 25+o"+( )
7. 4* , -&n|z-K ) 17. m>+144-( )
8. 16V-lfcr+( ) 18. o?+81-( )
9. 9-12o+( ) 19. J?+64+( )
10. 64m'+( )+9 20. 121+2>+( )
DigilzBd^y GoOgk
118 JUNIOR HIGH SCHOOL MATHEMATICS [X, j56
§ 66. Factors of Trinomial Squares.
Example 1. Factor x 3 +6xy+9y*.
x t +6xy+9y 1 ={x+Sy) t
Check. Multiply x+3y by x+3y.
Example 2. Factor 16x*-8x*y+y*.
16x*-8x*y+y>m(4x*-y)*
Check. Multiply 4**— y by 4a?— y.
Factor each of the following trinomials and check.
1. x*-2xy+y* 11. a*+46*-4a6
2. m*+2mn+n* 12. 64+m*-16»t
3. 16a*— 40a6+256* 13. Stf-SOa^St*
4. 4m , +4mn+n* 14. 16x*+40xy*+25y*
6. a*+x*+2ax 16. W+x*-8x
6. b I +m i ~2bm 16. l+a t -2a
7. 100+8»+20s 17. x*+2x 1 +l
8. 9x»+y»-6iy 18. 4o 1 -4o6 , +6*
9. x*+2x*y 1 +y* 19. m*+l-2m
10. a 4 -2a*6+b* 20. 49a 4 ft 4 +l-14o%*
21. 4o'+2a+l
Is 4o*+2o+l factorable under the type a t +2ab+b t 1
In what respect does it differ from the type?
In Exs. 22-31, some of the given trinomials cannot be
factored as squares of binomials, while others can be.
State in what respect each of the former differs from the
type.
22. a*+4<rf>+46* 24. x*-6xy+9y*
23. 16m*+4m+l 26. 9x*+6xy+4y>
UKinz.drvGoogle
XS571 FACTORS AND EQUATIONS 119
86. x*+2x*y+y* 99. a t -2ab+b*
27. M+x*-16x 30. x>+xy+y*
28. 16a; 1 +50xy+25tf* 31. a?-2ab+b t
32. &-2&y+xy*
If the monomial factor x is taken out, the other factor
is of the type o*— 2ab+b s , hence
x , -2xhj+xy t ^x(x t -2xy+y 1 )=x(x-y) t .
In Exs. 33-40, find the prime factors by removing the
monomial factor first.
S3. 3»*-6xj/+3y* 37. o'+afc*-2a s 6
34. 4ax*+4a J x+a* 38. x*y-4x t y t +4xv'
36. 12s*-36a;+27 39. 4a*+8ao+41» I
36. x"-2a?+a: 40. 9x 4 +9y*-18xy
S 57. Review of Types I-UJ.
EXERCISES
Find the prime factors of each of the following expres-
sions, and check.
1. m J -4n* 11. 25x*-16
2. mHm-n+4n* 12. 25x*+40xy+16y
S. m t +4n*+4mn 13. y*—y
4. 9i*-4i,* 14. y t -2y+l
5. S^-oiy+v* 15. 2y*-2y
6. a*-l 16. x'-xM-x
T. a*+2a+\ 17. (a-26)*-x»
8. e?-2a*+a 18. (2x-3)*-16
9. Sx'+Sxy+Sjf 19. (o*-10)*-36
10. 6*-66+9 20. (4p*-l)'-9
21. a 1 -2ab+b t -(?
Solution. o , -2a6+6*-e»=(o*-2a6+6 , )-c*
m(a-b+c)(a-b-c)
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120 JUNIOR HIGH SCHOOL MATHEMATICS (X, §57
22. 4a*-4ao+b 2 -25
Solution. 4a*-4a6+&*-25 = (2a-fc) ! -25
= (2a-o+5)(2a-o-5)
23. (o*-6o6+96 l )-«» 26. m«-4mra+4n*-25
24. (z*-6aH-9)-jf 27. 4a*-12a6+96 1 -16c i
25. (b*-10o+25)-16 28. (a*-2a*+l)-9
§ 86. Trinomials — The Product of Two Binomials
Having a Common Term.
Type IV. i 2 +(a+b)x+fl&.
x s +(a+o)x+<io=(x+fl)(z+o)
Example 1. Factor x*+8x+l5.
Solution. Since +15 is the product of the two num-
ber terms, their signs are like.
Since +8 is the sum of the two number terms, their signs
are plus, hence
xH-8x+15-(z+5)(a:+3)
Check. Multiply z+5 by z-f-3.
Example 2. Factor x i —8x+ 15.
Solution. Since + 15 is the product of the two number
terms, their signs are like.
Since — Sis the sum of the two number terms, their signs
are minus, hence
x t -Sx+15 = (x~5)(x~S)
Check. Multiply x— 5 by x— 3.
Example 3. Factor x t +2x— 15.
Solution. Since — 15 is the product of the two number
terms, their signs are unlike.
Since +2 is the sum of the two number terms, the sign
of the larger is plus, hence
s*+2r-15 = (x+5)(l-3)
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X.S58] FACTORS AND EQUATIONS 121
Check. Multiply x+5 by x— 3.
i Example 4. Factor a?— 2x— 15.
Solution. Since — 15 is the product of the two num-
ber terms, their signs are unlike.
Since —2 is the sum of the two number terras, the sign
of the larger is minus, hence
jr 2 -2i-15 = (x-5)(a:+3)
Check. Multiply x — 5 by x+Z.
EXERCISES
Factor each of the following trinomials and check.
s^+sv+e
11.
h'-14h-15
a'-Sa+lb
12.
t"+6t+5
o ! +2o-15
13.
m"+llm-60
wi* — wi — 6
14.
S'-25i,+ 150'
m'+m— 6
16
V-3b-M
i'+75+12
16.
xy-4xy+3
pM-p-12
17.
a*6 ! -afc-132
i*-9i+18
18.
<*p-Scd-9
e"-c-20
19.
y'+2y+l
10. *P+ 14d+33 20. a*-6a«-16
21. «*-a:-3
This trinomial cannot be factored under Type IV.
Why? In Exs. 22-31, some of the given trinomials can-
not be factored under Type IV, while others can be.
State why in each case.
22. o l -a-30 27. 6*+5b+6
23. m*+3»i-5 28. o s +66+8
24. a^-J-4x-t-5 29. x*+x-2
25. p l -3p+2 30. &+X+1
26. y*-5y-6 31. asH-3s+2
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122 JUNIOR HIGH SCHOOL MATHEMATICS [X, § 58
38. 8**-16x+18
If the monomial factor 3 is taken out, the other factor is
of Type IV.
aa?-15as+18»3(s , -6x+6)-3(*-3)(as-2)
In Exs. 33-40, find the prime factors.
33. atb+fab+Sb 37. 2a'-10a*+12o
34. <jy*-14ay+24a 38. a*x*-2a*x-15a*
36. 4P-8t-12 89. y 4 -13y»+36V
36. z»-&c*+5a; 40. 2x , +4a:+2
§ 59. Review of Types I-IV.
EXERCISES
Find the prime factors of each of the following expres-
sions and check.
1. 6a a +16a 9 a?-x
2. o*-4y» 10. 4n*+4
3. x*+12x+36 11. c*-4c-21
4. x*-2x-8 12. tfb+rab-lSb
6. 3x*+12x+6 13. a*-l
6. n*-4n+4 14. 3x*-3x-6
7. 4a I +4(i+l 15. a«-a*-6
8. 4x*~4xy+y t 16. 9a ! -18
§ 60. Trinomials — The Product of Any Two Binomials.
Type V. ai'+bx+c.
OI 1 + &I+C=(?X+?)(?X+?)
Example 1. Factor 4x 2 -lZx+ 10.
Solution. Since +10 is the product of the two num-
ber terms, their signs are like.
Since — 13s is the sum of the cross-products, the sign of
each number term is minus.
;, S ,:z K i:vC00gIe
FACTORS AND EQUATIONS
TxBTotm or 4i*
Factob* o» + 10
ix (2x
x° r [2x
{::»{::
Since the sum of the cross-products must be — 13x,
the factors are : 4x — 5
-13s
4*'-13a:+10=(4x-5)(x~2)"
Check. Multiply 4a;— 5 by x—2.
Example 2. Factor 6t/*— 5y— 4.
Solution. Since —4 is the product of the two number
terms, their signs are unlike.
Since — 5y is the sum of the cross-producte, the sign of
the greater cross-product is minus.
Factors or 8 b'
F.^„„
-*
f* J*
I""
{::«
-2
or
+2
{:;
Since the s
factors are ;
l of the cross-products must be — 5y, the
X
2y+i
-5y
6^-5j-4-(3 9 -4)(2j+1)
Check. Multiply 3y - 4 by 2y + 1 .
JigliZBdsy G00gle
124 JUNIOR HIGH SCHOOL MATHEMATICS [X, | 60
EXERCISES
Factor each of the following trinomials and check.
1. 3z'+5l+2
12. 3x ! +4z+l
2. 2x , -5x+2
13. 8^+221/+ 15
8. ijf-y-W
14. 8+22x+15x>
4. 6m s +7m-3
An*. (2+3z)(4+5»).
6. 6z*-ar-2
16. 2-a-21a>
«. 14o ! -39o+10
16. 15-r- 2r"
7. &r=- 101+3
17. 2-5l+2»*
8. 2t°+t- 15
18. 6m*-l$m+15
». 7i"-3l-4
19. 20a 2 -o-99
10. 3:r B +7:r.+2
20. 6-&t-4x»
11. 5»'-9»-2
21. 6s"+72+2
In Exs. 22-30, find the
prime factors.
22. 4P+221+10
27. 16x ! +l&r.-12
23. 20as-9x s -20x*
28. 15l , +21»+6
24. 10r"-5l— 75
28. 30»>-35l!+10
26. 7a^-x*-6x
SO. 15l , -21»+6
26. oz'-SSx+Se
§ 61. Summary of Factoring.
Type I. Monomial Factors in Polynomials.
am+bm—cm=m(a+b~c)
Type II. Binomials — The Difference of Two Squares.
a*-&*=(<H-t?)(a-&)
Type III. Trinomials — The Squares of Binomials.
a t ±2ab+b*=(a±b) t
Diqi! .zed sy G00gle
X, S 61]
FACTORS AND EQUATIONS
125
Type IV. Trinomials — The Product of Two Bi-
nomials Having a Common Term.
« , + (8+6)z+a*-(x+«)(x+fc)
Type V. Trinomials — The Product of Any Two
Binomials.
«r"+ta+c-(?i+?)(?i+?)
MISCELLANEOUS
Find the prime factors of
sions and check.
1. a , +2o ! +2a
2. m*-64y*
3. p<-ip+i
4. P-l-12
6. 6a'+7a-3
6. 3»'+5»+2
7. 3» ! +27a:+42
8. o'+36» ! +12o ! I
9. *■- 111+24
10. m'+llm-12
11. 100-9i"
12. «■- 171+72
13. J'- 16V- 100
14. 5o"+10o"l)'+30o'6<
15. l'-2«"-120
16. a*-a-240
17. y'-iy'-lSO
18. 12a'-23»6+106"
19. l-20»+75v'
20. 2n'-n*-3n
SI. zM-2zy-8»*
22. 1-1V
IK FACTOHIHG
each of the following expres-
23. ix'+ixy+y*
24. z—x*
28. *+4x
26. 6P-71-3
27. x<+y 4 -2zy
28. 3fl ! -12n-96
29. 6x ! -15«+6
30. ro ! - 14m -95
31. l-4o+4o*
32. l-2o+a s
88. l-15xj/+56*y
84. 7-3I-4* 1
35. a*— 4a
86. 2+2o*
37. a 8 -y«
38. 3z<-23* ! -36
39. (2a-o)"-c"
40. (3»+«)"-9
41. a 2 -6a+9-j/*
42. 4l"-12l!/+V-92'
43. z*-2z t y t +y*-W
44. (a+6) ! -3(o+6)+2
3, g ,i,z E d by Google
126 JUNIOR HIGH SCHOOL MATHEMATICS [X, j 62
§ 62. Quadratic Equations. Id Chapter V the equa-
tions solved were linear equations. Linear equations
are equations of the first degree ; that is, they involve only
the first power of the unknown number.
Quadratic equations are equations of the second de-
gree ; that is, they involve the second, but no higher, power
of the unknown number.
EXERCISES
In the following exercises :
(a) State the degree of each equation.
(b) Select the linear and the quadratic equations.
1. x-5=0 ■ 6. m»-16=0
2. x*+x=12 7. 5s*-9x=2
3. 3a-4 = 8. p"=12
4. o*-5a=14 9. #+5=0
5. 6x*-x«=2 10. P-&-10-0
§ 63. Solution of Quadratic Equations by Factoring.
To solve equations by factoring the following axiom
(Axiom A) is necessary. If the product of two or more
factors is zero, at least one of the factors is zero.
Example 1. Solve the equation x* — 5x=6, and check
the roots.
Solution.
® x»-5x = 6
®
x*-5x-6-0
®-6
®
(*-
-6)(x+l) =
®= (Factoring)
®
x-6-0, or x+1
-0
® by Ax. A
©
x=6, or x= -
-1
Check.
(6)<-5(6)i6 (-
-1)>
-5(-l)i6
36-30 = 6
+1+6-6
Ant. *-6, or -
J, S ,:z K i:vC00gIe
X, 1 63] FACTORS AND EQUATIONS 127
Example 2. Solve the equation a a +3o = 10a+18,
and check the roots.
Solution.
® o'+3o=10o+18
® a'-7a-lS-0
®- 10a- 18
® (o-9)(o+2)-0
©■(Factoring)
® a-9-0, ora+2-0
® by Ax. A
® a=9, ora=-2
Check.
(9) ! +3(9)il0(9)+18 (-2)M-3(-2)ilO(-2) + 18
81+27.1 90+18
+4-6i -20+18
108-108
-2--2
Ana. o=9, or —2.
Example 3. Solve the equation 4x+2 = 6x I +3x, and
check the roots.
® 4i+2-6V+3i
® 0-6x"-i-2
®-4l-2
® 0-(3i-2)(2i+l)
® 3*-2=0, or2*+l=0
©=s (Factoring)
® by Ax. A
® 3x-2,or2i=-l
® as-}, on- -J
Check.
4«)+2i6(*)>+3(f) 4(-«+2
A6(-i)'+3(-J)
}+2i6(t)+2 -2+2
i«(i)-|
4J = 4j
-0
Ana. *-4, or -J.
3,g,1 EE d by G00gle
128 JUNIOR HIGH SCHOOL MATHEMATICS [X, & 64
§ 64. Summary. To solve a quadratic equation by
factoring:
(1) Collect all the terms in one member (leaving the other
member zero).
(2) Factor that member.
(3) Write each factor equal to tero (Axiom A).
(4) Solve each equation thus formed.
Solve each of the following equations, and check the
roots.
1. rf-7s+13-0
18.
4m"-9 =
2. x l +3x=10
19.
5a*-10a =
S. j/M-8y=-7
20.
6!(<-l)-72
4. 2x ! -5x=-2
21.
60*+ir"+144-8z
5. 5x s =4x+l
22.
3s"-5x-2-0
6. 6x 2 -llx=2
23.
1%-63-t/"
7. j"-13y-S6
24.
3m*-6 7m
8. x s +100 = 20x
26.
0-2- -3a'
9. fc(*-ll) = -30
10. 10p-3 = 3p i
26.
x --x
7
11. a(a+10) = -24-4a
12. 2x*-4s = -40+14*
27.
?-l-4
2
13. m ! = 10m-25
28.
pi +3 p__j
14. x s — x=0 Ans.
15. a*-3a=0
x=0, or
3 "X = ^
16. y* = 5y
80.
s"-V»-i
17. c*-4 =
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X, S Ml FACTORS AND EQUATIONS 123
In checking, substitute each result in the statement of
the given problem.
1. Find two numbers whose difference is 9, and whose
product is 90. (Let n and n+9 be the two numbers.
Why?)
2. Find two numbers whose difference is 4, and whose
product is 12.
8. Find two numbers whose sum is 14, and whose
product is 33. (Let n and 14— n be the two numbers.
Why?)
4. Find two numbers whose sum is 10, and whose
product is 24.
6. The product of two consecutive integers is 20.
Find them. (Let n and n+1 be the two consecutive
integers. Why ?)
6. The product of two consecutive integers is 90.
Find them.
7. Find two consecutive integers the sum of whose
squares is 25.
8. Find two even consecutive integers the sum of
whose squares is 100.
9. Find a number whose square less 6 is equal to 5
times the number.
10. Find a number whose square increased by 8 is equal
to 6 times the number.
11. The square of a number exceeds the number itself by
56. Find the number.
IS. A rectangular room has an area of 240 sq. ft., one
side being 8 ft. shorter than the other. Find the dimen-
sions. Do both roots of the equation satisfy the problem?
ioogle
Pig. 23.
130 JUNIOR HIGH SCHOOL MATHEMATICS [X, 9 64
13. A picture, 8" by 12", is placed in
a frame of uniform width. If the area
of the frame is the same as the area of
the picture, what is the width of the
frame? (Fig. 23.) Do both roots of the
equation satisfy the problem 7
14. An open box is made from a
square piece of tin by cutting out a 4-inch square from
each corner and turning up the sides. How large is the
original square if the box con-
tains 64 cu. in.? (Fig. 24.) (Let
«=width of piece of tin.)
16. A fence 190 ft. long sur-
rounds a rectangular field that
contains 1800 sq. ft. Find the
dimensions of the field.
16. A ventilator with a rec- «■—#--•
tangular opening 20 in. high gives Fl °' M '
64 sq. in. more space than one with a square opening of
the same width. How wide is the ventilator?
17. A room is 15 feet square. On account of an error
in measuring the dimensions of the floor, an area of 361
square inches too large was obtained. What was the error
in the measurement (in inches) ? [The equation is :
(n+180)' = 180 3 +361.]
18. A room is 3.2 meters square. On account of an
error in measuring the dimensions of the floor, an area of
1276 square centimeters too small was obtained. What
was the error in the measurement (in centimeters) ?
i BV Google
CHAPTER XI
RADICALS AND ROOTS
§ 66. Rational and Irrational Numbers. For each of
the quadratic equations in Chapter VI, the values of the
unknown number were rational.
A rational number is an integer, or a number which is
the quotient of two integers, for example : 8,-7, 0.875,
f V25.
For many quadratic equations the values of the un-
known are irrational; for example, to find the side of a
square whose area is 45 sq. in., the equation is 45 — a*, in
which 8 = V45.
An irrational number is a number that is not rational;
for example : V2, vl2, "v^, ^Mj.
In these examples the symbol ■>/ is called the radical
sign. In the -$!>, the 3 is the index of the root and shows
that the cube root is required. When no index is written,
the root required is the square root.
Make a list of the rational numbers and of the irrational
numbers from among those that follow. Name the index
and root required for each irrational number. 5, —12,
Vjt, i, 0.305, Vl0, <fl, $, vTB", ^SO", 0.04, vT7, A.
VT6, </ r ¥7.
§ 66. Reduction of Radicals. When a radical is irra-
tional, it is often convenient not to evaluate it, but to re-
tain it in radical form. When this is to be done the radi-
cal should be reduced to its simplest form.
131
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132 JUNIOR HIGH SCHOOL MATHEMATICS [XI, J 66
A radical, involving square root only, is in its simplest
form when the number under the radical sign is integral,
and when the number contains no factor that is a perfect
square.
Example 1. Reduce V^SO to its simplest form.
Solution. V50 = V25X2 = 5V2. Arts. 5\^2.
Example 2. Reduce Vf to its simplest form.
Solution. V\ = V^=V-^x6 = ^Vq. Ans. \V&.
Example 3. Reduce \^ to its simplest form.
Solution,
"OOC*
^56c vouo "'~5fc v
Reduce each of the following radicals to its simplest
i. via
8. V1000
1«. Vj
2. V24
3. VJO
10. Vte*y
"•>i
4. VS?
8. Via
. 11. VS?
12. v'oOWo 1
»\f
6. V80
7. V200
13. Vj
14. Vj
18 J™5
v 27
In Exs. 19-30,
after reducing each radical, evaluate it.
Use the following values: V2*
= 1.414, V3- 1.732,
^5-2.236, and VlO-3.162.
19. V8
23. V20
27. VJ
20. Vis
24. V75
28. VA
21. V45
26. vT25
29. V|
22. V300
2«. VlOOO
30. V^
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XI, S 67] RADICALS AND ROOTS 133
§ 67. Multiplication of Radicals. The square root of a
number is one of the two equal factors of the number.
Hence, when you square the square root of a number, the
result must be the given number. For example, (v'2)*=2,
(V7)»=7, etc.
Example 1. Find the product : 2V3X5V3.
Solution. 2^3X5^3 = 10(>/3)> = 10x3«30.
Am. 30.
Example 2. Find the product : V3(2"\/3-5).
Solution.
V3(2V , 3-5)=2(V3)*-5v^=2X3-5v / 3 = 6-5V^.
Ana. 6-5V3.
BXERCISBS
In each of the following exercises, find the product.
1. VExVl 6. V5(2V5-1)
2. 3V2XV2 7. 2V3(V3-3)
8. 4VlOX3vlO 8. (2+V3)(2-V^)
4. vlSxVs 9. (2+V3)(2+V3)
5. V2(3-V2) 10. (VB-2)(V5-3)
In each of the following problems, leave the result in
radical form ; reduce the radical to its simplest form unless
required to evaluate it.
1. The sides of a right triangle are 4" and 2". Find
the hypotenuse. (See Ex. 7, page 26.)
2. The side of a square is 5". Find its diagonal.
3. The side of a square is 4". Find its diagonal.
4. The side of a square is 10". Find its diagonal.
B. The side of a square is a inches. Find its diagonal.
jigiiized by Google
134 JUNIOR HIGH SCHOOL MATHEMATICS [XI, |67
6. The side of an equilateral triangle is 10". (a) Find
its altitude. (See Ex. 14, page 41.) (&) Find its area.
7. The side of an equilateral triangle ia 4". (a) Find
its altitude. (o) Find its area.
8. The side of an equilateral triangle is 12". (a) Find
its altitude. (6) Find its area.
9. The side of an equilateral triangle is a inches,
(a) Find its altitude. (6) Find its area.
10. The sides of a right triangle are V% in. and V2 in.
Find its hypotenuse (to three figures).
11. The sides of a rectangle are V7 in. and V3 in.
Find its diagonal (to three figures).
12. The side of a square is 2 V2 in. Find its diagonal.
§ 68. Table of Squares and Square Roots. In Chapter
II you found the square root of a number by estimating
one of its two equal factors, and then by dividing the given
number by your estimate.
It is customary for those who need to evaluate square
roots, cube roots, etc., frequently, to consult tables.
On pages 288-290 you will find the squares of all num-
bers of three figures from 1.00 to 10.00. This table may
also be used for finding the square roots of numbers.
When using the table, make a mental estimate of your
result in order to check the location of the decimal
point.
A study of the examples on pages 285-287 will make it
clear how to use the table for finding the square and the
square root of any number of three figures.
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XI, |68]
RADICALS AND ROOTS
135
EXERCISES
Copy the following table and find the squares and square
roots to three figures using the table on pages 288-290.
Make the estimates for all the exercises first.
n™™
btfett-
Stnui
Est. Suuabb
Snu mi Root
1. 5.43
2. 54.3
3. 0.543
4. 1.27
5. 12.7
6. 127.
7. 0.127
8. 6.40
9. 64.0
10. 640.
11. 0.640
12. 0.0640
13. 3
14. 30
15. 300
PROBLEMS
Check the answer to each of the following problems in
the statement of the given problem. Record the answer
to three figures, when it is not exact.
1. The area of a square room is 192 sq. ft. What is
its length?
2. The sides of a right triangle are 8.00" and 11.0".
Find its hypotenuse. (Formula, Ex. 7, page 26.)
3. The sides of a right triangle are 22.4 cm. and 18.6
em. Find its hypotenuse.
4. The sides of a square are each 12.0". Find the
diagonal.
5. A trunk is 41.3" by 23.8" (inside). What is the
longest cane that can be placed in the bottom of it?
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136 JUNIOR HIGH SCHOOL MATHEMATICS [XI, J 68
6. The hypotenuse of a right triangle is 20" and one
side is 12". Find the other side.
Sold
(ton. The formula is
©
tf-c 1 -^
®
o»=( c +6)(c-6)
©■(Factoring)
®
o*-32X8
©■ (Substituting)
®
a* =266
®-
®
a -16
®v
7. The hypotenuse of a right triangle is 12.3" and
one side is 11.8". Find the other side.
8. A ladder is 38' long and just
reaches a window. If its foot is
13' from the building, how high is
the window?
9. Find the altitude of an
equilateral triangle whose side is
12.4" (Fig. 25).
10. Find the altitude of an
equilateral triangle whose side is 18.6
area.
11. Each side of a regular hexa-
gon is 8.6" long (Fig. 26). Find its
area.
(Suggestion. Find the area of
one triangle first. The triangles are
equilateral. Why ?)
12. Each side of a regular hexa-
gon is 27.6 cm. Find its area.
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XI, S 68] RADICAIS AND ROOTS 137
IS. A metal plate 10.3" square is required to be cut
from a circular plate. What is the diameter of the
smallest circular plate that can be used ?
14. What is the diameter of a circle whose area is 47.4
sq. in.?
15. The inside cross-sectional area of & water pipe is
1.52 sq. cm. What is its diameter?
(In Exs. 16-25, change the subject of the formula as
required.)
16. A-tt 1 , r-7
31. e'-o'+J', o-
17. t-W, t-1
18. V-ir'J, r. ?
2.5'
1». V-^h, r-7
SO. S-*rf", r-7
a. A-t^,.;
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CHAPTER XII
QUADRATIC EQUATIONS
§ 69. Types of Quadratic Equations. The equation
ax t +bx+c = is called a complete quadratic equation, be-
cause it contains one term involving x 1 , one term involv-
ing x, and a third term free from x.
Quadratic equations of the forms
xH-e-0
x 1 +bx =
are called incomplete quadratic equations, because either the
term involving x, or the term free from x is missing.
EXERCISES
From the following equations make a list of the complete
and of the incomplete quadratic equations.
1. 5xH-3;r~7=0 5. 7y ! -15=0
2. a?-7 = 6. 36 J +56+3=0
3. a^-5x-2=0 7. 12o s *0
4. 5a i ~a=0 8. m*-m =
§ 70. Solution of Incomplete Quadratic Equations.
Example 1. Solve x"=lQ, and check the roots.
Solution A (by factoring).
® i*=16
© x s -16=0 ®-16
© (x-4)(x+4)=0 ©^(Factoring)
© x-4=0, orx+4 = © by Ax. A
© 3 = 4, or x=— 4
From this solution you should note that the equation
x*— 16 = has two roots, namely +4 and —4.
138
is, Google
xii, s 7o] Quadratic equations 139
Solution B (by square root).
® a?=16
© a;=+4,or-4 ®V
Solution A shows that it is necessary to write both
+4 and —4 for the square root of 16.
Check. (+4) ! = + 16 (-4) s =+16
Ans. 3= +4, or— 4.
It follows that the square root of a number may be
either plus or minus. The plus square root is called the
principal root.
Example 2. Solve 3 ! -7 = 0, and check the roots.
Solution A (by factoring).
® x l -7=0
Since 7 is not a perfect square, its two equal factors are
irrational.
®{x~V7)(x+V7)=0 ®=(Factoring)
© x-V7 = 0,on+V7=0 ©by Ax. A
© I=+ V7 )0ra . == _V7
© x= +2.65, or x = -2.65
Solution B (by square root).
® x ! -7 =
© x* = 7 ©+7
© x=+2.65,or -2.65 ® = (Table)
Check. (Use Table of Squares.)
(+2.65)*-7=0 (-2.65) ! -7i0
7.02-7 = 7.02-7i0
Ans. x= +2.65, or -2.65.
The discussion of these two methods and the interpre-
tation of the checks follows :
J, S ,:z K i:vC00gIe
140 JUNIOR HIGH SCHOOL MATHEMATICS [XII, (70
Solution A shows that both the plus and the minus
values of V7 must be recorded. The check shows that
both values satisfy the equation.
Method B requires less labor, but it is essential that you
record both signs in your answer. The answer may be
written: z=±2.65.
Note. In the check the two members are not identical,
because the square root of 7 is only an approximate value. The
more figures used in the square root, the more nearly identical
would the members of the check become. When irrational
roots are evaluated, the two members in the "Check" will not
be identical.
Example 3. Solve the equation —-11, and check
the roots.
Solution.
® ap-ii
® 3j*=44 ®X4
® **= 14.7 ®+3
® i-±3.83 ®V~
Since the tables give the square roots of three-figure
numbers only, retain not more than three figures in ®
and ®.
Check. (Use Table of Squares.)
3(3.83)* ± n 3(-3.83)' j. 11
4
3X14.7 _L 11
4
11.0 = 11 11.0411
Ana. *=±3.83.
Note. The members of this check appear to be identical,
but if the fourth figure were written, they would differ, that is,
3><1M- 11.02.
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QUADRATIC EQUATIONS
Example 4. Solve the proportion = — , and check
the roots.
Solution.
12 = m
m = 32
®
® W^-a 2 ^) ©X32m
Check.
® 384 = ™* ©=
© ±19.6= in ®V~
J2_±l$£ 12 j —19.6
19.6 32 -19.6 32
0.612-0.612 -0.6124 -0.612
Am. m=±19.6.
Note. The equation — "S?' a a proportion, since it expresses
the equality of two ratios. In this proportion, m is called the
mean proportional between 12 and 32.
EXERCISES
Find, to three figures, two roots for each of the follow-
ing equations, and check each root.
1.
3a"- 96.6
10.
48-3.141*
2.
3.
1.
9-4.5m'
¥-17.5
11.
11.
p 20
5 d
d 8
5.
6.
4
3_2a'
5~ 4
5fe'+4-18
13.
It.
10_K
» 3
16 a
o = 6
7.
8.
10(o'+l)-18
2(3i>'-i)«4.66
15
3.5 d
d 12
9.
3
le.
2.4 1
x "10.6
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142 JUNIOR HIGH SCHOOL MATHEMATICS [XII, } 70
17. I™-J£ 19. «*-J
m 18 a 4
18 2™--* L12 = _6_
fc 10 6 0.61
PROBLEMS
Check the answers to each of these problems in the
statement of the given problem.
1. How long will it take a bomb to fall from an air-
plane 3000 feet high? (Use the formula s = 16(*.) Will
both values of ( satisfy the conditions?
2. How long will it take a bomb to fall from an air-
plane 2000 meters high ?
5. The horse-power of a 4-cylinder automobile engine
is 35. What is the diameter of its cylinders in inches?
(Formula, page 37, Ex. 5.)
4. Write the formula for the statement : The areas
of two similar surfaces have the same ratio as the squares of
any two corresponding lines.
The formula is :
AlJl
As V
Explain the meaning of each letter. (Use the formula
in Prob. 4 for Probs. 5-7.)
6. The areas of two circles are 40 sq. in. and 60 sq. in.
The diameter of the smaller is 7.14 in. Find the diameter
of the larger.
6. The areas of two similar triangles are 50 sq. in. and
30 sq. in. The altitude of the second is 15 in. Find
the altitude of the other.
7. The surfaces of two balls are 16 sq. cm. and 40 sq.
cm. What is the diameter of the smaller one, if the
diameter of the larger is 3.57 cm.?
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XII, 1 71] QUADRATIC EQUATIONS 143
§ 71. Completing the Square. In order to solve
quadratic equations of the type ax'+bx+c = 0, that have
irrational roots, you must consider again how to complete
a trinomial square. (See page 89.)
From (x-{-a) i =x i -\-2ax+a i you note that the third term
of the trinomial is the square of half the coefficient of x in
the second term. Hence, if you are given only the two
terms x i +2ax, you have to add a 2 in order to complete the
square.
Figure 27 is a geometric illustra-
tion of the incomplete square x s +2ax.
Example 1. Complete the square
ina^+10x.
Half the coefficient of x = 5.
5*=25.
Thecompletesquarei8i 2 +10a;+25. Fw - 27 -
Example 2. Complete the square in x 1 — 1#.
Half the coefficient of x = — £.
(-#>■=#*.
The complete square is x 1 — $£+$$.
EXERCISES
Complete the square in each of the following.
1. x i +Qx 7. x*-5x
2. x*+12x 8. x?+x
8. x*-$x 9. x i -x
4. x*-\0x 10. x*+llx
5. 3?+Zx 11. x'+fs
6. x*+7x 12. ar'+Js
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144 JUNIOR HIGH SCHOOL MATHEMATICS 1X11,171
13. i»-|a: 17. x*+*s
14. x'-^-x 18. i*-fc
15. a?+£r 18. x*+mx
16. **-fc SO. x*-bx
§ 72. Solution of Complete Quadratic Equations.
Example 1. Solve the equation tf+Qx-Z, and check
the roots.
Solution A (by completing the square and taking the
square root of each member).
© 3s s +6x = 3
® s*+6a;+9-12 ®-f3«
® x+3= ±3.46 ®y/ (Table)
® x=+3.46-3, or -3.46-3 ®-3
® 1 = 0.46, or -6.46 ®=
Solution B (by completing the square and factoring).
© 3 l +6x = 3
© ^+61-3-0 ®-3
® (x»+6i+9)-12-0 ®= (Completing the
square)
(t)(x+Z-Vi2)(_x+i+Vl2)~0 ®= (Factoring)
® i+3-Vl2=0, or
x+3+vT2«0 ® by Ax. A
® s-0.46-0, or
«+6.46 = ®=(Table, collect-
ing terms)
© i = +0.46, or* =-6.46
Solution B shows the necessity of using both + and
- for Vl2 in equation ®, Solution A.
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XII, 1 72] QUADRATIC EQUATIONS 145
Check.
(0.46)*+6(0.46)i3 (-6.46)*+6(-6.46)=l3
.212+2.76=L3 41.7-38.8£3
2.97=^3 2.913
Ana. «=0.46, or -6.46.
Why are the two numbers in these checks not identical ?
Example 2. Solve the equation 2i*-3a;=7, and check
the roots.
Solution. Before completing the square, divide by
the coefficient of x*.
® 2*=-3x = 7
® *-U-l
®+2
® z"-t*+A-l+A
©+({)'
® *=-»*+*-«
®-
4 4
®V
4 4
or»-*°«
4 4
®+|
© x-2.76, or
-1.26
®-
Check.
2(2.76)"-3(2.76)i7
2(-1.26)'-
-3(-1.26)i7
2(7.62)-8.28i7
2(1.59)+3.78i7
16.24-8.28-7
3.18+3.78J7
6.98i7
6.96 i 7
Am. x
-2.76, or -1.1
L
;,;,iz*d=vGoogI
146 JUNIOR HrGH SCHOOL MATHEMATICS (XII,! 73
§ 73. Summary. To solve a quadratic equation by
completing the square :
(1) Write the equation in the form x^+bx^c.
(2) Complete the square by adding the square of half
the coefficient of x to each member.
(3) Take the square root of each member of this equation;
write both the plus and the minus sign in front of the member
containing the number term only.
(4) Solve each of the first degree equations thus formed.
EXERCISES
Find two roots for each of the following equations by
completing the square, and check each root. (For the
equations in Exs. 1-16 the roots are rational.)
1. ^+2* =8 9. 2x*+3s=2
2. y»+4y=12 10. 3a*+5o=-2
S. x*-2x=\b 1L 2y*-y=2&
4. m*-6m=-8 12. 3&H-7b+2=0
5. &"+»-10 1 IS. 4a;>-4a;=-l
6. a s +a-12=0 U. 6p*=p+2
7. x*+$x-2=0 16. 2s* = 5;f-2
8- y*-iy = $ 18. 3m J +l=-4m.
(For the equations in Exs. 17-30 the roots are irrational.
Find each root to three figures.)
17. xM-4r=8 24. 3x»+4z=l
18. m ! +6m=3 28. 5y i -2y = 2
1». j,*-2j/=4 28. 2m*+3m = l
20. **-8a;=-14 27. 2a s = 5-5a
21. p ! = 2+4p 28. 4(^-8^+1=0
22. 5-6o = a* 29. R(R+1) = 3
28. l-x=x* 30. 5 = 2a(a+3)
81. Solve ax*+bx+c=Q for x by completing the square.
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XI I. 1 74] QUADRATIC EQUATIONS 147
§ 74. Solution of Quadratic Equations by the Formula.
The complete quadratic equation is ax*+bx+c=Q.
The solution of the equation ax i +bx+c = Q by complet-
ing the square is as follows :
® ax*+bx+c=Q
® ax*+bx=-c
a 4a 1 4a 1
®
®
®+(£)"
®=(Combining frac-
tious in sec-
ond member)
®V
x= ^ ± VV-4ae @ _±.
® r ^ ~b±Vb 1 -4ac ®= (Combining frac-
2a tions)
Hence the solutions of the equation
ax s +bx+c=0,
x=z b±V^E^ or - & ~ V ^ :
2a ' 2a
-b±Vb*-4ac
J, S ,:z K i:vC00gIe
®-
148 JUNIOR HIGH SCHOOL MATHEMATICS [XII, J74
Example 1. Solve 2x a +7a;+6=0 by the formula, and
check the roots.
Solution. ® 2x*+7s+6 =
In equation ®, a=+2, b=+7, and c=+6.
Substituting these values for a, b, and c in the formula,
-b±Vb t -4ac
X Ya
® a- -7*^49=48
® *-^±l,or.
4
® as- -If or -2 ®s
Check.
2(-i)*+7(-f)+6i0 2(-2)*+7(-2)+6=L0
|-^l+6=L0 8-14+6=0
0-0
Ana. x= — li, or— 2.
Example 2. Solve 2x ! +3as— 2 by the formula, and
check the roots,
Solution. ® 2a^+3x = 2
® 2x*+3a:-2 = ®-2
In equation ©, a— +2, fe = +3, and c= —2.
Substituting these values for a, b, and c in the formula,
— b±Vb*—4ac
®
©
®
2a
-3±V9+16
4
,=2+5 ,^3^5
4 4
®-
'i, »r -2
®-
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XII, 1 75) QUADRATIC EQUATIONS 149
Check. 2(*)*+3(±)-2 2(-2)*+3(-2)i.2
i+l-2 8-6«=2
Ana. x = \, or —2.
Example 3. Solve 5a?=2+2;r by the formula, and
check the roots.
Solution. (J) &r s =2+2a:
® 5^-21-2=0 ©-2-fcr
In equation ©, a=+5, 6= —2, and c= —2.
Substituting these values for a, b, and c in the formula,
„_— &±vV-4ac
2a
®
. 2±vT+40
* 10
®
2+6.63 2-6.63 ®«
10 ' 10
®
1-0.863, or -0.463 ®m
Check.
5(.863)",
l2+2(.863)
5(-.463)'i2+2(-.463)
5(.745),
12+1.726
5(.214)i2-.926
3.725.
13.726
1.070il.074
Am. 1-0.863, or -0.463.
§ 75. Summary. To solve a quadratic equation by the
formula:
(1) Write it in the form az?+bx+c = 0.
(2) Substitute the values for a, b, and c in the formula,
— b±Vb*-4dc
x = — .... — .
2a
(3) Simplify the values of x thus obtained.
is, Google
18. n--=li
150 JUNIOR HIGH SCHOOL MATHEMATICS [XII, j 75
EXERCISES
A. Find two roots for each of the following equations,
using the formula. Check each root.
1. x*+4s+3=0 „ usb+^J .
2. 2x 2 +x-28=0 R
9. x*-6x=-8 15 1+ g2 M
4. 6x*-x = 2 A
6. 2*»-ll*+2-0 w fa _ 8 _21
6. x*+6x=-9 o
T. y=-8!/+14 = 1? 3j/-10 _ y+120
8. a ! +6a=5 14 y
*. 3^-4^=1
10. 2m ! +5m = 5
11. 3x*-5=5x-s'+2 m+3 = -L-
IS. 6x*-7x-5 = 3aH-5 m-1
2 2 m 2x+l
B. Find (wo roots for each of the following equations :
(a) By factoring. (6) By completing the square, (c) By
the formula.
1. a*-7o+12=0 5. 3x 2 +10x+3 =
2. »H-3tf-10 6. 5^-6^=11
3. 6 2 +9&+14 = 7. 2x*+3x+l =
4. m a -6m=-9 8. 9a , -6o=-l
PROBLEMS
1. Find two numbers whose difference is 6 and whose
product is 135.
2. Find two numbers whose sum is 24 and whose
product is 95.
3. Find two consecutive integers whose product is 72.
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XII. 1 75] QUADRATIC EQUATIONS 151
1. One side of a right triangle is 7" longer than the
other side. The hypotenuse is 13". Find the two sides.
5. One side of a right triangle is 2 cm. longer than the
other side. The hypotenuse is \^34 cm. Find the two
sides.
6. One side of a right triangle is 5" less than the
hypotenuse. The other side is hVz". Find the hypote-
nuse.
7. A rectangle is 18" by 24". If the diagonal is to be
increased 4" in length, what will be the length of the
rectangle, the width remaining the same?
8. The area of a square in square feet and its perimeter
in inches are expressed by the same number. Find one side,
9. The area of a square in square centimeters and its
perimeter in centimeters are expressed by the same num-
ber. Find one side.
10. Around a rectangular flower bed, 12' by 16', there
is a border of turf which is everywhere of equal width, and
whose area is 4 times the area of the bed. How wide is
the turf to the nearest tenth of a foot?
11 « = s ( g - 1 ); a
- is the formula for finding the total num-
ber of different connections possible in a telephone ex-
change, given the number of subscribers s. If the total
number of connections in a certain exchange is 4851, find
the number of subscribers.
12. A photograph 8" by 10" is enlarged to twice its
original area. What are the dimensions of the enlarged
photograph to the nearest hundredth of an inch ?
13. A photograph 6 cm. by 8 cm. is enlarged to 2J
times its original area. What are the dimensions of the
enlarged photograph to the nearest millimeter?
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CHAPTER XIII
RATIO, PROPORTION, AND VARIATION
S 76. Ratio. A ratio is the expression of the quotient of
two quantities. The quantities must be of the same kind.
The ratio of a 12" line to a 6" line is if-, or 2 (an integer).
The ratio of a 5-lb. weight to a 12-lb. weight is ■& (a
common fraction). The ratio of the weight of a cubic foot
of ice (57.5 lb.) to the weight of a cubic foot of water
(62.5 lb.) is ~, or .92 (a decimal fraction). The ratio
62.5
of the diagonal of a square to one side is V2 (an irrational
number).
EXERCISES
1. Draw two lines of different lengths.
(a) Measure them to the nearest hundredth of an inch
and find the ratio of the first to the second.
(b) Measure them with a metric rule to the nearest
millimeter, and find the ratio of the first to the second.
(c) Test the two ratios obtained in (a) and (6) for equal-
ity.
2. Two weights are 12£ lb. and 7\ lb.
(a) Find the ratio of the second to the first.
(b) Express the weights in ounces, and find the ratio
of the second to the first.
(c) Test the two ratios obtained in (a) and (6) for
equality.
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XIII, S 76] RATIO, PROPORTION, VARIATION 163
3. Two pieces of iron weigh 7 lb. and 10 lb.
(a) Find the ratio of the first to the second.
(6) Express the weights in kilograms. (See page 46.)
Find the ratio of the first to the second,
(c) Test the two ratios for equality.
4. The areas of two rectangles are 45 sq. in. and 12 sq.
in. What is the ratio of the first to the second? Of the
second to the first ?
6. The volumes of two balls are 250 cu. cm. and 450
cu. cm. What is the ratio of the first to the second?
«. Two balls weigh 16J oz. and 7J oz. What is the
ratio of the first to the second ?
7. One line is 1.4 meters and another is 40 centimeters.
What is the ratio of the first to the second?
8. Two weights are 4£ lb. and 21 oz. Find the ratio
of the second to the first.
9. The area of a lot of land is 4000 sq. ft. The area
of the plan of this lot is 4 sq. in. What is the ratio of
the area of the lot to the area of the plan?
10. A square room is 40 ft. on each side. The plan of
the room, drawn to scale, is 2 in. on each side.
(a) Find the area of the room.
(6) Find the area of the plan.
(c) What is the ratio of the area of the room to the area
of the plan?
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154 JUNIOR HIGH SCHOOL MATHEMATICS [XIII, 577
§ 77. Proportion. A proportion expresses the equality
of two ratios. Hence a proportion is an equation each
of whose members is a ratio. Usually the ratios in a pro-
portion are common fractions. Exercises 35-60, pages 53
and 54, and Exercises 11-20, page 141, are proportions.
The first and last terms of a proportion are called the
extremes; the second and third terms are called the
means.
For example, in the proportion $=i$, 2 and 25 are the
extremes, and 5 and 10 are the means.
In the proportion 7 =-, these means are the same
o c
number. Then that number, 6, is called the mean pro-
portional between a and c.
EXERCISES
Select from Exercises 1-10 those statements that are
true proportions.
1. JiA «• tn
»• iift '• i-i
3- *A*
«■ &AA
2.1 5
5~ 12.5
JLi 5
2.4 4
, 3.6 , 1.8
•• *i* 10. T - T
11. Select from Exercises 1-10 the proportions that
contain a mean proportional and name the mean propor-
tional in each case.
12. Solve the proportions 11-20, page 141.
18. Separate 84 into two parts which are in the ratio
3 toll. (Let n= the larger part; form an equation at the two
ratios.)
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XIII, [77] RATIO, PROPORTION, VARIATION 155
14. Separate 240 into two parts which are in the ratio
of 7 to 5.
16. Find the mean proportional between 18.4 and 30.5.
16. Brass is an alloy consisting of two parts copper and
one part zinc. How many ounces of copper and zinc
are there in 2 pounds of brass ?
17. Gun metal consists of nine parts copper and one
part zinc. How many ounces of each are there in 22
ounces of gun metal?
18. The total area of land on the earth is to the total
area of water as 7 is to 18. If the total surface of the
earth is 197,000,000 square miles, find the number of
square miles of land and of water (to three figures).
19. A study of family budgets shows that on a salary
of $900 per year, $405 is required for food. At the same
rate how much would be required for food when the
salary is $1000 per year?
20. On a salary of $850 per year, $168 is required for
rent. At the same rate how much would be required for
rent when the salary is $950 per year (to the nearest
dollar)?
21. When one pound of substitutes must be purchased
with every four pounds of wheat flour, how many pounds
of substitutes must be purchased with ten pounds of wheat
flour?
22. When two pounds of substitutes must be purchased
with every three pounds of rye flour, how many pounds of
substitutes must be purchased with ten pounds of rye
flour (to the nearest pound) ?
23. A 1-lb. loaf of bread contains f as much nutrition
as 1 pound of rice. When a 1-lb. loaf of bread costs 10 cents
and 1 pound of rice costs 12 cents, which is the cheaper?
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156 JUNIOR HIGH SCHOOL MATHEMATICS [XIII, 6 78
§78. Variables in Arithmetic.
Example 1. When sugar is selling at 9 cents a pound,
2 pounds will cost 18 cents, 3 pounds will cost 27 cents,
4 pounds will cost 36 cents, etc. The total cost de-
pends upon the number of pounds purchased. Aa the
number of pounds («) increases, the cost (c) increases in
the same ratio. In this illustration the total cost and the
number of pounds are variables, and the price per pound
(p) is a constant. The total cost is said to vary directly as
the number of pounds. The relation may be expressed
by the formula
cnp.
Example 2. A sum of $200 is loaned, the annual rate
of interest being 6%. The interest for 1 month is $1,
for 2 months it is $2, etc. The interest that accumu-
lates depends upon the number of months. As the time
(() increases, the interest (i) increases in the same ratio.
In this illustration the interest and the time are variables;
the principal and the rate are constants. The interest
varies directly as the time. The relation is expressed by
the formula
i=prt.
Dependence of one quantity upon another is of frequent
occurrence. For example, the amount a man earns de-
pends upon the number of days that he works ; an agent's
commission depends upon the amount of goods that he
sells; the amount of the electricity used determines the
amount of the bill ; etc.
In the statement of a problem a letter that may represent
different numbers is called a variable, whereas a constant
represents only one number.
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XIII, ! 78] RATIO, PROPORTION, VARIATION 157
Illustrate direct variation from each of the topics in
arithmetic named in Exs. 1-8. Name the variables and
the constant. Write a formula using whatever letters
you think most suitable.
1. A mechanic is paid 60 cents per hour. Total
wages vary ••■ (?).
Solution. The mechanic's total wages vary directly
as the number of hours that he works. His total wages
and the time are variables, the wage per hour is the con-
stant. The relation may be expressed by a formula,
W=tw,
where W = total wages, t = number of hours, and w = wage
per hour.
2. Coal is selling at $9.50 per ton. Total cost varies
3. The profit is 25% of the cost. Profit varies ■■■(?).
4. The profit is 20% of the selling price. Profit
varies --■ (?).
6. An agent gete a commission of 5%. Commission
varies ■■- (?).
6. The tax rate is $17.50 per thousand valuation.
Taxes vary •■• (?).
7. The cost of electricity is 10 cents per kilowatt hour.
Total cost varies •■• (?).
8. The Third Liberty Bond pays interest at 4J%.
Total income varies ■■■ (?).
9. A certain quantity of flour costs $4.75. What will
be the cost of m times as much ?
10. Knowing how much a plumber will earn in 8 hours,
find out how much he will earn in n hours.
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158 JUNIOR HIGH SCHOOL MATHEMATICS [XIII, S78
11. In a certain city the taxes on a given piece of prop-
erty are $48. At the same rate, what will be the taxes
on another piece of property valued at b times the first ?
12. During a certain month your gas bill is $4.50.
The next month you use ^ as many cubic feet of gas.
4
What is your gas bill ?
13. Explain the statement: In making bread the
number of one-pound loaves varies directly as the number
of pounds of flour used.
14. Explain the statement : For a given state the
number of people per square mile varies directly as the
total population of the state.
§ 79. Variables in Geometry.
Exercises 1-4 illustrate direct variation. Name the
variables and the constant. Write the formula.
1. The area of a rectangle having a 16" base va-
ries ■■■(?).
Solution. The area varies directly as the height.
The area and the height are variables, the length is a
constant.
FoHMTJLA. A = 16ft.
2. The area of a triangle having a 10" altitude varies
••■(?).
3. The perimeter of a square varies --• (?).
4. The volume of a cylinder having a base containing
50 sq. in. varies ■•■ (?).
6. The area of a rectangle that has a given height
varies •■■ (?).
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XIII, § 79] RATIO, PROPORTION. VARIATION 169
6. Explain the statement: The circumference of a
circle varies directly with the diameter.
(a) Write the formula.
(6) Name the constant.
7. The diameter of one circle is three times that of
another.
(a) What is the ratio of their circumferences?
(b) What is the circumference of the larger, if that of
the smaller is 15"?
8. Explain the statement : The area of a rectangle
varies jointly as its base and its height.
Solution. The formula is : A = bk.
As b increases, A increases in the same ratio.
As h increases, A increases in the same ratio.
That is, A varies directly as b , also A varies directly as h.
Hence A varies jointly with 6 and k.
9. A rectangle has an area of 40 sq. in.
(o) What would be the area of a rectangle twice as long ?
(6) What would be the area of a rectangle twice as high ?
(c) What would be the area of a rectangle both twice
as long and twice as high ?
10. A rectangular field has an area of 1000 sq. ft. What
is the area of a field three times as long and twice as wide?
11. Explain the statement: The volume of a block
varies jointly with its length, its width, and its height.
12. A certain box contains 200 cu. in.
(a) How much would a box contain that is three times
as long?
(6) How much would a box contain that is half as high ?
(c) How much would a box contain that is twice as
wide?
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160 JUNIOR HIGH SCHOOL MATHEMATICS [XIII, 1 79
(d) How much would a box contain that is three times
as long, one half as high, and twice as wide?
13. Explain the statement : The area of a circle varies
directly as the square of its radius.
Solution. The formula is : A=irr t .
As r* increases, A increases at the same ratio ; that is,
A varies directly as H. The constant ratio is *■.
14. The radius of one circle is twice that of another
circle.
(a) The square of the radius of the first circle is how
many times the square of the radius of the second circle?
(6) What is the ratio of the areas ?
(c) The area of the smaller circle is 25 sq. in. What
is the area of the larger circle ?
IK. Two circles have radii of 2" and 6". What is the
ratio of their areas?
Solution. ® ^--
n* 1* 1
r** 3* 9
(T) Squared
A, 1
A,~9
© Substitution
Arts. £.
Two circles have radii of 5'
and 3". What is the
IS.
ratio of their areas?
17. The area of a circle varies with the square of its
diameter. The diameter of one circle is 4 times that of
another. What is the ratio of their areas?
18. The diameter of one circle is 2$ times that of
another. What is the ratio of their areas ?
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XIII, |801 RATIO, PROPORTION, VARIATION
§ 80. Variables in Science.
1. The distance that a train travels varies directly as
the time, given its average rate per hour.
(a) Write the formula. Name the variables and the
constant.
(b) The train travels 145 miles in a certain number of
hours. How far will it go in three times as many hours?
2. The distance that a body travels varies jointly
as the rate per hour and the number of hours. An air-
plane goes 140 miles in a certain number of hours.
(a) How far will it go in twice as many hours?
(b) If the rate is doubled, how far will it go in the given
number of hours ?
(c) How far will it go at twice the first rate and in three
times as many hours?
3. The weight of water varies directly as the volume.
A certain volume of water weighs 450 lb. What is the
weight of three times as much ?
4. The specific gravity of a substance varies directly
as the density. A cubic foot of steel weighs 490 lb. and a
cubic foot of pine weighs 25 lb. What is the ratio of their
specific gravities ?
D. The distance passed over by a falling body varies
directly as the square of the time. What is the ratio of
the distance passed over if the time is doubled ?
6. The current of electricity (in amperes) varies di-
rectly as the electromotive force (in volts). Doubling
the voltage will have what effect upon the current ?
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162 JUNIOR HIGH SCHOOL MATHEMATICS [XIII, 1 81
§ 81. Inverse Variation.
Example. An airplane travels 270 miles in 3 hours,
hence its average rate is 90 miles an hour. On another
trip it travels 270 miles in 5 hours, hence its average rate
is 54 mileB an hour. The greater the time required the
slower is the rate. The rate is said to vary inversely as
the time. The formula is : d = rt.
Letting d be a constant and giving larger and larger
values to (, you note that r becomes smaller and smaller.
EXERCISES
1. With a certain sum of money the number of articles
of the same kind that can be purchased varies inversely
as the price per article. Explain.
2. With a certain sum of money I can purchase 50
pounds of flour. If the price is doubled, how many pounds
can I purchase?
3. If a rectangle has a given area, its length varies
inversely as its width. Explain.
4. A rectangle 20' wide has a certain area. A second
rectangle has the same area but is four times as long ; how
wide is it?
6. The law of the lever states that the moment (weight
X weight arm) on one side of the fulcrum equals the
moment (forcexforce arm) on the other side of the ful-
crum. For a given moment, the force varies inversely as
its distance from the fulcrum. To raise a certain weight,
a 30-lb. force is required at a certain distance from the
fulcrum. How large a force would need to be applied at
a distance three times as far from the fulcrum?
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XIII, J 82] RATIO, PROPORTION, VARIATION 163
t. The volume of gas varies inversely as the pressure
upon it. The pressure upon a certain quantity of gas is
40 lb. per square inch. What will be the effect upon the
volume if the pressure is reduced to 20 lb. per square inch?
7. The intensity of the current of electricity in a cir-
cuit varies directly as the electromotive force and inversely
as the resistance. The relation is expressed by the for-
mula: /=p-
(a) What is the effect upon the intensity of doubling
the electromotive force?
(ft) What is the effect of doubling the resistance?
(c) The electromotive force of a current is increased
threefold and the resistance doubled. If the intensity
was 12 amperes before the change, what is it after the
change?
8. The intensity of light on an object varies inversely
as the square of the distance between the object and the
source of light. An object is moved from a distance of
1 foot from a lamp to a distance of 3 feet. What will be
the ratio of the intensity of light at the greater distance
to the intensity at the nearer distance ?
§ 82. Functions. If two variables are so related that
when a value of one is given, a corresponding value of
the other is determined, the second variable is called a
function of the first. In the formula c=ird, c is & function
of d. In the formula A =irr f , A is & function of r. In the
formula p = rb, p is & function of b, if r is constant. The
total cost of a quantity of sugar is a. function of the number
of pounds, if the price per pound is constant. In the equa-
tion y = 3x+2, y is a function of x. In the equation
i BV Google
164 JUNIOR HIGH SCHOOL MATHEMATICS [XIII, i 82
y may be expressed as a function of x, giving
Or x may be expressed as a function of y,
3x+2j/~4, y may be expressed as a function of x, giving
y 2
giving x =~r^- In the statement y = ■ ■ x , y is called
the dependent variable and x is called the independent
variable.
In the statement x = ~ ? , a; is the dependent variable
and y is the independent variable
Example. Given the equation 2x— 3y = 5, express x
as a function of y.
Solution. ® 2x— 3y=5
® 2x«5+3y ®+3y
® *-5^tt ®+ 2
An,. .-***.
EXERCISES
In Exs. 1-6, express x as a function of y.
1. x+y = 7 4. 2x-3y = 5
2. 2x-y = 8 5. 4x+y=7
3. 3x+4j/-5 6. 3x-2y =
In Exs. 7-12, express y as a function of x.
7. *+y=12 10. 2x-3y=0
8. 2x+y=7 11. 4x-5y = 7
9. 3j/-4x=10 12. 3x-7y=15
In Exs. 13-18, make x the dependent variable.
18. x 2 +y i =18 16. xy = 15
Am. x=±\/l8=?. 17. 3xM-2j/*=18
15. Sx'-y'-lS
18. xy+^-12
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XIII, S 83] RATIO, PROPORTION, VARIATION 165
§ 83. Graphs of linear Functions of Two Variables.
On pages 105-107 you plotted graphs of linear equations
containing two variables. In each case, before getting
pairs of values, you expressed one variable as a function
of the other. You noted that, when the equations were
of the first degree, the graph was a straight line, hence
only two pairs of values were essential for determining
its direction. You tabulated three pairs of values, the
purpose of the third pair being to check the accuracy of
the other two.
In Exercises 1-3 which follow, draw the graphs on the
same sheet of squared paper. Make a comparison of these
three graphs, noting the effect upon the graphs when cer-
tain constants are changed.
1. Given the functions y = x, y = 2x, y = 4x, and
y=—4z:
(a) Find three pairs of values for each function.
(b) Plot the graphs of the four functions on the same
axes.
(c) Through what point do all the graphs pass ?
(d) What change in the graph is produced by changing
the coefficient of x?
2. Given the functions y = x, y = x+2, y=x+i, and
(a) Find three pairs of values for each function.
(6) Plot the graphs of the four functions on the same
axes.
(c) What change in the graph is produced by changing
the constant ?
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166 JUNIOR HIGH SCHOOL MATHEMATICS [XIII, 5 83
3. Given the functions 2x+3y=5, 3x+2y = 5, and
(a) Find three pairs of values for each function.
(b) Plot the graphs of the three functions on the same
axes.
(c) What name can you give to these three equations ?
(See page 106, Ex. 1.)
4. The graph of the function y=kx, where fc is a con-
stant, passes through what point? [See Ex. 1 (c).]
8. The graphs of y = mx+b and y = mx+c are how
located, if m has the same value in each? [See Ex. 2 (c).[
§ 84. Graphs of Quadratic Functions of Two Variables.
Quadratic equations are equations of the second degree.
You will see that the graph of an equation of the second
degree is a curve, not a straight line ; hence you will need
to tabulate many pairs of values before plotting a quad-
ratic function.
Example 1.
Solution.
Plot the graph of the function y=x*.
*
-5
-4
-3
-2
-1
+1
+2
+3
+4
+5
y
25
16
9
4
1
1
4
9
16
26
The points are located in Fig. 28 and a smooth curve
is drawn through them. Figure 28 is the graph for the
squares of numbers from 1 to 10. From the graph, read
the squares of 6, -7, 4$, -5£, 8£.
(Suggestion. Leti-6, -7,4i,etc. Then locate, on the curve,
the to rre spoil ding value of y.)
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XIII, | 841 RATIO, PROPORTION, VARIATION 167
Example 2. Plot the graph of the function x'-f-y* = 25.
Solution. Express y as a function of x.
® ^+^=25
® V s -25-i» ®-J»
® y=±V25=x* ©V
In tabulating values for x and y, values of x are avoided
that render the quantity under the radical sign negative.
X
-5
-4
-3
-2
-1
+1
+2
+3
+4
+5
V
±3
±4
±4.5
±4.9
±6
±4.9
±4.5
±4
±3
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168 JUNIOR HIGH SCHOOL MATHEMATICS [XIII. 1 84
Figure 29 is the graph of a?+s*=*25, a circle.
Pio. 20.
EXERCISES
Plot the graphs of each of the following quadratic func-
tions. First express one variable as a function of the other,
then tabulate pairs of values.
1. *'+y e =16. (A circle)
2. 4r*+j/ s =4.
3. ^ = 81+1.
4. xy = 8.
6. 4x*-j^=4.
(An ellipse)
(A parabola)
(An hyperbola)
(An hyperbola)
i^v Google
CHAPTER XIV
FRACTIONS AND EQUATIONS
§ 86. Fractions. In algebra a fraction is an indicated
division, in which the dividend and the divisor are alge-
braic expressions. The dividend is called the numerator
of the fraction, and the divisor the denominator of the
fraction. The two are called the terms of the fraction.
The numerator of a fraction may have any value what-
ever, while the denominator may have any value excepting
zero.
The principles of fractions used in arithmetic apply
also in algebra.
One important principle is :
The numerator and the denominator of a fraction may be
multiplied by, or divided by, the same number without chang-
ing the value of the fraction. The resulting fraction is said
to be equivalent to the given fraction. For example :
(a\ m +n =j wtx+na: Both terms of the fraction are
x t x* multiplied by x.
(K\ 5s+5y _ a:+y Both terms of the fraction are
10a 2a divided by 5.
(c) j;1 ~y' = ( x +y)( :c ~y) = x +y Both terms of the
x % —xy x(x—y) x fraction are di-
vided by *— y.
§ 86. Addition and Subtraction of Fractions. As in
arithmetic, fractions can be added, or subtracted, only
when their denominators are the same, hence the first step
is to change all fractions into equivalent fractions having
the same denominator.
100
i BV Google
170 JUNIOR HIGH SCHOOL MATHEMATICS [XIV, f 86
Example 1. Combine- — h- — — into a single frac-
4a 6a 3a
tion.
Solution. l+A-l „_?_ + i0 __i_n.
4a 6a 3a 12a 12a 12a 12a
_ 2_ 4_3s*_2s_4 _ 3x*-2i-
fl? X s x* I s X s *"
EXERCISES
In each of the following exercises, combine the fractions
into a single fraction. Reduce the result, when possible.
i+i+i
a b c
1.
3,1 1
8 2 3
2.
3n , a 1
5 3 10
3.
1-2,1-1
5 2
1
3 + ^ 5_
y 5y lOy
5.
If I 2
Solution. —
1-1+1
f ab o'
o+6
4 o+6 (o+b)(a-6) (a+6)(o-i?)
(a+i.)-(a-t) _ 26
(a+6)(o-t) (o+5)(a-t)
An. 2il
■ (a+t)(a-6)
KigilizBd by G00gle
XIV. 1 86] FRACTIONS AND EQUATIONS 171
Note. It is advisable to keep the denominators in eaeh
step io the form of factors.
1
+~4 16.
»+l lf-1 ' *v* + l)
_? ?_ ,, _*
x-l x+1
« j. 2_
18.
' (o+5)(o-6) T a+i> ' (a+l)(a-l) (o+l)
16 q+fc q-E> 19 v+l g-1
« 6 ' »+2 s-2
if 1 -* lf+2y
Solution.
„t_4 ^ +2 j, („_ 2 )(„+2) y(y+2)
c » »- 2
»(V-2)<v+2) »to-2)(»+2)
^ y-t/+2 _ 2
- »fe-2)(s+2) »(»-2)(»+2)
2z
x»-l
1
1+1
3
2m-4
5
6m—
12
2
2a
<i-l
a'-l
x+1
1
*•
x-l
Atis.
»(»-2)(»+2)
i. -2—
"o+5 + o^6
o— 6 i
-i— i
•+6
i 3
2 x
1 x-2
r+3 xM-i-6
,. j(±L
(D-D'
»"-l
, m
2 ..
m , -m-12 m+3
3, g ,i EE d by Google
172 JUNIOR HIGH SCHOOL MATHEMATICS (XIV, J87
§ 87. Multiplication of Fractions. The product of two
or more fractions in arithmetic is found by finding the
product of the numerators for a new numerator and the
product of the denominators for a new denominator. It is
customary to reduce the resulting fraction to its simplest
form by applying the principle of fractions stated on
page 169.
In algebra the procedure is the same and it is also cus-
tomary to reduce the resulting fraction to its simplest form.
To find the product of two or more fractions :
(1) Find the product of the numerators for the numerator
of the result.
(2) Find the product of the denominators for the denomina-
tor of the result.
(3) Reduce the resulting fraction to its simplest form by
applying the principle of fractions stated on page 169.
' 3i>* 20
K 26~4b
!. Find the product : t^MyX+JL.
x?-y* I*
Note. When the division of both terms of a fraction is
shown by crossing out the common factors, as in Exs. 1 and
2, the process is called cancellation.
jigiiized by Google
XIV, j 87) FRACTIONS AND EQUATIONS 173
EXERCISES
In each of the following exercises find the product, and
reduce it.
15 8 2 ' a'+ab a-b.
3b 25xy x?-Zx ^-1-6
, 3x* 8a^_ 17 q*-4 „ q'+3o
' 2j' 9 ' a<+5a+6 a+1
6. =X=X-£ 18. ^ 't'Yo.y
n p m x+2y x t —Sxy+2y 1
6. ^X*X^ 19. ti2+l x ^±^-
y* 7? ay a 1 — a a 2 +a— 2
12a* 10a6 y3bc -- n*+n .. na;— n ^. g+1
56* 9a 1 4a* ' n*~n nx+n x—1
■m a'+4o v a*+3a
S1 * 0+3 X a+4
(l+y) ! x ! -xp
a y g+ft ya-fr a- 4as*-l „ j'+j
o+& a o+6 ' 2s s +3x+l 6xH-x-2
aj*(:e+2) „ s*+3x -. 1-x' .. 3^+3:
z x*-4 ' 3x*-2x-l x+1
n*-4 y q'-ie 2 - q*-16 .. a*+2o-15
tf(a-4) a*-2o ' 12-a-a* 25-a !
9tf'-4 6m* M 6a?-13x+6 ., 7+34x-5x*
4m" 3y-2 ' 15tf-7i-2 2x*-17*-21
Jigilized by GoOgle
174 JUNIOR HIGH SCHOOL MATHEMATICS [XIV, (88
§ 88. Division of Fractions. To find the quotient of
one fraction divided by another fraction :
(1) Invert Ike fraction which is the divisor.
(2) Find the product of the two fractions after the divisor
has been inverted.
Example 1. Find the quotient : ^+^-
Solution.
5o . 2a' m 5$ lP m 5b
96 T 3i ! it 60
6a
. Find the quotient :
o-l
a*-l
o'+3o+2
a ! +2a
Solution.
o-l . a"-l_
o-l .-e
+3a+2
0(4*2)
+1)(9^1)
'+2a
o*+3o+2 ' a a +2a a
(ojr-SKa+lA
»'-l
a
(a+1)
Arw
a
(o+D !
IQafy.&r*
o " a*
-2ap+fc'
•oi
BXKRCISES
In each of the following exercises find the quotient, and
reduce it.
o'+o» .
o'-o 1 ' o !
J-6i+9
i+2 ' i"+i-2
ro+n . ro'+wm
w'-mn * m*— n*
f-18o+80 !>'— 15t>+56
6'-5»-50 o=-7b
i^W+3,-35)
;, S ,:z K i:vC00gIe
XIV, 1 89] FRACTIONS AND EQUATIONS 175
§ 89. Complex Fractions. A complex fraction is a
fraction having a fraction in its numerator, or in its de-
nominator, or in both its numerator and denominator.
The complex fraction "may be written -
The complex fraction 2 may be written
KH°-;>
All complex fractions can be simplified by treating them
as exercises in division of fractions. However, it is often
shorter to multiply each term of the complex fraction at
once by that expression which is the least common multiple
of the denominators of the two terms.
Example 1
Simplify :
a
a
Solution.
a \ aj a 1 -
4-1
-1
Both terms of the complex fraction are multiplied by a.
i BV Google
176 JUNIOR HIGH SCHOOL MATHEMATICS [XIV, SS9
Example 2. Simplify: !+£
Solution.
i^ +1 _ (1+3;) fe +1 ) , x + l + , . 1+2,
' 1 j.S\ 1+x+x 1 1+x+x*
Both terms of the fraction are multiplied by 1+x.
Simplify each of the following complex fractions.
l H 8 £ 10. L_2
I" So
l+S I+1+-2-T
O X— 1
°* a x-1
Z*
5*
1-4+5
JigrzBd^y GoOgle
XIV, S 90] FRACTIONS AND EQUATIONS 177
§ 90. Fractional Linear Equations. In every chapter
in this course you have been solving equations, many of
which contained fractions.
In solving a fractional equation the first step is to get
rid of the fractions.
To get rid of the fractions :
(1) Multiply each member of (he equation by the least
common multiple (1. cm.) of the denominators.
(2) Divide the numerator and denominator of every frac-
tion by all factors common to both. This process is called
cancellation.
Example 1. A surveyor drives two stakes in the
ground 220 ft. apart. He wishes to drive a third stake so
that its distance from the farther stake divided by its dis-
tance from the nearer stake is 1.2. Find its distance from
each stake.
Solution. Let d=the distance of the third stake from
the nearer stake, then
220— d = its distance from the farther stake.
The equation is :
® 230^. 12
a
© &^1-1M ®Xd
© 220-2.2d ®+d
© 100-d ©+2.2
© 220-d-W
Chuck. W8-1-2
Am. 120 ft. and 100 ft.
KigilizBd by G00gle
178 JUNIOR HIGH SCHOOL MATHEMATICS (XIV, 190
Example 2. The denominator of a fraction exceeds
its numerator by 2. If I is added to both terms of the
fraction, the resulting fraction will be equal to f. Find
the fraction.
Solution. Let n = the numerator of the fraction, then
n+2=the denominator, and
The equation is :
w n+3 3
The 1. c. m. of the denominators is 3(n+3).
„ 3i!t!^±l)_3(n+3))<2 @X 3 (n+s)
sH^s 3
® 3(n+l)-2(»+3) ®- (Cancellation)
© 3n+3 = 2n+6 ©■(Parentheses re-
solving®, »-3 moved)
n+2-5
Chm - I+i-I Am - *•
Example 3. Solve the equation — ——=-■
Solution. © — ~^=|
The 1. c. m. of the denominators is 6x.
@ mx-\) ftK«-2) _fe 0x6x
® &(x-\)-2(x-2) = Zx ©= (Cancellation)
® 6s-6-2*+4 = 3s ® = (Parentheses re-
moved)
Solving ®, x-2
3, g ,i EE d by Google
XIV, { 90] FRACTIONS AND EQUATIONS 179
Check. ^-^il
2 3X2 2
i-O-J Ana. x-2.
Note. Equation ® may be obtained directly from © by
dividing the 1. c. m. of the denominators by each denominator
and multiplying the corresponding numerator by the quotient.
EXERCISES
Solve each of the following equations, and check each
answer.
1.
5-1
».
2-1
3.
i— »
5a
1.
!2-i-o
a
6.
1.8-|i
2j
6.
12-i-0
a
7.
5+1-8
V
S.
9 -3
9.
2.5 ,
l-»
L0.
S^T-«
11.
1 1
m+6 15
12.
1 _ 1
2i+l i+7
Il-
cm.: (2i+l)(«+7)]
ls.
5 7
o-l o+l
It.
7-6
1+3 1-2
15.
3 7.5
2y-2 2y-l
16.
-15— 2
3o+4
17.
M-*
18.
i+j-.i
3l 2x
19. i,i+_L+i.
K 30 40 15
B 4.5 T 1.5 T 13.5
JigilizBd by GoOgle
180 JUNIOR HIGH SCHOOL MATHEMATICS [XIV, !90
21. -51.
-7
4*
Solution. First multiply the numerator and denomi-
nator of the fraction by 6. (See page 175.)
®
it.
.7
®
6X3J_
6X4*~
■7
®-
®
21
■ 7
®-
Solving ®,
x =
J
Ana. x-\.
22. A = 12
^+3
fira
28.
i 8
23. iM_4
5-1
1»P
2
24. ?fc=44
74
*!-2
2 B . i-ia
29.
| 2
3* I'l
t+1
26. -L-tt
3
2| 12
i +1
27. =—=4
30.
3 °- i_i
2+^
[In Ex. 27, multiply both terms of the fraction by 4;
?»+i-4, etc.]
6; . 2 16
1+2 »-2"l"-4
JigilizBd by GoOgk
XIV, 5 90] FRACTIONS AND EQUATIONS 181
Solution. Before getting the 1. c. m. of the denomi-
nators, factor the third denominator. Then the 1. c. m.
is (*+2)(i-2).
w x+2 1-2 x?-i
» 6te*2)(x-2) . 2(x+2)(i^2) _16(i>|-i)(x^r)
®X(*+2)(x-2)
® 6(i-2)+2(x+2)-16 ®«(Cancella-
Solving®, 1 = 3 tion)
Am. as = 3.
(See Note following Example 3, page 179.)
31.
2i+3 + 2i-3"4i"-9
si.
2x+l 2x-l_ 8
2x-l 2i+l 4x"-l
34.
3-ro»_ 4 m+1
nt*— 1 m+1 m— 1
35.
8 4 16
B+3 R+l fi"+4B+3
36.
l + _2 8_
X z — l x*—x
37.
i_ + J_ + I_A +1
4K 3K 2 6K
38.
5 _ 7 _1_ 3
m+3 2wi+6 2 2(ro+3)
39.
26-1 36 , 4 „
6-1 P-rM-l
to.
a-4_6o'-20a-13 2o-15
2a-5 4a'-2o-20 2a+4
^Google
182 JUNIOR HIGH SCHOOL MATHEMATICS [XIV, S90
PROBLEMS
1. The sum of two numbers is 160. The quotient of
the larger divided by the smaller is 4. Find the numbers.
(Represent the numbers by n and 160— n.)
2. What number must be added to each term of the
fraction $■ to obtain a fraction equal to ■£? (The equation
is: 3±»«3 \
8+n 4 J
3. What number must be subtracted from each term
of the fraction £$ to obtain a fraction equal to j-?
4. What number added to both terms of the fraction
J will double the value of the fraction ?
6. Separate 42 into two parts whose ratio is J.
6. One half of a certain integer is £ of the sum of the
next two consecutive integers. Find the three integers.
7. The denominator of a fraction exceeds its numerator
by 5. If 3 is added to both the numerator and denomina-
tor, the resulting fraction will be f . Find the fraction.
8. The difference between two numbers is 60. If the
greater is divided by the less, the quotient is 7 and the re-
mainder is 6. Find the numbers.
9. One lot of land contains 1200 sq. ft. more than a
second lot. The ratio of their areas is J. Find their
areas.
10. There are two lots of land, the ratio of whose areas
is 1.25. The larger lot contains 750 sq. ft. more than, the
smaller. Find the area of each lot.
11. There are two lots of land, the ratio of whose areas
is 0.625. The smaller lot contains 360 sq. ft. less than the
larger. Find the area of each lot.
;, S ,:z K i:vC00gIe
XIV, { 90] FRACTIONS AND EQUATIONS 183
12. A can paint a certain building in 10 days. B can
paint it in 8 days. In how many days can they do it
working together?
Solution. A can paint -fa of the building in one day.
B can paint £ of the building in one day.
Let d = the number of days required for both of them
working together, then - = the fractional part they both
a
can do in one day. The equation is :
10 8 d
Solving the equation, d=4|
Check. They both can paint — , or --, of the building
4£ 40
in one day.
A+*=A Ana. 4$ days.
13. A can do a certain piece of work in 3 days and B in
4£ days. How long will it take them working together?
14. A can hoe a certain field in 4£ days, B in 5£ days, and
C in 5J days. How long will it take them working to-
gether?
16. A and B can paint a building in 12 days. A can
paint it in 20 days. How long would it take B to paint
it?
16. An oil tank can be filled by one pump in 8 hours,
or by another pump in 12 hours. How long will it take
to nil the tank if both pumps are working?
17. Two pumps working at the same time can fill an
oil tank in 6£ hours. One pump working alone can fill
it in 10J hours. How long will it take the other pump
working alone to fill it?
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184 JUNIOR HIGH SCHOOL MATHEMATICS [XIV, 1 91
§ 91. Formulas.
1. In a certain electric circuit an electromotive force
{E) of 16.0 volte produces a current (C) of 6.4 amperes.
Find the number of ohms of resistance {R).
Formula: C=^
2. In a certain electric circuit an electromotive force
of 8.5 volts produces a current of 2.5 amperes. The
internal resistance (r) is 1.4 ohms. Find the external
resistance (R).
Formula: C = R+
3. Five equal electric cells connected in series form
a battery for ringing bells. The internal resistance (r)
of each cell is 0.8 ohm, the voltage of each cell is 1.5.
Find the external resistance (R) in the circuit, if the cur-
rent strength is 1 ampere.
Formula : C = t- , where n = the number of cells.
R+nr
4. Six electric cells, connected in parallel, form a bat-
tery for automatic sparks for lighting gas. The internal
resistance of each cell is 0.4 ohm and the voltage of each
cell is 1.6. Find the external resistance in the circuit, if
the current strength is 8 amperes. Formula :
C = -^-
R+-
5. An electric circuit between two points is divided
into two branches, the resistance on one branch (r{) is
12 ohms, and the resistance on the other branch (r 2 ) is
8 ohms. Find the total resistance (R) of the divided cir-
cuit. Formula : _i_ li 1_
R T L r t
;, S ,:z K i:vC00gIe
XIV, §92] FRACTIONS AND EQUATIONS 185
6. In the formula for Ex. 5, ri=17.4 ohms, rz-24.3
ohms, R = ?
7. When a circuit is divided into three branches, the
formula is :
i.i+i+i
a ri Tt r,
Find R, if ri = 6.4 ohms, r s = 5.8 ohms, and r» = 7.9 ohms.
8. A formula used for certain work with mirrors and
lenses is :
i.x+i
F D, D.
Find D„ if F = 30" and D,=45".
9. FindF, if D,=30cm. and Z>,=- 15 cm.
10. FindD ( .if^=25cm.andZ) = 15cm.
§ 92. Transformation of Formulas.
EXERCISES
Change the subject of the formula in each of the follow-
ing exercises.
1. D-^, W-t
J. D-2 V-?
S. C=|, K-T
«. C-{(F-32),F-?
«. C-—.R-1
7.
C — —,E=t
R+nr
8.
9.
-e-t,c-?
360 c'
LO.
«. C--2-
11. £-§, V,-?
3, g ,i,z E d by Google
186 JUNIOB HIGH SCHOOL MATHEMATICS [XIV, 1 92
3. » =
r ±=2, 1- ?
17
4-M-.S-?
r-1
R n r t
4. «.
r— 1
IS
n ri r t
i
19.
i.i + i F , 7
5. 8 =
1 r=?
r— 1
F D, D,
6. V
-V (l+0.00365!,), 1,.
, 20.
i,J-+i, D) ,,
f D, D.
893
Fractional Linear Pairs.
Example. Solve the pair of equations :
(I)
a 26
o+l 26-3*
-0
(II)
3,5
o-l 6+2
-0
Solution.
(i)
o 26
=
a+1 26-3"
®
o(26-3)-26(o+l).
■0
(I)X(o+l)(26-3)
®
2o6-3o-2a6-26.
-0
®-
®
-3a-26-
■0
®-
(id
3,5
o-l 6+2*
=
©
3(1+2) +6(0-1).
.0
(II)X(o-l)(6+2)
©
36+6+5o-5.
■0
©-
©
5o+36.
.-1
®-l
Solving equations ® and ©, o= -
2, 6- +3.
Check.
(I)-
-2 2X3 jl
(II)
_3_+_6_i
2+1 2X3-3
-2-13+2
=1-1*0
X+5io
-1 3
-3 5
2-2-0
-1+1-0
Ant. a--2, 6-+3.
3,g,1 EE d by G00gle
XIV, S 93] FRACTIONS AND EQUATIONS
Solve the following pairs of equations, and check the
answers.
1.
S±S+%-9
6.
2o+36 ,
1+46
«=a_I!+9-o
3 10
4o = 26_
0-6 2
J.
o+6 o-6_7
2 3 "5
7.
% 5
» _ 4
0+6 , o— 6 = 5
3 4 _ 4
*+6 8
»-2 _ 3
3.
i-2_s+4
1-3 ji+3
x+l_g-ll
1+2 J/-12
8.
e+l_ ,
3e+2ii_ 2
d+6
'
*(9-4)-i(P+12)-
-6
"±2+8-20-2=*
26 _3o^26_ 3o+4
'
2+4n=5m
n+2 m-1
10. 1
1 , 1 _ 20
x—y x+y X s — y*
S+?_i!r3.3
5 K
PROBLEMS
In solving these problems, use two equations involving
two unknowns.
1. Find two numbers whose sum is 46, such that if the
greater is divided by the less, the quotient is 6£.
2. The value of a fraction is f . If 8 is added to the
numerator and 6 to the denominator, the value of the re-
sulting fraction is -$. Find the fraction.
i BV Google
188 JUNIOR HIGH SCHOOL MATHEMATICS [XIV, j 93
3. The sum of the terms of a fraction is 11. If £ is
added to the numerator and £ to the denominator, the
value of the resulting fraction is / T . Find the fraction.
4. If a certain number of two digits is divided by the
sum of its digits, the quotient is 7. The tens' digit ex-
ceeds the units' digit by 3. Find the number.
(Suggestion. Represent the digits by t and u, then
the number is I0t+u. Why ?)
6. If a certain number of two digits is divided by one
less than its tens' digit, the quotient is 19. The tens'
digit is 1 less than half the units' digit. Find the number.
§ 94. Fractional Quadratic Equations.
Example. Solve the equation 2±£— 2fc£— 1,
x— 7 x-b
Solution.
~ x+2_ s+5 ,
W x-7 x-5
® (*+2)(*-5)-(s-7)(»+8)-(*-7)(x-6)
®X(«-7)(s-5)
® x 1 -Zx-10-x t +2x+35=x t -12x+35
®= (Parentheses
removed)
® -:r-t-25=x*-12x+35 ®»
0=x 2 -llx+10
®+x-25
Solving equation © by factoring, x = l, or 10.
Check. i±?-l±5il lO+a.lO+S,.,
1-7 1-5 10-7 10-5
_3 6_j. I2_I<>il
-6-4 3 5
-i+f-1 4-3-1
Am. x=l, or 10.
3, g ,i,z E d by Google
XIV, i 94) FRACTIONS AND EQUATIONS
EZBBCISES
Solve each of the following equations, and check both
of the roots.
1-2
, *+3_ 4
4 *-3
a. 6-?-i
b
a. -K-_« + Itzi
y-1 6 y
s. "-^-e
a— 1 a*-l
4. 4-+?-3
s+l x
10. -?-H — — — 1=0
as+3 as»-9
o-2 o-2
u V | J/-3 5
y— 1 y*-l 4
6. **=3_3+.j2-
1— x 1— a:
12. _™_+™±l_24
m+1 m
PROBLEMS
1. The sum of a number and its reciprocal is 3£. Find
the number.
The reciprocal of 5 is J.
The reciprocal of a number n is —
The reciprocal of a number is 1 divided by that number.
The equation is : n-\ — =3£.
2. A number is 4.8 larger than its reciprocal. Find
the number.
8. A number exceeds its reciprocal by 2.1. Find the
number.
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190 JUNIOR HIGH SCHOOL MATHEMATICS [XIV, (M
1. Two thirds of a number is 3£ more than twice its
reciprocal. Find the number.
6. The ratio of the square of the smaller of two con-
secutive integers to the square of the larger is ^. Find
the integers.
6. One half the larger of two consecutive integers
added to the reciprocal of the smaller is 3$ . Find the
integers.
7. The ratio of two sides of a rectangle is f , and its
diagonal is 15". Find the sides.
8. The ratio of the two aides of a rectangle is ^, and
ita diagonal is 2.6". Find the sides.
9. The sum of two numbers is 20. The sum of the
first number and the reciprocal of the second is lfrj. Find
the numbers.
10. The numerator of a fraction exceeds its denomina-
tor by 2. The sum of the fraction and its reciprocal is
2^g. Find the fraction.
11. Solve the formula s = n V"~7 ' . f or n .
12. Solve the formula - = -£- form.
p c—m
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PART H. GEOMETRY
CHAPTER XV
LIKES AMD ANGLES
§ 96. Straight Lines. In the measurement and con-
struction of geometric figures you have used the straight-
edge (ruler), protractor, and compasses. The straight-
edge was used for drawing straight lines, the protractor
for measuring and drawing angles, the compasses for
drawing arcs of circles.
From the work with the straightedge, the following
properties of straight lines may he accepted as true :
(1) Only one straight line can be drawn through two points.
(2) Two straight lines can intersect in only one point.
(3) A straight line is the shortest line between two points.
These statements are three of the axioms of geometry.
The axioms of geometry are sometimes called postulates.
§ 96. Angles. An angle ( / ) is formed when two lines
meet at a point. Ite size ia determined by the amount of
rotation of a line in a plane about one of its points from one
position to another. The greater the amount of turning
the greater the angle.
In Fig. 30, the line is turned about from the position
OA to the position OB, forming
the angle AOB. is called
the vertex of the angle AOB,
and OA and OB the sides. In
reading an angle the vertex
letter is always read in the
middle {/.AOB).
191
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192 JUNIOR HIGH SCHOOL MATHEMATICS [XV, {96
If the line (Fig. 30) is turned about from the position
0A until it again takes the position OA, a revolution is
completed. A revolution is divided into 360 equal parts
called degrees (360°). Thus one degree is one three
hundred sixtieth of one revolution. The degree is sub-
divided into 60 equal parts, called minutes C). The
minute is divided into 60 equal parts, called seconds (").
An angle of 25 degrees, 36 minutes, and 12 seconds is
written 25° 36' 12".
A right angle is one fourth of a
revolution, or an angle of 90".
An acute angle is an angle less
An obtuse angle is an angle more
than 90° and less than 180°.
The bisector of an angle is the line
that divides the angle into two equal
J, S ,:z K i:vC00gIe
XV, |971
LINES AND ANGLES
COWSTRUCTIOHS
1. To construct a given angle at a given point on a given
line.
Carry out this construction as indicated in Fig. 35,
using ruler and compasses. (See Course II, page 100.)
C P p2>
Fig. 35.
2. To bisect a given angle. (See Course II, page 101.)
Cany out this construction as
indicated in Fig. 36.
§ 97. Perpendicular Lines. Perpendicular lines are
lines that form right angles with each other.
The perpendicular bisector of a given line is the line
that is perpendicular to the given line at its mid-point.
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194 JUNIOR HIGH SCHOOL MATHEMATICS [XV, |fi7
COKSTKUCTIORS
1. To construct the perpen-
dicular bisector of a given line.
Carry out this construction
as indicated in Fig. 38. (See
Course II, page 97.)
2. To construct a perpen-
dicular to a given line through
a given point.
Cany out these construc-
tions as indicated in Figs. 39
and 40. (See Course II, page 98.)
A. When the given point is on the line.
B. When the given point is not on the line.
>*
.yl j »
>^
;$
Facts Concerning Pehpbndicuiah Lines
(1) AU right angles are equal.
(2) Through a given point only one perpendicular can be
drawn to a given line.
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XV, SOT]
LINES AND ANGLES
195
§ 98. Complementary Angles. Complementary angles
are two angles whose sum is 90°. Either angle is called
the complement of the other. In Fig. 41,
® Zx+43-90
© ZH-43-90
Fig. 41.
® Z;c+43 = Zy+43 (Axiom 5. A quantity may be
substituted for its equal in an
equation. See page 199.)
® Zx- Ay ®-43 (Axiom 2, page 199)
In ®, Z x is the complement of 43° ; in ®, Z y is the
complement of 43°.
The complements of equal angles are equal.
§ 99. Supplementary Angles. Supplementary angles
are two angles whose sum is 180°. Either angle is called
the supplement of the other. In Fig. 42,
-\us°
y\ne*
® ZaH-115=180
® Zff+115-180
Fia. 42.
® Zs+115= Zy+115 (Axiom 5)
® Zx=Ay ®-115(Axiom2)
In®, Zxis the supplement of 115°; in®, Zj/isthe
supplement of 115°.
The supplanents of equal angles are equal.
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196 JUNIOR HIGH SCHOOL MATHEMATICS [XV, {99
1. What is the complement of an angle of 22°? Of
39°? Of 57°? Of 86°? Of 58° 24'?
2. What is the supplement of an angle of 34°? Of
69°? Of 101°? Of 128°? Of 154° 6'?
3. Construct the complement of an angle of 75° with
ruler and compasses. Check your construction with a
protractor.
4. Construct the supplement of an angle of 135° with
ruler and compasses. Check your construction with a
protractor.
5. Two angles are complementary and the greater
exceeds the smaller by 18". Find each angle.
6. Two angles are complementary and the greater
is twice the smaller. Find each angle.
7. The angles x and y are complementary and x=Zy.
Find each angle.
8. Two angles are supplementary and the greater
exceeds the smaller by 52°. Find each angle.
9. The angles x and y are supplementary and x^2y.
Find each angle.
10. The ratio of two complementary angles is $ . Find
each angle.
11. The ratio of two complementary angles is $. Find
each angle.
12. The ratio of two supplementary angles is $. Find
each angle.
13. Two angles are complementary and one is 6° more
than twice the other. Find each angle.
14. Two angles are complementary and one is 18° less
than five times the other. Find each angle.
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XV, { 101] LINES AND ANGLES 197
§ 100. Vertical Angles. When the sides of one angle
are the prolongations of the sides of another angle, these
angles are called vertical angles.
Thus in Fig. 43, A AOB and LCOD are vertical angles;
also Z BOC and £DOA are vertical angles. To construct
this figure, first draw the line AC. Then, using your pro-
tractor, draw BD, making LCOD equal to 38°.
How many degrees are there in Z AOD 1
How many degrees are there in LA0B1
How many degrees are there in LBOC1
How do the angles COD and AOB compare?
How do the angles BOC and AOD compare?
If one straight line intersects another straight line, the
vertical angles formed are equal.
§ 101. Adjacent Angles. Two
angles which have the same vertex
and a common side between them
are called adjacent angles. Thus
in Fig. 44, Z AOB and Z BOC
are adjacent angles, for they have * v> ' **'
the same vertex 0, and the common side OB between
them.
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198 JUNIOR HIGH SCHOOL MATHEMATICS [XV, 5 101
XZEKCISSS
*^
1. In Figs. 45-47, read the pairs of angles that are ad-
jacent. Read the vertical angles. Read the angles that
are complementary. Read the angles that are supple-
mentary. Read the angles that are both adjacent and
complementary at the same time. Read the angles that
are both supplementary and adjacent at the same time.
8. In Kg. 48, ZAOZ>=90°. What
is the value of each of the other angles ?
Explain.
3. In Fig. 49, Zm=« 35'
the values of Zn, Zx, and Ly.
Fio. 49.
1. In Fig. 50, the three straight lines
crossatO. If Za=42°and Za+Z/= 90°,
find the value of each of the other angles.
5. In the figure of Ex. 4, if La were 26°
and Z 6 were 90°, find the value of each
of the other angles.
8. The angle of elevation of the sun above the horizon,
as Z SO A, may be measured in the following way (Fig. 51).
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XV, 1 103]
LINES AND ANGLES
Hold a quadrant in a
vertical position, so that
a plumb line OP, which is
suspended from a pin at 0,
will fall upon the 90°
mark. The shadow of the
pin will then fall upon the
scale at C. Angle BOC
shows the angle of eleva-
tion of the sun. Explain.
§ 102. List of Axioms.
(1) If the same quantity is added to equal quantities, the
sums are equal.
(2) 1/ the same quantity is subtracted from equal quantities,
the remainders are equal.
(3) // equal quantities are multiplied by the same quantity,
the products are equal.
(4) If equal quantities are divided by the same quantity,
the quotients are equal.
(5) A quantity may be substituted for its equal in a given
equation.
§ 103. Symbols.
equals, or is equal to
^, not equal, or is not equal to
s, identical, or is identical to
~', similar, or is similar to
ss, congruent, or is congruent to
II , parallel, or is parallel to
The plural of any symbol representing a noun is obtained
by using the letter s. Thus, A for angles ; A for triangles.
X, perpendicular, or
is perpendicular to
Z, angle
A, triangle
I— I, parallelogram
, circle
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CHAPTER XVI
CONGRUENT TRIANGLES
§ 104. Classification of Triangles. A triangle (A) is a
plane figure inclosed by three straight lines.
A. Triangles (A) are divided into three groups accord-
ing to the lengths of their sides :
(a) Equilateral triangles, having all three sides equal.
(b) Isosceles triangles, having two sides equal.
(c) Scalene triangles, having no two sides equal.
S. Triangles are divided into three groups according
to their angles :
JiqrzBd^y GoOgk
XVI, 5 1071 CONGRUENT TRIANGLES 201
(a) Right triangles, having one angle right (90°).
(6) Obtuse triangles, having one angle obtuse.
(c) Acute triangles, having all three angles acute.
In an acute triangle, if all three angles are equal, the
triangle is called equiangular.
§ 106. Corresponding Angles and Sides. If two tri-
angles have the angles of the one respectively equal to the
angles of the other, the equal angles are called correspond-
ing angles, and the Bides opposite these angles are called
corresponding sides.
§ 106. Congruent Triangles. Congruent triangles are
triangles that can be made to coincide in all their parts.
The symbol, =«, means "congruent," or "is congruent
to," page 199.
§ 107. Morion. When we said that the size of an angle
was determined by the amount of rotation of a line in a
plane about one of ite points, the idea of motion was im-
plied ; when we used the compasses in the construction
of angles, perpendiculars, etc., the idea of motion was
also implied.
For the study of geometry it is necessary that you get
a clear idea of the three kinds of motion that are possible :
(1) Any figure in a plane may be imagined to slide along
the plane from one position to another.
(2) Any figure in a plane may be imagined to rotate in
the plane about any one of its points.
(3) Any figure in a plane may be imagined to make a com-
plete revolution about any line of the plane as an axis until
it comes into the plane again. This motion is sometimes
called overturning.
These three statements are the postulates of motion.
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202 JUNIOR HIGH SCHOOL MATHEMATICS [XVI, J 108
§ 108. Theorem I.* // two triangles have two sides
and the included angle of one equal respectively to two
sides and the included angle of the other, the two triangles
are congruent.
On a piece of tracing paper draw a straight line ; on
this line mark off with the compasses a line-segment
XY = 1.G".
At X draw an angle of 35°, and on the other side of Lz
mark off with the compasses XZ= 1.3".
Complete the AXYZ.
(a) Comparison of AXYZ with AABC.
Slide AX YZ along the page until point X falls on point A.
Rotate AXYZ about A until XY lies along AB.
Does point Y then fall on point Bt (XY=AB)
Does XZ then he along AC? (ZX= LA)
Does point Z then fall on point CI (XZ = AC)
Do the AXYZ and ABC coincide; that is, do they fit
exactly in all their parts ?
* A theorem is a geometric statement requiring proof.
J, S ,:z K i:vC00gIe
XVI, ! 108] CONGRUENT TRIANGLES 203
Since the AXYZ and ABC are precisely the same
shape and size, they are said to be congruent; that is,
AXYZ^AABC
(6) Comparison of AXYZ with ADEF.
Slide AXYZ along the page until point X falls on
point D.
Rotate AXYZ about D until XY lies along DE.
Does point Y then fall on point El (XY=DE)
Does XZ then lie along DF? (ZX= ZD)
Does point Z then fall on point F? ( XZ^DF )
Is AXYZ^ADEF?
Note. The symbol, ^, means "is not equal to," page 199.
(c) Comparison of AXYZ with AGHK. ■
Slide AXYZ along the page until point X falls on
point. G.
Rotate AXYZ about G until AT lies along GH.
Does point Y then fall on point H ? ( X F = Gff )
Does A"Z then lie along GK? (LX* LG)
IsAXYZ^AGHKI
(d) Comparison of AXYZ with ALMN.
Slide AXYZ along the page until point ^ falls on
point M.
Rotate A.XYZ about M until XK lies along ML.
Does point F then fall on point L ? Why ?
(Note that the two &. are now on opposite sides of ML.)
Overturn AXYZ on ML as an axis, bringing A XYZ
again into the plane of ALMN and on the same side of .
LM.
Does XZ then lie along MN? Why?
Does point Z then fall on point NJ Why?
Is AXYZ^ALMN?
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204 JUNIOR HIGH SCHOOL MATHEMATICS [XVI, 1 108
After superposing the AXYZ on each of the &ABC,
DEF, QHK, and LMN, what conclusion do you reach ?
State this conclusion as definitely as possible.
In the two A.ABC and LMN the equal parts are said
to be arranged in the opposite order.
§ 109. Proof by Superposition is the method of proving
the congruence of two figures by making them coincide.
This method of proof is used in fundamental propositions
only. In order to use it, there must be at least one pair
of equal angles in the figures being compared. You should
always begin by placing a line or angle on its equal part ;
then, by successive steps, trace the position of the rest of
the figure.
Note. In congruent triangles, the corresponding aides are
equal and the corresponding angles are equal.
1. In Fig. 55, ABCD is a square ;
E is the mid-point of the side AD.
Can you prove that the triangles
ABE and ECD are congruent?
Does this prove that EC=EBt
Why?
What kind of a triangle is EBC ?
a. In Fig. 56, ABCD is a rec-
tangle; AD = BC; AE=FB.
Can you prove that rt. ADAE
art. AFBCt
Can you prove that ED-FC?
Can you also prove that Zl= Z2?
E F B
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XVI, i 109]
CONGRUENT TRIANGLES
[XI
3. In Fig. 57, ABCD is a rec-
tangle; AE=FB; DM = MC.
Can you prove that ADEM z*
ACFM after what was proved in
Ex.2?
(Suggestion. 1 1+ L%- Z4+ Z2.)
Does#M = Fitf?
What kind of a A is EMFt
4. In Fig. 58, AO = OB, and
CO-OD.
Prove that AC-BD.
6. In Fig. 59, AB=AC; Zl =
Z2.
Can you prove that AABD as
AADC?
Does this prove that LB= £C1
Does this prove that BD=DC1
Does it follow that (Ac bisector of
the vertex angle of an isosceles triangle
bisects the base t
Note. The vertex angle (also called vertical angle) of a
angle is the angle opposite the base. The base of a triangle ii
side on which the figure is supposed to rest.
6. In Fig. 60, AD is ±BC, and
BE = EC; A is any point in AD,
and AC and AB are drawn.
Prove that AB=AC.
Does it follow that any point in the
perpendicular bisector of a line is equi-
distant from the extremities of the line f
Why?
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206 JUNIOR HIGH SCHOOL MATHEMATICS [XVI, S 109
7. If two lines bisect each other at right angles, prove
that any point in either is equidistant from the ends of
the other.
8. In a square ABCD, the points V, W, X, Y are the
mid-points of the consecutive sides AB, BC, CD, DA,
respectively. Prove that VW = WX=X¥=YV.
9. InFig.61,ABCDisarectangle,
and AF = BE. .
Can you prove that the &AFD and
EBCare congruent?
Does this prove that DF=EC7 '" ~ a ^
10. If D and E, C and F , in Fig. 61, are connected by
two straight lines, prove that ADAE 3s ACBF.
Does this prove that DE= CF ?
11. After DE and CF are drawn, in Fig. 61 (Ex. 10),
provethat ADEF g* ACFE. (Suggestion. ZDEFisz
supplement of Z DEA ; Z CFE is a supplement of Z BFC ;
then apply § 99.)
Does this also prove that DF=CE"!
12. In Fig. 62, ABCD is a ra
tangle.
Prove that ADAB ~ ACAB.
Does this prove that AC = DB1
Does this prove that the diagonals A
of a rectangle are equal?
Note. The diagonals of a rectangle are the lines joining the
opposite vertices.
IS. In Fig. 62, prove the following :
(1) that AACD s* ABDC,
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XVI, 1 109] CONGRUENT TRIANGLES
(2) that &BAD B ACDA,
(3) that AABC * ADCB.
14. In Fig. 63, to measure the
distance from A to P, draw ABX.
to AP; make OB = OA, and CBX
to OB at B ; then draw OP and ex-
tend it to meet CB at C.
Prove that CJ?- PA.
How could the distance AP be c
found?
15. In Fig. 64, to measure
the distance from M to JV,
measure AN and produce it
through A so that AB = AN ;
measure AM and produce it
so that AC = AM. n °- M
Prove that BC=MN.
How could you find the distance WJV ?
lfl. If equal segments, AB and AC, measured from the
vertex of the angle A, are laid off on the sides of the angle,
and if B and C are joined to any point M in the bisector
of the angle, prove that BM = CM.
Note. A segment is a limited portion of a straight line.
17. If equal segments measured from the vertex are
laid off on the equal sides of an isosceles triangle, prove
that the lines joining the ends of these segments to the
opposite ends of the base are equal.
18. If the equal sides of the isosceles triangle are ex-
tended through the vertex and the equal segments are laid
off on the sides extended, give the proof.
is, Google
208 JUNIOR HIGH SCHOOL MATHEMATICS (XVI, j 110
§ 110. Theorem II. // two triangles have two angles
and the included side of one equal respectively to two
angles and the included side of the other, the two triangles
are congruent.
On a piece of tracing paper draw a straight line ; on
this line mark off with the compasses a line-segment
XY- 1.6".
At X draw an angle of 35", and at Y an angle of 60°.
Complete the AXYZ.
(a) Comparison of AXYZ with AABC.
Slide AXYZ along the page until point X falls on
point A .
Rotate AXYZ about A until XY lies along AB.
Does point Y then fall on point B? Why?
Does XZ then lie along ACt Why?
Does YZ then lie along BCt Why?
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XVI, 1110] CONGRUENT TRIANGLES 209
Does point Z then fall on point C? Why?
Do the AXYZ and ABC fit exactly in all their parte?
Make a statement about the A XYZ and ABC.
(b) Comparison of AXYZ with ADEF .
Slide AXYZ along the page until point X falls on
point D.
Rotate AXYZ about D until XY lies along DE.
Does point Y then fall on point El Why?
Does YZ then lie along EFt Why?
Does XZ then lie along DF ? ( Z X ^ Z D)
Is AXYZszADEF?
(c) Comparison of AXYZ with AGHK..
Slide AXYZ along the page until point X falls on
point G.
Rotate AXYZ about G until XY lies along Gtf.
Does XZ then he along GK ? Why ?
Does point Y then fall on point H? (XY^GH)
Is AXYZsiAGHK?
(d) Comparison of AXKZ with ALMN.
Slide AXYZ along the page until point X falls on
point Af .
Rotate AX7Z about M until XK lies along ML. (Note
that the two A are now on opposite sides of ML.)
Does point Y then fall on point L? Why?
Overturn AXKZ on ML as an axis bringing AXYZ
again into the plane of ALMN and on the same side of
LM.
Does XZ then lie along M# ? Why ?
Does rZ then lie along LN? Why?
Is AXYZ =* ALMN?
After superposing the AXYZ on each of the &ABC,
DEF, GHK, and LMN, what conclusion do you reach ?
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210 JUNIOR HIGH SCHOOL MATHEMATICS [XVI. 1 110
EXERCISES
1. In Fig. 66, Zl-30", Z2-30",
Z3-45", Z4-45". Prove that
AABC a AACD.
2. In the parallelogram ABCD
(Fig. 67), AD-BC, AA-/.C,
Z 1 = Z4. Prove that &ADF a
ABEC.
3. In the parallelogram BCD A
(Pig. 68), Z1-Z4, Z2-Z3.
Prove that AB = DC and that
AD-BC.
4. In the triangle ABC (Fig.
AB=AC and Zl=Z2. Prove that
&ABD si AADC.
5. In Fig. 70, AB-AC, ZB-
ZC. Prove that AABDaAAEC.
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XVI, S HI] CONGRUENT TRIANGLES
211
6. To measure the distance (BD)
that a boat (B) is anchored from
the shore at D (Fig. 71), s boy
proceeds as follows. He marks off
the line DE, and by sighting to
the boat from D and E obtains the
A EDB and DEB. He then makes
LCDE^lEDB, LDEC=LDEB,
ADCE.
Where can he measure the distance equal to BD?
Explain.
§ 111. Theorem III. If two sides of a triangle are
equal, the angles opposite the equal sides are equal.
Fro. 71.
and completes the
In &ABC draw CD so that it bisects ZC.
Compare the two triangles thus formed ; that is
&.ACD
and CDB.
AC=BC
Why?
Z1-Z2
Why?
CDmCD
AACD a; &CDB
Why?
i.A-i.B
Why?
State your conclusion.
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212 JUNIOR HIGH SCHOOL MATHEMATICS (XVI, | 111
EXERCISES
1. In Fig. 73, AB -AC, and AB
and AC are extended to D and E
respectively. Provethat Z2= Z4.
2. In Fig. 74, Zl= Z2, and BD =
CE. Prove that ABDC*tABEC.
3. In Fig. 75, AB=AC; AD -
iAB; AE = iAC. Prove that
Z1=Z3; also Z2 = Z4.
4. The triangle ABC (Fig.
76) is an isosceles triangle; D
is the mid-point of BC ; £\ *"
= Z3. Prove that FD =ED.
6. The triangles ABC and DBC
(Fig. 77) are isosceles triangles.
Prove that Z.ABD= I ACT).
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CONGRUENT TRIANGLES
6. In Fig. 78, AB=AD, and
BC=CD; BD is drawn. Prove
that LABC= LADC.
7. In Fig. 79, AB = AC, and BD
=DC. Prove that IABD= LACD.
8. In Fig. 80, AC = AB, and
DC = DB. Prove that Zl= Z3.
9. In Fig. 81, Z1=Z2, and Z3
= Z 4. Prove that CD = BD.
10. To cut two converging beams by
a line AB which shall make equal angles
with them, a carpenter proceeds as fol-
lows : He places two squares against the
beams (Fig. 82), so that AO = BO.
Explain why line AB will make equal angles with the
two beams.
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214 JUNIOR HIGH SCHOOL MATHEMATICS [XVI, 1 112
§ 112. Theorem IV. // two triangles have three sides
of one equal respectively to three sides of the other, the
two triangles are congruent.
In &ABC and DEF, AC = DE, AB = DF, BC=EF.
Imagine ADEF to slide along the page until point D
comes on A.
Let ADEF rotate about A until DE coincides with AC.
In the resulting figure at the right, draw BF.
What kind of a A is ABF1
Z1=Z3 Why?
What kind of a A is CBF1
Z2=Z4 Why?
Z1+Z2=Z3+Z4 Why?
£B=LF Why?
AABCz* AACF Why?
AABC a ADEF Why?
State your conclusion.
Note. This theorem is not proved by superporition for the
reason that no angle value is known. If Z C were known to be
equal to LE, we could overturn A DEF on AC as an axis and
then EF would lie along CB. Since I C is not known to be
equal to LE, we do not know where EF would lie, henee we do
not attempt superposition.
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XVI, { 112] CONGRUENT TRIANGLES
EXERCISES
1. In Fig. 84, AB = AD, and
DC = BC. Prove that AC bisecta A
the A A and C.
2. In Fig. 85, ABC is an
isosceles triangle ; D is the
mid-point of BC ; AD is drawn.
Prove that AD bisects /A.
3. In Fig. 86, ABCD is a
parallelogram in which AD = BC,
and AB = CD. Prove that ZS
4. In Fig. 87, AC
= AD, and BC =
BD. Prove that
the line through A
and B bisects the
ACADaodCBD.
Fig. 87.
6. In the AABC and DEF (Fig. 88), AB = DE;
BC-EF; M and JV are mid-points of BC and EF respec-
tively; AM=DN. Prove that AABM =* ADEN. '
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216 JUNIOR HIGH SCHOOL MATHEMATICS (XVI, 1 112
6. Fasten three pieces of
wood together, by using a nail
at each corner, so as to form
a triangle (Fig. S9). Explain
why this frame is rigid.
Fig. 68.
7. Why is a roof sufficiently braced when a board is
nailed across each pair of rafters?
8. Why are the brace-rods on a bridge always ar-
ranged in triangular shapes ?
9. Explain why a carpenter can bisect
the angle A when he proceeds as follows
(Fig. 90): Lay off AB=AC. Place a
steel square so that BD = CD as shown.
Mark D, and then draw AD. Fia ^
10. A simple form of level may be constructed in the
following manner : Construct a frame as shown in Fig. 91,
with BA equal to BC, and forming the
isosceles triangle BDE. Mark Q the
mid-point of DE, Suspend a plumb bob
from a nail at B. Show that any object
upon which the feet A and C rest will
be level when the plumb line passes
through Q. Fig. 91,
11. On page 193, Ex. 1, you constructed with ruler and
compasses a given angle at a given point on a given
straight line. Prove that Z 1 (Fig. 35) is equal to L MAR ;
that is, that the construction is correct.
12. On page 193, Ex. 2, you bisected a given angle.
In Fig. 36, prove that LABX = ZXBS.
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XVI, 1 113] CONGRUENT TRIANGLES 217
§ 113. Theorem V. If two angles of a triangle are
equal, the sides opposite the equal angles are equal.
On a piece of tracing paper draw a straight line ; on
this line mark off with the compasses, a line-segment
XY=2". At X draw an angle of 50°, and at Y an angle
of 50°. Complete the AXYZ.
Slide AXYZ along the page until point X falls on
point B.
Rotate AXYZ about B until XY lies along BA, point
Y falling on point A. (Note that the two triangles are
on opposite sides of AB.)
Overturn AXYZ on AB as an axis, bringing AXYZ
again into the plane of A ABC,
Does point Z then fall on point C?
Is AXYZ^tAABC!
Can AXYZ be made to coincide with AABC if point
X is placed on point A and XY along ABt
Does this prove that AC-CB1
State your conclusion.
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218 JUNIOR HIGH SCHOOL MATHEMATICS [XVI, 1 113
2. In Fig. 94, IB= IC.
E is the mid-point of AB.
D is the mid-point of AC.
F is the mid-point of BC.
Prove that EF = FD.
3. In Fig. 95, AABC is equiangular.
E, F, and D are the mid-points of
the sides.
Prove that AEFD is equilateral.
Pio. 95.
4. In Fig. 96, ZC=90°; ZZ>=
K)°; Z1=Z3.
Prove that BC=BD.
Z2= Z4. Why?
B. In Fig. 97, A C and D are
rt. A ; AC=AD.
Prove that AADB »s AACB.
Note. ADBC is the plan for a kite,
whose axis is AB.
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XVr, ! 114] CONGRUENT TRIANGLES
219
§ 114. Thbobem VI. // two right triangles have the
hypotenuse and a side of one equal respectively to the
hypotenuse and a side of the other, the two right triangles
are congruent.
In the rt. AABC and DEF, the hypotenuse AB =
hypotenuse DE, and AC=DF.
Imagine the rt. ADEF to slide along the page until
the point D comes on A .
Let ADEF rotate about A until the hypotenuse DE
coincides with the hypotenuse AB. (This gives us the
figure at the right.)
In the figure at the right, ZC- ZF. Why?
DrawCF.
In AAFC, Zl= Z3 (AF-AC)
In ABFC, Z2= Z4 (Complements of equal angles)
BC-BF Why?
rt. AABC as rt. AABF Why?
rt. AABC =s rt. ADEF Why?
State your conclusion.
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220 JUNIOR HIGH SCHOOL MATHEMATICS [XVI. 1 114
EXERCISES
1. In Fig. 99, OC is ±BC; OA
is ±BA; OC = OA.
Prove that OB bisects the angle
ABC.
a. In the AABC (Fig. 100), FE
and FD are ± to AB and AC re-
spectively; FE=FD; F is the mid-
point of BC.
Prove that the AABC is isos-
celes.
3. In Fig. 101, prove that the
&ABC is isosceles, if the perpen-
diculars from the extremities of the
base to the opposite sides are equal.
4. Prove that every point in the
perpendicular bisector of a line is E.
equidistant from theendsof the line. Fia - 101,
6. Prove that the altitude upon the base of an isosceles
triangle bisects the base ; also the vertex angle.
5. To measure the distance from A to B across a river,
you may proceed as follows (Fig.
102) : Run the line AC at right
angles to AB and mark its mid-point
0. At C run the line CD at right
angles to AC. Locate D in line with
points and B. Then measure CD.
Prove that CD=AB.
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CHAPTER XVII
PARALLEL LINES AND PARALLELOGRAMS
§ 116. Parallel Lines. Parallel lines are lines that lie
in the same plane and do not meet, however far they are
extended.
Postulate. Two lines in the same plane perpendicular
to the same line are parallel (||).
CONSTRUCTIONS
1. To construct a line parallel
to a given line through a given
point.
Carry out this construction ■*
with ruler and compasses, as
shown in Fig. 103. (See Course
II, page 99.)
S. A draftsman lays off
parallel lines by moving his
T-square along the straight
edge of his drawing board as
shown in Fig. 104. Explain.
221
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222 JUNIOR HIGH SCHOOL MATHEMATICS [XVII, $115
3. A draftsman lays off
vertical parallel lines by
using a drawing triangle,
the T-square, and drawing
board as shown in Fig. 105.
Explain.
Pio. 106.
§ 116. Transversal. A line that
crosses two or more lines is called
a transversal of those lines.
In Fig. 106, EF is a transversal of
Fig. 106. the jj^a AB and CD
§ 117. Theorem I. If two parallel lines are crossed by
a transversal, then all of the acute angles formed are equal,
and all of the obtuse angles formed are equal.
B b
Using a protractor, verify in each figure the following
equalities :
Acute Angles Obtuse Angles
Z1=Z5 Z2=Z6
Z7=Z3 Z4=Z8
Z1-Z7 Z2=Z8
Z5=Z3 Z4=Z6
A\ and 5, A4 and S, A2 and 6, -43 and 7 are called
corresponding angles.
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CVII, S 117] PARALLEL LINES 223
A 4 and 6, A 3 and 5 are called aUernate-interior angles.
A 1 and 7, A 2 and 8 are called cdternote-exterior angles.
PiQ. 108.
Fig. 109.
1. In Fig. 108, Z 1 = 125°. Find
the number of degrees in each of
the other angles.
2. In Fig. 109, ^6 = 30°. Find
the number of degrees in each of the
other angles.
S. In Fig. 110, AB\\CD;
Zl-25 ; Z3 = 70°. Find the
number of degrees in each of q—
the remaining angles. Fiq. no.
4. A line crosses two parallel lines so that one obtuse
angle is 30° more than twice one of the acute angles. Find
the number of degrees in each angle in the figure.
5. A line crosses two parallel lines so that one of the
obtuse angles is 5° more than four times one of the acute
angles. Find the number of degrees in each angle in the
figure.
6. In Fig. Ill, AB[\CD;
Z6 = 100°; Z7=40°.
How many degrees are there
in each of the other angles? Fio. ill.
F
7. In Fig. 112, AB || DE;
BC\\EF.
Prove that £ABC= £DEF.
iA X^°
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224 JUNIOR HIGH SCHOOL MATHEMATICS [XVII, (117
8. In Fig. 113, AB || CD; GH A V^ *
Prove that MH-BK. c >M _jN "
' Tja. 118.
». In Pig. 114, AB\\CD; is L
the mid-point of LM.
Prove that is also the mid-point
.15, DE || BC through „ A
LA+ZB+ZC of B Z A c
of any other transversal through c — $f — ~
included between AB and CD. ¥ia - IW -
10. In Fig. 115, DE \\ BC through
A.
Prove that /.A+ZB+z
AABC= Z1+ Z2+ Z3. Fio. us.
§118. TheoremJI. The sum of the angles of a triangle
it equal to 180°, or two right angles.
Draw the line DE through the vertex C of AABC\\AB.
Number the angles as in the figure.
L\=LA Why?
Z2= LB Why?
Z1+ZC+Z2--?
ZA+ZC+ZB=?
State the conclusion.
Using the same AABG, can you prove this theorem
by drawing a line through the vertex B||AC? Try it.
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PARALLEL LINES
1. In Fig. 117, IC = 90°;
ZA=35°; £B=? The two
acute angles of a right triangle
are complementary. Why?
2. Prove that if two angles of one triangle are equal
to two angles of a second triangle, the third angles are
also equal.
3. If two right triangles have the hypotenuse and an
acute angle of the one equal respectively to the hypotenuse
and an acute angle of the other, the two triangles are
congruent. (Apply Ex. 2.)
1. In a certain right triangle one acute angle is twice
as large as the other. How many degrees are there in
each acute angle?
6. In a certain right triangle one acute angle is three
times as large as the other. How many degrees are there
in each acute angle?
6. In a certain isosceles triangle the vertex angle is
twice as large as a base angle. Find the number of
degrees in each angle of the isosceles triangle.
7. Find the value of each angle of an equilateral tri-
8. Prove that an exterior angle of a triangle is equal
to the sum of the two opposite interior angles.
(Suggestion. In Fig. 116, extend AB through B;
draw a line through B [[ AC.)
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226 JUNIOR HIGH SCHOOL MATHEMATICS (XVII, g 118
9. ABC (Fig. 118) is an equilateral
A ; CD J.AB.
Find the values of A 1 and 2.
Prove that AD = DB.
State your conclusion.
10. One angle of a triangle is 40°. t
The other two angles have the ratio p, , ng.
4 to 1. Find the other two angles,
and draw a figure to represent such a triangle. (The
equation is: A+4A+40 = 180.)
11. One angle of a triangle is 50°. The other two
angles have the ratio 1.5. Find the other two angles.
(The equation is : 4+1.5A+50-180.)
12. In an isosceles triangle, each of the angles at the
base is 10° more than twice the third angle.
(a) Find the three angles. (Represent the angles by
A, 2A+10, and 2A + 10. Why?)
(6) Draw the figure, making the base 2.6".
13. In a right triangle one acute angle is twice the
other; the shortest side is 1".
(a) Find the two acute angles.
(b) Draw the triangle, using the
shortest side for the base, with the
right angle at its left, as in &ABC,
Fig. 119.
(c) Draw another right triangle £—
with the same parts and place it in
the position of J\ADC, Fig. 119.
(d) Test the sides and angles of AABD. Name the
triangle.
(e) How does AB compare with BC ?
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XVII, E 1191 PARALLEL LINES 227
It follows that : In a right triangle if one acute angle is
twice the other, then the hypotenuse is twice the shortest side.
This is the 60°-30° right triangle used by the draftsman.
(See Fig. 121, page 228.)
§ 119. Theorem III. If two lines are crossed by a
transversal so that either of the following pairs of angles are
equal, the two lines are parallel :
(a) A pair of alternate-interior angles; or
(b) A pair of corresponding angle*
j£*
K £_
* m — 2^— '"
Fro. 120.
The transversal EF crosses the two lines AB and CD
so that,
(a) a pair of alternate-interior angles are equal ; that is,
Z3=Z5.
is the mid-point of KL.
NM is drawn through 01.CD.
AOML^AONK Why?
ZOML=a rt. Z Why?
£KNO= LOML Why?
AB || CD (Postulate, page 220.)
(b) a pair of corresponding angles are equal ; that is,
Z1=Z5.
Z1=Z3 Why?
AB |1 CD Why?
State your conclusion.
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228 JUNIOR HIGH SCHOOL MATHEMATICS |XVII, { 119
1. Prove that if a line is perpendicular to one of two
parallel lines, it is perpendicular to the other one also.
2. Figure 121 shows a draftsman's triangle which is
used for drawing perpendicular lines and parallel lines.
If the hypotenuse is 8", how long is the shortest side?
How long is the third side?
8. In order to draw a line parallel to a line I (Fig. 122)
through a point P, a draftsman places a drawing triangle
so that one side coincides with I, and the other side passes
through P. He then lays a ruler against the side of the
triangle that passes through P, and finally slides the tri-
angle along the edge of the ruler, until one vertex of the
triangle comes to P. Why will a line drawn along the
side of the triangle, originally in coincidence with I, be
the parallel to I through P? ,
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PARALLEL LINES
§120. Classification of Quadrilaterals. A quadrilateral
is a plane figure inclosed by four straight lines.
Quadrilaterals are divided into three groups :
Parallelograms, having opposite sides parallel.
Trapezoids, having only two sides parallel.
Trapeziums, having no two sides parallel.
Quadrtiateraii
• I \ WiamboUl\
Parallelograms are divided into two groups :
(a) Rectangles, having all angles right angles.
A square is an equilateral rectangle.
(b) Rhomboids, having no angles right angles.
A rhombus is an equilateral rhomboid.
§ 121. Parallelograms. A parallelogram is a quadri-
lateral whose opposite sides are parallel.
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230 JUNIOR HIGH SCHOOL MATHEMATICS (XVII, S 122
§ 122. Theorem IV. In any parallelogram, either
diagonal divides it into two congruent triangles, and the
opposite sides of the parallelogram are equal.
Fig. 124.
In EJABCD draw diagonal DB.
Compare the AABD with ABCD.
Z4 of AABD= /2 of ABDC Why?
Zl of AABD= Z3 of ABDC Why?
BD=BD
AABD^ADBC Why?
What follows as to AS and CD1 AD and BCJ
EXERCISES
1. Prove that the opposite angles of a parallelogram
are equal.
2. Prova that parallel lines which are included between
parallel lines are equal.
S. Prove that the sum
of any two consecutive /
angles of a parallelogram /
is equal to 180° (Fig. £-
125). ' F»o. 126.
(Suggestion. Z 1 = Z C. Why?)
4. Prove that if one angle of a parallelogram is a right
angle all the angles are right angles.
5. Prove that if two adjacent sides of a parallelogram
are equal, all of its sides are equal.
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XVII, S 1231 PARALLEL LINES 231
§ 123. Theorem V. The diagonals of a parallelogram
bisect each other.
Fig. 126.
In C3ABCD draw the diagonals AC and BD.
Call their point of intersection 0.
Compare AABO with ACDO.
AB = DC Why?
Z1=Z4 Why?
Z2=Z3 Why?
AABO** ACDO Why?
BO = OD Why?
AO=OC Why?
State your conclusion.
Could you have used AADO and BCO? Try them.
EXERCISES
1. Prove that the diagonals of a rectangle are equal.
2. Prove that the diagonals of a square bisect each
other at right angles.
(Suggestion. Each diagonal is the hypotenuse of an
isosceles right triangle.)
3. In Fig. 127, XY, DE, and
BC are parallel lines.
D is the mid-point of AB.
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232 JUNIOR HIGH SCHOOL MATHEMATICS [XVII, i 123
Prove that E is the mid-point of AC.
Draw DF \\ AC. Then prove,
(a) that AADE&ABFD,
(o) that DF-AE,
(c) that EC-DF,
(<i) thatAE-£C. Why?
1. Using Fig. 127, prove that DE-iBC.
§ 124. Theorem VI. 7/ a quadrilateral has each pair
of its opposite sides equal, it is a parallelogram.
ABCD is a quadrilateral in which AB-
■DC,
and AT)
• BC.
Draw the diagonal BD.
AABDssABCD
Why?
£ABD=£BDC
Why?
AB\\CD
Why?
IADB- Z.DBC
Why?
AD\\BC
Why?
ABCD is a O.
Why?
State your conclusion.
If two sides of a quadrilateral are equal and parallel,
the figure is a parallelogram.
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XVII, 5 126] PARALLEL LINES 233
§ 126. Polygons. A polygon is a plane figure inclosed
by any number of straight lines. A polygon of five sides
is a pentagon; of six sides, a hexagon; of seven sides, a
heptagon; etc.
§ 126. Theorem VII, The sum of all the angles of a
polygon is equal to (n— 2)180°, when n equals the number of
sides of the polygon.
In the quadrilateral (I) there are two triangles, or
(n-2)A.
In the pentagon (II) there are three triangles, or (n — 2) A .
In the hexagon (III) there are four triangles, or (n— 2) A.
In the heptagon (IV) there are five triangles, or
(n— 2)i,etc.
The sum of the angles of all of the triangles in each
polygon is equal to the sum of the angles of that polygon.
What is the sum of the angles of a triangle? Then
what follows? State your conclusion.
EXERCISES
1. Express the sum of all the angles of a polygon of
7 sides ; of 9 sides ; of 10 sides ; of 12 sides.
2. How many degrees are there in each angle of an equi-
angular quadrilateral? Pentagon? Hexagon? Octagon?
Decagon?
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CHAPTER XVIII
CIRCLES
§ 127. Definitions. A circle is a curve all points of
which are equally distant from a point within called the
center. This curve is often called the circumference of
the circle.
An arc of a circle is any portion of
its circumference (Fig. 130).
The radius of a circle is a straight
line drawn from the center to any
point in its circumference.
A diameter is a straight line drawn
through the center of the circle and
terminating in the circumference.
A chord is a straight line joining any two points of the
circumference.
A semicircle is one half of a circle.
A central angle is the angle between any two radii.
§ 128. Central Angles. In Fig. 131, the central angle
AOB is said to intercept {cut off) the arc AB ; while the
arc A B is said to subtend (stretch across)
the angle AOB.
A circle is generated when a seg-
ment of a line turns in a plane about
one end as a point through a complete
revolution. The circumference is de-
scribed by the other end of the line.
Any two positions of the line-seg- SM. 181.
234
i ^dsyGoogle
ment, or radius, form a central angle. Hence it follows
that:
(1) In the same circle (or equal circles) if two central
angles are equal, they intercept equal arcs, for the radius,
while turning through equal angles, describes equal arcs
of the circumference.
(2) In the same circle (or equal circles) if two arcs are
equal, they subtend equal central angles, for the radius,
while describing equal arcs of the circumference, turns
through equal angles.
§ 129. Theorem I. // the diameter of a circle is per-
pendicular to a chord, it bisects that chord.
In the circle O, the diameter CD is ± to the chord AB
BbE.
Draw the radii OA and OB.
Prove that rt. &AOE =* rt. AEOB.
AE=EB Why?
State your conclusion.
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236 JUNIOR HIGH SCHOOL MATHEMATICS [XVIII, f 129
1. Draw a circle having a 10" radius to the scale of
10 to 1. In the circle draw a chord 16" long, and draw
a line from the center of the circle perpendicular to the
chord. Find the distance from the center to the chord.
(Suggestion. Draw a radius to the end of the chord.
The radius is the hypotenuse of a right triangle.)
3. In a circle whose radius is 12 cm., a chord 9 cm.
long is drawn. How far is the chord from the center ?
3. In a 16" circle (diameter =16") a chord is at a dis-
tance of 3" from the center. How long is the chord ?
4. In a 24 cm. circle a chord is
9 cm. from the center. How long
is the chord?
6. In Fig. 133, the chords AB
and AC make equal angles with the
radius OA. OM and ON are X to
AB and AC, respectively. ~- -g
Prove, AM=AN; also, AB=AC.
Fio. 133.
§ 130. Tangents. A tangent to a circle is a straight
line of unlimited length that touches the circle at only one
point. This point is called the point of contact.
In Fig. 134, AB is tangent to the circle 0, at A . A is the
point of contact.
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XVIII, i 1311
CIRCLES
237
§ 131. Theorem II. A line perpendicular to a radius
of a circle at its extremity is tangent to the circle at that
point.
Ib OA the shortest line from to BC? Why?
Are all points on the circle at the distance OA from ?
Why?
Can any point other than A in the line BC, be on the
circle ?
State your conclusion.
EXERCISES
The radius of a circle is 8" (Fig.
A/* N.
^
136). The tangent to the circle from -
A is 6". Find the distance (OA) from
the center to the end of the tangent.
2. The distance from the center of a
circle to a point outside is 10 cm. and Fw " 138,
the radius of the circle is 5 cm. Find the length of the
tangent from that point.
3. A point is 8" from the center of a 12" circle. Find
the length of the tangent from that point.
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238 JUNIOR HIGH SCHOOL MATHEMATICS [XVIII, § 131
4. At a point on a circle construct a tangent to the
circle.
(Sugoebtion. Draw the radius to the given point.)
6. Prove that two tangents drawn to a* circle at the
extremities of a diameter are parallel. (See Postulate,
page 221.)
6. If two tangents are drawn to a circle from an ex-
ternal point, prove that they are equal (Fig. 137). Draw
AO, CO, and BO. What kind of triangles are formed?
Are these triangles congruent ?
7. In Fig. 137, Ex. 6, prove that LBAO- 10AC.
8. To locate the center of circular objects the machin-
ist uses an instrument called a center square (Fig. 138).
The center square consists of
two anna AK and AN fori
an angle NAK. Through the
vertex of this angle there j
a steel blade which bisects the
angle. If the arms are adjusted
to touch the circular object, will this blade pass through
the center of the object? Why?
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XVIII, S 133] CIRCLES 239
§ 132. Measurement. The numerical measure of a
quantity is the number of times that it contains a given
unit of measure.
For example, (a) A line is 5 feet long. The unit of meas-
ure is the foot, and the numerical measure is the number 5.
(6) The area of a floor is 24 square yards. The unit of
measure is the square yard, and the numerical measure is
the number 24.
(c) A revolution contains 360 degrees. The unit of
measure is the angle of one degree, and the numerical
measure is the number 360.
(d) A circle contains 360 degrees. The unit of measure
is the arc of one degree, and the numerical measure is the
number 360.
With the units of measure named in (c) and (d), a revo-
lution has the same numerical measure as the circle.
In generating a circle, as the radius turns through a
given angle, the arc described is the same fractional part
of the circle that the given angle is of a revolution ; hence
A central angle has the same numerical measure as its
intercepted arc. That is, a central angle of 45° intercepts
an arc of 45°.
§133. Inscribed angles. An inscribed angle is an angle
whose vertex is on the circle and whose sides are chords.
An angle is inscribed in an arc
when its vertex is on the arc and its
sides pass through the extremities of
the arc.
In Fig. 139, Z 1 is an inscribed
angle. It is inscribed in the arc
ABC.
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240 JUNIOR HIGH SCHOOL MATHEMATICS |XVIII, 5 134
§ 134. Theorem III. An inscribed angle has the
same numerical measure as one half its intercepted arc.
Case I. In Fig. 140, /.ABC is an inscribed angle,
having the diameter BC for one of its sides. Draw the
radius AO.
LA0C = Z1+ Z2 (An exterior angle of a triangle is
equal to the sum of the two op-
posite interior angles. Ex. 8,
page 225.)
Z1=Z2 Why?
/ AOC = Z 1 + L 1 (Substitution)
t/A0C=/l Why?
A AOC has the same numerical measure as arc AC.
Why?
Z 1 has the same numerical measure as \ the arc AC.
Why?
But, since /ABCm / \, /.ABC has the same numeri-
cal measure as J the arc AC.
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Cabb II. In Fig. 141, the center of the circle is within
the sides of the inscribed angle ABC. Draw BD, the
diameter through B.
Z 1 has the same numerical measure as £ the arc AD.
(Case I)
Z2 has the same numerical measure as b the arc DC.
Why?
Adding the numerical measures of the angles and of the
arcs, /.ABC has the same numerical measure as £ the arc
AC.
Case: III. In Fig. 142, the center of the circle is with-
out the sides of the inscribed angle ABC. Draw BD, the
diameter through B.
LABD has the same numerical measure as £the arc
ACD. Why?
Z CBD has the same numerical measure as £ the arc
CD. Why?
Subtracting the numerical measures of the angles and
of the arcs, Z.ABC has what numerical measure?
State the conclusion about the measure of any inscribed
angle.
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242 JUNIOR HIGH SCHOOL MATHEMATICS [XVIII, } 134
EXERCISES
1. An angle inscribed In a semi-
circle is a right angle. In Fig.
143, how many degrees are there
inarcAZ>C? Z.ABC=W. Why?
2. In Fig. 144, P is a point
outside the circle whose center
is 0. PK is constructed tangent P"
to the circle from P.
Fio. l«.
Note. PO is the diameter of the circle used ia the oonatniction
of the tangent PK.
Prove that PK is tangent to the circle at K.
8, Pattern makers use a carpenter's square to deter-
mine a semicircle. The square is placed as in Fig. 145.
If the heel of the square touches
the bottom of the hole for all posi-
tions of the square, while the sides
rest against the edges of the hole,
then the hole is a semicircle.
Why?
4. Show how to locate the diameter of
a given circle by applying a rectangular
sheet of paper to the circle (Fig. 146).
How would you locate the center of the
circle?
Fta.Ua.
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XVIII, S 134) CIRCLES 243
5. A surveyor desires to lay out a line at 90° to the line
AB (Fig. 147). He sets a stake at a convenient point P,
50 feet from a stake at B. With ,, „
one end of the 50-foot tape at P, A —Wr ;
he describes an arc with the other ^\ /
end, thus locating a stake at C. Vji-,
Keeping one end still at P, with the \
other end he locates a stake at D,
so that C, P, and D are in the
same straight line. Why is BD perpendicular to BC?
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CHAPTER XIX
SIMILAR TRIANGLES
§ 136. Similar Triangles. Similar triangles are tri-
angles whose corresponding angles are equal and whose
corresponding sides are in proportion.
§ 136. Theorem I. // two triangles have their cor-
responding angles equal, they are simitar.
■&1
Draw a triangle (I) having angles of 40°, 60°, and 80°,
making the longest side 1.6" long. Draw another triangle
(II) having angles of 40°, 60°, and 80°, making the longest
side 1.2" long.
Letter the triangles as in Fig. 148.
Measure a, and a, and find the ratio of a L to a s -
In a similar way find the ratio of foi to b t , and c t to c 3 .
How do these ratios compare in value ?
Are the triangles similar ?
In similar triangles, the pairs of corresponding sides are
opposite the pairs of equal angles.
244
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XIX, i 137] SIMILAR TRIANGLES
245
EXERCISES
1. Two triangles have two angles of one equal to two
angles of the other. Are the third angles equal? Are
the triangles similar 7
2. Two right triangles have an acute angle of one equal
to an acute angle of the other. Are the remaining acute
angles equal? Are the right triangles similar?
§ 137. Summary.
(1) Two triangles are similar when the three angles of
one equal the three angles of the other.
(2) Two triangles are similar when* two angles of one
equal two angles of the other.
(3) Two right triangles are similar when an acute angle
of one equals an acute angle of the other.
EXERCISES
1. The triangles in Fig. 149 are similar : hence-=^ = -
a b c
In Fig. 149, 1 = 1.8", y=2A", «=1.5", and 0-1.2".
(Scale 2 to 1.) Find b and c.
The equations a
Fia. 149.
i L8=L1
1.8 2.4 .1.8
L2 = T' "* U = c
Test your answers by measuring b and c.
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246 JUNIOR HIGH SCHOOL MATHEMATICS (XIX, 1 137
Using Fig. 149 for reference, copy the following table
and fill in the required values. (Estimate your results
first.)
-
»
'
-
b
-
2.
8"
9"
11"
16"
a.
3.5"
5"
7-fi"
15"
4
IS em.
3 era.
4 era.
5 om.
ft,
8"
17"
18"
4"
6.
5'
6'
7'
8'
7.
9 cm.
12 om.
13 om.
18 om.
8.
5.2 cm.
3.5 otn.
6.4 om.
8.7 om.
9. A vertical rod 6' long casts a shadow 8' long. How
high is a tree that casts a 40' shadow? (See Fig. 150.)
A ray of sunlight from the top of each object locates
the end of its shadow and right triangles are formed.
Why are the triangles right triangles ?
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XIX, i 137) SIMILAR TRIANGLES 347
At the same hour the angles formed by the sun's rays
must be equal. What do you know about the right
triangles? What do you know about the corresponding
sides of these triangles? State an equation for finding
h and solve it.
Note. The angle A is called the angle of elevation of the
10. The shadow east by a 4' vertical rod is 5J', when
the shadow cast by a church spire is 220'. How high is
the spire?
11. Estimate the height of your school building. By
the method in Ex. 9, find the height to the nearest foot.
Find the per cent error in your estimate.
12. A monument casts a shadow 118' long, when a
yardstick casts a 40" shadow. Find the height of the
monument to the nearest foot.
13. Estimate the heights of poles and buildings and
test your estimates by the method in Ex. 9. Bring into
class at least two such problems worked out.
14. The height of a certain hill is desired. At a place
A (Fig. 151), the angle of elevation of the top of the hill
is 30°. At a place B along
a level road, 300 feet nearer
the base of the hill, the angle „
of elevation of the top is 45°. j^x /?) ]/
fi'b
Draw a plan to the scale of
feet in the height of the hill. Fi°- isi.
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248 JUNIOR HIGH SCHOOL MATHEMATICS [XIX, S 137
15. The distances between two forts, I'\ and F 2 , within
the enemy's lines, also their distances from the two points
of observation, A and B, are
desired. Certain angles are
measured and found to be as
indicated in Fig. 152. The
distance AB is 1500 feet.
Draw a plan to the scale of
400 feet to one inch. Find
from the plan the distances ^^
Fi.Fi, AF lt AFi, BFi, BF t .
§ 138. Tangent Ratio. On pages 246-247 we found the
heights of objects by measuring their shadows. A more
practical method will now be shown.
It is as follows : measure the angle of elevation of the
top of the object and the horizontal distance from the
point of observation to the base of the object ; read the
tangent ratio of the angle of elevation from a table of
tangents (see page 291) * compute the height of the ob-
ject by substituting the values of tan A and 6 in the
formula given below.
In Fig. 153, a is the segment of a tangent, to a
circle whose radius is b, which is cut off by the central
angle A. The ratio of this segment to the radius, -, is
called the tangent ratio of the
angle A.
The formula is written :
tan A = *
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XIX S 138] SIMILAR TRIANGLEB 249
For measuring angles the engineer uses an instrument
called a transit (Fig. 154). With this instrument angles
in the vertical plane can be measured on the vertical
circle by turning the telescope up and down ; angles in
the horizontal plane can be measured on the horizontal
plate by turning the upper part of the transit to the
right and left.
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250 JUNIOR HIGH SCHOOL MATHEMATICS [XIX, 1138
1. In Fig. 155, three right triangles are drawn, all
having the same acute angle. / i 4=40°.
Copy the following table. In Fig. 155, measure the
lines designated in the table, to the nearest hundredth of
an inch. Fill in the remaining values, and find the tan-
gent ratios to two decimal places.
a, = 7
a,-?
o,-?
6, = ?
6,-r
b,
Average ■ ?
2. Draw a right triangle having I. A =40° and mark o
for the side opposite /. A, and b for the other side. Meas-
ure a and b to the nearest hundredth of an inch. Find
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XIX, { 139] SIMILAR TRIANGLES 251
the ratio of a to 6 and express it as a decimal to two
places. Test your ratio and the average obtained in Ex. 1
for equality.
However large the right triangle is, if the Z A - 40°,
the ratio ~ will always have the same value.
b
On page 291, a table of tangents of various angles is
given to three decimal places for angles from 1° to 89°.
Compare your values of tangent 40° (Exs. 1 and 2) with
the value given in the table.
Note. The study and use of this ratio and other similar
ratios is called Trigonometry. The table on page 291 also gives
the values of two of the other ratios, sines and cosines.
g 139. Table of Tangents. The angle of elevation
(or depression) of an object is the angle that a line from
the point of observation to the object makes with the
horizontal line in the same vertical plane. It is called
the angle of elevation if the object is above the eye of the
observer ; it is called the angle of depression if the object
is below the eye of the observer.
In Fig. 156, if the observer is at A
and the object is at B, Z 1 is the angle
of elevation. If the observer is at B
and the object is at A, /. 2 is the angle
of depression. Note that Z2= A\. Fiq. im.
Example 1. A tree casts a shadow 45 ft. long on a
horizontal plain, when the sun is 38° above the horizon.
Find the height of the tree.
(Suggestion. The angle of elevation of the sun is 38°.)
The solution with diagram (Fig. 157) follows on
page 262.
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252 JUNIOR HIGH SCHOOL MATHEMATICS [XIX, 1 139
Solution. Make 6 repre-
sent 45', using a scale of 20'
to 1". Construct ZA = 38°;
erect a X to b at the end of b
(Fig. 157). Measure o with a
ruler; change the measure-
ment to feet and record it for
your estimate. Substitute 38
® tan 38 = — (Est. a = 35')
45
® 0-781 = £ ®=(Table)
45
® 35.1 = a ©X45
Check. Compare 35.1 with the measured length of a.
Ans. 35.1'.
Example 2. A 9-foot pole casts a shadow 7.6' long.
Find the angle of elevation of the sun within one degree.
Solution. Make 6 represent 7.6' ;
draw a _L to b at the end of b, mak-
ing it represent 9' (Fig. 158). Meas-
ure I. A with a protractor and record
the result. Substitute 9 for a and
7.6 for 6 in the formula, tan A = -■ --,
b U 6-7.^—1 C
Fio. 168.
® tan A = y~ (Est. A - 50°).
® tan A = 1.184 ®=
® A =50 ®- (Table)
Check. Compare 50 with the measured value of LA.
Ans. 50°.
J, S ,:z K i:vC00gIe
SIMILAR TRIANGLES
For each of the following exercises, first draw a diagram
to a convenient scale ; then measure the required part for
your estimate.
1. At a horizontal distance 423' from the base of a
tower the angle of elevation of the top is 57°. Find the
height of the tower.
2. A vertical pole 244/ high casts a horizontal shadow
43A/ long. Find the angle of elevation of the sun.
Note. In exercises where the angle is required, find the
angle to within one degree.
3. At a horizontal distance of 137' from the foot of a
steeple the angle of elevation of its top is 60°. Find the
height of the steeple.
4. From the top of a cliff that rises 215' out of the water
the angle of depression of a boat is 23°. Find the distance
of the boat from the foot of the cliff.
5. A tower 37.5 meters high is situated on the bank of
a river. The angle of depression of an object on the oppo-
site bank is 32°. Find the width of the river.
6. The angle of elevation of an airplane at a point A
on level ground is 64°. The point M on the ground
directly under the airplane is 225 yd. from A. How high
is the airplane?
7. The beam of a searchlight on a tower shines directly
on a boat.' The angle of depression is 37° and the search-
light is 420 meters above the horizontal. How far is the
boat from the tower?
8. From the top of a cliff 145' high, on one side of a
river, the angle of depression of an object on the bank oppo-
site is 24°. How wide is the river?
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254 JUNIOR HIGH SCHOOL MATHEMATICS [XIX, §139
9. The angle of elevation of the sun is 42°. How tall
is a tree that casts a shadow on level ground 72' long?
10. A distance MN along n
the bank of a river (Fig. 159)
is found by measurement to
be 127'. A tree at on the
other bank is directly oppo-
site M , so that M is at right
angles to MN. The angle
AfJVOis37°. Find the width
of the river at M .
Fra. 159.
inclined plane if it
11. Find the angle of elevation of an
rises 18' in a horizontal distance of 40'
12. A railroad incline rises 20' in every 100' along the
horizontal. Find the angle of elevation of the road.
13. The grade of a railroad is 2J per cent ; that is, it
rises 2£' for every horizontal distance of 100'. Find the
angle of elevation.
14. Find the angle of elevation of a 3.5 per cent grade.
15. A railway grade is 200' to the mile. What is the
angle of elevation of the road bed?
16. Find the angle be-
tween the rafter and the span
in a £ pitch roof. (The pitch
is the ratio of the "rise"
to the "span." S ~
160.)
17. Find the angle between the rafter and the span in
a i pitch roof.
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XIX, } 140) SIMILAR TRIANGLES
255
§ 140. Theorem II. In any right triangle, the altitude
upon the hypotenuse divides the triangle into two right tri-
angles which are similar to the given triangle and to each
other.
Fra. 161.
In the rt
imAB.
AABC, CD is the altitude
upon the hypote-
Comparison of triangles :
(i)
AI with AABC.
LA-ZA
L\~/.B
Why?
AI~AABC
Why?
(2)
All with AABC.
£B=ZB
{2-ZA
Why 7
AII-AABC
Why?
(3)
AI with All.
/.A-L2
Zl-ZB
AI~AII
Why?
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256 JUNIOR HIGH SCHOOL MATHEMATICS [XIX, § 141
§ 141. Theorem III. In any right triangle, the alti-
tude upon the hypotenuse is a mean proportional between
the segments of the hypotenuse; also, either side of the right
triangle is a mean proportional between the hypotenuse and
the segment adjacent to that side.
Fw. 102.
Parts of AI compared with parts of All.
Z 1 = £ 3 (Both complements of L 2)
Z 4 = Z 2 (Both complements of Z 1)
m corresponds to h.
h corresponds to n.
Then, f=~ Why?
State your conclusion.
Parts of Al compared with parts of given triangle.
m b
Then,
State your conclusion.
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SIMILAR TRIANGLES 267
1. Prove that a (Fig. 162) is a mean proportional be-
tween the hypotenuse and the segment adjacent to a.
2. In a right triangle whose hypotenuse is 25", one
segment of the hypotenuse made by the altitude upon it
is 5". Find the altitude.
3. The hypotenuse of a right triangle is 13 cm. and the
altitude upon the hypotenuse is 6 cm.
(a) Find the segments of the hypotenuse.
(b) Find each side of the given right triangle.
4. The hypotenuse of a right triangle is 18.4", and the
altitude upon the hypotenuse is 7.2".
(a) Find the segments of the hypotenuse.
(b) Find each side of the given right triangle,
5. The diameter of a circle is 40", and the perpendicu-
lar from a point in the circumference upon the diameter
is 12". Find the segments of the
diameter formed by the perpendic- /"" \c
ular (Fig. 163). f ^^^hk .
(Suggestions. h±AB; m&ndn Ay^H — 2 fa\s
are the two segments of AB; the ^ *
chords AC and BC are drawn. \ J
What kind of a triangle is ABC? ita~iB3
What follows?)
6. The diameter of a circle is 13.2 cm. and the per-
pendicular from a point in the circumference upon the
diameter is 4.2 cm.
(a) Find the segments of the diameter.
(b) Find the lengths of the chords from the point in
the circumference to the ends of the diameter.
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258 JUNIOR HIGH SCHOOL MATHEMATICS [XIX, { 142
§ 142. Theorem IV. In any right triangle, the square
of the hypotenuse equals the sum of the squares of the
turn sides.
FlQ. 154.
Id the given right triangle, the altitude is drawn upon
the hypotenuse ; m and n are the segments of the hypote-
nuse ; a and b are the sides of the right triangle.
®
c_a
a n
Why?
®
cn = a i
®Xan
®
c_ b
b m
Why?
®
em=b*
®x6m
®
cn+n»=a'
+¥
©+®
®
c(n+m)=a?
+V
© (Fac
®
<?=a i
+b*
® (Sub
(?) (Substitution of c for n-f-m)
State your conclusion.
This theorem is known as the Pythagorean Theorem.
It is thought that Pythagoras proved this proposition by
proportion in a manner similar to the above proof, but
of this we are not certain. This theorem can also be
proved by means of comparing the areas of similar plane
figures constructed on the sides of the right triangle.
The theorem is not only the most famous theorem of all
geometry; but, as you have seen, it is one of the most
useful in the solution of problems.
is, Google
SIMILAR TRIANGLES
1. Builders, using a ball of twine and a 10-ft. pole,
apply the Pythagorean Theorem in staking out the founda-
tion of a building in the following way.
A string is stretched between two stakes set at A and B
(Fig. 165). This line determines the location of one wall.
If there is to be a square corner
at B, BC is measured off equal to
i ft. A 10-ft. pole is placed with ^^v\
its end at C and held in the posi- X,
tion shown in the figure. (CD
represents the pole.) On another &
string BN, fastened to stake B, *•■ 165 -
BD is measured off equal to 6 ft. This string is held by
a man at N, who brings the string BN so that it just
touches the end of the pole at D. A stake is then set at
N and the string is fastened to it.
Show that the angle B must be a right angle ; that is,
that the corner is "square" at B.
2. Another method of running one line at right angles
to a second line is known as the "rope-stretching" method.
To use this method you proceed as follows: Divide a
string into segments in the ratio of 3, 4, and 5 ; join the
ends and stretch the string taut, after pins have been put
between the segments. Prove that the triangle formed
will be a right triangle. If a 50-ft. tape is to be used,
what are the segments into which it can be divided con-
veniently in order to use this method?
3. A square having a side 12.5" long is inscribed in a
circle. Find the diameter of the circle.
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260 JUNIOR HIGH SCHOOL MATHEMATICS (XIX, i 142
4. The diagonal of a rectangle is 45". If the length
of the rectangle is twice its width, find ita dimensions.
5. The length of a rectangle exceeds its width by 2"
and the diagonal is 14". Find the dimensions of the rec-
tangle.
6. The diagonal of a rectangle is 6.1" and the ratio
of its length to its width is 12. Find its dimensions.
7. Find the height of an isosceles triangle whose pe-
rimeter is 25 cm. and whose base is half one of the equal
sides.
8. Find the diameter of a circle circumscribing a
square whose side is 3^2".
9. Find the diameter of the round rod from which a
square rod 1" on each Bide can be cut.
10. Find the diagonal of a
cube whose edge is 2". (Sug-
gestion. In Fig. 166, first
find AB ; then, using triangle
ABC, find AC.)
__ — —
^-"-^
A
/
/
/
\
11. A kite is inscribed in a circle
whose diameter is 6' (Fig. 167). The
length of one of the shorter sides is 3'.
Find the length of one of the longer
sides, and the perimeter of the kite.
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XIX, 1 142] SIMILAR TRIANGLES 261
12. A certain 60°-30° draftsman's triangle has a hy-
potenuse of 12". Find the length of the two sides. (See
Ex. 2, page 228.)
18. The bases of an isos- j" so *1 <
celes trapezoid are 30" A {\ ■
and 20" ; one of the equal / j j Y^»
sides is 13". Find the / j | \ v
height of the trapezoid >-, jq» ' ,,'
(Fig. 168). Bo. 168.
3,g,1 EE d by GoOgle
CHAPTER XX
MENSURATION
§ 143. Area. The area of a surface is the number
showing how many square units of a given kind it con-
5 144. Theorem I. The area of a rectangle is equal
to the product of its base and height.
Fig. 169.
The rectangle ABCD contains 133 squares, since there
are 8 rows of 17 squares each.
The number of squares thus formed is called the area
of the rectangle, hence
The area of rectangle A BCD = 136 square units.
If we let A stand for the number of square units in the
area of a rectangle, and b and h stand for the number of
linear units, of the same kind, in the base and height,
respectively, then
A (the number of square units) = 6xA square units;
that is,
A-bh
is the formula for the area of a rectangle.
Note. The square units most commonly used are the follow-
ing : the square inch, the square foot, the square yard, the square
centimeter, and the square meter.
262
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MENSURATION
1. Construct on squared paper a rectangle in which
6 = 1.2 in., and ft=0.50 in. Count the number of unit
squares and name the unit. Check your result numeri-
cally by substituting the values of b and h in the
formula.
2. Construct on squared paper a square on a line 1.4
in. long. Count the number of unit squares and name the
unit. Check your result numerically by substituting the
values of b and k in the formula for the area of the rec-
tangle.
3. In a square whose side is s, show by a geometric
diagram that its area can be expressed by the formula,
A =aX8, or «*.
4. In a rectangle, where 6 = ^ and h=V2, we shall
assume that the formula for the area still holds true ; that
is, that A = V'3XV / 2 = V6.
Using the following pairs of values for b and h, deter-
mine the number value of A in each case. (The method
of Approximate Products, pages 13-17, may be used to
advantage in these computations.)
b
1.4
1.41
1.414
1.4142
1.41421
h
1.7
1.78
1.732
1.7320
1.73205
A
7(2 figs.)
?(3 flgs.)
?(4 figs.)
?(5 figB.)
7(6 figs.)
From an inspection of these products, it seems reason-
able to conclude that they constantly approach the exact
area of the rectangle ; that is, VE, which is 2.44940.
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264 JUNIOR HIGH SCHOOL MATHEMATICS [XX, [ 144
6. Find the area of a rectangle with each of the follow-
ing pairs of dimensions.
b
3.8"
5.6"
v'a"
V3"
7.4 cm.
h
2.6"
0.82"
V?"
Ve"
3.8 em.
A
7
?
?
J
?
Note. Since the base 6 and the height h of any rectangle
may be measured by whole numbers to any required degree of
accuracy, if a sufficiently small unit of measure is chosen, it
follows that any rectangle may be thought of as consisting of
k rows of b square units each ; that is, that the area is measured
by bh to any required degree of accuracy.
6. Prove that any two rectangles are in the same ratio
as the products of their bases and heights.
Let A i and A s = the areas of the rectangles,
bi and bj=the bases, and
hi and ftj = the heights, then
Ai=Mi Why?
A s =Mj Why?
A_, Mi
A."m«
A V M L
7. Prove that any two rectangles having equal heights
are in the same ratio as their bases.
Aj Mi
A," Mi
But hi = ht, hence what follows?
8. Prove that any two rectangles having equal bases
are in the same ratio as their heights.
Why?
Ans.
(From Ex. 6)
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MENSURATION
§ 116. Theorem £1. The area of a parallelogram it
equal to the product of its bate and height.
Fra. 170.
In C3ABCD, b is the base and A is the height.
Draw AFXCD produced.
ABGF is a rectangle.
AF = BG
AD = BC
rt. AAFD a rt. ABGC
ABCF-rt. AAFD = ABCF-rt. ABGC
£7ABCZ>=rectangle ABGF
OABCD = bh
State your conclusion.
Why?
Why?
Why?
Why?
Why?
Why?
Why?
EXERCISES
1. Prove that any two parallelograms are in the same
ratio as the products of their bases and heights. (Sug-
gestion. See Ex. 6, page 264.)
2. Prove that any two parallelograms having equal
bases and heights are equal in area.
3. Prove that any two parallelograms having equal
bases are in the same ratio as their heights.
4. Prove that any two parallelograms having equal
heights are in the same ratio as their bases.
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266 JUNIOR HIGH SCHOOL MATHEMATICS [XX, 1 146
§ 146. Theohem III. The area of a triangle is equal
to one half the product of its base and height.
In AABC, b is the base and ft is the height.
Complete the tDABCD by drawing AD\\BC and CD\\AB
EJABCD = bk Why?
AABC=*AACD Why?
AABC=yOABCD Why?
&ABC=ibh. Why?
State your conclusion.
1. Prove that any two triangles are in the same ratio
as the products of their bases and heights.
2. Prove that any two triangles having equal bases
and heights are equal in area.
. 3. Prove that any two triangles having equal bases
are in the same ratio as their heights.
4.. Prove that any two triangles having equal heights
are in the same ratio as their bases.
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XX, i 147) MENSURATION 267
§ 147. Theorem IV. The area of a trapezoid is equal
to one half the product of the sum. of its bases by its height.
T
i \.
Fio. 173.
In trapezoid ABCD, b, and 5i are the bases and h is the
height. Draw the diagonal AC.
Al = $M Why?
AII-4M Why?
AI+AII = il>ifc+£iiA Why?
Trapezoid ABCD=ih(bi+b t ) Why?
State your conclusion.
Note. The area of a polygon of 4 or more sides can be found
in the following way : draw the longest diagonal of the polygon ;
then draw perpendioulara to the diagonal from the other vertices
of the polygon ; find the aum of the areas of all the triangles and
trapezoids thus formed. (See Fig. 194, page 279.)
§ 148. Regular Polygons. A regular polygon is a
polygon that is both equilateral and equiangular.
In a regular polygon the center is equidistant from the
vertices and equidistant from the sides.
The distance' from the center to one side is often called
the short radius ; the distance from the center to one ver-
tex is often called the long radius.
The perimeter of a polygon is the sum of all its sides.
Diqi^Bdsy GoOgle
268 JUNIOR HIGH SCHOOL MATHEMATICS (XX, 1 149
§ 149. Theorem V. The <
equal to one half the product &
radius.
tea of a regular -polygon is
its perimeter and the short
In the regular polygon ABCDEF, draw the long radii
AO, BO, CO, DO, EO, and FO.
This gives as many triangles as the polygon has sides.
Draw a short radius.
Then, AO=B0=CO=>DO = EO=FO. Why?
Also, ZAOB= ZBOC=ZCOD = LDOE, etc. Why?
Represent the perimeter by p, the side by a, and the
short radius by r.
The triangles AOB, BOC, COD, etc., are congruent
isosceles triangles. Why ?
AAOB=irs Why?
ABOC=$rs, etc. Why?
AAOB+ ABOC+ ACOD+etc. =$r8+$rs+ira+etc.
But polygon ABCDEF is equal to the sum of all the
triangles.
Hence, polygon ABCDEF - ir(s+s+s+etc.) or
Polygon ABCDEF^rp. Why?
State your conclusion.
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XX, S 160] MENSURATION 269
§ 160. Theorem VI. In any right triangle, the square
on the hypotenuse is equal to the sum of the squares on
the two sides.
The square on AC+the square on BC= 144+256
= 400 (square units)
AI+AII+AIII+AIV+AV=96+96+96+96+16
= 400 (square unite)
State your conclusion.
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270 JUNIOR HIGH SCHOOL MATHEMATICS [XX, 5 150
Note. The Pythagorean Theorem was proved on page 258
by means of proportion ; that form of proof is called an algebraic
proof. The method of proof used here, — that is, by a com-
parison of the areas of plane figures, constructed on the sides
of the right triangle, — is called a geometric proof. There are
many other ways of proving this theorem by the geometric
method, but the method used here is probably one of the sim-
plest.
1. Prove that the square on one side of a right tri-
angle is equal to the square on the hypotenuse minus the
square on the other side,
2. Construct a square equal to the sum of two given
squares. Carry out the construction as shown in Fig. 175,
using ruler and compasses.
Fio. 175.
(Suggestion. On a given line lay off a segment a;
erect a perpendicular at one end of a and on this perpen-
dicular lay off b; join the other ends of the segments
a and b.)
Prove that the square on c is the required square.
3. Construct a square equal to the difference of two
given squares.
(Suggestion. Reverse the construction in Ex. 2 as
follows : after erecting a perpendicular at one end of a,
use the other end of a as a center, and with a radius c
describe an arc cutting the perpendicular.)
i BV Google
XX, 5 151) MENSURATION 271
§ 161. Theorem YII. The ratio of the circumference
of a circle to its diameter is a constant.
Using a tape measure, find the distance around a circular
disk to the nearest tenth of an inch ; also, measure its di-
' ameter to the same degree of accuracy.
Divide the number of inches in the circumference by the
number of inches in the diameter. What is the ratio?
In a similar way, measure the circumferences and di-
ameters of several circular objects ; such as, an ash can,
a pail, a half dollar, a bicycle wheel, etc. Find the ratio
of the circumference to the diameter in each case.
State your conclusion.
Note. If your measurements were very accurate, you would
find the ratio to be nearly 3.1416. This mixed decimal is com-
monly expressed by the symbol r (read pi).
EXERCISES
1. Prove that the circumference of a circle is equal to
r times its diameter.
2. Prove that the circumference of a circle is equal
to 2t times its radius.
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272 JUNIOR HIGH SCHOOL MATHEMATICS [XX, f 152
§ 152. Theorem VIII. The area of a circle is equal
to one half its circumference times its radius.
Cut the surface inclosed by a circle into any number
of equal parts, say 16, as shown in Fig. 177. Fit them
together as in Fig. 178.
Figure 178 resembles a parallelogram, having for its
base one half the circumference of the circle, and for its
height the radius of the circle.
For the parallelogram, we have
A = bh square units.
Hence, for the circle, we have
A—\er square units.
3ig.1iz.ed by GoOgk
MENSURATION
1. Letting c — the circumference of a circle and r=
the radius, write a formula for the area, A.
2. Prove that the area of a circle may be expressed by
the formula : A = rr 1 .
3. In Fig. 179, count
the number of small
squares and parts of
squares in one quarter
of the circle ; multi-
ply the total number
of whole squares that
you get by 4. Com-
pare your result with
the statement in Ex.
Fra. 179.
4. Prove that the area of a circle may be expressed
by the formula : A =%xcP.
5. Prove that the areas of two circles are in the same
ratio as the squares of their radii ; as the squares of their
diameters.
6. In Fig. 180, the shaded part is
a sector of a circle. It is the same
fractional part of the area of the circle
as its angle is of 360°. How would
you find the area of a Bector?
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CHAPTER XXI
FORMULAS OF MENSURATION
§ 163. Formulas for the Mensuration of Plane Figures.
In formulas (1) to (5), A = area; p = perimeter; 6, bi, bt,
= bases; h = height; and s = side.
(1) Rectangle: A = bh p = 2b+2h
(2) Square: A = s? p=As
(3) Parallelogram: A = bh
(4) Triangle: A = \bh
(5) Trapezoid: A = \h(b 1 +b 2 )
In formulas (6) and (7), A = area, c = circumference,
r = radius, d = diameter, and x = 3.1416 (3.14 to three
(6) Circle: c=ird, orc = 2irr
(7) Circle : A = -or*, or A = \trtF
(8) Angles of a plane triangle:
Z4+ZB+ZC=180°,
where A, B, and C = the angles of the triangle.
(9) Sides of a right triangle:
where c = hypotenuse, and a and b = the sides.
§ 154. Solids. A rectangular
block is a solid bounded by six
rectangles, called faces (Fig. 181).
274
X-]
y
i
s
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XXI, ! 1541 FORMULAS OF MENSURATION 275
A cube is a rectangular block all of whose
faces are squares (Fig. 182).
A right prism is a solid bounded by two
congruent polygons, called bases, and by
rectangles, called lateral faces. The height, or
altitude, is the perpendicular distance between
the bases (Fig. 183).
A regular pyramid is a solid
bounded by a regular polygon,
called the base, and congruent
isosceles triangles, called lateral
faces, meeting at a common
point called the vertex (Fig. 184).
The height, or altitude, is the
perpendicular distance from the
vertex to the" base (ft in Fig. 184). The slant height
is the height of one of the lateral faces {OM in Fig. 184).
A frustum of a regular pyramid
is the solid included between its base
and a section made by a plane par-
allel to the base.
The height, or altitude, is the per-
pendicular distance between the bases r '°' 100 '
(ft in Fig. 185). The slant height is the height of one
of the lateral faces (MN in Fig. 185). The lateral faces
are congruent isosceles trapezoids.
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276 JUNIOR HIGH SCHOOL MATHEMATICS [XXI, { 154
A right circular cylinder is the solid
formed when a rectangle is turned through
a complete revolution about one of its
sides as an axis.
The height, or altitude, is the perpen-
dicular distance between the bases (h in
Fig. 186).
A right circular cone is the solid
formed when a right triangle is
turned through a complete revolu-
tion about one of its perpendicular
sides as an axis.
The height, or altitude, is the per-
pendicular distance from the vertex ,
to the base (ft in Fig. 187). The
slant height is the distance from the '
vertex to any point hi the circumference of the base (OM
in Fig. 187).
A frustum of a right circular cone is the portion of a
right circular cone included between the base and a section
made by a plane parallel to the base.
The base of the cone and the parallel section are to-
gether called the bases of the frustum.
The height, or altitude, of the frus-
tum is the perpendicular distance
between the bases (ft in Fig. 188).
The slant height is the distance from
a point in the circumference of the
upper base to the corresponding
point in the circumference of the lower base (MN in
Fig. 188).
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XXI, i 155] FORMULAS OF MENSURATION
A sphere is a round solid
inclosed by a surface, all points
of which are equidistant from
a point within called the
center (Fig. 189).
Fro. ISO.
§ 166. Formulas for the Mensuration of Solids. The
volume of a solid is the number showing how many cubic
units of a given kind it contains.
In Fig. 190, the rectangular
block consists of 4 layers, each
made up of 15 cubic units, each
equal to M.
Hence, the volume of the
block = 60 cubic units. XTJ
In formulas (1) to (5), V=
volume ; S = area of lateral
surface; B, B L , Bi=areas of bases;
of bases; fc=height; I = length (1), or slant height (4) and
(5) ; and e = edge.
(1) Rectangular block : V=tuth
(2) Cube: F=e*
(3) Right prism: V=Bh S=ph
(4) Regular pyramid: V=\Bh S=\pt
(5) Frustum of regular
pyramid : V= |h(Bi+B*+ V5IF,)
S-foi+pM
Fig. 190.
■ Pi Pi) ps = perimeters
3,g,1 EE d by GoOgle
278 JUNIOR HIGH SCHOOL MATHEMATICS [XXI, S 155
In formulas (6) to (9), V"=volume; S=area of curved
surface; r, n, r s = radii; A = height; I = slant height; and
x = 3.1416 (3.14 to three figures).
(6) Right circular
cylinder : V= irr t h S = 2irrft
(7) Right circular
cone : V = [nr 1 /! S = irrt
(8) Frustum of right
circular cone: V=JirA(ri i +rj*+rirj) S=ir(ri+r 2 )f
(9) Sphere: V=|irr» S = tir7 2
1. On a map drawn to a scale of 60 mi. to the inch,
what area is represented by a strip 3" long and 2" wide?
2. Find the cross-sectional i ^ j
area of an excavation for a t
railroad bed, if it is 6' wide \
at the top, 5' wide at the [^
bottom, and 1.5' deep. How f, g . !91
many cubic yards of earth
are removed in a section of 100 yards? (See Fig. 191.)
3. Find the area of a right triangle if its hypotenuse
ia 15 cm. and one side is 9 cm.
4. Find the area of a right triangle if its hypotenuse
is 5V^" and one side ia 5".
~7f¥'
5. Find the area of an isosceles right triangle whose
hypotenuse ia 10V2".
J, S ,:z K i:vC00gIe
XI, (Uf XXI, { 156] FORMULAS OF MENSURATION 279
6. Find the cross - sectional [
area of the Z-bar, with dimen-
sions as shown in Fig. 192.
!•— 3.*i'-*
Fiq. 192.
7. Find the cross-sectional area of the channel-ire
with dimensions as shown in Fig. 193.
8. Figure 194 is the plan of a held. Make a plan to
the scale of 50' to 1" ; draw the diagonal and draw the
altitudes of the triangles
as shown in the figure.
Measure the lines neces-
sary for finding the areas
of the triangles ABC and
ADC. Find the approxi-
mate number of square feet
in the area of the field. FlQ ' 194 '
9. The perimeters of an equilateral triangle, of a square,
of a regular hexagon, and of a circle are each equal to 60
cm. Find the area of each and arrange them in order of
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280 JUNIOR HIGH SCHOOL MATHEMATICS [XXI, f 155
10. How many revolutions does a 40" automobile wheel
make in going one mile ?
11. Four circular boiler plates 24" in diameter are cut
from a plate 48" by 48".
(a) Find the number of square inches of stock wasted,
(t) Find the per cent of stock wasted.
12. Find the area of a sector of a 10-cm. circle if the
angle of the sector is 30°. (See Ex. 6, page 273.)
IS. In Fig. 195, the diameter of the circle is 12" and
angle A0B~ 60°.
(a) Find the length of the arc
AB.
(b) Find the area of the sector
AOB. '
(c) Find the area of the triangle
AOB.
(d) Find the area inclosed by the
chord AB and its arc. **■ 185
14. The diameter of a piston of an engine is 22". The
pressure on the end of the piston is 100 lb. per square
inch. Find the total pressure on the end of the piston.
15. The inside diameter of a cer-
tain water pipe is 10". The pipe
is 1" thick. Find the cross-sec-
tional area of the metal in the pipe |
(Fig. 196).
16. Findtheweightof4-ft.length
of the pipe in Ex. 15. The density of
the metal is 0.26 lb. per cubic inch. "°- ,m
{Suggestion. To get the volume, multiply the cross-
sectional area of the metal by the length of the pipe.)
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XXI, 1 1551 FORMULAS OF MENSURATION
281
17. Neglecting overlapping, how many square feet of
tin are required to make a stovepipe 8' long and 6" in
diameter ?
18. Find the total area (curved surface and two ends)
of a tin can, 4" in diameter and 6" high.
19. How many ends for cans of the dimensions given
in Ex. 18 can be cut from a sheet of tin 40" by 48"?
How many square inches are wasted ? What per cent is
wasted?
20. A plate glass window is 8' by 10' and J" thick.
Find its weight, the density of the glass being 172 lb. per
cubic foot.
21. How many square feet of lead will be required to
cover the bottom and sides of a rectangular taiik, 3.5 ft.
deep, 4.2 ft. long, and 3.5 ft. wide? What will be the
weight of the lead if it is 0.05 in. thick and a cubic inch
weighs 0.41 lb. ?
22. The Great Pyramid in Egypt is about 450 ft. high
and its base is a
square about 764 ft.
on each side. Find
the approximate
number of cubic
yards in its volume.
23. Figure 197 is
the plan of a pave- ■ :
ment. What vol-
ume of cement \"
thick is required to
cover it?
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282 JUNIOR HIGH SCHOOL MATHEMATICS [XXI, 1 155
24. A cylindrical pail is 45 cm. deep and 15 cm. in diam-
eter (inside). Find its contents in cubic centimeters.
How many liters will it hold ?
25. A cast-iron bar is 2.8 cm. in diameter and 2.5 m.
long.
(a) Find its volume in cubic centimeters.
(6) A piece 12 mm. long is cut off. What per cent of
the bar is left ?
26. Find the weight of a certain steel cone 6 in. in diam-
eter and 5 in. high. A cubic inch of the steel weighs
0.28 lb.
27. Find the weight in kilograms of a certain iron ball
11 cm. in diameter. The density of the iron is 7.2 grams
per cubic centimeter.
28. Figure 198 is the plan of " ty+r
a hemispherical head for a steel
bolt whose diameter is 1+/'. Find
the weight of the head, if a cubic
inch of the steel weighs 0.28 lb.
29. The area of a circle in
square inches and its circumfer-
ence in inches are expressed by the
samenumber. Find its diameter.
30. The volume of a cylinder in cubic centimeters and
the area of its curved surface in square centimeters are ex-
pressed by the same number. Find its diameter.
31. Cylindrical cans of a certain brand of condensed
milk are shipped to France in boxes containing 2 layers
of 12 cans each. Before the boxes are nailed up, the
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XXI, i 155] FORMULAS OF MENSURATION 283
spaces between the cans are filled with wheat. The diam-
eter of a can is 3". The box is 12" long, 9" wide, oj"
deep (inside dimensions). If one bushel of wheat occupies
approximately { cu. ft., how many bushels of wheat can
be sent in 10,000 such boxes?
32. In building a retaining wall of concrete the propor-
tion of gravel, sand, and cement is 5 to 3 to 1. The wall
is 50 ft. long, 4 ft. high, and 10 in. thick. How many
cubic feet of each kind of material is needed, if there is an
allowance of 10% for waste?
33. Find the weight per linear foot of lead pipe of 1J"
bore (inside diameter) and ^" thick. A cubic inch of
the lead weighs 0.41 lb.
34. A cylindrical pail is 20 centimeters in diameter
(inside). How deep must it be to hold 5 liters?
SB. A cubic inch of copper is drawn into a wire 5 mm.
in diameter. Find the length of the wire in meters.
36. A disk 10 cm. in diameter and 0.8 cm. thick is
turned out of a piece of stock 12 cm. square and 1 cm.
thick. How much is wasted in shavings? What per
cent is wasted ?
37. A conduit is made of concrete (shaded part in Fig.
199).
(o) Find the total cross-sectional
area of the entire figure.
(b) Find the total cross-sectional
area of the shaded part.
(c) Find the number of cubic feet
of concrete needed for 400 yards of ]
this conduit. Fro. 199.
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284 JUNIOR HIGH SCHOOL MATHEMATICS [XXI, J 155
38. A galvanized iron pail, in the shape of a frustum
of a cone, has the dimensions given in Fig. 200.
(a) Find the length of AB.
(b) Find the number of
square inches in the bottom.
(c) Find the number of
square inches in the conical
part.
(d) Find the number of cubic
inches in the contents.
(e) How many gallons will
the pail hold?
Fig. 200.
39. Figure 201 represents a certain type of milk can.
(a) Find the area of the bottom.
(b) Find the area of the two cylindrical surfaces.
(c) Find the length of AB.
(d) Find the area of the surface
of the frustum.
(e) Find the total surface of
the entire can (without cover).
(/) Find the total number of
square inches of tin sheeting needed
to make the can, allowing 10% for
the seams.
(ff) Find the volume of the
larger cylindrical part.
(A) Find the volume of the
frustum.
(i) Howmany quarts will the can
hold, to within a tenth of a quart? Fm - 201 -
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TABLES 07 SQUARES AND SQUARE ROOTS
§ 166. Squares of Numbers. The table on pages 288-
290 gives the squares of all three-figure numbers from 1.00
to 10.00. The first two digits of the number are in the
column n, and the third digit is in the row beginning with
u. The square is at the intersection of the row and column
on which the number is located. For example, 4.76* =
22.66.
To find the square of a number less than 1 or more
than 10, it is necessary to make the number fit the table.
35 ! =(3.5X10) S = 12.25X100 = 1225; hence, when you
divide the given number by 10 to make it fit the table,
the square of the given number is 100 (10 2 ) times the
square found in the table.
.. /3 : 5\*_12.25_
\io)~
tiply the given number by 10 to make it fit the table, the
square of the given number is — - ( — - 1 times the square
found in the table.
Example 1. Find the square of 48.6.
,1* ...,„„
10
(h) Finding the square of 4.86 in the table, 4.86'=23.62.
(c) Making the square 23.62 fit the number 48.6,
48.6 1 - (4.86 X 10)* = 23.62 X 100 - 2362
Ant. 2362.
is, Google
0.35»=
/3.46y_11.97
I, 10 J 100 "
286 JUNIOR HIGH SCHOOL MATHEMATICS [APP.,5 156
Example 2. Find the square of 0.346.
Solution, (a) Making 0.346 fit the table,
0.346=^
(h) Finding the square of 3.46 in the table, 3.46*= 11.97.
(c) Making the square 11.97 fit the number 0.346,
0.1197
Ans. 0.1197.
§ 157. The Square Roots of Numbers. The table on
pages 288-290 gives the square roots of numbers from 1
to 100. To find the square root of a number, look in
the table for the square nearest the given number. This
square will be at the intersection of a row and a column.
The first two digits of the square root are in column n in
that row in which the square is located and the third
digit is in the row beginning with n in that column in
which the square is located. For example, the square
root of 11.16 is 3.34 ; the square root of 14 is 3.74.
To find the square root of a number less than 1 or more
than 100, it is necessary to make the number fit the table.
I— ) = ttt:; hence, in order to make a square larger
\10/ 100
than 100 fit the table, it is necessary to divide the square
by 100 (or 100*, etc). Then the required square root is
10 (or 10*, etc.) times the number found in the table.
(10n)*=100n*; hence, in order to make a square less
than 1 fit the table, it is necessary to multiply the square by
100 (or 100*, etc.). Then the required square root is tV (or
— , etc.) times the number found in the table.
3,g,1 EE d by GoOgk
APR, S 157] SQUARES AND SQUARE ROOTS 287
Example 1. Find the square root of 392.
Solution, (a) Making the square 392 fit the table,
^?=3.92
100
(b) Finding the square root of 3.92 in the table,
V592-1.98
(c) Making the square root 1.98 fit the square 392,
V502-V'LOOX3.92-10X 1.98-19.8 .ins. 19.8.
Example 2. Find the square root of 3920.
Solution, (a) Making the square 3920 fit the table,
^=39.2
100
(6) Finding the square root of 39.2 in the. table,
^393 = 6.26
(c) Making the square root 6.26 fit the square 3920,
V3920=V100X39.2 = 10X6.26 = 62.6 An*. 62.6.
Example 3. Find the square root of 0.455.
Solution, (a) Making the square 0.455 fit the table,
0.455X100=45.5
(&) Finding the square root of 45.5 in the table,
V45l> = 6.75
(c) Making the square root 6.75 fit the square 0.455,
v / O455.=0.675 Am. 0.675.
Example 4. Find the square root of 0.0455.
Solution, (a) Making the square 0.0455 fit the table,
0.0455X100 = 4.55
(6) Finding the square root of 4.55 in the table,
vT55 = 2.13
(c) Making the square root 2.13 fit the square 0.0455,
V0.0455- 0.213. Ana. 0.213.
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§ 168. TABLE I-SQUARES OF NUMBERS
FROM 1.00 TO 10.00
1.000 1.020 1.040 1.061 1.082
1.210 1.232 1.254 1.277 1.300
1.440 1.464 1.488 1.513 1.538
102 1.124 1.145 1
2.250 2.280 2.310 2.341 2.372
2.560 2.592 2.624 2.657 2.690
2.890 2.924 2.958 2.993 3.028
3.240 3.276 3.312 3.349 3.386
3.610 3.64S 3.686 3.725 3.764
4.000 4.040 4.080 4.121 4.162
4.410 4.452 4.494 4.537 4.580
4.840 4.884 4.928 4.973 5.018
5.290 5.336 5.382 5.429 5.476
5.760 5.808 5.856 5.905 5.954
6.250 6.300
6.760 6.812
7.290 7.344
7.840 7.898
8.410 6.468
6.350 6.401 6.452
6.864 6.917 6.970
7.398 7.453 7.508
7.952 8.009 8.066
8.526 8.585 8.644
1(1.2'
10.30 10.37 10.43 10.50
10.89 10.96 11.02 11.09 11.18
11.56
12.25 12.32
12.96 13.03
13.89 13.76
14.44 14.52
15.21 15.29
1.70 11.76 1
12.39 12.46 12.53
13.10 13.18 13.25
13.84 13.91 13.99
14.59 14.87 14.75
15.37 15.44 15.52
16.24 1
17.06 1
17.89 1
18.75 1
19.62 1
19.54
20.43 20.52 20.61
20.25
21.16 21.25 21.34 21.44 21.53
22.09 22.18 22.28 22.37 2
23.04 23.14 23.23 23.33 2
24.01 24.11 24.21 24.30 2
1.613 1.638 1.664
.434 2.465 2.496 2.528
.756 2.789 2.822 2.856
! 4.285 4.326 4.368
i 4.709 4.752 4.796
: 5.153 5.198 5.244
> 5.617 5.664 5.712
: 6.101 6.150 6.200
■. 6.605 6.656 6.708
. 7.129 7.182 7.236
I 7.673 7.728 7.784
I 8.237 8.294 8.352
! 8.821 8.880 8.940
.364 9.425 9.486 9.548
.97 12.04 12.11 12.18
40 13.47 13.54 13.6
14 14.21 14.29 14.36
.90 14.98 15.05 15.13
15.78 15.84 15.92
48 16.56 14.65 16.73
31 17.39 17.47 17.56
8.23 18.32 18.40
.79 20.88 20.98 21.07
.72 21.81 21.9
.66 22.75 22.85 22.94
23.72 23.81 23.91
.60 24.70 24.80 24.90
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APP.. i 1581 SQUARES AND SQUARE ROOTS
-
1 1 8 4
5 6 7 8 9
0.0
5.2
5.3
H5
5.6
5.7
58
5.9
60
6.1
6.2
6 3
8.4
6.1
6ft
67
66
6.9
T.O
7.1
7.2
7.3
74
7.6
7.6
7.7
7.8
7.9
80
8.1
8.2
83
8.4
B.6
8.6
8.7
8.8
8.9
25.00 25.10 25.20 25.30 25.40
26.01 26.11 26.21 26.32 26.42
27.04 27.14 27.25 27.36 27.46
28.09 28.20 28.30 28.41 28.52
29.18 29.27 29.38 29.48 29.59
30.26 30.36 30.47 30.58 30.60
31.36 31.47 31.68 31.70 31.81
32.49 32.60 32.72 32.83 32.95
33.64 33.76 33.87 33.99 34.11
34.81 34.93 35.05 35.16 35.28
36.00 36.12 36.24 36.36 36.48
37.21 37.33 37.45 37.58 37.70
38.44 38.56 38.69 38.81 38.94
39.69 39.82 39.94 40.07 40.20
40.96 41.09 41.22 41.34 41.47
42.25 42.38 42.51 42.64 42.77
43.66 43.69 43.82 43.96 44.09
44.89 45.02 45.16 45.29 45.43
46.24 46.36 46.51 46.66 46.79
47.61 47.76 47.89 48.02 48.16
49.00 49.14 49.28 49.42 49.56
60.41 50.55 50.69 60.84 60.96
51.84 51.98 62.13 52.27 52.42
53.29 53.44 53.58 53.73 53.88
54.76 54.91 55.06 65.20 65.35
56.25 66.40 66.55 56.70 66.85
57.76 57.91 58.06 68.22 58.37
69.29 69.44 59.60 59.75 69.91
60.84 60.99 61.15 61.31 61.47
62.41 62.57 62.73 62.88 63.04
64.00 64.16 64.32 64.48 64.64
65.61 65.77 65.93 68.10 68.26
67.24 67.40 67.57 67.73 67.90
68.89 69.06 69.22 69.39 69.56
70.66 70.73 70.90 71.06 71.23
72.25 72.42 72.59 72.76 72.93
73.96 74.13 74.30 74.48 74.65
75.69 75.86 76.04 76.21 78.39
77.44 77.62 77.79 77.97 78.15
79.21 79.39 79.57 79.74 79.92
25.50 26.60 25.70 25.81 25.01
26.52 26.63 26.73 26.83 26.94
27.58 27.67 27.77 27.88 27.98
28.62 26.73 28.64 28.94 29.06
29.70 29.81 29.92 30.03 30.14
30.80 30.91 31.02 31.14 31.26
31.92 32.04 32.15 32.26 32.38
33.06 33.18 33.29 33.41 33.52
34.22 34.34 34.48 34.67 34.69
35.40 35.52 35.64 35.76 35.88
36.60 36.72 36.34 36.97 37.09
37.82 37.96 38.07 38.19 38.32
39.06 39.19 39.31 39.44 39.58
40.32 40.45 40.68 40.70 40.83
41.60 41.73 41.86 41.99 42.12
42.90 43.03 43.16 43.30 43.43
44.22 44.36 44.49 44.62 44.76
45.56 45.70 45.83 45.97 46.10
46.92 47.06 47.20 47.33 47.47
48.30 48.44 46.58 46.72 48.66
49.70 49.84 49.98 60.13 50.27
51.12 51.27 51.41 51.55 51.70
52.56 62.71 52.85 53.00 53.14
64.02 64.17 64.32 54.46 54.61
55.60 55.65 56.80 55.95 56.10
57.00 57.15 57.30 57.46 57.61
58.52 58.66 58.83 68.96 59.14
60.06 60.22 60.37 60.53 60.68
61.62 61.78 61.94 62.09 62.25
03.20 63.36 63.52 63.68 63.64
64.80 64.96 65.12 65.20 65.45
86.42 66.59 66.75 66.91 67.08
68.08 68.23 68.39 68.55 68.72
69.72 89.89 70.06 70.22 70.39
71.40 71.57 71.74 71.91 72.08
73.10 73.27 73.44 73.62 73.79
74.82 75.00 75.17 75.34 75.52
76.58 76.74 76.91 77.09 77.26
78.32 78.50 78.68 78.85 79.03
i BV Google
290 JUNIOR HIGH SCHOOL MATHEMATICS [APP.. 5158
-
1 1 I 4
■ •789
90
9.1
0.9
9.3
94
9.5
96
9.7
98
9.9
81.00 81.18 81.36 81.64 81.72
82.81 82.99 83.17 83.36 83.64
84.64 84.82 85.01 86.19 85.38
86.49 86.68 86.86 87.05 87.24
88.36 88.55 88.74 88.92 89.11
90.25 90.44 90.63 90.82 01.01
92.16 92.35 92.54 92.74 92.93
94.09 94.28 94.48 94.67 94.87
96.04 96.24 96.43 96.63 96.83
98.01 98.21 98.41 98.00 08.80
81.90 82.08 82.26 82.45 82.63
83.72 83.91 84.09 84.27 84.46
85.66 85.75 85.93 86.12 86.30
87.42 87.61 87.80 87.98 88.17
89.30 89.49 89.68 89.87 90.06
91.20 91.39 91.58 91.78 91.97
93.12 93.32 93.51 93.70 93.90
95.06 95.26 95.45 95.65 95.84
97.02 97.22 97.42 97.61 97.81
99.00 99.20 99.40 99.60 99.80
TABLE II — RECIPROCALS OP NUMBERS
FROM I TO B.9
Jt
.1 .1 J
A .• .«
.T
M »
1
1.000
0.909
0.833
0.769
0.714 0.667 [0.625
0.588
0.666 0.526
1
0.500
0.478
0.456
0.435
0.417 0.400 0.386
0370
0.357 0.345
a
0,333
0.323
0.313
0.303
0.294 0.286 0.278
0.270
0.263 0.256
4
0.250
0.244
0.238
0.233
0.227 0.222 0.217
213
0.208 0.204
it
6
0.167
0.164
0.161
0159
0.156 0.164 0152
0.149
0.147 0.146
7
0.143
0.141
0.139
0.137
0.135 0.133 0.132
0130
0.128 127
K
0.125
0.123
U.12-2
0.120
0.119 0.118 0.116
0.11a
0.114 0.112
9
0.111
0.110
0.109
0.10H
0.106 0.106 0.104
0.103
0.102 0.101
3ig.1iz.ed by GoOgk
TABLE III— TRIGONOMETRIC RATIOS
(Tin abbreviation hyp m
(op
nnt
Cohih*
T.NDKMT
(op
n C
T.«M»t
i/hyp)
(ifl/hjp)
<«t-p/«ij;
/hyp) (Ml
/fc'jT)
(°pp/»hi>
0°
000
1.000
.000
M°
707
707
1.000
1°
01T
-037
719
68B
1.036
2°
035
.999
47°
731
682
1.072
3°
082
.052
48°
743
669
1.111
4°
070
.998
.070
49°
75S
656
9°
087
.996
.087
»0°
766
643
1.192
105
.999
61"
777
1.235
7°
122
.123
52°
788
1.280
139
.990
53°
799
602
1.327
9°
156
.988
.158
54°
809
588
1.376
10°
174
.98B
.176
•6°
819
574
1.428*
1!)1
.982
.194
56°
559
1.483
12°
208
.B13
57"
839
545
1.540
J3°
22fl
.974
.231
848
530
1.600
242
.970
.243
59°
857
1.664
15°
25!)
.966
.268
60°
866
500
1.732
16°
276
.961
.287
«1°
876
485
1.804
17°
292
.956
.306
62°
883
469
1.881
18°
309
.951
63°
891
454
19°
326
.946
.344
64°
899
2.060
80°
342
.940
..364
68°
906
423
2.145
21°
353
.934
.384
66°
914
407
2.246
22°
175
.927
.404
921
391
2.356
23°
391
.424
68°
927
375
2.476
24°
407
.914
.445
69°
934
358
. 2.605
86°
423
.906
.466
70°
940
843
2.747
26°
.899
71°
46
326
2.904
27°
454
.891
-610
72°
951
309
3.078
28°
469
.883
.532
966
292
3.271
29°
485
.875
.551
961
276
3.487
30°
.866
.577
78"
259
3.732
31°
515
.857
.601
970
242
4.011
32°
530
.848
77°
974
225
4.331
.649
978
.1)8
4.70B
559
329
.675
982
191
5.145
35°
.174
.819
.700
80°
985
174
5.671
588
.727
81°
156
6.314
37°
602
.799
.754
82°
990
139
7.116
016
.788
.781
993
li!
8.144
.777
84°
996
106
40°
643
.766
SB"
996
087
11.430
41°
rwifi
.755
86°
998
070
14.301
42"
ra
.743
.900
87°
999
052
19.081
43°
.731
88 a
999
oris
28.636
695
.719
.966
89" 1
XH7
67.290
U°
707
.707
1.000
90" l
000
000
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TABLE IV— IMPORTANT NUMBERS
A. Units of Length
English Uifin Mbtbio Units
12 Inches (in.) = 1 foot (ft.) 10 millimeters — 1 centimeter (cm.)
8 feet = 1 yard (yd.) (mm.)
5J yards = 1 rod (rd.) 10 centimeters = 1 decimeter (dm.)
S20roda ■-■ 1 mile (mi.) 10 decimeters = 1 meter (m.)
10 meters = 1 dekameter (Dm.)
10OO meters = 1 kilometer (Em.)
English to Mbtbio Metric to English
lin. =8.5400 cm. lcm. =0.3987 in.
lft. = 80.480cm. 1 m. = 39.37 in. = 8.2809 ft.
1 mi. = 1.0008 Km. 1 Em. = 0.6214 mi.
B. Units of Area or Surface
1 square yard = 9 square feet = 1206 square Inches
1 acre (A.) = 160 square rods = 4840 square yards
1 square mile = 640 acres = 102400 square rods
C. Units of Measurement of Capacity
Dry Measure Liquid Mb a bum
2 pints (pt.) = 1 quart (qt.) 4 gills (gi.) = 1 pint (pt.)
8 quarts = 1 peck (pk. ) 2 pints = 1 quart (qt.)
4 pecks = 1 bushel (bu.) 4 quarts = 1 gallon (gal.)
1 gallon = 281 on. in.
D. Metric Units to English Units
1 liter = 1000 cu. cm. =61.02 on. in. s 1.0667 liquid quarto
1 quart = .94680 liter = 946.30 on. cm.
1000 grams = 1 kilogram (Eg.) = 2.2046 pounds (lb.)
1 pound = .458583 kilogram = 468.69 grams
B. Other Numbers
t = ratio of circumference to diameter of a circle
= 8.14169266
1 radian = angle subtended by an arc equal to the radius
= 67° 17' 44".8 = 67°.2957705 = 180°/x
1 degree = 0.01745829 radian, or r/180 radians
Weight of 1 cu. ft. of water - 63.425 lb.
i BV Google
Addition and subtraction of frac-
tions, 169
Addition of monomials, 73
□f polynomials, 74
of signed numbers, 64, 66
Algebraic expressions, 70
Angle, 191, acute, 192
bisector of, 192
central, 234, 239
inscribed, 239
obtuse, 192
right, 192
vertex of, 206
Angles, adjacent, 197
alternate-exterior, 223
13
complementary, 195
corresponding, 222
supplementary, 195
vertical, 197
Approximate products,
quotients, 18
Arc, 234
of one degree, 239
Areas of plane figures, 262, 26S,
266, 267, 268, 272
Axes, 63. 106
Axioms, list of, 50, 199
Base of a power, 71
Binomials, 114, 124
Braoes, 75
Brackets, 75
Cancellation, 172, 177
Capacity, 47
Centimeter, 43
Checks, 3, 7, 19, 23, 25, 51, 74, 78,
96, 105, 113, 135, 140
Chord, 234
Circle, 234, 272
Circumference of a circle, 234
Classification of quadrilaterals, 229
of triangtea, 200
Coefficient, 70
Completing the square, 143
Complex fractions, 176
Cone, 276, 278
Constant, 156
Cube, 276, 277
Cubic centimeter, 47
Cylinder, 276, 278
Degree, 192
Degree of a polynomial, 72
Denominator, 169
Diagonals of a rectangle, 206
Diameter, 234
Division of fractions, 174
of monomials, 95
of polynomials, 96, 97
of signed numbers, 94
Draftsman's triangle, 227, 228
Equations, 50, 100, 161
Explanation of symbols, 7, 52, 199
Exponent, 71
Extremes, 154
Factors, 70, 113
Formulas of mensuration, 34, 274,
277,278
of science and industry, 37
Fractional equations, 177, 186, 188
Fractions, 169
Frustum of a cone, 276, 278
pyramid, 276, 277
Fulcrum, 31
Function, 163
Gram, 46
Graphs of equations, 60, 105, 165,
Identity, 60
Inconsistent equations, 106
Independent e<
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Index of a root, 131
Pi (*■), 20, 271, 274
Inverse variation, 162
Point of contact, 236
Irrational numbers, 131
Polygon, 233, 265
Polynomial, 70
Kilogram, 46
Positive numbers, 63
Kilometer. 43
Postulates, 191, 201, 221
Kite, 218
Power of a number, 71
Prime factor, 113
Lateral area, 277, 278
Prime number, 113
Law of signs, in division, 94
Principal root, 139
in multiplication, 81, 82, 84
Prism, 275, 277
Lever, 31
Products of two binomials, 90, 93
Like terms, 73
Proof by superposition, 204
Linear equations, 107, 126
Proportion, 154
Linear measure, 43
Pyramid, 275, 277
Liter, 47
Pythagorean theorem, 258
Lone radius, 267
Quadratic equations, 126, 138, 144,
146, 147
Quadrilateral, 229
Means, 154
Measure of a central angle, 239
of an inscribed angle, 240
Radical sign, 131
Meter, 43
Radius, 234
Metric measures, 43. 292
Ratio, 1, 152
Millimeter, 43
Rational numbers, 131
Minuend, 76
Ratios as per cents, 2
Minute, 192
Rectangle. 229. 262
Monomial factors, 113, 124
Rectangular block, 274, 277
Motion, 201
Reduction of radicals, 131
Multiplication, constructions for,
Regular polygon. 267
85,89
Regular pyramid, 275
of fractions, 172
Revolution, 192, 239
of monomials, 84
Rhomboid, 229
of polynomials, 85, 86
Rhombus, 229
of radicals, 133
Root of a number, 71
of signed numbers, 82
of an equation, 51
Negative numbers, 63
Second, 192
Numerator, 169
Sector of a circle, 273
Numerical measure, 239
Segment of a line, 207
Semicircle, 234
Order of operations, 100
Short radius, 267
Signed numbers, 63
Parallel lines, 221
Similar terms, 73
Parallelogram, 229, 265
Special products, 89
Parentheses, 75
Specific gravity, 48
Pentagon, 233
Sphere, 277, 278
Per cent error. 3
Square, 220
Percentage formula, 7
Square root, 23, 286
Perimeter of a polygon, 267
Squares of numbers, 28S
Perpendicular lines, 193
Straight lines, 191
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Subtraction, 76
of monomials, 76
of polynomials, 78
Subtrahend, 76
Summary of factoring, 124
Symbols, 7, 52, 199
Tables, appendix, 288-292
Tangent ratio, 248
Terms of a fraction, 169
Three-figure accuracy, 16
Transformation of formulas,
186
Transit, 249
Transversal, 222
Trapezium, 229
Trapesoid, 229, 267
Triangle, acute, 201
bisector of vertex angle, 205
Triangle, equiangular, 201
equilateral, 200
isosceles, 200
obtuse, 201
right, 201
scalene, 200
Triangles, congruent, 201
Types of quadratic equations, 138
Variables in arithmetic, 156
in geometry, 158
in science, 161
Volume, 277
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GEOMETRY
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plane and Solid Geometry. Goth, ia°, ill., ix and 331 page*
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The authors of this book believe that in the study of geometry
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Another feature of the book is the careful selection and arrange-
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The editor of the series to which this book belongs was a member
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The first book presents the elements of algebra,
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The early exercises are simple and well graded. They
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The manual is made up in the size of the standard slip-
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