1 Find the square root of 144.

SOLUTION. We first resolve the number into its prime factors, then taking one of every two equal factors we have the square root of 144 equal to 2X2X8=12.

NOTE -We have marked the factors taken with a little star, and it will be well for the student to do the same in his solutions.

OPERATION.

2)144

*2)72

2)36

*2)18

3)9

*3

425. There are two methods of explaining the general process of extracting square and cube roots, called the Geometrical or Synthetic Method, and the Algebraic or Analytic Method.

NOTE TO TEACHERS.-I. The geometrical method is placed at the close of Evolution, but teachers who prefer may omit the analytic method and turn over and give their pupils the geometrical method, and then return and apply it to the problems given under the analytic method.

II. With young pupils who have a difficulty in understanding evolution, it will be well to drill them upon the method of doing the work, not requiring them to give the explanation until they are better prepared to understand it.

ANALYTIC METHOD OF SQUARE ROOT.

426. The Analytic Method of square root is so called because it analyzes the number into its elements, and derives the process of evolution from the law of involution.

1. Extract the square root of 1225.

SOLUTION.-Since the square of a number consists of twice as many figures as the number itself, or twice as many less one, the square root of 1225 consists of two places, and hence consists of tens and units, and the given number consists of the square of the tens, plus twice the tens into the units, plus the square of the units, as is indicated in the second operation in the margin.

The greatest number of tens whose square is contained in 1225 is 3 tens; squaring the tens

and subtracting we have 325, which equals twice the tens into the units, plus the square of the units. Now since twice the tens into the units is much greater than units into units, 325 must consist principally of twice the tens into the units, hence if we divide by twice the tens we can ascertain the units. Twice the tens equals 30 X 2=60; dividing, we find the units to be 5; now finding twice the tens into the units, plus the square of the units, or what is the same, twice the tens, plus the units, both multiplied by units, which equals (60+5) X5=325, and subtracting, nothing remains. Hence the square root of 1225 is 3 tens and 5 units, or 35.

principally of twice the hundreds. into the tens; hence if we divide by twice the hundreds we can ascertain the tens, and then proceed as before.

3

10'49 76(324

9

62

149

644

124

2576

2576

II. In practice we generally ab- OPERATION AS IN PRACTICE. breviate the work by omitting the ciphers and condensing the other parts, merely preserving the trial divisors and true divisors, as is indicated in the margin. Here 6 is the first trial divisor and 62 the first true divisor. 64 is the second trial, and 644 the second true divisor.

427. From the above solution and remark we derive the following

f

RULE.-I. Commence at units place and separate the number into periods of two figures each.

II. Find the greatest number whose square is contained in the left hand period, place it to the right as a quotient and its square under the left hand period, subtract and annex the next period to the remainder.

III. Double the root found and place it on the left for a trial divisor, divide the dividend exclusive of the right hand figure by it, the quotient will be the second figure of the root.

IV. Annex the second figure of the root to the trial divisor for the true divisor, multiply the result by the second figure of the root, subtract and bring down the next period for the next dividend.

V. Double the root now found, find the third figure of the root as before, and thus proceed until all the periods have been used.

NOTES.-I. If the product of any true divisor by the corresponding figure of the root exceeds the dividend, the figure of the root must be diminished by a unit.

II. When a dividend, exclusive of the right hand figure, will not contain the trial divisor, place a cipher in the root and at the right of the trial divisor, then bring down the next period and proceed as before.

III. To extract the square root of a decimal, we point off the decimal into periods of two figures each, counting from units place, and proceed as with whole numbers, the reason of which may be easily seen.

IV. To find the square root of a common fraction, it is evident that we extract the square root of both numerator and denominator. When these terms are not perfect squares, the shortest way is to reduce the fraction to a decimal and extract the root. When a number is not a perfect square, annex ciphers and find the root on to decimals.

APPLICATIONS OF SQUARE ROOT.

428. Square root is extensively used in treating of geometrical figures, such as squares, right angled triangles, etc.

1. A man owns a farm in the form of a square which contains 10 acres; how many rods in length or breadth is it? Ans. 40rd. SOLUTION. Reduce acres to square rods, and extract the square root. 2. A man has a square lot of land containing 1440 acres; how many rods in length or breadth ? Ans. 480 rods.

3. A man has a rectangular board 128in. long and 32in. wide, from which he makes a square table as large as possible; required its length, no allowance being made for sawing. Ans. 64in.

4. A general trying to mass his army into a solid square of 80 on each side, found he lacked 500 men to fill out each square; how many men in his army? Ans. 5900.

5. A general attempting to draw his army of 9480 men into a square found he had 71 men over; required the number of men in rank and file.

Ans. 97.

6. What would it cost to enclose a square lot, containing 160 acres, with a fence costing at the rate of $4 per rod for fencing? Ans. $2560.

7. A general drew up his army of 27175 men in three grand divisions in the form of three equal squares, and found he had 168 over in the first, 132 in the second, and lacked 200 in the third; what was the number of men in the side of each square?

Ans. 95.

SUG. Find the number in the three squares, divide by 3 and take the root.

S

RIGHT ANGLED TRIANGLES.

429. A Right Angled Triangle is a triangle which has

one right angle.

430. The side opposite the right angle 1s called the Hypothenuse, one of the other sides is the Base, the other the Perpendicular. In right angled triangles we have the following principles from geometry:

PRIN. I.—The square of the hypothenuse equals the sum of the squares of the other two sides.

PRIN. II.-Hence the square of either side equals the square of the hypothenuse diminished by the square of the other side.

EXAMPLES FOR PRACTICE.

1. The two sides of a right angled triangle are 51 and 68 inches respectively; required the hypothenuse.

SOLUTION. Hypothenuse=1/512+ 682 = 1/7225 — 85, Ans.

2. The hypothenuse of a right angled triangle is 115, the baso 92; what is the perpendicular?

Ans. 69. 3. A ladder 65ft. long is placed against a house so that its foot is 25ft. from the house; how high does it reach? Ans. 60ft. 4. A rectangular lot of land is 1080 rods long and 810 rods broad; what is the distance between two opposite corners? Ans. 1350 rods. 5. Two vessels sail from the same port, one sails north 3 miles an hour, the other west 4 miles an hour; how far are they apart in 2 days? Ans. 240 miles.

6. A ladder 82ft. long stands close against a building; how far must it be drawn out at the bottom that the top may be lowered 2 feet? Ans. 18ft.

7. A pole was broken 52ft. from the top, and fell so that the end struck 39ft from the foot; required the length of the pole. Ans. 117ft.

8. A ladder 130ft. long, with its foot in the street, will reach on one side to a window 78ft. high, and on the other to a window 50ft. high; what is the width of the street? Ans. 224ft.