The greatest number of tens whose square is contained in 2025 is 4 tens. Let A, Fig. 1, represent a square whose sides are 40 units, its area will be 402, or 1600 square units. Subtracting 1600 from 2025, we find remaining a surface containing 425 square units. By inspection we find this surface to consist principally of the two rectangles B and C, Fig. 2, each of which is 40 units long, and since they nearly complete the square, their area is nearly 425 units; hence if we divide 425 by their length, we will find their width. The length of both is 40x2=80; dividing 425 by 80, we find their width to be 5 units. Adding the length of the little corner square D, Fig. 3, whose sides are 5 units, we find the entire length of the surface remaining after the removal of the square A, is 80+5=85 units, and multiplying this by the width, we find the whole area of the remainder to be 85X5= 425 square units. Subtracting 425 square units from the square units left after subtracting 1600 square units, nothing remains, therefore the side of the square whose area is 2025 square units is 45 units; hence the square root of 2025 is 45.

NOTES.-1. When there are three figures in the root, by the analytic method we use the formula for three terms; by the geometrical method, after removing the first rectangles and small square, we have two rectangles and a small square remaining, which we remove as before.

2. In practice, we determine the number of figures in the root by pointing off the number into periods of two figures each, beginning at the right; we also abbreviate the work by omitting ciphers and condensing the other parts, preserving only the trial and true divisors. For illustration see solution in the margin.

3. This can also be explained by building up the square instead of separating it into its parts, for which see Manual.

Rule.-I. Begin at units, 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 at the right as a quotient, subtract its square from the left hand period, and annex the next period to the remainder for a dividend.

III. Double the root found and place it at the left for a TRIAL DIVISOR; divide the dividend, excluding the right hand term, by this divisor; the quotient will be the second term of the root.

IV. Annex the second term of the root to the trial divi

sor for the TRUE DIVISOR, multiply the result by the second term of the root, subtract the product from the dividend, and bring down the next period for the next dividend.

V. Double the root now found for a second TRIAL DIVISOR, find the third term of the root as before, and thus proceed until all the periods have been used.

NOTES.-1. If the product of a true divisor by a term of the root exceeds the dividend, the term must be diminished by a unit.

2. When a cipher occurs in the root, annex a cipher to the trial divisor, bring down the next period, and proceed as before.

3. The square root of a common fraction is evidently the square root of each term. When these terms are not perfect squares, reduce the fraction to a decimal, and extract the root. When a number is not a perfect square, annex periods of ciphers and carry the root on to decimals.

4. By squaring 1, .1, .01, etc., we see that the square of a decimal contains twice as many decimal places as the decimal, hence to extract the square root of a decimal, we point off the decimals into periods of two figures each, counting from the decimal point, and proceed as in whole numbers.

Extract the square root of

12

1

.12 = = .01 .012=

=.0001

35

Ans. . 21. 4907 0025.

74

Ans. 2. 22. 89526.025681. Ans. 299.209.
Ans. 518. 23. .i00.

Ans. .3163859+.

Ans. .58. 24. 642521104. Ans. 25348. Ans. .29. 25. 185383635844. Ans. 430562. Ans. .035. 26. 4122544464025..

15. .0841.

16. .001225.

17. .099856.

Ans. .316.

18. .061009.

Ans. 247. 27. 77531660905535929.

19. 364. Ans. 6.036923+.

Ans. 2030405.

Ans. 278445077.

CONTRACTIONS IN SQUARE ROOT.

811. When the square root is to be extracted to many places of decimals, the work may be shortened by the following method:

Rule.-Find, as usual, more than one-half the terms of

the root, and then divide the last remainder by the last divisor, using the contracted method, as in Art. 278.

1. Extract the square root of 10.

APPLICATIONS OF SQUARE ROOT.

812. The Applications of Square Root to problems involving geometrical figures are extensive.

813. The Side of a square is equal to the square root of its area.

1. A man owns a square lot containing 25 hectares; how many meters does its side measure?

SOLUTION. The 25 hectares equal 250000 sq. meters; extracting the square root, we have 500 meters.

2. I own a square lot containing 7 acres; what is the length of one of its sides? Ans. 33.466+ rods.

3. A man owns a rectangular lot containing 20 acres, whose length is twice its breadth; what is the distance around it? Ans. 240 rods.

4. What will it cost to enclose a rectangular lot containing 12 hectares, whose length is 3 times its breadth, at the rate of 25 cents a meter? Ans. $400.

5. A cabinet maker has a board 26 ft. 3 in. long and 2 ft. 11 in. wide; what is the largest square table he can make out of it, no allowance being made for sawing?

Ans. 8 ft. 9 in.

6. If it cost $600 to inclose a farm 96 rods long and 54 rods wide, how much less will it cost to enclose a square farm of equal area with the same kind of fence?

Ans. $24.

7. A general drew up his army of 38000 in three grand divisions in the form of three equal squares, and found he had 354 over in the first, 414 in the second, and lacked 400 in the third; what was the number of men in the side of each square? Ans. 112 men.

RIGHT-ANGLED TRIANGLES.

814. A Right-angled Triangle is a triangle which has one right angle.

815. The Base of a triangle is the side on which it stands; as AB.

816. The Perpendicular is the side which forms the right angle with the base; as BC.

817. The Hypothenuse is the side opposite the right angle; as AC.

Hypothenuse.

A

Base.

B

Perpendicular.

818. The Principles of right-angled triangles are as

follows:

PRINCIPLES.

1. The square of the hypothenuse equals the sum of the squares of the other two sides.

2. Hence, the square of either side equals the square of the hypothenuse diminished by the square of the other side.

NOTE.-The smallest integers which can express the relation of the three sides of a right-angled triangle are 3, 4, and 5. We may have an infinite number of right-angled triangles with their sides in this relation. Other integral relations of sides are as follows: 5, 12, 13; 8, 15, 17; 20, 21, 29. These are obtained by substituting in the formula (2rs)2+(s2—r2) 2 =(s2+r2)2, in which r is less than s.

1. The hypothenuse of a right-angled triangle is 230, and perpendicular 138; required the base.

SOLUTION. The base = √/2302—1382-184, Ans.

2. A rectangular lot containing 103.68 ares is twice as long as wide; required the distance between its opposite Ans. 72/5 meters.

corners.

3. A ladder leaning against a house reaches 72 feet, its foot being 30 feet from the house; what is the length of the ladder? Ans. 78 ft.

4. Two rafters, each 35 feet long, meet at the ridge of a roof 15 feet above the attic floor; what is the width of the house? Ans. 63.2454+ ft.

5. Two ships sail from the same port, one going due north, 8 miles an hour, and the other due east, 6 miles an hour; how far are they apart in three days? Ans. 720 miles.

6. A ladder 78 feet long stands close against a building; how far must it be drawn out at the foot, that the top may be lowered 6 feet? Ans. 30 ft.

7. A tree was broken 51 ft. from the top, and fell so that the end struck 24 feet from the foot; required the length of the tree. Ans. 96 feet.

8. A ladder 60 feet long, standing with its foot in the street, will reach on one side to a window 23 ft. high, and on the other to a window 37 ft. high; what is the width of the street? Ans. 102.65 ft.

9. A light-house was built upon a rock; if the distance from a point of observation to that point of the rock on a level with the eye is 620 meters, to the top of the rock is 846 meters, and to the top of the light-house 900 meters, what is the height of the light-house? Ans. 76.78 meters.

10. Required the distance between the lower corner and the upper opposite corner of a room 60 ft. long, 32 ft. wide, and 51 ft. high. Ans. 85 ft.