period 6, (hundreds,) and bringing down the next period by the side of the remainder, making 225, as before.

3d. The square A is now to be enlarged by the addition of the 225 remaining yards; and, in order that the figure may retain its square form, it is evident, the addition must be made on two sides. Now, if the 225 yards be divided by the length of the two sides, (20+20= 40,) the quotient will be the breadth of this new addition of 225 yards to the sides c d and b c of the square A.

But our root already found, = 2 tens, is the length of one side of the figure A; we therefore take double this root, 4 tens, for a divisor.

The divisor, 4, (tens) is in reality 40, and we are to seek how many times 40 is contained in 225, or, which is the same thing, we may seek how many times 4 (tens) is contained in 22, (tens,) rejecting the right hand fig ure of the dividend, because we have rejected the cipher in the divisor. We find our quotient, that is, the breadth of the addition, to be 5 yds.; but, if we look at Fig. II., we shall perceive that this addition of 5 yards to the two sides does not complete the square; for there is still wanting, in the corner D, a small square, each side of which is equal to this last quotient, 5; we must, therefore, add this quotient, 5, to the divisor, 40, that is, place it at the right hand of the 4, (tens,) making it 45; and then the whole divisor, 45, multiplied by the quotient, 5, will give the contents of the whole addition around the sides of the figure A, which, in this case, being 225 yards, the same as our di vidend, we have no remainder, and the work is done. Consequently, room, 25 yards on a side, which 625

Fig. II. represents the floor of a square yards of carpeting will exactly cover. The proof may be seen by adding together the several parts of the figure, thus:

The square A contains 400 yards.

The figure B .......... 100

.....

........ C

100

Or we may prove it by involution, thus:-25 X 25 = 625, as before.

From this example and illustration, we derive the following general

RULE

FOR THE EXTRACTION OF THE SQUARE ROOT.

I. Point off the given number into periods of two figures each, by putting a dot over the units, another over the hundreds, and so on. These dots show the number of figures of which the root will consist.

11. Find the greatest square number in the left hand period, and write its root as a quotient in division. Subtract the square number from the left hand period, and to the remainder bring down the next period for a dividend.

III. Double the root already found for a divisor; seek how many times the divisor is contain ed in the dividend, excepting the right hand figure, and place the result in the root, and also at the right hand of the divisor; multiply the divisor, thus augmented, by the last figure of th root, and subtract the product from the dividend, to the remainder bring down the next period for a new dividend.

IV. Double the root already found for a new divisor, and continue the operation as before, until all the periods are brought down.

Note 1. If we double the right hand figure of the last divisor, we shall have the double of the root.

Note 2. As the value of figures, whether integers or decimals, is determined by their distance from the place of units, so we must always begin at unit's place to point off the given number, and, if it be a mixed number, we must point it off both ways from units, and if there be a deficiency in any period of decimals, it may be supplied by a cipher. It is plain, the root must always consist of so many integers and decimals as there are periods belonging to each in the given number.

EXAMPLES FOR PRACTICE.

2. What is the square root of 10342656 ?

¶ 108. In this last example, as there was a remainder, after bringing down all the figures, we continued the operation to decimals, by annexing two ciphers for a new period, and thus we may continue the operation to any assigned degree of exactness; but the pupil will readily perceive that he can never, in this manner, obtain the precise root; for the last figure in each dividend will always be a cipher, and the last figure in each divisor is the same as the last quotient figure; but no one of the nine digits multiplied into itself produces a number ending with a cipher; therefore, whatever be the quotient figure, there will still be a remainder.

11. What is the square root of 3?
12. What is the square root of 10?
13. What is the square root of 184'2?
14. What is the square root of 1

Ans. 1'73

Ans. 3'16

Ans. 13'57

Note. We have seen, (T 105, ex. 9,) that fractions are squared by squar. ing both the numerator and the denominator. Hence, it follows that the square root of a fraction is found by extracting the root of the numerator and of the denominator. The root of 4 is 2, and the root of 9 is 3.

Ans, 3.

Ans. . Ans..

When the numerator and denominator are not exact squares, the fraction may be reduced 40 a decimal, and the approximate root found, as directed above.

1. What is involution? 2. What is understood by a power? 3. the first, the second, the third, the fourth power? 4. What is the index, or exponent? 5. How do you involve a number to any required power? 6. What is evolution? 7. What is a root? 8. Can the precise root of all numbers be found? 9. What is a surd number? 10. a rational? 11. What is it to extract the square root of any number? 12. Why is the given sum pointed into periods of two figures each? 13. Why do we double the root for a divisor? 14. Why do we, in dividing, reject the right hand figure of the dividend ? 15. Why do we place the quotient figure to the right hand of the divisor? 16. How may we prove the work? 17. Why do we point off mixed numbers both ways from units? 18. When there is a remainder, how may we continue the operation? 19. Why can we never obtain the precise root of surd numbers? 20. How do we extract the square root of vulgar fractions?

EXERCISES.

1. A general has 4096 men; how many must he place in rank and file to form them into a Bquare 7 Ans, 64. 2. If a square field contains 2025 square rods, how many rods does it measure on each side? Ans. 45 rods. Ans. 75.

3. How many trees in each row of a square orchard containing 5625 trees? 4. There is a circle whose area, or superficial contents, is 5184 feet: what will be the length of the side of a square of equal area? 518472 feet, Ans.

5. A has two fields, one containing 40 acres, and the other containing 50 acres, for which B offers him a square field containing the same number of acres as both of these; how many rods must each side of this field measure? Ans. 120 rods.

6. If a certain square field measure 20 rods on each side, how much will the side of a square field measure containing 4 times as much? 20X20X4=40 rods, Ans. 7. If the side of a square be 5 feet, what will be the side of one 4 times as large? - 9 times as large? 16 times as large? 25 times as large ?- 36 times as large? Answers, 10 ft.; 15 ft.; 20 ft.; 25 ft.; and 30 ft. 8. It is required to lay out 288 rods of land in the form of a parallelogram, which shall le twice as many rods in length as it is in width. Note. If the field be divided in the middle, it will form two equal squares.

Ans. 24 rods long, and 12 rods wide. 9. I would set out, at equal distances, 784 apple trees, so that my orchard may be 4 times as long as it is broad; how many rows of trees must I have, and how many trees in each row? Ans. 14 rows, and 56 trees in each row. 10. There is an oblong piece of land, containing 192 square rods, of which the width is as much as the length; required its dimensions. Ans. 16 by 12.

11. There is a circle whose diameter is 4 inches; what is the diameter of a circle 9 times as large ?

Note. The areas or contents of circles are in proportion to the squares of their diameters, or of their circumferences. Therefore, to find the diameter required, square the given dianc ter, multiply the square by the given ratio, and the square root of the product will be the diameter required. 4 X 4X 912 inches, Ans. 12. There are two circular ponds in a gentleman's pleasure ground; the diameter of the less is 100 feet, and the greater is 3 times as large; what is its diameter ? Ans. 1732 feet. 13. If the diameter of a circle be 12 inches, what is the diameter of one as large? Ans. 6 inches.

T 109. 14. A carpenter has a large wooden square; one part of it is 4 feet long, and the other part 3 feet long; what is the length of a pole which will just reach from one end to the other?

Hypotenuse.

A

Perpendicular.

Note. A figure of 3 sides is called a triangle, and, if one of the corners be a square corner, or right angle, like the angle at B in the annexed figure, it is called a right-angled triangle, of which the square of the longest side, A C, (called the hypote nuse,) is equal to the sum of the squares of the other two sides, A B and B C. 4216, and 329; then, 9+16= 5 feet, Ans. 15. If, from the corner of a square room, 6 feet be measured off one way, and 8 feet the other way, along the sides of the room, what will be the length of a pole reaching from point to point?

Base.

B

Ans. 10 feet. 16. A wall is 32 feet high, and a ditch before it is 24 feet wide; what is the length of a ladder that will reach from the top of the wall to the opposite side of the ditch ? Ans. 40 feet.

17. If the ladder be 40 ft. and the wall 32 ft., what is the width of the ditch? 18. The ladder and ditch given, required the wall.

Ans. 24 ft.
Ans. 32 ft.

19. The distance between the lower ends of two equal rafters is 32 ft. and the height of the ridge above the beam on which they stand is 12 ft.; required the length of each rafter. Ans. 20 ft. 20. There is a building 30 ft. in length and 22 ft. in width, and the eaves project beyond the wall 1 foot on every side; the roof terminates in a point at the centre of the building, and is there supported by a post, the top of which is 10 ft. above the beams on which the rafters rest; what is the distance from the foot of the post to the corners of the eaves; and what is the length of a rafter reaching to the middle of one side? a rafter reaching to the middle of one end? and a rafter reaching to the corners of the eaves? Answers, in order, 20 ft.; 15'62ft.; 18'86+ft.; and 22'36+ ft. 21. There is a field 800 rods long and 600 rods wide; what is the distance between two opposite corners ? Ans. 1000 rods.

22. There is a square field containing 90 acres; how many rods in length is each side of the field; and how many rods apart are the opposite corners? Answers, 120 r.; and 169'7+ r. 23. There is a square field containing 10 acres; what distance is the centre from each corner ? Ans. 28 28+ rods.

EXTRACTION OF THE CUBE ROOT.

T 110. A solid body, having six equal sides, and each of the sides an exact square, is a CUBE, and the measure in length of one of its sides is the root of that cube; for the length, breadth, and thickness of such a body are all alike; consequently, the length of one side, raised to the 3d power, gives the solid contents. (See ¶ 36.)

Hence it follows, that extracting the cube root of any number of feet is finding the length of one side of a cubic body, of which the whole contents will be equal to the given number of feet. 1. What are the solid contents of a cubic block, of which each side measures 2 feet 7 Ans. 232 X 2 X 2 = 8 fect. 2. How many solid feet in a cubic block, measuring 5 feet on each side?" 3. How many feet in length is each side of a cubic block, containing 125 solid feet?

Note. The root may be found by trial.

4. What is the side of a cubic biock, containing 64 solid feet? 216 solid feet? -512 solid feet?

Ans.3255 feet

27 solid feet? Answers, 4 ft.; 3 ft.; 6 ft.; and 8 ft. 5. Supposing a man has 13824 feet of timber, in separate blocks of I cubic foot each; he wishes to pile them up in a cubic pile; what will be the length of each side of such a pile?

It is evident, the answer is found by extracting the cube root of 13824; but this number is so large, that we cannot so easily find the root by trial as in the former examples. We will endeavor, however, to do it by a sort of trial; and,

1st. We will try to ascertain the number of figures of which the root will consist. may do by pointing the number off into periods of three figures each, ( 107, ex. 1.)

This we

Pointing off, we see, the root will consist of two figures, a ten and a unit. Let us, then, seek for the first figure, or tens of the root, which must be extracted from the left hand period, 13, (thousands.) The greatest cube in 13 (thousands) we find by trial, or by the table of powers, to be 8, (thousands,) the root of which is 2, (tens ;) therefore we place 2 (tens) in the root. The root, it will be recollected, is one side of a cube. Let us, then, form a cube, (Fig. 1.) each side of which shall be supposed 20 feet, expressed by the root now obtained. The contents of this cube are 20 X 20 X 20 8000 solid feet, which are now disposed of, and which, consequently, are to be deducted from the whole number of feet, 13824. 8000 taken from 13824 leave 5824 feet. This deduction is most readily performed by subtracting the cubic number, 8, or the cube of 2, (the figure of the root already found,) from the period 13, (thousands,) and bringing down the next pe riad by the side of the remainder, making 5824, as before.

2d. The cubic pile A D is now to be enlarged by the addition of 5824 solid feet, and, in order to preserve the cubic form of the pile, the addition must be made on one half of its sides, that is, on 3 sides, a, b, and c. Now, if the 5824 solid feet be divided by the square contents of these 3 equal sides, that is, by 3 times, (20 X 20400) 1200, the quotient will be the thickness of the addition made to each of the sides a, b, c. But the root, 2, (tens,) already found, is the length of one

of these sides; we therefore square the root, 2, (tens,) = 20 X 20 = 400, for the square contents of one side, and multiply the product by 3, the number of sides, 400 × 3 = 1200; or, which is the same in effect, and more convenient in practice, we may square the 2, (tens,) and multiply the product by 300, thus, 2 x 2 = 4, and 4 X 300 1200, for the divisor, as before.