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1. If 529 feet of boards be laid down in a square form, what will be the length of the sides of the square ? or, in other words, what is the square root of 529?

From what was shown (264), wc know the root must consist of two fig. ures, in as much as 629 consists of two periods. Now to understand the method of ascertaining these two figures, it may be well to consider how the square of a root consisting of two figures is formed. For this pur

pose we will take the number 23, and 23

square it. By this operation, it appears. 23

that the square of number cousisting

of tens and units is inade up of the 9 square of units.

squarc of the units, plus twice the pro60 ) twice the product of duct of the tens, by the units, plus the 60 | the tens by units. square of the tens. See this exhibited 400 square of the tcns. in figure F. As 10x10=100, the square

of the tens can never make a part of the 529 square of 23.

two right hand figures of the whole square. Hence the square of the tens is always contained in the second peri

ol, or in the 5 of the present example. 4 00

The greatest square in 5 is 4, and its

root %; hence, we conclude, that the 1 29

tens in the root are 2–20, and 20X20

400. But as the square of the tens can never contain significant figures below hundreds, we need only write the square of the figure denoting tens under : second period. From what precedes it appears that 400 of the 529 fee f boards are now disposed of

in a square fori

. E measuring 20 feet on each 20 ft

side, and that 1 feet are to be added to this square in suchi

nner as not to alter its form; und in order to this, the additions must be E

muude upon two des of the square, E—20+

-10 feet. Nif 129, the number of feet to 20

be added, be divided by 40, the length of the 1 20

additions, or, dropping the cipher and $, if 12 be

divided by 4, the quotient will be the width of 400 ft.

the additions; and as 4 in 12 is had 3 times, we conclude the addition will be 3 feet wide, and 40X3120 feet, the quantity added upon

the two sides. But since these additions are no longer than the sides of the square, E, there must be a deficiency at

the corner, as exhibited in F, whose sides 20 ft.

3 ft. are equal to the width of the additions, 20X3-60.

or 3 feet, and 3X3=9 feet, required to fill out the corner, so as to complete the square. The whole operation may be arranged as on the next page, where it will be

20 ii.



that we first find the root

of the greatest square in the left hand 20x20=400

period, place it in the form of a quotient, subtract the square from the period and to the remainder bring down the next period, which we divide, omitting the right hand figure, by double the root,

and place the quotient for the second 23 ft.

figure of the root; and the square of this

23 ft.

20 ft. 20x3=60.

529 ( 23

figure being necessary to preserve the form of the square, by filling the corner,

we place it at the right of the divisor, in 43 ] 129

place of the cipher, which is always un129

derstood there, and then multiply the

whole divisor by the last figure of the 23x23529 proof.

root. As we may conceive every root

to be inade up of tens and units, the above reasoning may he applied to any number whatever, and may be given in the following gcueral

RULE. 267. Distinguish the given numbers into periods ; find the root of the greatest square number in the left hand period, and place the root in the manner of a quotient in division, and this will be the highest figure in the root required. Subtract the square of the root already found from the left hand period, and to the remainder bring down the next period for a dividend. Double the root already found for a divisor; seek how many times the divisor is contained in the dividend (excepting the right hand figure), and place the result for the next figure in the root, and also on the right of the divisor. Multiply the aivisor by the figure in the root last found; subtract the proauct from the dividend, and to the remainder bring down the next period for a new dividend. Double the root now found for a divisor, and proceed, as before, to find the next ligure of the root, and so on, till all the periods are brought down.

of ?

Ans. g.

QUESTIONS FOR PRACTICE. 1. What is the square root

6. What is the square root of 529? 2. What is the square root

Ans. .64549+. of 2? Ans. 1.4:49+

Reduce to a decimal and The decimals are found by an then extract the root (130). bexing pairs of ciphers continually to the remainder for a new dividend.

7. What is the square root in this way a surd root may be obained to any assigned degree of of 1?

8. What is the square root 3. What is the square root

of 143? of 182.25 ? Ans. 13.5.

4. What is the square root 9. An army of 567009 men of .0003272481 ?

are drawn up in a solid body, Ans. .01809.

in form of a square; what is Hence the root of a decimal is the number of men in rank greater than iis powers.

and file ?

Ans. 753. 5. What is the square root of 5499025?

Ans. 2:345. 10. What is the length of


Ans. 14

the side of a square, which I aineter of a circle 4 times as shall contain an acre, or 160 large ?

Ans. 24. rods? Ans. 12.619+ rods.

Circles are to one another as the 11. The area of a circle is squares of their diameter ; therefore 234.09 rods; what is the length square the given diameters, multiply of the side of a square of or divide it by the given proportion, equal area?

as the required diameters is to be

greater or less than the given diamAns. 15.3 rods. eter, and the square root of the

pro12. The area of a triangle duct, or quotient, will be the diamis 44944 feet; what is the

eter required ? length of the side of an equal 14. The diameter of a circle aquare ? Ans. 212 feet.

is 121 feet; what is the diam13. The diameter of a circle i eter of a circle une half as is 12 inches; what is the di- | large ? Ans. 85.5+ feet.

268. Having two sides of a right angled triangle given to find the other side.

RULE.—Square the two given sides, and if they are the two sides which include the right angle, that is, the two shortest sides, add them together, and the square root of the sum will be the length of the longest side ; if not, the two shortest; sub- · tract the square of the less from that of the greater, and the square root of the remainder will be the length of the side required. (See demonstration, Part I. Art. 68.)

QUESTIONS FOR PRACTICE. 1. In the right angled tri If A B be 45 inches, and angle, A B C, the side A C is A C 36 inches, what is the 36 inches, and the side B. C, / length of B C? 27 inches; what is the length A B2=45X45=2025 of the side A B?

A C2=36X36=1296

B C2=

729 BC=729=27 in. Ans.

If A B=45, B C=27in., what is the length of A C?

A B2=452, B C2=272, A base.

C2 and A C=1296336in.

A C2=36X36=1296
B C2=27X27= 729



A B2=

2025 A B=\/2025=45 in. Ans.

2. Suppose a man travel east and one on the other side of 40 miles (from A to C), and the street, 21 feet from the then turn and travel north 30 | ground; what is the width of miles (from C to B); how far the street? is he from the place (A) where

Ans. 56.64+ feet. he started ? Ans. 50 miles. 5. A line 81 feet long, will

3. A ladder 48 feet long exactly reach from the top of will just reach from the oppo a fort, on the opposite bank of site side of a ditch, known to a river, known to be 69 feet be 35 feet wide, to the top of broad; the height of the wall a fort ; what is the height of is required. the fort? Ans. 32.8+ feet.

Ans. 42.426 feet. 4. A ladder 40 feet long, 6. Two ships sail from the with the foot planted in the same port, one goes due east same place, will just reach a 150 miles, the other due north window on one side of the 252 miles; how far are they street 33 feet from the ground, I asunder? Ans. 293.26 miles.

269. To find a mean proportional between two numbers. RULE.-Multiply the two given numbers together, and the square root of the product will be the mean proportional soughta

QUESTIONS FOR PRACTICE. 1. What is the mean propor

2. What is the mean pro tional between 4 and 36 ? portional between 49 and 64 ? 36X4=144 and 144=12

Ans. 56. Ans.

3. What is the mean proThen 4 : 12 :: 12 : 36. portional between 16 and 64?

Ans. 32.



270. To extract the cube root of a given number, is to find a number which, multiplied by its square, will produce the given number, or it is to find the length of the side of a cube of which the given number expresses the content.

1. I have 12167 solid feet of stone, which I wish to lay up in a cubical pile; what will be the length of the sides ? or, in other words, what is the cube root of 12167 !

By distinguishing 12167 into periods, we find the root will consist of two figures (265). Since the cube of tens (264) can contain no significant figures less than thousands, the cube of the tens in the root must be found in the left hand period. The greatest cube in 12 is 8, whose root is %;

12167 (23 root but the value of 8 is 8000, and the 234X2X2=8

2 is 20, that is, 8000 feet of the

stone will make a pile measuring 20 29X30042X30=1260)4167 feet on each side, and (12167–

8000–) 4167 feet remain to be add. 1200X33600

ed to this pile in such a manner as 63X3X3= 510

to continue it in the form of a cube. 3X3X3 27

Now it is obvious that the addition

must be made upon 3 sides; and 4167

each side being 20 feet square, the

surface upon which the additions must be made will be (20X20X3_2X2X300=) 1200 feet, but when these additions are made, there will evidently be three deficiencies along the lines where these additions come together, 20 feet long, or (20X32X30=-) 60 feet, which must be filled in order to continue the pile in a cubic form. Thus the points upon which the additions are to be made, are (1200-+60=) 1260 feet and 4167 feet, the quantity to be added divided by 1260, the quotient is (4167-1260) 3, which is the thickness of the additions, or the other figure of the root. Now if we multiply the surface of the three sides by the thickness of the additions, the product (1200X3), 3600 feet, is the quantity of stone required for those additions. Then to find how much it takes to fill the deficiencies along the line where these additions come together, since the thickness of the additions upon the sides is 3 feet, the auditions here will be 3 feet square, and 60 feet loug, and the quantity of stone added will be (60X3X3=) 540 feet. But after these additions there will be a deficiency of a cubical form, at the corner, between the ends of the last mentioned additions, the three dimensions of which will be just equal to the thickness of the other additions, or 3 feet, and cubing 3 feet we fiind (3X3X3=) 27 feet of stone required to fill this corner, and the pile is now in a cubic form, measuring 23 feet on every side, and adding the quantities of the additions upon the sides, the edges, and at the corner together, we find them to amount to (36007-5407-27=) 4167 feet, just cqual to the quantity remaining of the 12167, after taking out 8000. To illustrate the foregoing operation, make a cubic block of a convenient size to represent the greatest cube in the left hand period. Make 3 other square blocks, each equal to the side of the cube, and of an indefinite thickness, lo represent the additions upon the three sides, theu 3 other blocks, each equal in length to the sides of the cube, and their other dimensions equal to the thickness of the square blocks, to represent the additions along the edges of the cube, and a small cubic block with its dimensions, each equal to the thickness of the square blocks, to fill the space at the corner.. These, placed together in the manner described in the above operation, will render The reason or each step in the process perfectly clear. The process may bo summed up in the following


271. 1. Having distinguished the given number into periods, of three figures each, find the greatest cube in the left hand period, and place its root in the quotient. Subtract the cube from the left hand period, and to the remainder bring down the next period for a dividend. Multiply the square of the quotient by 300, calling it the triple square, and the quotient by 30, call. ing it the triple quotient, and the sum of these call the divisor.

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