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figure being necessary to preserve the form of the square, by filling the corner, we place it at the right of the divisor, in place of the cipher, which is always understood there, and then multiply the whole divisor by the last figure of the root. As we may conceive every root to be made up of tens and units, the

23x23 529 proof.

above reasoning may be applied to any number whatever, and may be given in the following general

529 [23

43] 129
129

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 divisor by the figure in the root last found; subtract the product 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 figure of the root, and so on, till all the periods are brought down.

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the side of a square, which | ameter of a circle 4 times as shall contain an aere, or 160 Ans. 24. rods? Ans. 12.649+ rods.

large?

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

1. In the right angled triangle, A B C, the side A C is 36 inches, and the side B C, 27 inches; what is the length of the side A B?

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B

hypothenuse.

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; subtract 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.)

Circles are to one another as the squares of their diameter; therefore square the given diameters, multiply or divide it by the given proportion, as the required diameters is to be greater or less than the given diameter, and the square root of the product, or quotient, will be the diameter required?

QUESTIONS FOR PRACTICE.

perpendicular.

14. The diameter of a circle is 121 feet; what is the diameter of a circle one half as Ans. 85.5+ feet.

base.

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

2025

A B2-
A B/2025 45 in. Ans.

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2. Suppose a man travel east 40 miles (from A to C), and then turn and travel north 30 miles (from C to B); how far is he from the place (A) where he started? Ans. 50 miles.

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3. A ladder 48 feet long will just reach from the opposite side of a ditch, known to be 35 feet wide, to the top of a fort; what is the height of the fort? Ans. 32.8+ feet.

4. A ladder 40 feet long, with the foot planted in the same place, will just reach a window on one side of the street 33 feet from the ground,

and one on the other side of the street, 21 feet from the ground; what is the width of the street?

Ans. 56.64+ feet. 5. A line 81 feet long, will exactly reach from the top of a fort, on the opposite bank of a river, known to be 69 feet broad; the height of the wall is required.

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 sought.

1. What is the mean proportional between 4 and 36? 36X4 144 and 144-12 Ans.

Then 4: 12 :: 12: 36.

Ans. 42.426 feet. 6. Two ships sail from the same port, one goes due east 150 miles, the other due north 252 miles; how far are they asunder? Ans. 293.26 miles.

QUESTIONS FOR PRACTICE.

2. What is the mean proportional between 49 and 64 ? Ans. 56. 3. What is the mean proportional between 16 and 64? Ans. 32.

EXTRACTION OF THE CUBE ROOT.

ANALYSIS.

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 2,

12167 (23 root but the value of 8 is 8000, and the 2 is 20, that is, 8000 feet of the stone will make a pile measuring 20 feet on each side, and (121678000) 4167 feet remain to be added to this pile in such a manner as to continue it in the form of a cube. Now it is obvious that the addition must be made upon 3 sides; and each side being 20 feet square, the surface upon which the additions

must be made will be (20×20×3-2×2×300) 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 (1200×3=), 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 additions here will be 3 feet square, and 60 feet long, and the quantity of stone added will be (60×3×3) 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 find (3×3×3=) 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 (3600+540+27) 4167 feet, just equal 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, to represent the additions upon the three sides, then 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 to fill the space at the corner. These, placed together in the manner described in the above operation, will render the reason of each step in the process perfectly clear. The process may be summed up in the following

23=2×2×28

22X300+2×30=1260)4167

1200X3 3600 60X3X3 540 3X3X3 27

4167

RULE.

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, calling it the triple quotient, and the sum of these call the divisor.

Seek how often the divisor may be had in the dividend, and place the result in the quotient. Multiply the triple square by the last quotient figure, and write the product under the dividend; multiply the triple quotient by the square of the last quotient figure, and place this product under the last; under these write the cube of the last quotient figure, and call their sum the subtrahend. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend, with which proceed as before; and so on, till the whole is finished.

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2. A bullet 3 inches in diameter weighs 4 lbs. what is the weight of a bullet 6 inches in diameter?

6. What is the cube root of 27054036008? Ans. 3002. 7. What is the cube root of 138?

Ans.

27

8. What is the cube root of

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1520

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Ans.

9. What is the cube root of 436036824287? Ans. 7583.

272. Solids of the same form are in proportion to one another as the cubes of their similar sides or diameters.

1. If a bullet, weighing 72 | 3×3×3=27 and 6×6×6—216 lbs. be 8 inches in diameter, Then 27: 4 :: 216. what is the diameter of a bulAns. 32 lbs. let weighing 9 lbs. ? 3. If a ball of silver 12 inch72:83:9: 64 Ans. 4 in. in diameter be worth $600, what is the worth of another ball, the diameter of which is 15 inches?

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Ans. $1171.87+.

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