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), we know the root must consist of two figures, in as much as 29 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

23 23

9 square of units.

60 twice the product of 60 the tens by units. 400 square of the tons.

529 square of 23.

5 29 [20

4. 00

1 29

pose we will take the number 23, and square it. By this operation, it appears. that the square of a number consisting of tens and units is inade up of the square of the units, plus twice the product of the tens, by the units, plus the square of the tens. See this exhibited in figure F. As 10X10-100, the square of the tens can never make a part of the two right hand figures of the whole square. Hence the square of the tens is always contained in the second period, or in the 5 of the present example. The greatest square in 5 is 4, and its root 2; hence, we conclude, that the tens in the root are 2-20, and 20×20= 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 precedes it appears that 400 of the 529 fee

20 ft

E

in a square for
side, and that 1
square in such
und in order to
ade upon two

second period. From what of boards are now disposed of E measuring 20 feet on each feet are to be added to this nner as not to alter its form; > this, the additions must be des of the square, E-204-40 feet. Nif 129, the number of feet to be added, be divided by 40, the length of the additions, or, dropping the cipher and 9, if 12 be divided by 4, the quotient will be the width of the additions; and as 4 in 12 is had 3 times, we conclude the addition will be 3 feet wide, and 40×3 120 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 a

23 ft.

20

20

400 ft.

20 ft. 20X360.

F

20X20 400

23 ft.

20 ft.

the corner, as exhibited in F, whose sides are equal to the width of the additions, or 3 feet, and 3×3-9 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 seen, that we first find the root of the greatest square in the left hand 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 figure of the root; and the square of this

529 [ 23

43] 129

129

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

23X23529 proof.

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 figure of the root, and so on, till all the periods are brought down.

the side of a square, which shall contain an acre, or 160 rods? Ans. 12.649+ rods.

11. The area of a circle is 234.09 rods; what is the length of the side of a square of equal area?

Ans. 15.3 rods.

12. The area of a triangle Is 44944 feet; what is the length of the side of an equal aquare? Ans. 212 feet.

13. The diameter of a circle is 12 inches; what is the di

ameter of a circle 4 times as
large?
Ans. 24.

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?

14. The diameter of a circle is 121 feet; what is the diameter of a circle une half as 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.)

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.

QUESTIONS FOR PRACTICE.

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

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,

23 2X2X28

22X300+2×30=1260)4167

1200X3 3600

6×3×3 540
3X3X3= 27

4167

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 (20X3230) 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 blocks, 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

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.