7. A trec, 140 ft. high, is in the center of a circular island 100 ft. in diameter; a line 600 feet long, reaches from the top of the tree to the further shore: what is the breadth of the stream, the land on each side being of the same level? Ans. 533.43+ ft.

8. A room is 20 ft. long, 16 ft. wide, and 12ft. high: what is the distance from one of the lower corners to the opposite upper corner? Ans. 28.28+ft.

ART. 291. Since the area or superficial contents of a square equals the square of one of its sides, (Art. 87), hence, the

RULE FOR FINDING THE SIDE OF A SQUARE EQUAL IN AREA

TO ANY GIVEN SURFACE.

Extract the square root of the given area; the root will be the side of the required square.

1. The superficial contents of a circle are 4096: what the side of a square of equal area?

Ans. 64. 2. A square field measures 4 rd. on each side: what the length of one side of a square field having 9 times as many sq. rd.? Ans. 12 rd.

3. There are 43560 sq. ft. in 1 A.: what is each side of a square, containing 1 A, A, A?

4

Ans. 208.71+ft.; 147.58+ft.; and 104.35+ft.

4. A man has 2 fields; 10 A. and 121 A.: find the side of a sq. field equal in area to both.

Ans. 60 rd.

EXTRACTION OF THE CUBE ROOT.

ART. 292. To extract the cube root of a number, is to resolve it into THREE equal factors; or, to find a number which, when multiplied by itself twice, will produce the given number.

Thus, 4 is the cube root of 64, because 4X4X4=64. Roots. 1, 2, 3, 4, 5, 6, 7, 8, 9. Cubes. 1, 8, 27, 64, 125, 216, 343, 512, 729.

REVIEW.-291. How find the side of a square equal in area to any given surface? 292. What is it to extract the cube root of a number?

ART. 293. From Art. 278, it follows that the cube root of a number expresses the side of a cube whose solid contents are equal to the given number.

Hence, extracting the cube root, is finding the side of a cube when its solid contents are known; or, arranging a given number of cubes, so as to form the largest cube possible.

ART. 294. From Art. 280, it follows that,

The cube root of 1000000 is 100; and so on: hence,

The cube root of a number between 1 and 1000, consists of one figure; between 1000 and 1000000, of two figures; between 1000000 and 1000000000, of three, &c.: hence,

RULE FOR POINTING.-If a dot (.) be placed over every 3d figure of any given number, beginning with units, the number of dots will denote the number of figures in the cube root.

ART. 295. 1. Extract the cube root of 13824; or, suppose 13824 cubic blocks, each 1 in. long, 1 in. wide, and 1 in. thick, are to be arranged in the form of a cube.

in 13 (thousand) is 8 (thousand), the cube root of which is 2 (tens), which place on the right, as in extracting the square root.

Subtract the cube of 2 (tens), which is 8 (thousand), from the given number, and 5824 remain.

While solving this example by figures, attend to arranging the cubic blocks. After finding that the cube root of the given number

REVIEW.-292. Of what numbers are the nine digits the cube roots? 293. What does the cube root of a number express ?

294. What the cube root of a number between 1 and 1000? Why? Of a number between 1000 and 1000000? Why? What the rule for pointing?

will contain two places of figures, (tens and units,) and that the figure in the tens' place is 2, form a cube, A, 20 (2 tens) inches long, 20 in. wide, and 20 in. high; this cube will contain (Art. 92) 20 X 20 X 20=8000 cu. in.; take this sum from the whole number of cubes, and 5824 cu. in. are left, which correspond to the number 5824 in the numerical operation.

It is obvious that to increase the figure A, and at the same time preserve it a cube, the length, breadth, and hight, must each receive an equal addition.

Then, since each side is 20 in. long, square 20, which gives 20×20=400, for the number of sq. in. in each face of the cube; and since an addition is to be made

to three sides, multiply the 400 by 3, which

FIG. 1.

20 wide

20 high

A

gives 1200 for the number of square inches in the 3 sides.

This 1200 is called the TRIAL DIVISOR; because, by means of it, the thickness of the additions may be determined.

By examining Fig. 2 (or the blocks, see Note 6), it will be seen, that after increasing each of the three sides equally, there will be required 3 oblong solids, C, c, c, of the same length as each of the sides, and whose thickness and hight are each the same as the additional thickness; and also a cube, D, whose length, breadth, and hight, are each the same as. the additional thickness.

Hence, the solid contents of the first three rectangular solids, the three oblong solids, and the small cube, must together be equal to the remaining cubes (5824).

Now find the thickness of the additions. It will always be something less than the number of times the trial divisor (1200) is contained in the dividend (5824). FIG. 2.

By trial we find 1200 is contained 4 times in 5824. Place the 4 in the quotient, and proceed to find the contents of the different solids: these added together, make the number to be subtracted, called the subtrahend.

The solid contents of the first three additions, B, B, B, are found, (Art. 93)

B

D

by multiplying the number of sq. in. in the face by the thickness;

now there are 400 sq. in. in the face of each, and 400 × 3=1200 sq., in. in one face of the three; then multiplying by 4, (the thickness,) gives 4800 cu. in. for their contents.

The solid contents of the three oblong solids, C, c, c, are found (Art. 93) by multiplying the number of sq. in. in the face by the thickness; now there are 20 X 4 = 80 sq. in. in one face of each, and 80 X 3240 sq. in. in one face of the three; then multiplying by 4, (the thickness,) gives 960 cu. in. for their contents.

Lastly, find the contents of the small cube, D, by multiplying its length (4) by its breadth (4), and that product by the' thickness (4); this gives 4X4 × 4 = 64 cu. in.

If the solid contents of the several additions be added together, their sum, 5824 cu. in., will be the number of small cubes remaining after forming the first cube, A.

ADDITIONS.

B, =4800 cu. in. 960 "

B, B,

C, c,

c,=

D, =

64"

Sum 5824

Hence, when 10824 cu. in. are arranged in the form of a cube, each side is 24 in.; that is, the cube root of 13824 is 24.

In finding the solid contents of the additions, in each case the last multiplier is the thickness.

To produce the same result more conveniently, find the area of one face of each of the additional solids, then the sum of the areas, (as in the numerical operation,) and multiply it by the common thickness.

The sum of the areas of one face of each of the additional solids, is termed the COMPLETE DIVISOR. Thus,

In the preceding operation, 1456 is the complete divisor.

NOTE. As the 1st figure of the root is always in the tens' place with regard to the 2d, annex to it a cipher before it is squared; or, omit the cipher, and multiply the square by 300 instead of 3.

REVIEW.-295. How obtain the 1st figure of the root? Why square it? Why multiply by 3? What is the product called? Why? 295. How obtain the 2d figure of the root? Why multiply the 1st figure by the 2d? Why multiply their product by 3? Why square the 2d figure of the root? How find the subtrahend? What is the complete divisor? 295. NOTE. Why is a cipher annexed to the 1st figure of the root before squaring? If the cipher is omitted, what must be done? What if the vipher is omitted in multiplying the 1st figure of the root by the 2d?

3d Bk.

19

For the same reason, annex a cipher to the figure first obtained, before multiplying it by the 2d (the thickness):

Or, omit the cipher, and multiply by 30 instead of 3.

Thus, 20 x 20 × 3 =1200

2 X 2 X 300 = 1200

20 × 4 × 3 = 240

Or, 2 X 4 X

20:

= 240

*2. What is the cube root of 1728?

3. Find the cube root of 413493625.

EXPLANATION.-By the rule for pointing (Art. 294), the root will contain 3 figures. Find the 1st and 2d figures of the root as in the preceding examples. Then consider 74 as so many tens, and find the 3d figure in the same manner as the 2d was obtained. 4. Find the cube root of 515.849608

EXP.-After obtaining the 1st figure, and bringing down the 20 period, we find the trial divisor is not contained in the dividend;