19. The distance between the lower ends of two equal rafters is 32 feet, and the height of the ridge, above the beam on which they stand, is 12 feet; required the length of each rafter. Ans. 20 feet.

20. There is a building 30 feet in length and 22 feet 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 feet 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.; 15'62 ft.; 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 ? what is the

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 rods; and 1697 rods. 23. There is a square field containing 10 acres; what distance is the centre from each comer? Ans. 28'28 + rods.

EXTRACTION OF THE CUBE
ROOT.

¶ 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 T 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 uumber of feet.

1. What are the solid contents of a cubic block, of which each side measures 2 feet? Ans. 232 X2 X2=8 feet. 2. How many solid feet in a cubic block, measuring 5 feet on each side? Ans. 5 125 feet.

3. How many feet in length is each side of a cubic block, containing 125 solid feet? Ans. 125 5 feet. Note. The root may be found by trial.

feet? feet?

27 solid feet?

4. What is the side of a cubic block, containing 64 solid 216 solid feet? 512 solid Answers, 4 ft.; 3 ft.; 6 ft.; and 8 ft. 5. Supposing a man has 13824 feet of timber, in separate blocks of 1 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 endeavour, 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. This we may do by pointing the number off into periods of three figures each (T 107, ex. 1.)

OPERATION.

13824(2

8

5824

FIG. I.

20

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 8000 feet, Contents. 20X20X20 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

20

400

20

the period 13, (thousands,) and bringing down the next period 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 20=400) 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 X 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.

=

=

The divisor, 1200, is con

OPERATION-CONTINUED. tained in the dividend 4 times;

consequently, 4 feet is the thickness of the addition made to each of the three sides, a, b, c, and 4 X 1200 4800, is the solid feet contained in these additions; but, if we look at Fig. II., we shall perceive, that this addition to the 3 sides does not complete the cube; for there are deficiencies in the 3 corners n, n, n. Now the length of each of these deficiencies is the same as the length of each side, that is, 2 (tens) 20, and their width and thickness are each equal to the last quotient figure, (4); their contents, therefore, or the number of feet required to fill these deficiencies, will be found by multiplying the square of the last quotient figure, (42) =16, by the length of all the deficiencies, that is, by 3 times

the length of each side, which is expressed by the former quotient figure, 2, (tens.) 3 times 2 (tens) are 6 (tens) = 60; or, what is the same in effect, and more convenient in practice, we may multiply the quotient figure, 2, (tens,) by 30, thus, 2 X 30= 60, as before; then, 60 X 16=960, contents of the three deficiencies n, n, n.

Looking at Fig. III., we perceive there is still a deficiency in the corner where the last blocks meet. This deficiency is a cube, each side of which is equal to the last quotient figure, 4. The cube of 4, therefore, (4 × 4 X 464,) will be the solid contents of this corner, which in Fig. IV. is seen filled.

Now, the sum of these several additions, viz. 4800 + 960645824, will make the subtrahend, which, subtracted from the dividend, leaves no remainder, and the work is done.

Fig. IV. shows the pile which 13824 solid blocks of one foot each would make, when laid together, and the root, 24, shows the length of one side of the pile. The correctness of the work may be ascertained by cubing the side now found, 243, thus, 24 X 24 X 24 = 13824, the proved by adding together

given number; or it may be
the contents of all the several parts, thus,

addition to the sides a, b, and c, Fig. 1. addition to fill the deficiencies n, n, n, Fig. II. addition to fill the corner e, e, e, Fig. IV.

contents of the whole pile, Fig. IV., 24 feet on

From the foregoing example and illustration we derive the following

RULE

FOR EXTRACTING THE CUBE KOOT.

I. Separate the given number into periods of three figures each, by putting a point over the unit figure, and every third figure beyond the place of units.

II. Find the greatest cube in the left hand period, and pu its root in the quotient.

III. Subtract the cube thus found from the said period, and to the remainder bring down the next period, and call this the dividend.

IV. Multiply the square of the quotient by 300, calling it. the divisor.

V. Seek how many times the divisor may be had in the dividend, and place the result in the root; then multiply the divisor by this quotient figure, and write the product under the dividend.

VI. Multiply the square of this quotient figure by the former figure or figures of the root, and this product by 30, and place the product under the last; under ail write the cube of this quotient figure, and call their amount the subtrahend.

VII. 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.

Note 1. If it happens that the divisor is not contained in the dividend, a cipher must be put in the root, and the next period brought down for a dividend.

Note 2. The same rule must be observed for continuing the operation, and pointing off for decimals, as in the square root.

Note 3. The pupil will perceive that the number which we call the divisor, when multiplied by the last quotient figure, does not produce so large a number as the real subtrahend; hence, the figure in the root must frequently be smaller than the quotient figure.