360. The side of a square equal in area to any given surface, is found by extracting the square root of the given surface.

OBS. To find the dimensions of a rectangular field, equal in area to a given surface, when its length is double, triple, or quadruple, &c., of its breadth, find the square root of 1,,, of the given surface, and this will be the width; and the width being doubled, tripled, or quadrupled, as the case may be, will be the 'ength.

55. What is the side of a square, whose area is equal to that of a circle which contains 225 sq. yds.? Ans. 15 yds. 56. A general has 906304 soldiers: how many must he place in rank and file to form them into a square?

57. A man bought a square tract of land containing 3840 acres: how many rods square is the tract?

58. He afterwards divided his tract into four equal and square farms what is the length of one of their sides?

59. A man having a garden 465 yards square, wished to extend it so as to make it 9 times as large: how many yards square will it then be?

60. What is the side of a square, whose area is equal to that of a triangle containing 576 sq. ft.?

61. The length of a rectangular field containing 80 acres, is twice its breadth: what are its length and breadth?

62. The breadth of a rectangular farm containing 160 acres, isits length: what are its length and breadth?

63. What is the side of a square equal in area to a rectangular field 32 rods long and 18 rods wide?

361. The areas of all similar figures are to each other as the square of their similar sides or dimensions. (Leg. IV. 25, 27. V. 10.)

64. If a pipe 1 inch in diameter will fill a cistern in 60 minutes, how long will it take a pipe 2 inches in diameter to fill the same cistern?

65. If a gate 9 inches in diameter will empty a mill-pond in 16 hours, how large must a gate be to empty it in 4 hours?

QUEST.-360. How do you find the side of a square equal in area to any given surface? Obs. How find the dimensions of a rectangular field equal in area to a given surface, when its length is double, triple, &c., of its breath? 301. In what ratio are the areas of similar figures to each other.

66. If one side of the base of a triangular pyramid measuring 16 square feet, is 20 inches in length, what is the length of a side of a similar pyramid, which measures 36 square feet?

67. A man owns a building lot containing 20 square rods in the shape of a right-angled triangle, the perpendicular of which is 20 yards in length: what is the perpendicular of a a similar lot, which contains 30 square rods?

362. A mean proportional between two numbers is found by multiplying the given numbers together, and extracting the square root of the product. (Art. 320. Obs. 1.)

68. What is the mean proportional between 9 and 16? Solution.-16 x 9-144; and 144-12. Ans.

Find the mean proportional between the following numbers:

EXTRACTION OF THE CUBE ROOT.

363. To extract the cube root is to find a factor which being multiplied into itself twice, will produce the given number. (Art. 344.)

1. What is the side of a cubical block containing 27 solid feet? Suggestion.-Let the given block be represented by the adjoining cubical figure, each side of which is divided into 9 equal squares, which we will call square feet. Now, since the length of a side is 3 feet, if we multiply 3 into 3 into 3, the product 27, will be the solid contents of the cube. (Art. 154. Obs. 3.) Hence, if we reverse the pro

3x3x3=27.

QUEST.-362. How find a mean proportional between two numbers? What is it to extract the cube root ?

363

cess, i. e. if we resolve 27 into three equal factors, one of these factors will be the side of the cube. (Art. 344. Obs.) Ans. 3 ft.

2. A man wishes to form a cubical mound containing 15625 solid feet of earth: what is the length of its side?

1. We first separate the given number Into periods of three figures each, by placing a point over the units' figure, then over thousands. This shows us that the root must have two figures, (Art. 342. Obs. 3,) and thus enables us to find part of it at a time.

Operation. 15625(25

8

7625 div.

Divisor.
1200

300

25

1525 7625

2. Beginning with the left hand period, we find the greatest cube of 15 is 8, the root of which is 2. Placing the 2 on the right of the given number, we subtract its cube from the period, and to the remainder bring down the next period for a dividend. This shows that we have 7625 solid feet to be added to the cubical mound already found.

3. We square the part of the root thus found, which in reality is 20, for since there is to be another figure annexed to it, the 2 is tens; then multiplying its square 400 by 3, we write the product on the left of the dividend for a trial divisor; and finding it is contained in the dividend 5 times, we place the 5 in the root.

4. We next multiply 20, the root already found, by 5, the last figure placed in the root; then multiply this product by 3 and write it under the divisor. We also write the square of 5, the last figure placed in the root, under the divisor, and adding these three results together, multiply their sum 1525 by 5, and subtract the product from the dividend. The answer is 25.

PROOF.-Multiply the root into itself twice, and if the last product is equal to the given number, the work is right.

.

Thus, 25 x 25 x 25=15625.

OBS. The simplest method of illustrating the process of extracting the cube root to those unacquainted with algebra and geometry, is by means of cubical blocks. A set of these blocks contains 1st, a cube, the side of which is usually about 1 in. square; 2d, three side pieces about in. thick, the upper and lower base of which is just the size of a side of the cube; 3d, three corner pieces, whose ends are in. square, and whose length is the same as that of the side pieces; 4th, a small cube, the side of which is equal to the end of the corner pieces. It is desirable for every teacher and pupil to have a set. If not conveniently procured at the shops, any one can easily make them for himself.

364. From the preceding illustrations and principles, we derive the following general

RULE FOR EXTRACTING THE CUBE ROOT.

I. Separate the given number into periods of three figures each, placing a point over units, then over every third figure towards the left in whole numbers, and over every third figure towards the right in decimals.

II. Find the greatest cube in the first period on the left hand; then placing its root on the right of the number, subtract the cube from the period, and to the remainder bring down the next period for a dividend.

III. Square the part of the root thus found with a cipher annexed to it; multiply this square by 3, and place the product on the left of the dividend for a trial divisor; find how many times it is contained in the dividend, and place the result in the root.

IV. Multiply the root previously found with a cipher annexed by this last figure placed in it, then multiply this product by 3, and write the result under the divisor; under this result write also the square of the last figure placed in the root.

V. Finally, add these results to the trial divisor; multiply the sum by the last figure placed in the root, and subtract the product from the dividend. To the right of the remainder bring down the next period for a new dividend; find a new divisor as before, and thus proceed till the root of all the periods is found.

OBS. 1. When there is a remainder, periods of ciphers may be added, and the figures of the root thus obtained will be decimals.

2. If the right hand period of decimals is deficient, this deficiency must be supplied by ciphers.

3. When there are decimals in the given example, find the root as in whole numbers; then point off as many decimal figures in the answer, as there are per riods of decimals in the given number.

4. If the trial divisor is not contained in the dividend, place a cipher in the root, also two ciphers on the right of the divisor, and bring down the next period.

5. In finding the cube root of a common fraction, first reduce the fraction to its lowest terms, then extract the root of its numerator and denominator.

When either the numerator or denominator is not a perfect cube, the fraction should be reduced to a decimal, and the root of the decimal be found as above. A mixed number should be reduced to an improper fraction.

QUEST.-364. What is the first step in extracting the cube root? The second ? Third Fourth? Fifth? How is the cute root proved?

DEMONSTRATION BY CUBICAL BLOCKS.

1. The reason for dividing the number into periods of three figures, is two fold: First, it shows us how many figures the root will contain: Second, it enables us to find part of it at a time. Now, placing the large cube upon a table or stand, let it represent the greatest cube in the left hand period, which in the example above is 8, the root of which is 2. We subtract this cube from the left hand period, and to the remainder bring down the next period, in order to find how many feet remain to be added. In making this addition, it is plain the cube must be equally Increased on three sides; otherwise its sides will become unequal, and it will then cease to be a cube. (Art. 154. Obs. 2.)

2. The object of squaring the part of the root already found with a cipher annexed, is to find the area of one side of the cube. (Art. 153. Obs. 3.) The cipher is annexed because the part of the root thus found, denotes tens of the order next following. We multiply its square by 3, because the additions are to be made to three of its sides; and, dividing the dividend by this product shows the thickness of these additions. Now placing one of the side pieces on the top, and the other two on two adjacent sides of the cube, they will represent these additions. 3. But we perceive there is a vacancy at three corners, each of which is of the same length as the root already found, or the side of the cube, viz: 20 ft., and the breadth and thickness of each is 5 ft., the thickness of the side additions. Placing the corner pieces in these vacancies, they will represent the additions netessary to fill them. The object of multiplying the root already found by the figure last placed in it, is to obtain the area of a side of one of those additions; we then multiply this area by 3, to find the area of a side of each of them.

4. We find also another vacancy at one corner, whose length, breadth, and thickness are each 5 ft., the same as the thickness of the side additions. This vacancy therefore is cubical. It is represented by the small cube, which being placed in it, will render the mound an exact cube again. The object of squaring the 5, or the figure last placed in the root, is to find the area of a side of this cubical vacancy. We now have the area of one side of each of the side additions, the area of one side of each of the corner additions, and the area of one side of the cubical vacancy, the sum of which is 1525. We next multiply the sum of these areas by the figure last placed in the root, in order to find the cubical contents of the several additions. (Art. 154. Obs. 3.) These areas are added together, and their sum multiplied by the last figure placed in the root, for the sake of finding the solidity of all the additions at once. The result would obvi ously be the same, if we multiply them separately, and then subtracted the sum of their products from the dividend.

3. What is the cube root of 1728? 4. Cube root of 13824?

6. Cube root of 571787? 8. Cube root of 2?

5. Cube root of 373248?
7. Cube root of 1953125?
9. Cube root 2357947691 ?

QUEST.-Dem. Why separate the given number into periods of three figures each? Why subtract the greatest cube from the left hand period? Why square the part of the root already found? Why annex a cipher to it? Why multiply its square by 3? Why divide the dividend by this product? Why multiply the root already fourd by the last figure placed in it? Why multiply this produet by 3? Why square the figure last placed in the root? Why multiply the sum of these areas by the last figure placed in the root ?