Problem V. A line 36 yards long will exactly reach from the top of a fort to the opposite bank of a river, known to be 24 yards broad. The height of the wall is required 7 36x36=1296 ; and 24×24=576. Then, 1296–576=720, and 4/720–26-83+yards, the Answer. Problem VI. The height of a tree growing in the centre of a circular island 44 feet in diameter, is 75 feet, and a line stretched from the top of it over to the hither edge of the water, is 256 feet. What is the breadth of the stream, provided the land on each side of the water be level ? Problem VII. Suppose a ladder 60 feet long be so planted as to reach a window 37 feet from the ground, on one side of the street, and without moving it at the foot, will reach a window 23 feet high on the other side ; I demand the breadth of the street 7 Ans. 102.64 feet, Problem VIII. Given the difference of two numbers, and the difference of their squares, to find the numbers. RULE.—Divide the difference of the squares by the difference of the numbers, and the quotient will be their sum. Then proceed by Prob. 4, p. 57. ExAMPLEs. 1. The difference of two numbers is 20, and the difference of their squares is 2000; what are the numbers ? 2. Said Harry to Charles, my father gave me 12 apples more than he gave brother Jack, and the difference of the squares of our separate parcels was 288 : Now, tell me how many he gave us, and you shall have half of mine. Harry's share 18 Y - Ans. }}. share 6 EXTRACTION OF THE CUBE ROOT. A CUBE is any number muliplied by its square. To extract the cube root, is to find a number which being multiplied into its square, shall produce the given number. RULE. 1. 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. 2. Find the greatest cube in the left hand period, and put its root in the quotient. 3. Subtract the cube thus found, from the said period, and to the remainder bring down the next period, and call this the dividend. o 4. Multiply the square of the quotient by 300, calling it the triple square, and the quotient by 80, calling it the triple quotient, and the sum of these call the divisor. 5. Seek how often the divisor may be had in the dividend, and place the result in the quotient. 6. Multiply the triple square by the last o figure, and write the product under this dividend; multiply the square of the last quotient figure by the triple quotient, and place this product under the last ; under all, set the cube of the last quotient figure and call their sum the subtrahend. 7. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend, with 4 which proceed as before, and so on till the whole be finished.* *The process for extracting the cube root may be illustrated in the same manner as that for the square root. Take the same number 37, and multiply as before, collecting the twice 21 into one sum, as they belong to the same place, and the operation will be simplified, 373=50653. , 293. What is a Cube 2–294. What is the method of extracting the cube root of a given number 3 295. I wish you to illustrate the process under this rule, by one of the examples given. Note.—The same rule must be observed for continuing the operation and pointing for decimals, as in the square root. EXAMPLES. 1. Required the cube root of 436036824287. 436036824287(7583 the root. 1st Divisor=14910)93036=1st Dividend. 73500 3×32 ) 189 . . =3×32 ×7 3×302 )18900–3×302 ×7 3×3 ) 441 . =3x3x72 3×30 ) 4410=3×30x72 As 27 or 27000 is the greatest cube, its root is 3 or 30, and that part of the cube is exhausted by this extraction. Collect those terms which belong to the same places, and we have 32 x7=63, and 2x32 x7=126, and 63--126–3X32 x7=189; and 2x3x72= 294 and 3x72=147, and 294+147=441=3x3x72 for a dividend, which divided by the divisor, formed according to the rule, the quotient is 7, for the next figure in the root. And it is evident, on inspecting the work, that that part of the cube not exhausted is composed of the several products which form the subtrahend, according to the rule.— The same may be shown in any other case, and the universality of the rule hence inferred. The other method of illustration, employed in the square root, is equally applicable in this case. 37=30+7, and 30+72=302-H2x30×7+72 303+2×302 ×7+30x72 373=50653=303+3x302x1+3x30x7s-H78 (30+7=37 Divisor 3x302-H3x30 )3x302 ×7+3x30x72 +73 div. 3×302 ×7+3x30x72--73 subtrahend. It is evident that 303 is the greatest cube. When its root is extracted, the next three terms constitute the dividend; and the several products formed by means of the quotient or second figure in the root, are precisely equal to the remaining parts of the power whose root was to be found. pHMONSTRATION OF The REASON AND NATURE OF THE RULE. A block of wood, or any solid body, having six equal sides, all exactly square, is a cube; the root of which is the measure in length of one of its sides. To gain a distinct understanding of the subject, let the scholar provide himself with little blocks of wood, and build them up into a cubick form, according to the rule. First make a cubick block of any given size, and mark it with the letter A. Then make three other blocks of a square form, of an indefinite thickness, but all equal to each other, each of which will just cover one side of the block A, and mark them B, C and D. Place these blocks on three adjoining sides of the block A, when there will be deficiencies at the three points where the blocks B, C and D meet, These deficiencies must be filled with three other blocks, each of which must be just equal in length to one side of the block A, and mark these blocks with E, F and G. When the blocks E, F, G are put in their places, there will be a deficiency at the place where the ends of these blocks meet. This deficiency must be filled with another block, which mark H. To illustrate the rule, take the following number. 10648(2 In the first place I seek the greatest cube in the left hand period, and place its root, 2, in the quotient. The cube of 2 is 8, which I place under the left hand period, and subtract it therefrom, which Heaves a remainder of 2. Now as there are two periods in the given number, there must be 2 figures in the root, consequently, 2, in the quotient, does not express 2, merely, but 20 ; and the cube of 20 is 8000, which 8, under the period 10, represents; thus 8000 of the parts of 10648, are disposed of into a cubick body, the length of each side of which is equal to 20 of those parts, and to render the explanation more plain, we will consider these parts as cubick feet, so that each side of this body is 20 feet square, and this body we will have represented by the block A. Now as each side of this block is 20 feet square, there are 400 feet on each side of it. Now 8000 feet are disposed of in this block, consequently, there are 2648 cubick feet to be added to the block A, in such a manner, that its cubick form will be preserved. To do this, the additions must be made to three sides of the block, and these additions are represented by the blocks B, C and D, each of which containing 400 feet, the sum of the whole is 1200. Thus it is evident, that if there were 1200 feet more, there would be just enough to cover three sides of the block A ; and it is to find the contents of these three sides, that the rule directs to “multiply the square of the quotient by 300." The square of the quotient shows the superficial contents of one side of the block A. viz. 400, for 2 in the quotient is in reality 20, and 20X20-400, and it is because the cipher is not annexed to the quotient figure, that we . |