In finding the root of a mixed number, reduce the mixed number to an improper fraction, and find the root of the numerator and denominator, and then reduce the fraction to a whole or mixed number. 10. What is the square root of 74 Ans. 811. Ans. 6. Ans. 3,25. Ans. 4,5. Ans. 8,5. Ans. 3. Ans. 625. Ans. 875. Ans. ,0625. Ans. ,003. Ans. ,04. Ans. 6,004. Ans. 3,03. Ans. 4,08. Ans. 325. Ans. 3,25. Having shown the scholar how sums are performed in the square root, I wish him now to attend carefully to the following explanation of the rule, that he may understand the nature of the process, and thus be able to give the why in relation to the several steps of the rule. Suppose a man is to receive 3249 square rods of land in one square lot; what will be the length of one side of the field? 3249 First point the sum to distinguish the periods. I point it off into periods of two figures each, because the square of a number contains twice as many figures as the root, or one less than twice as many. If the left-hand period contain two figures, then the given sum contains twice as many figures as the root. The periods serve to show how many figures will be contained in the root. Having found the number of figures that will be contained in the root, find the greatest square of the left-hand period, and place its root at the right hand of the given sum. Now, because the root will contain two figures, 324915 the first figure (5) in the root is in the place of tens. This figure (5) represents one side of a square, which is found by multiplying 5 tens by 5 tens. 324915 25 749 When we have multiplied 5 tens by 5 tens, we find the product to be 250 tens, or 2500. If 50 r. represent one side of a square lot, that whole lot will contain 2500 rods, which may be represented by this figure, each of which sides may be supposed to be 50 rods long. We 2500 50 have now disposed 2500 rods of land into a square form; but there still remain to be disposed of, 749 rods more. 324915 25 749 As we know the length of each side of the square, we know how long our additions must be. And in order to preserve the square form of the lot, we must make additions on two sides of the field, so as to lengthen all the boundary lines equally. Our next inquiry is, What must be the width of our additions? We have already shown, (XXI. SQUARE AND CUBIC MEASURE,) that when the length and the square contents of a surface are given, we can find the width by dividing the square contents by the length. The square rods to be added to the above lot, are 749. The length of additions is 50×2=100. Therefore, if we divide 749 by 100, the quotient will be the width of the additions. 25 But because the 5 is in place of tens, and 3249 57 the cipher is omitted, therefore I must omit the cipher in the divisor, and one figure in the dividend. Now, if I make additions on two sides, the length of each addition 107|749 749 350 2 2500 will be 50 rods, its width 7, and its square contents 350 rods, according to the accompanying figure. The deficiency in the upper corner is 7 rods one way, and 7 rods the other; or each side of the deficiency is equal 7 to the width of the additions. Therefore, by placing the width of the addition at the right hand of the divisor, and multiplying it by itself, the square contents of this deficiency is found. This deficiency requires 49 rods, which complete the square form of the lot. 7 PROOF.-The whole length of one side of the lot multiplied by itself, will give the area, according to XXI. SQUARE AND CUBIC MEASURE. Or the parts of the lot may be added thus: 2500 350 350 49 3249 EXAMPLES FOR PRACTICE. 1. What is the length of one side of a square field which contains 7056 rods? Ans. 84. 2. A general has 15625 soldiers; how many must he place in rank and file to form them into a square? Ans. 125. 3. There is a circular field which contains 1406,25 square feet; what must be the length of the side of a square field which will contain the same number of feet? Ans. 37,5. The square of the longest side of a right-angled triangle is equal to the sum of the squares of the other two sides. 4. The top of a castle from the ground is 45 yards high, and is surrounded with a ditch 60 yards broad; how long must a line be to reach from the outside of the ditch to the top of the castle? Ans. 75 yards. Circles are to each other as the squares of their diameters 5. There is a circle whose diameter is 4 inches; what is the diameter of a circle three times as large? Ans. 6,928X. due 6. Two ships sail from the same port; one goes north 45 leagues, and the other due west 76 leagues. How far are they apart? Ans. 88,32 leagues. 7. If a pipe 6 inches in diameter will conduct off a certain quantity of water in 4 hours, in what time will three pipes, each 4 inches in diameter, conduct off double the quantity? Ans. 6 hours. 8. If a pipe whose diameter is 1,5 inches, fill a cistern in 5 hours, in what time will a pipe whose diameter is 3,5 inches, fill the same? Ans. 55m. 6 sec. XXVI. Cube Root. When a number has been multiplied by itself twice, we obtain a product which is called a cube. When a number is considered in relation to such a product, it is called a cube root. To find the cube root of any number, therefore, is to find a number which, multiplied by itself twice, will produce the given cubic number. TO FIND THE CUBE ROOT. RULE.-Beginning with the place of units, distinguish the sum into periods of three figures each, by placing a dot over units, thousands, and so on. If the sum contain decimals, place a dot over thousandths, millionths, and so on. Find, by trial, the greatest cube in the left-hand period, and place its root in the quotient. Subtract the cube thus found from the left-hand period, and to the remainder annex the second period, and call them the first dividend. Multiply the square of the root by 3, to this product add the product of the root multiplied by 3, call the amount the divisor, by which divide the first dividend, and place the quotient in the root. Multiply three times the square of the root (except the last figure in the root) by the last figure in the root, and place the product under the first dividend. Multiply the square of the last quotient figure by the former figure, or figures, of the root, and this product multiply by 3, and its product place under the first dividend. Under all write the cube of the last quotient figure, and the amount of these three products subtract from the first dividend, and to the remainder bring down the third period, and call them the second dividend, with which proceed as before, until the whole root is found. Note. In squaring the root, it must be remembered that the value of the root must be expressed in a denomination corresponding with the period, the root of which you are finding. If you are finding the root of the second period from the right hand, reckon your root as tens; if that of the third period, call your root hundreds, omitting one cipher in the root for every period at the right hand of that period, the root of which you are seeking. Call the first right-hand period, reckoning from units, a unit period, the next lefthand period tens period, and so on. 1. What is the cube root of 1953125? 1953125 125, Ans. |