THEOREM SECOND. The base and hypothenuse given to find the RULE.-Square the base and hypothenuse; subtract the least from the greatest; extract the root of the remainder, the root is the answer, or perpendicular, Examples. 1. What is the height of the eaves of a house that requires a ladder 23:34 feet long to reach from the ground to the same; allowing the bottom of the ladder to stand 12 feet from the bottom of the house? RULE.-Square the hypothenuse and perpendicular subtract the less from the greater, extract the root of the remainder; the root is the answer, or base. Example. 1. Hypothenuse 23:31; perpendicular 20; required the base. Ans. 11.9†, or 12 ft. CUBE ROOT. DEFINITION. A cube is the third power of any number; that is, any number multiplied by itself, and that product again by the same number, produces a Cube; thus 9x9 81x9739 is a Cube; and 9 is the root. EXTRACTION OF THE CUBE ROOT. The extraction of the cube root is the finding of a number which has been multiplied three times by itself. RULE. Separate the number into periods of three figures each,beginning in units place; thus,943213427; find the greatest cube in the first period, place the . root in the quotient, and its cube under the first pe riod; subtract it therefrom, and to the remainder bring down the next period, which call the dividend, or resolvend; find a divisor by multiplying the square of the quotient by 300, seek how many times the divisor is contained in the resolvend, and put the answer in the quotient. To find the Subtrahend. Multiply the divisor by the last quotient figure, and place the product under the last dividend (units under units, &c.) then square the last quotient figure, and multiply the square by the preceding figures in the quotient, and the last product multiply by 30, and place this product also under the dividend (units under units, &c.) then cube the last quotient figure, and place its cube under the dividend also (units under units, &c.) add the three numbers (which have placed under the dividend) together, and subtract their sum from the last dividend; and to the remainder bring down the next period of figures, and proceed as before, until all have been brought down, and the quotient is the root. PROOF-Cube the quotient, or root, add in the remainder (if there is any) and the sum will be like the given cube if the work is right. Examples. 1. The cube root of 33076161 is required. 3×3×327)33076161(321 root Ans. 27 1st Divis. 3×3×3002700)6076 1st dividend or resolvend. 2d Divis. 32×32×300 307200)308161 second dividend.. NOTE. In the above example I first found the greatest cube that was in the first period, viz. 27, and its root was 3,. which I put in the quotient, and the cube I put under the first period, and subtracted it therefrom; and to the remainder (6) I brought down the next period of figures, viz. 076; which completed my first dividend or resolvend. Then for the first divisor, I squared the quotient or root, and multiplied the square by 300, which gave me 2700 for the first divisor; I found my divisor was contained in the dividend but twice, which I put in the quotient.-Then for the subtrahend I multiplied my last divisor, viz. 2700, by the last figure in the quotient, viz. 2, which product I put under the last dividend; and I also squared the last figure in the quotient, viz. 2, and multiplied its square by the preceding figures in the quotient, viz. 3, and that product again by 30 and put this last product under the dividend; I then cubed the last figure in the quotient, viz. 2, and put its cube also under the dividend; I then found the sum of the three last numbers (standing under the dividend) and subtracted their sum from the dividend; and to the remainder, viz. 308, I brought down the next and last period, viz, 161, and proceeded as before. 2. The cube root of 1728000 is required. Ans. 120. 3. The cube root of 997002999 is required. Ans. 999. 4. The cube root of 367061696 is required. Ans. 716. NOTE. When there is a remainder after the periods of f ures are brought down; annex periods of cyphers (three at once) and point the root for decimals, and proceed as far as you please in decimals; or you may find a denominator to the remainder by the following Rule, viz. Square the root and mul tiply the square by 3: then multiply the root by 3; add the 2 products together, and their sum is a denominator to the remainder, which fraction must be annexed to the root to make it complete. CASE II. When there are decimals annexed to integers. RULE.-Prepare the decimals that a point may fall upon units in the integers, then proceed as in case first Examples 1. The cube root of 1780-360128 is required. Ans. 12:12. 2. The cube root of 1815 848 is required. Ans. 12.2 3. The cube root of 1-442897 is required. CASE III. Ans. 1.13. To extract the cube root of a vulgar fraction. RULE.-Extract the root of the numerator, for a new numerator; and the root of the denominator, for a new denominator; if the fraction be a surd, that is, one whose root cannot exactly be found, reduce it to a decimal and extract the root. CASE IV. To extract the cube root of mixed numbers. RULE. Reduce the vulgar fraction to a decimal, annex it to the integers, and proceed as in case second. CUBE ROOT APPLIED. CASE I. Cube root applied in finding the solidity of globular figures, by having the diameter and solidity of one given. RULE-Globes are in proportion to one another as the cubes of their diameters; therefore cube the diameter of the given globe, and also of the required globe; and then say as the cube of the diameter of the given globe, is to its solidity; so is the cube of the diameter of the required globe to its solidity. Examples. 1. If a cannon ball 6 inches in diameter weighs 25 lb.; I demand the weight of another of the like metal, whose diameter is 3 inches. 6X6X6 216 the cube of the given diameter. 3X3X3 27 the cube of the required diameter. As 216: 25lb. :: 271b.: 312flb. Ans. 2. If a ball of silver 12 inches in diameter is worth $600; I demand the value of another ball, whose diam eter is 15 inches. CASE II. $1171.87 Ans. Having one side of a cubical figure given, to find the side of another, that will contain 2, 3, 4, 5 or 6 times as much as the given one. RULE.-Cube the given side, then multiply the cube by the number mentioned in the question, (if three times as large, multiply by 3, &c.) the cube root of the prod uct will be the side required, |