From these illustrations we have the following 744. RULE.—I. Separate the number into periods of two figures each, by placing a point over every second figure, beginning with the units figure. II. Find the greatest square in the left-hand period and place its root on the right. Subtract this square from the period and annex to the remainder the next period for a dividend. FII. Double the part of the root found for a trial divisor, and find how many times this divisor is contained in the dividend, omitting the right-hand figure. Annex the quotient thus found both to the root and to the divisor. Multiply the divisor thus completed by the figure of the root last obtained, and subtract the product from the dividend. IV. If there are more periods, continue the operation in the same manner as before. In applying this rule be particular to observe: 1. When there is a remainder after the last period has been used, annex periods of ciphers and continue the root to as many decimal places as may be required. 2. We separate a number into periods of two figures by beginning at the units place and proceeding to the left if the number is an integer, and to the right if a decimal, and to the right and left if both. 3. Mixed numbers and fractions are reduced to decimals before extracting the root. But in case the numerator and the denominator are perfect powers, or the denominator alone, the root may be more readily formed by extracting the root of each term separately, ·Find the square root to three decimal places: 37. What is the length of a square floor containing 9025 square feet of lumber? 38. A square garden contains 237169 square feet; how many feet in one of its sides ? 39. How many yards in one of the equal sides of a square acre? 40. An orchard containing 9216 trees is planted in the form of a square, each tree an equal distance from another; how many trees in each row? 41. A triangular field contains 1966.24 P. What is the length of one side of a square field of equal area? 42. Find the square root of 2, of 5, and of 11, to 4 decimal places. 43. Find the square root of, f, and of, to 3 decimal places. CUBE ROOT. PREPARATORY PROPOSITIONS. 746. PROP. I.—ANY PERFECT third power may be represented to the eye by a cube, and the number of units in the side of such cube will represent the THIRD or CUBE ROOT of the given power. Represent to the eye by a cube 343. (1) 1. We can suppose the number 343 to represent small cubes, and we can take 2 or more of these cubes and arrange them in a row, as shown in (1). 2. Having formed a row of 5 cubes, as shown in (1), we can arrange 5 of these rows side by side, as shown in (2), forming a square slab (2) containing 5 x 5 small cubes, or as many small cubes as the square of the number of units in the side of the slab. 3. Placing 5 such slabs together, as shown in (3), we form a cube. Now, since each slab contains 5x5 small cubes, and since 5 slabs are placed together, the cube in (3) contains 5 x 5 x 5, or 125 small cubes, and hence represents the third power 125, and each edge of the cube represents to the eye 5, the cube root of 125. We have now remaining yet to be disposed of 343-125, or 218 small cubes. 4. Now, observe, that to enlarge the cube in (3) so that it may contain the 343 small cubes, we must build the same number of tiers of small cubes upon each of three adjacent sides, as shown in (4). Observe, also, that a slab of small cubes to cover one side of the cube in (3) must contain 5x5 or 25 small cubes, as shown in (4), or as many small cubes as the square of the number of units in one edge of the cube in (3). (4) Hence, to find the number of cubes necessary to put one slab on each of three sides of the cube in (3), we multiply the square of its edge by 3 giving 52x35x5x3 = 75 small cubes. 5. Having found that 75 small cubes will put one tier on each of three adjacent sides of the cube in (3), we divide 218, the number of (5) small cubes yet remaining, by 75, and find how many such tiers we can form. Thus, 218÷75= 2 and 68 remaining. Hence we can put 2 tiers on each of three adjacent sides, as shown in (5), and have 68 small cubes remaining. 6. Now, observe, that to complete this cube we must fill each of the three corners formed by building on three adjacent sides. Examine carefully (6) and observe that to fill one of these three corners (6) we require as many small cubes as is expressed by the square of the number of tiers added, multiplied by the number of units in the side of the cube to which the addition is made. Hence we require 22 x 5 or 20 small cubes. And to fill the three corners we require 3 times 22 x 5 or 60, leaving 68-60 or 8 of the small cubes. 7. Examine again (5) and (6) and observe that when the three corners are filled we require to complete the cube as shown in (7), another cube whose side contains as many units as there are units added to the side of the cube on which we have built. Consequently we require 23 or 2 x 2 x 2 = 8 small cubes.. Hence we have formed a cube containing 343 small cubes, and any one of its edges represents to the eye 5+2 or 7 units, the cube root of 343. From these illustrations it will be seen that the steps in finding the cube root of 343 may be stated thus: We assume that 343 represents small cubes, and take 5 as the length of the side of a large cube formed from these. Hence we subtract the cube of 5 = 1. We observe it takes 52 x 3 = 75 to put one tier on three adjacent sides. Hence we can put on FIRST STEP. SECOND STEP. 2. We have now found that we can add 2 units to the (3) For the cube in the corner last formed 28 8 = 343 125 75)218(2 =218 Hence the cube root of 343 is 5+2 = 7. 747. Observe, that the number of small cubes in the cube (7) in the foregoing illustrations, are expressed in terms of 5+2; namely, the number of units in the side of the first cube formed, plus the number of tiers added in enlarging this cube; thus: In this manner it may be shown that the cube of the sum of any two numbers is equal to the cube of each number, plus 3 times the square of the first multiplied by the second number, plus 3 times the square of the second multiplied by the first number. Hence the cube of any number may be expressed in terms of its tens and units; thus, 74 = 70+4; hence, (70+4)=703+3 times 702 × 4+3 times 42 x 70+43 = 405224. Solve each of the following examples, by applying the foregoing illustrations: 1. Find the side of a cube which contains 729 small cubes, taking 6 units as the side of the first cube formed. 2. Take 20 units as the side of the first cube formed, and find the side of the cube that contains 15625 cubic units. 3. How many must be added to 9 that the sum may be the cube root of 4096? Of 2197? Of 2744? 4. Find the cube root of 1368. Of 3405. Of 2231. Of 5832. 5. Express the cube of 83 in terms of 80 +3. 6. Express the cube of 54, of 72, of 95, of 123, of 274, in terms of the tens and units of each number. |
