FIG. 5. represents the form of the solid after the three edge blocks have been placed in their proper places. We observe that our cube is still incomplete, as there is one corner to be filled by the block marked D in Fig. 3. Each of the dimensions of this block is evidently 3 ft., and hence its volume is 3 x 3 x 3 = 3 = 27 cu. ft. Placing this block in its proper position in Fig. 4 we have our complete and enlarged cube as represented by Fig. 5. This cube is composed of 2. Three rectangular slabs each 20 × 20 × 3; 8000 cu. ft. 3600 cu. ft. 3. Three rectangular blocks each 20 × 3×3; 540 cu. ft. 4. A cube whose edge is 3 ; or 3×3×3=33 = SOLUTION. 12167 (23 8 27 cu. ft. 12167 cu. ft. EXPLANATION. - We first separate the power into two periods, 12 and 167. 22 × 300 = 1200 4167 2 x 30 x 3 = 180 32 = 9 1389 4167 Since the cube of units produces units, tens, or hundreds, and the cube of tens can produce no lower order than thousands, the cube root of 12167 must be units and tens, or 2 figures; and the first figure, or tens, must be the cube root of some cube in 12. 8 is the largest exact cube in 12, and 2 is its cube root ; and we conclude that 2, or 2 tens, is the first figure of the root. Subtracting the cube of 2 tens from 12 and bringing down the next period, or what is the same thing, subtracting the cube of 20, 8000, from 12167, we have left 4167. This is to be placed on 3 faces, but these faces are each 20 ft. square. They have, therefore, 20 × 20 × 3, or 1200 ft. of surface. Dividing 4167 by 1200, the number of ft. in this surface, we obtain its thickness, which is at least 3 ft. But we must make provision also for the three edge-pieces which are each 20 × 32, and the corner cube which is 33. These seven parts are equal in volume to (202 × 3)×3+ (20 × 3 × 3) x 3 + (3)2 × 3 = (202 × 3 + 20 × 3 × 3+32) × 3 = (1200+180+9 = 1389) × 3 = 4167; this subtracted from 4167 leaves no remainder, and the work is complete. EXAMPLE 2. - Extract the cube root of 12812904. we find their thickness 4. We then add to 232 × 300, 23 × 30 × 4 (=230 × 3 × 4), for the length and breadth of the three-edge pieces; and add also 4a for length and breadth of corner cube; and then multiply the sum by 4, the thickness; and since the product is 645904, there is no remainder, and the work is complete. Since the cube of tenths produces thousandths, and the cube of hundredths, millionths, when we extract the cube root of a decimal, that decimal must have the form of thousandths, millionths, &c. Thus, the cube root of 3.5 equals the cube root of 3.500, or 3.500000, &c. 513. RULE.-Separate the power into periods of three figures each, beginning at the decimal point. For the first figure of the root, write the cube root of the greatest exact cube in the first, or left-hand period. Subtract the cube of this first figure of the root from the first period, and to the remainder annex the next period. Regard the result as a dividend. For a divisor take 300 times the square of the first figure of the root. Find how often the product is contained in the dividend, and write the quotient as the second figure of the root. Complete the divisor by adding to it 1st. The product of the first figure of the root by 30 times the last figure of the root ; 2d. The square of the last figure of the root. Multiply the divisor thus increased by the last figure of the root; subtract the product from the dividend, and to the remainder annex the next period for a new dividend. Form, in the same manner, successive divisors, and find corresponding figures of the root; being careful to complete each divisor by the addition of 30 times the product of the last figure by the preceding figures of the root, and also the square of the last root figure. When any product is greater than the dividend erase the figure that produced it, and with a figure of less value recalculate the additions to the trial divisor until the product is small enough for subtraction. When any trial divisor is not contained in the dividend, annex a cipher to the root, two ciphers to the trial divisor, bring down the next period to the right of the dividend, and proceed as before. The cube of a common fraction is the cube of the numerator written over the cube of the denominator; therefore, the cube root of a common fraction is the cube root of the numerator over the cube root of the denominator. 514. It is demonstrated in Geometry that All similar solids are to each other as the cubes of their like dimensions. We make a few applications of this principle, referring the pupil to the Section on Mensuration for definitions of pyramids, cones, &c. PROBLEMS. 1. How many times will a cube whose edge is 3 in. contain one whose edge is 1 in. ? Ans. 27 times. 2. If an apple 1 in. in diameter is worth 1, how much is Ans. 84. one 2 in. in diameter worth? 3. A teamster who had a cart-bed 4 × 6 × 1 ft., made another in which each dimension was increased 1. How does the former compare with the latter? 4. Of two similar pyramids, one has a height of 2 ft. and the other of 5 ft. How do they compare in volume? Ans. 23:53. 5. A Winchester bushel is represented as a cylindrical vessel 18+ in. in diameter, and 8 in. deep. What would be the dimensions of a similar vessel that would hold 2 bushels? Ans. Dia., 23.308; depth, 10.08 in. |