It will be seen from the periods that the root will have one integral and two decimal places, and therefore the decimal-point must be placed in the root as soon as one figure of the root is obtained. 280. If the given number be not a perfect cube, ciphers may be annexed, and a value of the root may be found as near to the true value as we please. Extract the cube root of 1250.6894. Since 300 is not contained in 200, the next figure of the root will 281. The following method very much shortens the work in long examples. Extract the cube root of 5 to five places of decimals. After the first two figures of the root are found, the next trial divisor is obtained by bringing down the sum of the 210 and 49 obtained in completing the preceding divisor, then adding the three lines connected by the brace, and annexing two ciphers to the result. It is seen at a glance that, when the trial divisor is increased by 3 times the 17 tens of the root, it will be greater than 87000; so that O is placed in the root, and 3× 17002 is obtained by annexing two ciphers to the 86700. Again: the trial divisor is obtained by bringing down the sum of the 45900 and 81, which was obtained in completing the preceding divisor, then adding the three lines connected by the brace, and annexing two ciphers to the result. The last two figures of the root are found by division. The rule in such cases is, that two less than the number of figures already obtained may be found without error by division, the divisor to be employed being three times the square of the part of the root already found. 282. The cube root of a common fraction is found by taking the cube roots of the numerator and denominator; but, if the denominator be not a perfect cube, it is best to reduce the fraction to a decimal, and then extract the root. 17. The liter contains 61.027 cu. in. Find the side of a cube containing a liter. 18. The edges of a rectangular solid are 154 ft. 11 in., 70 ft. 7 in., 53 ft 1 in. Find the edge of a cube equivalent to it. The square of (30 + 5) = 302 + 2 (30 × 5) + 52. ? 266. The 302 may be represented by a square (Fig. 1) 30 in. on a side. The 2(30 × 5) may be represented by two strips 30 in. long and 5 in wide, of Fig. 2, which are added to two adjacent sides of Fig. 1. The 52 may be represented by the small square of Fig. 3 required to make Fig. 2 a complete square. Fig. 1. Fig. 2. Fig. 3. In extracting the square root of 1225, the large square, which is 30 in. on a side, is first removed, and a surface of 325 sq. in. remains. This surface consists of two equal rectangles, each 30 in. long, and a small square whose side is equal to the width of the rectangles. The width of the rectangles is found by dividing the 325 sq. in. by the sum of their lengths, that is, by 60, which gives 5 in. Hence the entire length of the surfaces added is 30 in. + 30 in. +5 in. 65 in., and the width is 5 in. = Therefore the total area is (65 × 5) = 325 sq. in. 8392. The cube of (30 + 5) = 303 + 3 (302 × 5) + 3 (30 × 52) + 53. The 303 may be represented by a cube whose edge is 30 in. (Fig. 1). The 3 (302 × 5) may be represented by three rectangular solids, each 30 in. long, 30 in. wide, and 5 in. thick, to be added to three adjacent faces of Fig. 1. The 3(3052) may be represented by three equal rectangular solids, 30 in. long, 5 in. wide, and 5 in. thick, to be added to Fig. 2. The 53 may be represented by the small cube required to complete the cube of Fig. 3. Fig. 1. Fig. 2. Fig. 3. Fig. 4. In extracting the cube root of 42875, the large cube (Fig. 1), whose edge is 30 in., is first removed. There remain (42875 – 27000) cu. in. 15875 cu. in. The greater part of this is contained in the three rectangular solids which were added to Fig. 1, and which are each 30 in. long and 30 in. wide. The thickness of these solids is found by dividing the 15875 cu. in. by the sum of the three faces, each of which is 30 in. square; that is, by 2700 sq. in. The result is 5 in. There are also the three rectangular solids which are added to Fig. 2, and which are 30 in. long and 5 in. wide; and a cube which is added to Fig. 3, and which is 5 in. long and 5 in. wide. Hence the sum of the products of two dimensions of all these solids is For the larger rectangular solids, 3 (30 × 30) sq. in. 2700 sq. in. = 450 sq. in. (5 × 5) sq. in. H 25 sq. in. 3175 sq. in. This number multiplied by the third dimension gives (5 × 3175) cu. in. = 15,875 cu. in. 283. In bodies of the same shape, two. Two corresponding lines are in the same ratio as any other The ratio of two corresponding surfaces is the square of the ratio of two corresponding lines. The ratio of two corresponding volumes is the cube of the ratio of two corresponding lines. Conversely, The ratio of two corresponding lines is the square root of the ratio of two corresponding surfaces, and the cube root of the ratio of two corresponding volumes. Ex. 162. 1. The volume of a rectangular solid is 1728 cu. in. The volume of a similar solid is 3375 cu. in. Find the ratio of two corresponding edges. 2. The surface of a solid is 600 sq. in. What is the surface of a similar solid whose edges are twice as great? 3. If the volumes of two similar solids be 100 cu. in. and 1000 cu. in. respectively, find the ratio of their heights to the nearest thousandth of an inch. 4. If two hills have the same shape, and one is 2700 ft. high, while the other is 3600 ft. high, find the ratio of their surfaces, and also the ratio of their volumes. 5. A bushel measure and a peck measure are of the same shape. Find the ratio of their heights. 6. The surfaces of two hills having the same shape are as 25 16. Find the ratio of their heights. 7. Of two similar solids, the volume of the larger is 181 of that of the smaller. Find the ratio of their heights; find also the ratio of their bases. 8. The equatorial diameter of the earth is 7926 mi. Find that of Venus whose volume is 0.953 of the volume of the earth. |