This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1893 Excerpt: ...LM, whose altitude is the units' figure of the root. Now the altitude of a rectangle is equal to its area divided by its base. 30 30 M N Since the base of each of the rectangles P and Q is 30 in., the base of the rectangle LM is something more than 60 in. If we divide the area of LM, 396 sq. in., by its approximate ...
Read More
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1893 Excerpt: ...LM, whose altitude is the units' figure of the root. Now the altitude of a rectangle is equal to its area divided by its base. 30 30 M N Since the base of each of the rectangles P and Q is 30 in., the base of the rectangle LM is something more than 60 in. If we divide the area of LM, 396 sq. in., by its approximate base, 60 in., we obtain something more than 6 in. as the approximate altitude. If, now, we make trial of 6 in. as the altitude of the rectangle, the base iJVis 60 in. + 6 in., or 66 in.; and multiplying this by the altitude, 6 in., the result is 396 sq. in. But this is just the area of the irregular figure of Fig. 2. We then conclude that the units' figure of the root is 6; whence, the required root is 30 + 6, or 36. The above process is exactly in accordance with the Rule of Art. 200. Cube Root. Let it be required to find the cube root of 13824. Let AB be a cube containing 13824 cu. in. To find its edge in inches. Since a cube whose edge is 20 in. contains 8000 cu. in., and a cube whose edge is 30 in. contains 27000 cu. in., the edge of the given cube must be between 20 and 30 in. Thus the tens' figure of the root is 2. Removing from the given cube a cube whose edge is 20 in., there remains an irregular solid CD, whose volume is 13824--'8000, or 5824 cu. in. Removing from CD the three solids E, F, and Q-, there remains an irregular solid, shown in Fig. 3, composed of three rectangular parallelopipeds, H, K, and L, and a cube. The solids E, F, G, H, K, and L, and the cube M, may be arranged as shown in Fig. 4, forming an irregular solid NP, whose altitude is the units' figure of the root. Now the altitude of this solid is equal to its volume divided by the area of its base. Since the area of the base of each of the solids E, F, and G is 202, or 4...
Read Less
