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This problem is difficult to explain, and the difficulty is increased by some numerical mistakes. The most probable explanation that I can give is that the author is undertaking to determine the areas of certain sections of an isosceles triangle. The drawing suggests that 14 is the common length of the two equal sides and that the sections are made by two lines parallel to the base. One of these lines seems to bisect the two equal sides and perhaps it was intended that the other should bisect the parts next to the base, so that the lengths of the sides of the three sections would be 31⁄2, 31⁄2, and 7. The length of the shorter dividing line is put down as 21⁄4, and this would make the base 41⁄2 and the longer dividing line 31⁄41⁄8. In his figure the author puts down 6 as the length of both of these lines, but in his calculations he seems to take 41⁄2 for the base.
He first undertakes to determine the area of the largest section. Apparently he intended to multiply the base, 41⁄2, by the height or side 31⁄2. This would give him 151⁄21⁄4 setat. He recognizes that the area thus determined would be too large, because the top is less than the base. He does not seem able to find the length of the upper line, but proposes arbitrarily to take away Mo of the area that he has obtained. In this way, if he had finished his solution, he would have had, finally, as the area, 141⁄8 setat 5 cubit-strips.[1] This taking away of 1⁄10 reminds us of his solution of Problem 28 and of his method of determining the area of a circle by taking away 1⁄2 of the diameter. In Problem 82 we shall have an example where he obtains a quantity somewhat smaller than a given quantity by taking away 1⁄10 of 2⁄3 of it.
The true area, if 31⁄2 were the height, would be easily found. AB and FG being 41⁄2 and 21⁄4.
| AB + FG will be 61⁄21⁄4 | ||
| Its 1⁄2, DE | 31⁄41⁄8 | |
| Adding AB we get | 71⁄21⁄41⁄8 | |
| Its 1⁄2 | 31⁄21⁄41⁄81⁄16 |
one-half of the sum of the bases of the trapezoid ABED. Multiplying by 31⁄2 we have
| \ | 1 | 31⁄21⁄41⁄81⁄16 |
| \ | 2 | 71⁄21⁄41⁄8 |
| \ | 1⁄2 | 11⁄21⁄41⁄81⁄161⁄32 |
| Total | 131⁄21⁄41⁄32 | |
- ↑ Actually the author multiplied 41⁄2 by 1, 2, 1⁄2 and 1⁄4, checking the 1 and 2. Then he put down as his total 51⁄21⁄8, which is the sum of the first and last of these partial products. Finally he put down as 1⁄10, 11⁄41⁄8 setat 10 cubit-strips. This is very nearly 1⁄10 of 15 1⁄21⁄4, and much too large to be 1⁄10 of his 51⁄21⁄8.
