Page:The Rhind Mathematical Papyrus, Volume I.pdf/110

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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 3$1/undefined$, 3$1/undefined$, and 7. The length of the shorter dividing line is put down as 2$1/undefined$, and this would make the base 4$1/undefined$ and the longer dividing line 3$1/undefined$$1/undefined$. In his ﬁgure the author puts down 6 as the length of both of these lines, but in his calculations he seems to take 4$1/undefined$ for the base.

He ﬁrst undertakes to determine the area of the largest section. Apparently he intended to multiply the base, 4$1/undefined$, by the height or side 3$1/undefined$. This would give him 15$1/undefined$$1/undefined$ 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, 14$1/undefined$ setat 5 cubit-strips. This taking away of $1/undefined$ reminds us of his solution of Problem 28 and of his method of determining the area of a circle by taking away $1/undefined$ 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 $1/undefined$ of $1/undefined$ of it.

The true area, if 3$1/undefined$ were the height, would be easily found. AB and FG being 4$1/undefined$ and 2$1/undefined$.

one-half of the sum of the bases of the trapezoid ABED. Multiplying by 3$1/undefined$ we have