The papyrus states the problem for a field of 10 khet by 2 khet, and these numbers are in the figure, but the solution is for 10 khet by 1 khet, or 1,000 cubits by 100 cubits. Multiplying these numbers together gives 100,000 square cubits. Dividing this by 100 gives 1,000 cubit-strips, strips 1 cubit wide and 1 khet long.
Problem 50
Example of a round field of diameter 9 khet. What is its area?
Take away 1⁄9 of the diameter, namely 1; the remainder is 8. Multiply 8 times 8; it makes 64. Therefore it contains 64 setat of land.
Do it thus:
| 1 | 9 | |
| 1⁄9 | 1; |
this taken away leaves 8
| 1 | 8 | |
| 2 | 16 | |
| 4 | 32 | |
| \ | 8 | 64. |
Its area is 64 setat.
Problem 51
Example of a triangle of land. Suppose it is said to thee, What is the area of a triangle of side[1] 10 khet and of base 4 khet?
Do it thus:
| 1 | 400 | |
| 1⁄2 | 200 | |
| 1 | 1,000 | |
| 2 | 2,000. |
Its area is 20 setat.
Take 1⁄2 of 4, in order to get its rectangle. Multiply 10 times 2; this is its area.
The author seems to put the reckoning before the explanation. In the reckoning he puts down the base as 400 and the side as 1,000; that is, he expresses these lengths in cubits. Dividing 400 by 2 he gets 200 and 1,000 as the dimensions of the equivalent rectangle. Then to obtain the area expressed as so many cubit-strips he multiplies 1,000, not by 200, but by 2, as if he thought of the rectangle as made up of 1,000 pairs of cubit-strips. Finally, he writes down 2, that is, 20 setat (2 ten-setat), as the standard form of expressing the result.
- ↑ For a discussion of the question whether this word should be side or altitude see Introduction, pages 36—37, and Bibliography, pages 132—l34.
