| 1 | 8 | |
| \ | 1⁄2 | 4 |
| \ | 1⁄4 | 2 |
| Total | 1⁄21⁄4 |
Take 1⁄21⁄4 of 7 this is a cubit.
| 1 | 7 | |
| \ | 1⁄2 | 31⁄2 |
| \ | 1⁄4 | 11⁄21⁄4. |
The result is 5 palms 1 finger. This is its seked.
As in Problem 43, the two numbers given are interchanged in the statement in the papyrus, or we might say that the side of the base and the altitude are interchanged; and then once in the solution "altitude" is written for "side of the base."
The original contains two or three words at the end of the solution that are uncertain. See Literal Translation.
The next problem is the inverse of the preceding and Eisenlohr counted the two together as 59. Indeed, the author may have regarded them as one problem, since none of the second is written in red. In order to hold to the numbering of Eisenlohr I have followed Feet in calling it 59B.
Problem 59B
If the seked of a pyramid is 5 palms 1 finger per euhit and the side of its base 12 cubits long, what is its altitude?
Multiply 5 palms 1 finge-r doubled, which is 101⁄2, so as to get 1 cubit; a cubit is 7 palms. 2⁄3 of 101⁄2 is 7; therefore 2⁄3 of 12, which is 8, is the altitude.
Problem 60
If a pillar (?)[1] is 30 cubits high and the side (diameter?) of its base 15 cubits, what is its seked?
Take 1⁄2 of 15; it is 71⁄2. Multiply 30 so as to get 71⁄2; the result is 1⁄4.
This is the seked.
The working out:
| 1 | 15 | ||
| \ | 1⁄2 | 7 | 36 |
| 1 | 30 | ||
| 1⁄2 | 15 | ||
| \ | 1⁄4 | 71⁄2. |
- ↑ For a discussion of the meaning of this word see Peet, page 101
