| 1 | 7 | |
| 1⁄2 | 31⁄2 | |
| 1⁄5 | 11⁄31⁄15 | |
| 1⁄50 | 1⁄101⁄25 |
The seked is 51⁄25 palms.
Problem 57
If the seked of a pyramid is 5 palms 1 finger per cubit and the side of its base 140 cubits, what is its altitude?
Divide 1 cubit by the seked doubled, which is 101⁄2. Multiply 101⁄2 so as to get 7, for this is a cubit: 7 is 2⁄3 of 101⁄2. Operate on 140, which is the side of the base: 2⁄3 of 140 is 931⁄3. This is the altitude.
{{smaller block|In this inverse problem and in 59B the author doubles the seked instead of taking 1⁄2 of the side of the base, and instead of dividing the seized doubled by 7 and dividing the side of the base by the result, he divides 7 by the seked doubled and multiplies the side of the base by the result, which amounts to the same thing.
Problem 56
If a pyramid is 931⁄2 cubits high and the side of its base 140 cubits long, what is its seked?
Take 1⁄2 of 140, which is 70. Multiply 931⁄3 so as to get 70. 1⁄2 is 462⁄3, 1⁄4 is 231⁄3. Make thou 1⁄21⁄4 of a cubit. Multiply 7 by 1⁄21⁄4. 1⁄2 of 7 is 31⁄2, 1⁄4 is 11⁄21⁄4. together 5 palms 1 finger. This is its seked.
The working out:
| 1 | 931⁄3 | |
| \ | 1⁄2 | 462⁄3 |
| \ | 1⁄4 | 231⁄3 |
| Total | 1⁄21⁄4. |
Make thou 1⁄21⁄4 of a cubit; a cubit is 7 palms.
| 1 | 7 | |
| 1⁄2 | 31⁄2 | |
| 1⁄4 | 11⁄21⁄4 | |
| Total | 5 palms 1 finger. |
This is its seked.
Problem 59
If a pyramid is 8 cubits high and the side of its base 12 cubits long, what is its seked?
Multiply 8 so as to get 6, for this is 1⁄2 of the side of the base.
