This result expressed in standard form (see Introduction, page 33) is 131⁄21⁄4 setat31⁄8 cubit—strips. The result obtained by the author‘s method exceeds this by 63⁄100 setat, or a little less than 3%.
He next determines the area of the apex. To do this he multiplies 7 by 21⁄4 and divides by 2, getting 71⁄21⁄41⁄8 setat as this area.
Finally, to determine the area of the middle section he would simply have to subtract the areas of the other two sections from the area of the whole triangle.
Problems 54 and 55 are exactly alike. although the first is given in the papyrus more briefly and with some words missing. I will explain them together.
Problem 54
What equal area: should be taken from 10 fields if the sum of these areas is to be 7 setat?
Multiply 10 so as to get 7.
| 1 | 10 | |
| \ | 1⁄2 | 5 |
| \ | 1⁄5 | 2 |
| Total | 1⁄21⁄5[1] |
Expressed as parts of a setat and cubit—strips this is 1⁄2 setat 10 cubit—strips.
Proof.
| 1 | 1⁄21⁄8 | setat | 71⁄2 | cubit—strips | |
| \ | 2 | 11⁄41⁄8 | " | 21⁄2 | " |
| 4 | 21⁄21⁄4 | " | 5 | " | |
| \ | 8 | 51⁄2 | " | 10 | " |
| Total | 7 | " |
Problem 55
What equal areas should be taken from 5 fields the sum of these areas is to be 3 setat?
Multiply 5 so as to get 3.
| 1 | 5 | |
| 1⁄2 | 21⁄2 | |
| 1⁄10 | 1⁄2 | |
| Total | 1⁄21⁄10. |
Expressed as parts of a setat and cubit—strips this is 1⁄2setat 10 cubit-strips.
- ↑ In Problem 4 this is given as 2⁄31⁄30. See notes to that problem.
