In his explanation he uses the number of khet in the base and the number of khet in the side, and says that 10 times 2 will give the area.
A similar problem is given on a fragment of the Golenishchev papyrus (see Bibliography under Tsinserling, 1925).
Problem 52
Example of a cut—off (truncated) triangle of land. Suppose it is said to thee, What is the area of a cut-off triangle of land of 20
khet in its side, 6 khet in its base, 4 khet in its cut—off line?
Add its base to its cut-off line; it makes 10. Take 1⁄2 of 10, that is 5. in order to get its rectangle. Multiply 20 times 5; it makes 10 (10 ten-setat). This is its area.
Do it thus:
| 1 | 1,000 | |
| 1⁄2 | 500 | |
| \ | 1 | 2,000 |
| 2 | 4,000 | |
| \ | 4 | 8,000 |
| Total | 10,000. | |
Its area is 100 setat (10 ten-setat).
This solution is very much like the preceding. Taking 1⁄2 of the sum of the bases he gets 500 cubits or 5 khet as the base of the equivalent rectangle with its side 2,000 cubits. Multiplying 2,000 by 5 he gets 10,000 cubit-strips for the area. Finally be expresses this in setat (or ten-setat). In the papyrus 20 was written where 10 should have been.
Problem 53
Areas of sections of a triangle.
| \ | 1 | 41⁄2 | setat | ||
| \ | 2 | 9 | " | ||
| \ | 1⁄2 | 21⁄4 | " | ||
| Total | 151⁄21⁄4 | " | |||
| 1⁄10 | 11⁄2 | " | 71⁄2 | cubit-strips. | |
| Its 1⁄10 taken away leaves the area | 141⁄8 | " | 5 | " | |
| 1 | 7 | " | |||
| \ | 2 | 14 | " | ||
| 1⁄2 | 31⁄2 | " | |||
| \ | 1⁄4 | 11⁄21⁄4 | " | ||
| Total | 151⁄21⁄4 | " | |||
| 1⁄2 | 71⁄21⁄41⁄8 | " | |||
