CHAPTER II. GEOMETRY
SECTION I
Problems 41-46. Problems of Volume
Problem 41
Find the volume of a cylindrical granary of diameter 9 and height 10.
Take away 1⁄9 of 9, namely, 1; the remainder is 8. Multiply 8 times 8; it makes 64. Multiply 64 times 10; it makes 640 cubed cubits. Add 1⁄2 of it to it; it makes 960, its contents in khar. Take 1⁄20 of 960, namely 48. 4800 hekat of grain will go into it.
Method of working out:
| 1 | 8 | 2 | 16 | 4 | 32 | \ | 8 | 64. | 1 | 64 | \ | 10 | 640 | \ | 1⁄2 | 320 | ||||
| Total | 960 | |||||||||||||||||||
| 1⁄10 | 96 | |||||||||||||||||||
| \ | 20 | 48. | ||||||||||||||||||
As explained in the Introduction (pages 35-36), the author, in order to obtain the volume of the cylinder, subtracts from the diameter its 1⁄9, squares the remainder, and multiplies by the altitude. In this way he finds that 640 is the contents of the granary in cubed cubits. As a khar is 2⁄3 of a cubed cubit he adds to 640 its 1⁄2, which gives him 960 as the number of khar in the granary. Finally, dividing by 20, he obtains 48 as the number of hundreds of quadruple hekat. (See Introduction, page 32.)
Problem 42
Find the volume of a cylindrical granary of diameter 10 and height 10.
Take away 1⁄9 of 10, namely 11⁄9;the remainder is 82⁄31⁄61⁄18. Multiply 82⁄31⁄61⁄18 times 82⁄31⁄61⁄18; it makes 791⁄1081⁄324. Multiply 791⁄1081⁄324
