Page:The Rhind Mathematical Papyrus, Volume I.pdf/51

EGYPTIAN GEOMETRY

The author of this papyrus was able to determine the areas of rectangles, triangles, and circles, and the volumes of cylinders and prisms, and he knew that in a right triangle the relation of the lengths of two sides determines one of the angles. In the papyrus he takes up problems of volume, then problems of area, and, finally, problems involving the relative lengths of the sides of a triangle.

Determination of Volumes

In the problems of volume, 41-46, the author calculates the amount of grain that can be stored in certain spaces or bins of given dimensions, and the dimensions of bins that will contain given amounts of grain. Rather strangely he takes up the case of cylinders first and then that of rectangular parallelepipeds. As to the latter, all that I need to say is that he multiplies the three dimensions together. Problem 44 is an example of the direct calculation of such a volume, the three dimensions being all equal to 10. Problem 45 is the inverse of 44, and 46 is a problem of the same kind as 45. It is to be noticed that the author takes first the inverse of a numerical problem that he has worked out directly, and therefore he knows the answer.^[1] Problem 45 as stated has only the volume given. Taken as the inverse of 44 we should consider it as involving the taking of a cube root. What he does is to take two of the dimensions as each equal to 10, and to find the third. Problem 46 is not the inverse of a problem already solved, but here also he assumes two of the dimensions as each equal to 10 and gets the third dimension, which in this case is not the same number, but is one-third of it, or 3+¹⁄₃, so that the bin is a rectangular parallelepiped but not a cube.

In Problems 41-43 the author has given the dimensions of a cylindrical body to find the volume. Thus in 41 the diameter of the cylinder is given as 9 and the altitude as 10. In order to obtain the area of the base the author subtracts from the diameter its ¹⁄₉, squares the remainder, and obtains 64 as the area of the circle. Multiplying this by 10 he gets 640 as the contents of the granary in cubed cubits.

Thus we see that in this problem he gives the relation between the

1. In the same way 58 is the inverse of 57 and 59B is the inverse of 59.