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Beyond the discovery of the volume of a square pyramid, the problem of finding the slope and height of a square pyramid can be found in the Rhind Mathematical Papyrus. [10] The Babylonian mathematicians also considered the volume of a frustum, but gave an incorrect formula for it. [11]
The fourteenth problem of the Moscow Mathematical calculates the volume of a frustum. Problem 14 states that a pyramid has been truncated in such a way that the top area is a square of length 2 units, the bottom a square of length 4 units, and the height 6 units, as shown. The volume is found to be 56 cubic units, which is correct. [1]
In problem 41 Ahmes computes the volume of a cylindrical granary. ... Consider a right regular square pyramid ... Problem 56 is the first of the "pyramid problems" or ...
The volume of a cylinder was taken as the product of the base and the height, however, the volume of the frustum of a cone or a square pyramid was incorrectly taken as the product of the height and half the sum of the bases. The Pythagorean rule was also known to the Babylonians. [21] [22] [23]
The Egyptians knew the correct formula for the volume of such a truncated square pyramid, but no proof of this equation is given in the Moscow papyrus. The volume of a conical or pyramidal frustum is the volume of the solid before slicing its "apex" off, minus the volume of this "apex":
Image of Problem 14 from the Moscow Mathematical Papyrus. The problem includes a diagram indicating the dimensions of the truncated pyramid. Several problems compute the volume of cylindrical granaries (41, 42, and 43 of the RMP), while problem 60 RMP seems to concern a pillar or a cone instead of a pyramid.
A square pyramid of cannonballs at Rye Castle in England 4900 balls arranged as a square pyramid of side 24, and a square of side 70. The cannonball problem asks for the sizes of pyramids of cannonballs that can also be spread out to form a square array, or equivalently, which numbers are both square and square pyramidal. Besides 1, there is ...
A triangular-pyramid version of the cannonball problem, which is to yield a perfect square from the N th Tetrahedral number, would have N = 48. That means that the (24 × 2 = ) 48th tetrahedral number equals to (70 2 × 2 2 = 140 2 = ) 19600. This is comparable with the 24th square pyramid having a total of 70 2 cannonballs. [5]