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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":
The truncated pyramid is a pyramid cut off by a plane; ... The volume of a pyramid is the one-third product of the base's area and the height.
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 the Moscow Mathematical Papyrus, Egyptian mathematicians demonstrated knowledge of the formula for calculating the volume of a truncated square pyramid, suggesting that they were also acquainted with the volume of a square pyramid, but it is unknown how the formula was derived.
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. It is rather small and steep, with a seked (slope) of four palms (per cubit). [10]
A pyramid with side length 5 contains 35 spheres. Each layer represents one of the first five triangular numbers. A truncated triangular pyramid number [1] is found by removing some smaller tetrahedral number (or triangular pyramidal number) from each of the vertices of a bigger tetrahedral number.
Rectangular (Cuboid): Several problems in the Moscow Mathematical Papyrus (problem 14) and in the Rhind Mathematical Papyrus (numbers 44, 45, 46) compute the volume of a rectangular granary. [13] Truncated pyramid (frustum) Frustum: The volume of a truncated pyramid is computed in MMP 14. [8]
The volume of a tetrahedron can be obtained in many ways. It can be given by using the formula of the pyramid's volume: =. where is the base' area and is the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apices to the opposite faces are inversely proportional to the areas of ...