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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 right pyramid is a pyramid whose base is circumscribed about a circle and the altitude of the pyramid meets the base at the circle's center; otherwise, it is oblique. [12] This pyramid may be classified based on the regularity of its bases. A pyramid with a regular polygon as the base is called a regular pyramid. [13]
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 section IV.3 of the Lahun Mathematical Papyri the volume of a granary with a circular base is found using the same procedure as RMP 43. 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]
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":
to calculate the volume of a cylindrical grain silo with a diameter of 9 cubits and a height of 10 cubits. Give the answer in terms of cubic cubits. Furthermore, given the following equalities among other units of volume, 1 cubic cubit = 3/2 khar = 30 heqats = 15/2 quadruple heqats, also express the answer in terms of khar and quadruple heqats.
This is a list of volume formulas of basic shapes: [4]: 405–406 ... Pyramid – , where is the base's area ...
Volume of Pyramid; Volume of Frustum — The 14th problem of the Moscow Mathematical Papyrus 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.