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The formula for the volume of a pyramidal square frustum was introduced by the ancient Egyptian mathematics in what is called the Moscow Mathematical Papyrus, written in the 13th dynasty (c. 1850 BC): = (+ +), where a and b are the base and top side lengths, and h is the height.
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. [10] [11] Problem 14 of the Moscow Mathematical Papyrus computes the volume of a truncated pyramid, also known as a frustum.
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]
The volume of a pyramid was recorded back in ancient Egypt, where they calculated the volume of a square frustum, suggesting they acquainted the volume of a square pyramid. [26] The formula of volume for a general pyramid was discovered by Indian mathematician Aryabhata, where he quoted in his Aryabhatiya that the volume of a pyramid is ...
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.
The formula for the volume of a frustum of a paraboloid [23] [24] is: V = (π h/2)(r 1 2 + r 2 2), where h = height of the frustum, r 1 is the radius of the base of the frustum, and r 2 is the radius of the top of the frustum. This allows us to use a paraboloid frustum where that form appears more appropriate than a cone.
In general, the volume of a pyramid is equal to one-third of the area of its base multiplied by its height. [8] Expressed in a formula for a square pyramid, this is: [9] =. Many mathematicians have discovered the formula for calculating the volume of a square pyramid in ancient times.
I tried studying a limiting case, and it seems incompatible with the formula presented here. I am considering a regular hexagon with side length L, so its area is 3*sqrt(3)*L^2/2. For a frustum with side lengths L and 0 and perpendicular height 0, we expect its total surface area to be twice that of the regular hexagon, 3*sqrt(3)*L^2.