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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 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 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]
Given that is the base's area and is the height of a pyramid, the volume of a pyramid is: [25] =. 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 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.
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]
Circles: Problem 48 of the RMP compares the area of a circle (approximated by an octagon) and its circumscribing square. This problem's result is used in problem 50, where the scribe finds the area of a round field of diameter 9 khet. [8] Hemisphere: Problem 10 in the MMP finds the area of a hemisphere. [8] Volumes:
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. [19] [20] [21]