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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]
Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities; List of volume formulas – Quantity of three-dimensional space
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. [30] 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 ...
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.
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]
A triaugmented triangular prism with edge length has surface area [10], the area of 14 equilateral triangles. Its volume, [10] +, can be derived by slicing it into a central prism and three square pyramids, and adding their volumes.
If the areas of the two parallel faces are A 1 and A 3, the cross-sectional area of the intersection of the prismatoid with a plane midway between the two parallel faces is A 2, and the height (the distance between the two parallel faces) is h, then the volume of the prismatoid is given by [3] = (+ +).