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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.
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.
The base regularity of a pyramid's base may be classified based on the type of polygon: one example is the star pyramid in which its base is the regular star polygon. [28] The truncated pyramid is a pyramid cut off by a plane; if the truncation plane is parallel to the base of a pyramid, it is called a frustum.
A square frustum, with volume equal to the height times the Heronian mean of the square areas. The Heronian mean may be used in finding the volume of a frustum of a pyramid or cone. The volume is equal to the product of the height of the frustum and the Heronian mean of the areas of the opposing parallel faces. [2]
This is a list of volume formulas of basic shapes: [4]: 405–406 Cone – 1 3 π r 2 h {\textstyle {\frac {1}{3}}\pi r^{2}h} , where r {\textstyle r} is the base 's radius Cube – a 3 {\textstyle a^{3}} , where a {\textstyle a} is the side's length;
Such a formula would be needed for building pyramids. In the next problem (Problem 57), the height of a pyramid is calculated from the base length and the seked (Egyptian for slope), while problem 58 gives the length of the base and the height and uses these measurements to compute the seked.
The surface area of a right square pyramid can be expressed as = +, where and are the areas of one of its triangles and its base, respectively. The area of a triangle is half of the product of its base and side, with the area of a square being the length of the side squared.
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]