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In geometry, a frustum (Latin for 'morsel'); [a] (pl.: frusta or frustums) is the portion of a solid (normally a pyramid or a cone) that lies between two parallel planes cutting the solid. In the case of a pyramid, the base faces are polygonal and the side faces are trapezoidal .
For a regular n-gonal bifrustum with the equatorial polygon sides a, bases sides b and semi-height (half the distance between the planes of bases) h, the lateral surface area A l, total area A and volume V are: [2] and [3] = (+) () + = + = + + Note that the volume V is twice the volume of a frusta.
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The lateral surface area of a right circular cone is = where is the radius of the circle at the bottom of the cone and is the slant height of the cone. [4] The surface area of the bottom circle of a cone is the same as for any circle, . Thus, the total surface area of a right circular cone can be expressed as each of the following: Radius and ...
For a cube the lateral surface area would be the area of the four sides. If the edge of the cube has length a, the area of one square face A face = a ⋅ a = a 2. Thus the lateral surface of a cube will be the area of four faces: 4a 2. More generally, the lateral surface area of a prism is the sum of the areas of the sides of the prism. [1]
A bi-conic nose cone shape is simply a cone with length L 1 stacked on top of a frustum of a cone (commonly known as a conical transition section shape) with length L 2, where the base of the upper cone is equal in radius R 1 to the top radius of the smaller frustum with base radius R 2. = +
The tenth problem of the Moscow Mathematical Papyrus asks for a calculation of the surface area of a hemisphere (Struve, Gillings) or possibly the area of a semi-cylinder (Peet). Below we assume that the problem refers to the area of a hemisphere. The text of problem 10 runs like this: "Example of calculating a basket.
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.
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