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If the edges connecting bases are perpendicular to one of its bases, the prism is called a truncated right triangular prism. Given that A is the area of the triangular prism's base, and the three heights h 1, h 2, and h 3, its volume can be determined in the following formula: [14] (+ +).
An augmented triangular prism with edge length has a surface area, calculated by adding six equilateral triangles and two squares' area: [2] +. Its volume can be obtained by slicing it into a regular triangular prism and an equilateral square pyramid, and adding their volume subsequently: [ 2 ] 2 2 + 3 3 12 a 3 ≈ 0.669 a 3 . {\displaystyle ...
To get the surface area of a triangular prism, you need to find the base area(0.5*bh) of the triangle. This is known as A1 in the following formula. The rectanges are known as A2, A3, and A4 in this formula. The formula for an equilateral triangular base in the prism is: A1×2+A2×3. The formula for an isosceles triangular base in the prism is:
The mensuration of polyhedra includes the surface area and volume. An area is a two-dimensional measurement calculated by the product of length and width; for a polyhedron, the surface area is the sum of the areas of all of its faces. [11] A volume is a measurement of a region in three-dimensional space. [12]
The surface area of a right prism is: +, where B is the area of the base, h the height, and P the base perimeter. The surface area of a right prism whose base is a regular n-sided polygon with side length s, and with height h, is therefore: = +.
An elongated triangular pyramid with edge length has a height, by adding the height of a regular tetrahedron and a triangular prism: [4] (+). Its surface area can be calculated by adding the area of all eight equilateral triangles and three squares: [2] (+), and its volume can be calculated by slicing it into a regular tetrahedron and a prism, adding their volume up: [2]: ((+)).
An elongated triangular orthobicupola with a given edge length has a surface area, by adding the area of all regular faces: [2] (+). Its volume can be calculated by cutting it off into two triangular cupolae and a hexagonal prism with regular faces, and then adding their volumes up: [ 2 ] ( 5 2 3 + 3 3 2 ) a 3 ≈ 4.955 a 3 . {\displaystyle ...
The elongated triangular bipyramid is constructed from a triangular prism by attaching two tetrahedrons onto its bases, a process known as the elongation. [1] These tetrahedrons cover the triangular faces so that the resulting polyhedron has nine faces (six of them are equilateral triangles and three of them are squares), fifteen edges, and eight vertices. [2]
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