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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 ...
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.
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 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: = +.
The surface area of an elongated triangular bipyramid is the sum of all polygonal face's area: six equilateral triangles and three squares. The volume of an elongated triangular bipyramid V {\displaystyle V} can be ascertained by slicing it off into two tetrahedrons and a regular triangular prism and then adding their volume.
The basic 3-dimensional element are the tetrahedron, quadrilateral pyramid, triangular prism, and hexahedron. They all have triangular and quadrilateral faces. Extruded 2-dimensional models may be represented entirely by the prisms and hexahedra as extruded triangles and quadrilaterals.
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 ...