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3D model of regular octahedron. The surface area of a regular octahedron can be ascertained by summing all of its eight equilateral triangles, whereas its volume is twice the volume of a square pyramid; if the edge length is , [12] =, =. The radius of a circumscribed sphere (one that touches the octahedron at all vertices), the radius of an ...
Every two ideal polyhedra with the same number of vertices have the same surface area, and it is possible to calculate the volume of an ideal polyhedron using the Lobachevsky function. The surface of an ideal polyhedron forms a hyperbolic manifold , topologically equivalent to a punctured sphere, and every such manifold forms the surface of a ...
In geometry, a Platonic solid is a convex, regular polyhedron in three-dimensional Euclidean space.Being a regular polyhedron means that the faces are congruent (identical in shape and size) regular polygons (all angles congruent and all edges congruent), and the same number of faces meet at each vertex.
Graphs of surface area, A against volume, V of the Platonic solids and a sphere, showing that the surface area decreases for rounder shapes, and the surface-area-to-volume ratio decreases with increasing volume. Their intercepts with the dashed lines show that when the volume increases 8 (2³) times, the surface area increases 4 (2²) times.
The surface area of a polyhedron is the sum of the areas of its faces, for definitions of polyhedra for which the area of a face is well-defined. The geodesic distance between any two points on the surface of a polyhedron measures the length of the shortest curve that connects the two points, remaining within the surface.
Given the edge length .The surface area of a truncated tetrahedron is the sum of 4 regular hexagons and 4 equilateral triangles' area, and its volume is: [2] =, =.. The dihedral angle of a truncated tetrahedron between triangle-to-hexagon is approximately 109.47°, and that between adjacent hexagonal faces is approximately 70.53°.
The surface area of a rhombicuboctahedron can be determined by adding the area of all faces: 8 equilateral triangles and 18 squares. The volume of a rhombicuboctahedron V {\displaystyle V} can be determined by slicing it into two square cupolas and one octagonal prism.
To find the volume and surface area of a pentagonal hexecontahedron, denote the shorter side of one of the pentagonal faces as , and set a constant t [1] = + (+) + + () Then the surface area ( A {\displaystyle A} ) is: A = 30 b 2 ⋅ ( 2 + 3 t ) ⋅ 1 − t 2 1 − 2 t 2 ≈ 162.698 b 2 {\displaystyle A={\frac {30b^{2}\cdot (2+3t)\cdot {\sqrt ...