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Circumscribed sphere

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In geometry, a circumscribed sphere of a polyhedron is a sphere that contains the polyhedron and touches each of the polyhedron's vertices. The word circumsphere is sometimes used to mean the same thing. When it exists, a circumscribed sphere need not be the smallest sphere containing the polyhedron; for instance, the tetrahedron formed by a vertex of a cube and its three neighbors has the same circumsphere as the cube itself, but can be contained within a smaller sphere having the three neighboring vertices on its equator. All regular polyhedra have circumscribed spheres, but most irregular polyhedra do not have all vertices lying on a common sphere, although it is still possible to define the smallest containing sphere for such shapes.

The radius of sphere circumscribed around a polyhedron P is called the circumradius of </math>P</math> .

Volume

V=\frac{\pi}{48}\left(\frac{abc}{A}\right)^3
V=\pi\frac{a^3b^3c^3}{48(\sqrt{\frac{(a^2+b^2+c^2)^2}{16}-\frac{a^4+b^4+c^4}{8}})^3}
V=\frac{\pi}{6}s^3
V=\frac{4\pi}{3}\left(\left(1-\frac{2}{n}\right)\tan\left(\frac{180}{n}\right)\right)^3s^3

The volume of sphere circumscribed around a polyhedron is therefore:

V=\frac{\pi}{6}\left(\frac{s}{\sin\left(\frac{180}{n}\right)}\right)^3

Surface area

SA=\frac{\pi}{4}\left(\frac{abc}{A}\right)^2
SA=4\pi\frac{a^2b^2c^2}{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}
SA=\pi s^2
SA=4\pi\left(\left(1-\frac{2}{n}\right)\tan\left(\frac{180}{n}\right)\right)^2s^2

The surface area of sphere circumscribed around a polyhedron is therefore:

SA=\pi\left(\frac{s}{\sin\left(\frac{180}{n}\right)}\right)^2

Circumscribed hemisphere

Volume

V= \pi\frac{a^3b^3c^3}{98A^3}
V= \pi\frac{a^3b^3c^3}{98(\sqrt{\frac{(a^2+b^2+c^2)^2}{16}-\frac{a^4+b^4+c^4}{8}})^3}
V= s^3\frac{\pi}{12}
V= s^3\pi\frac{2(n-2)^3}{3n^3}tan^3(\frac{180}{n})

The volume of the hemisphere circumscribed in a polyhedron is therefore:

V= s^3 \frac{\pi}{12sin^3(\frac{180}{n})}

Surface area

SA= \pi\frac{a^2b^2c^2}{8A^2}
SA= \pi\frac{2a^2b^2c^2}{(a^2+b^2+c^2)^2-2(a^4 + b^4 +c^4)}
SA= s^2\frac{\pi}{2}
SA= s^2\pi2(1-\frac{2}{n})^2tan^2(\frac{180}{n})

The surface area of the hemisphere circumscribed in a polyhedron is therefore:

SA= s^2 \frac{\pi}{2sin^2(\frac{180}{n})}

See also

External links


eo:Ĉirkaŭskribita sfero

pl:Kula opisana

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