# 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 P.

## VolumeEdit

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

The volume of sphere circumscribed around a polyhedron is therefore:

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

## Surface areaEdit

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

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

$SA= s^2 \frac{4\pi}{4sin^2(\frac{180}{n})}$

## Circumscribed hemisphereEdit

### VolumeEdit

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

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

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

### Surface areaEdit

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

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

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