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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 $P$ .

## Volume

\begin{align} V&=\frac{\pi}{48}\left(\frac{abc}{A}\right)^3\\ V&=\frac{4\pi}{3}\left(\frac{abc}{\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}}\right)^3\\ V&=\frac{\pi}{6}s^3\\ V&=\frac{4\pi}{3}\left(\left(1-\frac2n\right)\tan\left(\tfrac{180}{n}\right)\cdot s\right)^3 \end{align}

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

\begin{align} A&=\frac{\pi}{4}\left(\frac{abc}{A}\right)^2\\ A&=4\pi\frac{(abc)^2}{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}\\ A&=\pi s^2\\ A&=4\pi\left(\left(1-\frac2n\right)\tan\left(\tfrac{180}{n}\right)\cdot s\right)^2 \end{align}

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

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

## Circumscribed hemisphere

### Volume

\begin{align} V&=\frac{\pi}{98}\left(\frac{abc}{A}\right)^3\\ V&=\frac{32}{49}\pi\left(\frac{abc}{\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}}\right)^3\\ V&=\frac{\pi}{12}s^3\\ V&=\frac{2\pi}{3}\left(\left(1-\frac2n\right)\tan\left(\tfrac{180}{n}\right)\cdot s\right)^3 \end{align}

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

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

### Surface area

\begin{align} A&=\frac{\pi}{8}\left(\frac{abc}{A}\right)^2\\ A&=2\pi\frac{(abc)^2}{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}\\ A&=\frac{\pi}{2}s^2\\ A&=2\pi\left(\left(1-\frac2n\right)\tan\left(\tfrac{180}{n}\right)\cdot s\right)^2 \end{align}

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

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