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
{\displaystyle P}
circumradius of
P
{\displaystyle P}
Volume [ ]
V
=
π
48
(
a
b
c
A
)
3
V
=
4
π
3
(
a
b
c
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
4
+
b
4
+
c
4
)
)
3
V
=
π
6
s
3
V
=
4
π
3
(
(
1
−
2
n
)
tan
(
180
n
)
⋅
s
)
3
{\displaystyle \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
=
π
6
(
s
sin
(
180
n
)
)
3
{\displaystyle V=\frac{\pi}{6}\left(\frac{s}{\sin\left(\frac{180}{n}\right)}\right)^3}
Surface area [ ]
A
=
π
4
(
a
b
c
A
)
2
A
=
4
π
(
a
b
c
)
2
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
4
+
b
4
+
c
4
)
A
=
π
s
2
A
=
4
π
(
(
1
−
2
n
)
tan
(
180
n
)
⋅
s
)
2
{\displaystyle \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
=
π
(
s
sin
(
180
n
)
)
2
{\displaystyle A=\pi\left(\frac{s}{\sin\left(\frac{180}{n}\right)}\right)^2}
Circumscribed hemisphere [ ] Volume [ ]
V
=
π
98
(
a
b
c
A
)
3
V
=
32
49
π
(
a
b
c
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
4
+
b
4
+
c
4
)
)
3
V
=
π
12
s
3
V
=
2
π
3
(
(
1
−
2
n
)
tan
(
180
n
)
⋅
s
)
3
{\displaystyle \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
=
π
12
(
s
sin
(
180
n
)
)
3
{\displaystyle V=\frac{\pi}{12}\left(\frac{s}{\sin\left(\tfrac{180}{n}\right)}\right)^3}
Surface area [ ]
A
=
π
8
(
a
b
c
A
)
2
A
=
2
π
(
a
b
c
)
2
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
4
+
b
4
+
c
4
)
A
=
π
2
s
2
A
=
2
π
(
(
1
−
2
n
)
tan
(
180
n
)
⋅
s
)
2
{\displaystyle \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
=
π
2
(
s
sin
(
180
n
)
)
2
{\displaystyle A=\frac{\pi}{2}\left(\frac{s}{\sin\left(\tfrac{180}{n}\right)}\right)^2}
See also [ ] External links [ ]
eo:Ĉirkaŭskribita sfero
pl:Kula opisana
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