# How to compute the volume of the polyhedron with vertices at centre of a cube?

The centers of the faces of a cube are also the vertices of polyhedron. How to Compute the ratio of the volume of the polyhedron to that of the cube containing it?

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The dual to the cube is the octahedron, which is made up of two pyramids glued together. Draw a picture of the appropriate cross sections to derive the properties of these pyramids. If the side length of the cube is $2$ then the lengths of the edges of the octahedron are $\sqrt{2}$, so the area of the base of the pyramids is $2$ and the height is $1$ (half the cube). So the volume of the two pyramids is $$\frac{2}{3}(\text{area of base}) \cdot (\text{height})= \frac{4}{3} .$$ The volume of the cube is $8$ so that the ratio is $\frac{1}{6}$.