Elongated Triangular Orthobicupola - Volume

Volume

The volume of J35 can be calculated as follows:

J35 consists of 2 cupolae and hexagonal prism.

The two cupolae makes 1 cuboctahedron = 8 tetrahedra + 6 half-octahedra. 1 octahedron = 4 tetrahedra, so total we have 20 tetrahedra.

What is the volume of a tetrahedron? Construct a tetrahedron having vertices in common with alternate vertices of a cube (of side, if tetrahedron has unit edges). The 4 triangular pyramids left if the tetrahedron is removed from the cube form half an octahedron = 2 tetrahedra. So


The hexagonal prism is more straightforward. The hexagon has area, so

Finally

V_{J_{35}} = 20 V_{tetrahedron} + V_{prism} = \frac{5 \sqrt{2}}{3} + \frac{3 \sqrt{3}}{2}

numerical value:

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