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
numerical value:
Read more about this topic: Elongated Triangular Orthobicupola
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