Elongated triangular orthobicupola

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 ${\displaystyle {\frac {1}{\sqrt {2}}}}$, 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

${\displaystyle V_{tetrahedron}={\frac {1}{3}}V_{cube}={\frac {1}{3}}{\frac {1}{{\sqrt {2}}^{3}}}={\frac {\sqrt {2}}{12}}}$

The hexagonal prism is more straightforward. The hexagon has area ${\displaystyle 6{\frac {\sqrt {3}}{4}}}$, so

${\displaystyle V_{prism}={\frac {3{\sqrt {3}}}{2}}}$

Finally

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

numerical value:

${\displaystyle V_{J_{35}}=4.9550988153084743549606507192748}$

The elongated triangular orthobicupola forms space-filling honeycombs with tetrahedra and square pyramids.[2]

• Eric W. Weisstein, Elongated triangular orthobicupola (Johnson solid) at MathWorld.