# Chirorhombidodecahedron

(Redirected from Rhombidodecahedron)
Chirorhombidodecahedron
Rank3
TypeUniform
SpaceSpherical
Notation
Bowers style acronymKred
Elements
Components6 pentagonal prisms
Faces30 squares, 12 pentagons
Edges30+60
Vertices60
Vertex figureIsosceles triangle, edge lengths (1+5)/2, 2, 2
Measures (edge length 1)
Circumradius${\displaystyle \sqrt{\frac{15+2\sqrt5}{20}} ≈ 0.98672}$
Volume${\displaystyle \frac{3\sqrt{25+10\sqrt5}}{2} ≈ 10.32286}$
Dihedral angles4–4: 108°
4–5: 90°
Central density6
Related polytopes
ArmySemi-uniform Ti
RegimentKred
DualCompound of six pentagonal tegums
ConjugateGreat chirorhombidodecahedron
Topological properties
OrientableYes
Properties
SymmetryH3+, order 60
ConvexNo
NatureTame
Discovered by{{{discoverer}}}

The chirorhombidodecahedron, rhombidodecahedron, kred, or compound of six pentagonal prisms is a uniform polyhedron compound. It consists of 30 squares and 12 pentagons, with one pentagon and two squares joining at a vertex.

Its quotient prismatic equivalent is the pentagonal prismatic hexateroorthowedge, which is eight-dimensional.

## Vertex coordinates

The vertices of a chirorhombidodecahedron of edge length 1 are given by all permutations of:

• ${\displaystyle \left(±\sqrt{\frac{5-2\sqrt5}{20}},\,±\sqrt{\frac{5+2\sqrt5}{20}},\,±\sqrt{\frac{5+2\sqrt5}{20}}\right),}$

Plus all even permutations of:

• ${\displaystyle \left(0,\,±\sqrt{\frac{5-\sqrt5}{40}},\,±\sqrt{\frac{5+\sqrt5}{8}}\right),}$
• ${\displaystyle \left(±\sqrt{\frac{5-\sqrt5}{40}},\,±\sqrt{\frac{5+\sqrt5}{40}},\,±\sqrt{\frac{5+\sqrt5}{10}}\right).}$

## Related polyhedra

This compound is chiral. The compound of the two enantiomorphs is the disrhombidodecahedron.