# Parabiaugmented truncated dodecahedron

Parabiaugmented truncated dodecahedron Rank3
TypeCRF
SpaceSpherical
Notation
Bowers style acronymPabautid
Elements
Faces10+10+10 triangles, 10 squares, 2 pentagons, 10 decagons
Edges10+10+10+10+10+10+20+20+20
Vertices10+10+10+20+20
Vertex figures10 isosceles trapezoids, edge length 1, 2, (1+5)/2, 2
20 irregular tetragons, edge length 1, 2, 1, (5+5)/2
40 isosceles triangles, edge lengths 1, 2+2, 2+2
Measures (edge length 1)
Volume$\frac{515+251\sqrt5}{12} ≈ 89.68776$ Dihedral angles3–4 join: $\arccos\left(-\sqrt{\frac{23+3\sqrt5}{30}}\right) ≈ 174.34011^\circ$ 3–4 cupolaic: $\arccos\left(-\frac{\sqrt3+\sqrt{15}}{6}\right) ≈ 159.09484^\circ$ 3–10 join: $\arccos\left(-\sqrt{\frac{65-2\sqrt5}{75}}\right) ≈ 153.94242^\circ$ 4–5: $\arccos\left(-\sqrt{\frac{5+\sqrt5}{10}}\right) ≈ 148.28253^\circ$ 3–10 tid: $\arccos\left(-\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 142.62263^\circ$ 10–10: $\arccos\left(-\frac{\sqrt5}{5}\right) ≈ 116.56505^\circ$ Central density1
Number of external pieces52
Level of complexity24
Related polytopes
ArmyPabautid
RegimentPabautid
DualParabirhombirhombistellated triakis icosahedron
ConjugateParabiaugmented quasitruncated great stellated dodecahedron
Abstract & topological properties
Flag count480
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
Symmetry(I2(10)×A1)/2, order 20
ConvexYes
NatureTame

The parabiaugmented truncated dodecahedron is one of the 92 Johnson solids (J69). It consists of 10+10+10 triangles, 10 squares, 2 pentagons, and 10 decagons. It can be constructed by attaching two pentagonal cupolas to two opposite decagonal faces of the truncated dodecahedron.

## Vertex coordinates

A parabiaugmented truncated dodecahedron of edge length 1 has vertices given by all even permutations of:

• $\left(0,\,±\frac12,\,±\frac{5+3\sqrt5}{4}\right),$ • $\left(±\frac12,\,±\frac{3+\sqrt5}{4},\,±\frac{3+\sqrt5}{2}\right),$ • $\left(±\frac{3+\sqrt5}{4},\,±\frac{1+\sqrt5}{2},\,±\frac{2+\sqrt5}{2}\right),$ • $±\left(±\frac12,\,\frac{15+13\sqrt5}{20},\,3\frac{5+\sqrt5}{10}\right),$ • $±\left(±\frac{1+\sqrt5}{4},\,\frac{25+13\sqrt5}{20},\,\frac{25+\sqrt5}{20}\right),$ • $±\left(0,\,\frac{10+9\sqrt5}{10},\,\frac{15+\sqrt5}{20}\right).$ 