# Gyroelongated pentagonal rotunda

Gyroelongated pentagonal rotunda Rank3
TypeCRF
SpaceSpherical
Notation
Bowers style acronymGyepro
Elements
Faces5+5+5+5+10 triangles, 1+5 pentagons, 1 decagon
Edges5+5+5+10+10+10+10+10
Vertices5+5+5+5+10
Vertex figures10 rectangles, edge lengths 1 and (1+5)/2
10 irregular pentagons, edge lengths 1, 1, 1, 1, (1+5)/2
10 Isosceles trapezoids, edge lengths 1, 1, 1, (5+5)/2
Measures (edge length 1)
Volume$\frac{45+17\sqrt5+10\sqrt{2-2\sqrt5+2\sqrt{650+290\sqrt5}}}{12} ≈ 13.66705$ Dihedral angles3–3 join: $\arccos\left(\sqrt{\frac{5-2\sqrt5}{15}}\right)+\arccos\left(-\sqrt{\frac{11+4\sqrt5-2\sqrt{50+22\sqrt5}}{3}}\right) ≈ 174.43432°$ 3–3 antiprismatic: $\arccos\left(\frac{1-\sqrt{10+2\sqrt5}}{3}\right) ≈ 159.18651°$ 3–5 join: $\arccos\left(\frac{\sqrt5}{5}\right)+\arccos\left(-\sqrt{\frac{11+4\sqrt5-2\sqrt{50+22\sqrt5}}{3}}\right) ≈ 158.68159°$ 3–5 rotundaic: $\arccos\left(-\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 142.62263°$ 3–10: $\arccos\left(-\sqrt{\frac{11+4\sqrt5-2\sqrt{50+22\sqrt5}}{3}}\right) ≈ 95.24664°$ Central density1
Related polytopes
ArmyGyepro
RegimentGyepro
DualPentapentarhombidecapentagonal hemitrapezohedron
ConjugateGyroelongated pentagrammic rotunda
Abstract & topological properties
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryH2×I, order 10
ConvexYes
NatureTame

The gyroelongated pentagonal rotunda is one of the 92 Johnson solids (J25). It consists of 5+5+5+5+10 triangles, 1+5 pentagons, and 1 decagon. It can be constructed by attaching a decagonal antiprism to the decagonal base of the pentagonal rotunda.

If a second rotunda is attached to the other decagonal base of the antiprism, the result is the gyroelongated pentagonal birotunda.

## Vertex coordinates

A gyroelongated pentagonal rotunda of edge length 1 has the following vertices:

• $\left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,\sqrt{\frac{5+2\sqrt5}{5}}+h\right),$ • $\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5-\sqrt5}{40}},\,\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$ • $\left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$ • $\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{25+11\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}+H\right),$ • $\left(±\frac{3+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}+H\right),$ • $\left(0,\,-\sqrt{\frac{5+2\sqrt5}{5}},\,\sqrt{\frac{5+\sqrt5}{10}}+H\right),$ • $\left(±\frac12,\,±\frac{\sqrt{5+2\sqrt5}}2,\,H\right),$ • $\left(±\frac{3+\sqrt5}4,\,±\sqrt{\frac{5+\sqrt5}8},\,H\right),$ • $\left(±\frac{1+\sqrt5}2,\,0,\,H\right),$ • $\left(±\frac{\sqrt{5+2\sqrt5}}2,\,±\frac12,\,-H\right),$ • $\left(±\sqrt{\frac{5+\sqrt5}8},\,±\frac{3+\sqrt5}4,\,-H\right),$ • $\left(0,\,±\frac{1+\sqrt5}2,\,-H\right),$ where H = $\sqrt{\frac{-4-2\sqrt5+\sqrt{50+22\sqrt{5}}}8}$ is the distance between the decagonal antiprism's center and the center of one of its bases.