# Gyroelongated pentagonal cupola

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Gyroelongated pentagonal cupola
Rank3
TypeCRF
SpaceSpherical
Notation
Bowers style acronymGyepcu
Coxeter diagramoxo10sox&#xt
Elements
Faces5+5+5+10 triangles, 5 squares, 1 pentagon, 1 decagon
Edges5+5+5+10+10+10+10
Vertices5+5+5+10
Vertex figures5 isosceles trapezoids, edge lengths 1, 2, (1+5)/2, 2
10 irregular pentagons, edge lengths 1, 1, 1, 1, 2
10 isosceles trapezoids, edge lengths 1, 1, 1, (5+5)/2
Measures (edge length 1)
Volume${\displaystyle \frac{5+4\sqrt5+5\sqrt{2-2\sqrt5+2\sqrt{650+290\sqrt5}}}{6} ≈ 9.07333}$
Dihedral angles3–3 antiprismatic: ${\displaystyle \arccos\left(\frac{1-\sqrt{10+2\sqrt5}}{3}\right) ≈ 159.18651°}$
3–4 cupolaic: ${\displaystyle \arccos\left(-\frac{\sqrt3+\sqrt{15}}{6}\right) ≈ 159.09484°}$
4–5: ${\displaystyle \arccos\left(-\sqrt{\frac{5+\sqrt5}{10}}\right) ≈ 148.28253°}$
3–3 join: ${\displaystyle \arccos\left(\sqrt{\frac{5+2\sqrt5}{15}}\right)+\arccos\left(-\sqrt{\frac{11+4\sqrt5-2\sqrt{50+22\sqrt5}}{3}}\right) ≈ 132.62401°}$
3–4 join: ${\displaystyle \arccos\left(-\sqrt{\frac{5+\sqrt5}{10}}\right)+\arccos\left(-\sqrt{\frac{11+4\sqrt5-2\sqrt{50+22\sqrt5}}{3}}\right) ≈ 126.96412°}$
3–10: ${\displaystyle \arccos\left(-\sqrt{\frac{11+4\sqrt5-2\sqrt{50+22\sqrt5}}{3}}\right) ≈ 95.24664°}$
Central density1
Related polytopes
ArmyGyepcu
RegimentGyepcu
DualPentadeltadecapentagonal hemitrapezohedron
ConjugateGyroelongated retrograde pentagrammic cupola
Abstract & topological properties
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryH2×I, order 10
ConvexYes
NatureTame

The gyroelongated pentagonal cupola is one of the 92 Johnson solids (J24). It consists of 5+5+5+10 triangles, 5 squares, 1 pentagon, and 1 decagon. It can be constructed by attaching a decagonal antiprism to the decagonal base of the pentagonal cupola.

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

## Vertex coordinates

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

• ${\displaystyle \left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,\sqrt{\frac{5-\sqrt5}{10}}+H\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5-\sqrt5}{40}},\,\sqrt{\frac{5-\sqrt5}{10}}+H\right),}$
• ${\displaystyle \left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,\sqrt{\frac{5-\sqrt5}{10}}+H\right),}$
• ${\displaystyle \left(±\frac12,\,±\frac{\sqrt{5+2\sqrt5}}2,\,H\right),}$
• ${\displaystyle \left(±\frac{3+\sqrt5}4,\,±\sqrt{\frac{5+\sqrt5}8},\,H\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}2,\,0,\,H\right),}$
• ${\displaystyle \left(±\frac{\sqrt{5+2\sqrt5}}2,\,±\frac12,\,-H\right),}$
• ${\displaystyle \left(±\sqrt{\frac{5+\sqrt5}8},\,±\frac{3+\sqrt5}4,\,-H\right),}$
• ${\displaystyle \left(0,\,±\frac{1+\sqrt5}2,\,-H\right),}$

where H = ${\displaystyle \sqrt{\frac{-4-2\sqrt5+\sqrt{50+22\sqrt{5}}}8}}$is the distance between the decagonal antiprism's center and the center of one of its bases.