Pentagonal orthocupolarotunda

{{Infobox polytope The pentagonal orthocupolarotunda is one of the 92 Johnson solids (J32). It consists of 5+5+5 triangles, 5 squares, and 1+1+5 pentagons. It can be constructed by attaching a pentagonal cupola and a pentagonal rotunda at their decagonal bases, such that the two pentagonal bases are in the same orientation.
 * type=CRF
 * img=Pentagonal orthocupolarotunda 2.png
 * 3d=J32 pentagonal orthocupolarotunda.stl
 * off=Pentagonal orthocupolarotunda.off
 * dim = 3
 * obsa = Pocuro
 * faces = 5+5+5 triangles, 5 squares, 1+1+5 pentagons
 * edges = 5+5+5+5+10+10+10
 * vertices = 5+5+5+10
 * verf = 5 isosceles trapezoids, edge lengths 1, $\sqrt{2}$, (1+$\sqrt{5}$}/2, $\sqrt{2}$
 * verf2 = 10 rectangles, edge lengths 1 and (1+$\sqrt{5}$)/2
 * verf3 = 10 irregular tetragons, edge lengths 1, $\sqrt{2}$, 1, (1+$\sqrt{5}$)/2
 * coxeter = xoxx5ofxo&#xt
 * army=Pocuro
 * reg=Pocuro
 * symmetry = H2×I, order 10
 * volume = $$5\frac{11+5\sqrt5}{12} ≈ 9.24181$$
 * dih = 3–4 cupolaic: $$\arccos\left(-\frac{\sqrt3+\sqrt{15}}{6}\right) ≈ 159.09484°$$
 * dih2 = 4–5: $$\arccos\left(-\sqrt{\frac{5+\sqrt5}{10}}\right) ≈ 148.28253°$$
 * dih3 = 3–5 rotundaic: $$\arccos\left(-\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 142.62263°$$
 * dih4 = 3–4 join: $$\arccos\left(-\frac{\sqrt{15}-\sqrt3}{6}\right) ≈ 110.95106°$$
 * dih5 = 3–5 join: $$\arccos\left(-\sqrt{\frac{5-2\sqrt5}{15}}\right) ≈ 100.81237°$$
 * smm = Yes
 * dual = Pentadeltodecadeltopentadelto-pentarhombic icosipentahedron
 * conjugate = Retrograde pentagrammic orthocupolarotunda
 * conv=Yes
 * orientable=Yes
 * nat=Tame}}

If the cupola and rotunda are joined such that the bases are rotated 36º, the result is the pentagonal gyrocupolarotunda.

Vertex coordinates
A pentagonal orthocupolarotunda of edge length 1 has vertices given by the following coordinates:


 * $$\left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,\sqrt{\frac{5+2\sqrt5}{5}}\right),$$
 * $$\left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,\sqrt{\frac{5+2\sqrt5}{5}}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5-\sqrt5}{40}},\,\sqrt{\frac{5+2\sqrt5}{5}}\right),$$
 * $$\left(0,\,-\sqrt{\frac{5+2\sqrt5}{5}},\,\sqrt{\frac{5+\sqrt5}{10}}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{25+11\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}\right),$$
 * $$\left(±\frac{3+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}\right),$$
 * $$\left(±\frac12,\,±\frac{\sqrt{5+2\sqrt5}}{2},\,0\right),$$
 * $$\left(±\frac{3+\sqrt5}{4},\,±\sqrt{\frac{5+\sqrt5}{8}},\,0\right),$$
 * $$\left(±\frac{1+\sqrt5}{2},\,0,\,0\right),$$
 * $$\left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,-\sqrt{\frac{5-\sqrt5}{10}}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5+\sqrt5}{40}},\,-\sqrt{\frac{5-\sqrt5}{10}}\right),$$
 * $$\left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,-\sqrt{\frac{5-\sqrt5}{10}}\right).$$

Related polyhedra
A decagonal prism can be inserted between the two halves of the pentagonal orthocupolarotunda to produce the elongated pentagonal orthocupolarotunda..

The pentagonal orthocupolarotunda also has a connection with the regular icosahedron, being a partial Stott expansion of a pentagonal-symmetric faceting of the icosahedron.