Elongated pentagonal orthocupolarotunda

{{Infobox polytope The elongated pentagonal orthocupolarotunda, or epocuro, is one of the 92 Johnson solids (J40). It consists of 5+5+5 triangles, 5+5+5 squares, and 1+1+5 pentagons. It can be constructed by inserting a decagonal prism between the two halves of the pentagonal orthocupolarotunda.
 * type=CRF
 * img=Elongated pentagonal orthocupolarotunda 2.png
 * 3d=J40 elongated pentagonal orthocupolarotunda.stl
 * off=Elongated pentagonal orthocupolarotunda.off
 * dim = 3
 * obsa = Epocuro
 * faces = 5+5+5 triangles, 5+5+5 squares, 1+1+5 pentagons
 * edges = 5+5+5+5+5+5+10+10+10+10
 * vertices = 5+5+5+10+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 trapezoids, edge lengths 1, $\sqrt{2}$, $\sqrt{2}$, $\sqrt{2}$
 * verf4 = 10 irregular tetragons, edge lengths 1, (1+$\sqrt{5}$)/2, $\sqrt{2}$, $\sqrt{2}$
 * coxeter = xoxxx5ofxxo&#xt
 * army=Epocuro
 * reg=Epocuro
 * symmetry = H2×I, order 10
 * volume = $$5\frac{11+5\sqrt5+6\sqrt{5+2\sqrt5}}{12} ≈ 16.93602$$
 * dih = 3–4 rotundaic join: $$\arccos\left(-\sqrt{\frac{10+2\sqrt5}{15}}\right) ≈ 169.18768°$$
 * dih2 = 3–4 cupolaic: $$\arccos\left(-\frac{\sqrt3+\sqrt{15}}{6}\right) ≈ 159.09484°$$
 * dih3 = 4–5 join: $$\arccos\left(-\frac{2\sqrt5}{5}\right) ≈ 153.43495°$$
 * dih4 = 4–5: $$\arccos\left(-\sqrt{\frac{5+\sqrt5}{10}}\right) ≈ 148.28253°$$
 * dih5 = 4–4 prismatic: 144°
 * dih6 = 3–5: $$\arccos\left(-\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 142.62263°$$
 * dih7 = 3–4 join: $$\\arccos\left(-\sqrt{\frac{10-2\sqrt5}{15}}\right) ≈ 127.37737°$$
 * dih8 = 4–4 join: $$\arccos\left(-\sqrt{\frac{5-\sqrt5}{10}}\right) ≈ 121.71747°$$
 * smm = Yes
 * dual = Pentadeltodecadeltorthodecatetragopentarhombirhombic triacontapentahedron
 * conjugate = Rlongated retrograde pentagrammic orthocupolarotunda
 * conv=Yes
 * orientable=Yes
 * nat=Tame}}

Vertex coordinates
An elongated pentagonal orthocupolarotunda of edge length 1 has vertices given by the following coordinates:
 * $$\left(±\frac12,\,±\frac{\sqrt{5+2\sqrt5}}{2},\,±\frac12\right),$$
 * $$\left(±\frac{3+\sqrt5}{4},\,±\sqrt{\frac{5+\sqrt5}{8}},\,±\frac12\right),$$
 * $$\left(±\frac{1+\sqrt5}{2},\,0,\,±\frac12\right),$$
 * $$\left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,\frac{1+2\sqrt{\frac{5+2\sqrt5}{5}}}{2}\right),$$
 * $$\left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,\frac{1+2\sqrt{\frac{5+2\sqrt5}{5}}}{2}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5-\sqrt5}{40}},\,\frac{1+2\sqrt{\frac{5+2\sqrt5}{5}}}{2}\right),$$
 * $$\left(0,\,-\sqrt{\frac{5+2\sqrt5}{5}},\,\frac{1+2\sqrt{\frac{5+\sqrt5}{10}}}{2}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{25+11\sqrt5}{40}},\,\frac{1+2\sqrt{\frac{5+\sqrt5}{10}}}{2}\right),$$
 * $$\left(±\frac{3+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{40}},\,\frac{1+2\sqrt{\frac{5+\sqrt5}{10}}}{2}\right),$$
 * $$\left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,-\frac{1+2\sqrt{\frac{5-\sqrt5}{10}}}{2}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5+\sqrt5}{40}},\,-\frac{1+2\sqrt{\frac{5-\sqrt5}{10}}}{2}\right),$$
 * $$\left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,-\frac{1+2\sqrt{\frac{5-\sqrt5}{10}}}{2}\right),$$