Gyroelongated pentagonal cupolarotunda

The gyroelongated pentagonal cupolarotunda, or gyepcuro, is one of the 92 Johnson solids (J47). It consists of 5+5+5+5+5+5+5 triangles, 5 squares, and 1+1+5 pentagons. It can be constructed by attaching a pentagonal cupola and a pentagonal rotunda to opposite bases of the decagonal antiprism.

Vertex coordinates
A gyroelongated pentagonal cupolarotunda of edge length 1 has the following vertices:
 * (±1/2, –$\sqrt{5}$, $\sqrt{5}$+H),
 * (±(1+$\sqrt{2}$)/4, $\sqrt{5}$, $\sqrt{2}$+H),
 * (0, $\sqrt{5}$, $\sqrt{2–2√5+2√650+290√5}$+H),
 * (±(1+$\sqrt{(5–2√5)/15}$)/4, $\sqrt{(11+4√5–2√(50+22√5)/3}$, $\sqrt{10+2√5}$+H),
 * (±(3+$\sqrt{3}$)/4, –$\sqrt{15}$, $\sqrt{5}$+H),
 * (0, –$\sqrt{(11+4√5–2√(50+22√5)/3}$, $\sqrt{(5+√5)/10}$+H),
 * (±1/2, ±$\sqrt{5+2√5)/15}$/2, H),
 * (±(3+$\sqrt{(5+2√5)/15}$)/4, ±$\sqrt{(11+4√5–2√(50+22√5)/3}$, H),
 * (±(1+$\sqrt{(5+√5)/10}$)/2, 0, H),
 * (±$\sqrt{(11+4√5–2√(50+22√5)/3}$/2, ±1/2, –H),
 * (±$\sqrt{(5+2√5)/20}$, ±(3+$\sqrt{(5+2√5)/5}$)/4, –H),
 * (0, ±(1+$\sqrt{5}$)/2, –H),
 * (–$\sqrt{(5+√5)/40}$, ±1/2, –$\sqrt{(5+2√5)/5}$–H),
 * ($\sqrt{(5+√5)/10}$, ±(1+$\sqrt{(5+2√5)/5}$)/4, –$\sqrt{5}$–H),
 * ($\sqrt{(25+11√5)/40}$, 0, –$\sqrt{(5+√5)/10}$–H).

where H = $\sqrt{5}$/2 is the distance between the decagonal antiprism's center and the center of one of its bases.