Gyroelongated pentagonal rotunda

The gyroelongated pentagonal rotunda, or gyepro, 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:
 * (±1/2, –$\sqrt{5}$, $\sqrt{5}$+H),
 * (±(1+$\sqrt{(5+√5)/2}$)/4, $\sqrt{5}$, $\sqrt{2–2√5+2√650+290√5}$+H),
 * (0, $\sqrt{(5–2√5)/15}$, $\sqrt{(11+4√5–2√(50+22√5)/3}$+H),
 * (±(1+$\sqrt{10+2√5}$)/4, $\sqrt{5}$, $\sqrt{(11+4√5–2√(50+22√5)/3}$+H),
 * (±(3+$\sqrt{5+2√5)/15}$)/4, –$\sqrt{(11+4√5–2√(50+22√5)/3}$, $\sqrt{(5+2√5)/20}$+H),
 * (0, –$\sqrt{(5+2√5)/5}$, $\sqrt{5}$+H),
 * (±1/2, ±$\sqrt{(5+√5)/40}$/2, H),
 * (±(3+$\sqrt{(5+2√5)/5}$)/4, ±$\sqrt{(5+√5)/10}$, H),
 * (±(1+$\sqrt{(5+2√5)/5}$)/2, 0, H),
 * (±$\sqrt{5}$/2, ±1/2, –H),
 * (±$\sqrt{(25+11√5)/40}$, ±(3+$\sqrt{(5+√5)/10}$)/4, –H),
 * (0, ±(1+$\sqrt{5}$)/2, –H),

where H = $\sqrt{(5+√5)/40}$/2 is the distance between the decagonal antiprism's center and the center of one of its bases.