Gyroelongated pentagonal rotunda

The gyroelongated pentagonal rotunda 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: where H = $$\sqrt{\frac{-4-2\sqrt5+\sqrt{50+22\sqrt{5}}}8}$$is the distance between the decagonal antiprism's center and the center of one of its bases.
 * $$\left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,\sqrt{\frac{5+2\sqrt5}{5}}+h\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5-\sqrt5}{40}},\,\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$$
 * $$\left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,\sqrt{\frac{5+2\sqrt5}{5}}+H\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{25+11\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}+H\right),$$
 * $$\left(±\frac{3+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}+H\right),$$
 * $$\left(0,\,-\sqrt{\frac{5+2\sqrt5}{5}},\,\sqrt{\frac{5+\sqrt5}{10}}+H\right),$$
 * $$\left(±\frac12,\,±\frac{\sqrt{5+2\sqrt5}}2,\,H\right),$$
 * $$\left(±\frac{3+\sqrt5}4,\,±\sqrt{\frac{5+\sqrt5}8},\,H\right),$$
 * $$\left(±\frac{1+\sqrt5}2,\,0,\,H\right),$$
 * $$\left(±\frac{\sqrt{5+2\sqrt5}}2,\,±\frac12,\,-H\right),$$
 * $$\left(±\sqrt{\frac{5+\sqrt5}8},\,±\frac{3+\sqrt5}4,\,-H\right),$$
 * $$\left(0,\,±\frac{1+\sqrt5}2,\,-H\right),$$