Efficient construction of a dodecahedron This is essentially a manufacturing question: Is it possible to build a dodecahedron using 12 pentagonal panels that are all identical and can be joined mechanically? I am trying to come up with a solution that does not require separate parts.
 A: Let $a$ be the edge length of each of 12 (regular) pentagonal panels. Then the inscribed radius $\color{blue}{r_{i}}$ of each pentagonal panel is $$\color{blue}{r_{i}}=\frac{a}{2}\cot \frac{\pi}{5}$$$$\color{blue}{=\frac{a}{2}\sqrt{\frac{5+2\sqrt{5}}{5}}}$$ 
in a dodecahedron, the radius $R_{i}$ of the inscribed spherical surface touching all 12 identical pentagons is given as $$\color{blue}{R_{i}=\frac{a}{2}\sqrt{\frac{25+11\sqrt{5}}{10}}}$$ Now assume that the dihedral angle between any two consecutive pentagonal faces (panels) meeting at the common edge is $\color{blue}{\theta}$ & drop a perpendicular to any of 12 pentagonal faces (panels) to obtain a right triangle in    which we have $$\tan\frac{\theta}{2}=\frac{R_{i}}{r_{i}}$$ $$=\frac{\frac{a}{2}\sqrt{\frac{25+11\sqrt{5}}{10}}}{\frac{a}{2}\sqrt{\frac{5+2\sqrt{5}}{5}}}$$ $$=\sqrt{\frac{25+11\sqrt{5}}{2(5+2\sqrt{5})}}=\frac{1+\sqrt{5}}{2}$$  $$\implies\color{blue}{ \theta=2\tan^{-1}\left(\frac{1+\sqrt{5}}{2}\right)\approx 116.5650512^o}$$
It is clear that each two pentagonal panels (faces), out of 12, are inclined with each other at an equal diahedral angle $\theta$ at the common edge.   
Thus by mechanically joining (either by gluing or by welding) all 12 pentagonal panels at an equal angle $\theta\approx 116.5650512^o$ at all 30 edges each edge is shared (common) by two adjacent pentagonal panels (faces), the dodecahedron can be easily constructed/manufactured.  
