Edge length of a Dodecahedron Good morning, 
If I have a $12$ sided regular pentagonal structure - that is, a Dodecahedron - and the widest point is $3.5m$ in diameter, what is the length of an edge (if they are all the same).
Regards, 
Connor
 A: The radius $R$ of the spherical surface passing through all 20 identical vertices of a dodecahedron with edges length $a$ is given by generalized formula (derived in HCR's Formula for platonic solids)$$\bbox[4pt, border: 1px solid blue;]{R=\frac{\sqrt{3}(\sqrt{5}+1)a}{4}} $$ $$\implies a=\frac{4R}{\sqrt{3}(\sqrt{5}+1)}$$$$\implies \color{purple}{a=\frac{R(\sqrt{5}-1)}{\sqrt{3}}}$$ Since $R=\frac{3.5}{2}$ hence the edge length of dodecahedron is $$a=\frac{(3.5)(\sqrt{5}-1)}{2\sqrt{3}}$$$$=\frac{R(\sqrt{5}-1)}{\sqrt{3}}\approx 1.248877314 \space m$$
A: All Platonic solids except the tetrahedron are centrally symmetric, meaning they are preserved under reflection through the origin. (ref: wiki http://en.wikipedia.org/wiki/Platonic_solid#Symmetry_groups). So diameter of a dodecahedron = diameter of the circumscribed sphere.
Radius of circumscribed sphere and the side of a pentagon in dodecfahedron are related by $R \approx 1.4 a$. (ref. http://en.wikipedia.org/wiki/Dodecahedron).
So $a \approx \frac{R}{1.4} = \frac{D}{2.8} = 1.25m$
