How can I determine the radius of a dodecahedron? I am making a dodecahedron that needs to fit inside of a sphere. The sphere has a diameter of 56mm. What is largest possible measurement of one segment of a pentagon side of a dodecahedron that would fit inside the sphere? How do I determine this?

 A: The answers provided here are correct. Let me explain the formula.
By joining together the diagonals of the pentagons on the dodecahedron we build a cube. For reference use the following figure:

Let us focus in the blue pentagon. The diagonal separates the pentagon
into a triangle and a trapezoid. The other 11 diagonals (chosen here. Note
that you could form up to 5 cubes by selecting different sets of diagonals)  form a cube. The vertices of the cube lie in a sphere which is the same sphere
circumscribing the dodecahedron. So, to find the radius of that sphere is just half of the diagonal of the cube.
It is a well known fact (see here Pentagon ) that the length of a diagonal $d$ of a pentagon of side length $\ell$ is given by
\begin{equation}
 d = \phi \ell
\end{equation}
where $\phi=(1+ \sqrt{5})/2$ is the golden ratio.
Now, the diagonal of the cube is
\begin{equation}
  D = \sqrt{ d^2 + d^2 + d^2} = \sqrt{3} d = \sqrt{3} \phi \ell.
\end{equation}
So the radius of the sphere is 
\begin{equation}
  r = \frac{D}{2} = \frac{\sqrt{3} \, \phi \ell}{2} = \frac{\sqrt{3} (1 + \sqrt{5}) \ell }{4}.
\end{equation}
A: If you know the side $\ell$ of the pentagons, the radius of the circumcribing sphere is $r=\tfrac{\ell}{4}(\sqrt{15}+\sqrt{3})$.This is given in Wikipedia (which is, I guess, where you got the picture from :) ) and I recall having read it being deduced in a book written by Coxeter.
A: The radius $R$ of the sphere circumscribing the dodecahedron, having edge length $a$, is given here as follows $$\color{blue}{R=\frac{\sqrt{3}(\sqrt{5}+1)a}{4}}$$ $$\implies a=\frac{(\sqrt{5}-1)R}{\sqrt{3}}$$ 
Now, substituting the radius $R=\frac{56}{2}=28\space mm$, the edge length of the dodecahedron $$a=\frac{(\sqrt{5}-1)(28)}{\sqrt{3}}\approx 19.98203703 \space mm$$
A: If the measure for the Dodecahedron’s diameter is 56mm then the measure for the edge of the Dodecahedron can be found by first dividing the measure for the diameter of the Dodecahedron 56mm into the square root of 3 = 1.732050807568877 and then divide the larger part of the division into the Golden ratio of 1.6180339887555 and the result will be the edge of the Dodecahedron for the correct Sphere = 19.98mm.
