Constructing the Dodecahedron

Let $e$ be the edge length. Then

$e = 10$

The height of the Dodecahedron can be written as the sum

$h = h_1 + h_2$

where $h_1$ is the height due to the rotated altitude of a pentagon.

(from base to apex) , and $h_2$ is the height from the other half of the Dodecahedron due to the rotate edge (that is connected to one to the two vertices of the edge about which rotation occurs)

$h_1 = e ( \sin 72^{\circ} + \sin (72^{\circ} + 72^{\circ}) )$

$= e ( \sin 72^{\circ} + \sin 144^{\circ} )$

and

$h_2 = e sin 72^{\circ}$

Finally, the altitude of the Dodecahedron is given by

$A = h \sin 63.435^{\circ}$

This angle is the angle of rotation of the pentagons from the initial relaxed planar position (see problem "3-D Analytic Geometry" )

Therefore,

$A = e (2 \sin 72^{\circ} + \sin 144^{\circ} ) \sin 63.435^{\circ}$

Substituting the value of $e = 10$ gives

$A = 22.27$

Let's consider a pentagon ABCDE with "base" DC and vertex "A".

Let's call M the middle point of AE, N the middle point of AB, T the middle point of DC and H the middle point of MN.

Now we know that in a Dodecahedron the angle between 2 adiacent faces is $\angle 26.565^\circ$ (from the previous problem 3-D Analytic Geometry).

At this point we can observe the fact that $(AT-AH) \cdot cos(26.565)$ gives half of the height of the dodecahedron.

So $AH=5 \cdot cos(54)$ (angles in a regular pentagon measure 108 degrees);

$AT=\sqrt{AD^2-DT^2}$ = $\sqrt{ 2 \cdot 10^2-2 \cdot 10^2 \cdot cos(108)-5^2}$

At the end: $(\sqrt{2 \cdot 10^2-2 \cdot 10^2 \cdot cos(108)-5^2}-5 \cdot cos(54)) \cdot cos(26.565) \cdot 2= 22.270337$

In general, the distance between the opposite faces of a regular dodecahedron having edge length $a$ is given by the general formula (for detailed analysis, go through HCR's formula for all five regular polyhedrons (platonic solids) )

$\frac{(3+\sqrt5)a}{4\sin36^\circ}\approx 2.227032729a$

As per given problem, the distance between the opposite faces of a regular dodecahedron having edge length $10$ units is $=\frac{(3+\sqrt5)10}{4\sin36^\circ}\approx 22.27032729\ units$

$N[ \ \ 10 \text{Subtract}\text{@@} \\ \ \ \ \ \text{Table}[\text{If}[\text{And}\text{@@}\text{Table}[q[[1]]=\text{qv},\{\text{qv},q\}],q[[1]],\text{Nothing}], \\ \ \ \ \ \ \ \{q,\text{Table}[\text{ToRadicals}[\text{PolyhedronData}[\text{Dodecahedron},\text{Vertices}][[v]][[3]]], \\ \ \ \ \ \ \ \ \{f,\text{PolyhedronData}[\text{Dodecahedron},\text{Faces}]\},\{v,f\}]\}]] \Longrightarrow \\ 22.2703272882321$

$20 \sqrt{\frac{5}{8}+\frac{11}{8 \sqrt{5}}}$

Quick translation, Using Wolfram polyhedron data, find the top and bottom of the dodecahedron, subtract the lower $z$ coordinate from the upper $z$ coordinate and multiple by 10 as the edge length in the data is $1$ .