<s>[INST] Dodecahedron volume How can I derive the volume of a regular dodecahedron? A. P. Kiselev's Stereometry suggests that we cut it into a cube and congruent "tent-like" solids. I'm finding some difficulty to compute the volume of these "tent-like" solids.
 [/INST] First, note that if $s$ is the side length of the dodecahedron, then the side length of the cube is $$c = \frac{s}{2} \csc \frac{\pi}{10} = s \left(\frac{1 + \sqrt{5}}{2}\right).$$
Now consider looking at the tent solid from above, it looks something like this:

where the dotted line $x$ denotes the slant height of the triangular face. Since $x, c/2, s$ form a right triangle, we can compute $x$ using the Pythagorean Theorem to be $$x = \sqrt{s^2 - \left(\frac{c}{2}\right)^2} = s\sqrt{1 - \left(\frac{1 + \sqrt{5}}{4}\right)^2} = s\sqrt{\frac{5 - \sqrt{5}}{8}}.$$
It is also easy to see that $$y = \frac{c - s}{2} = \frac{1}{2}s \left(\frac{1 + \sqrt{5}}{2} - 1 \right) = s \left(\frac{\sqrt{5} - 1}{4} \right)$$
Letting $h$ be the vertical height of the tent, note $y, h, x$ form a right triangle, so we have $$h = \sqrt{x^2 - y^2} = s\sqrt{\frac{5 - \sqrt{5}}{8} - \left(\frac{\sqrt{5} - 1}{4} \right)^2} = \frac{s}{2}.$$
We can now compute the volume as follows: cut the volume into three pieces along the dashed lines. The left and right pieces can be joined to form a pyramid whose base is a $2y$ by $c$ rectangle and whose height is $h$, and the piece in the middle is a prism whose base is a triangle with base $c$ and height $h$ and whose height is $s$.
Can you take it from here?
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