Here's an exercise to get you started.
Start with a cube. Now build a "roof" on the top face by placing a ridge line parallel to one set of edges, but shorter than those edges, directly above the cube, and making some connections. If your cube vertices are $(\pm1, \pm 1, \pm 1)$, then the endpoints of your "ridge line" for the upper $xy$-face (i.e., the $z = 1$ face) would be something like
$$
(-a, 0, b) \\
(a, 0, b)
$$
where $b$ is a little larger than $1$, and $a$ is a bit smaller,
with "new" edges from $(-1, -1, 1)$ and $(-1, 1, 1)$ to the first of these, and from $(1, -1, 1)$ and $(1, 1, 1)$ to the second. For now, pick $a$ to be about $0.6$, just do get yourself started drawing a picture. Also treat the ridge line as a "new" edge. So you have five new edges for this top face.
Erect a parallel "ridge line" beneath the $z = -1$ face, and on the four vertical sides, erect four more ridges. Do it so that the ridges on any two adjacent faces of the cube are perpendicular rather than parallel.
Now forget the edges of the original cube. If you pick $a$ just right, it'll turn out that the faces formed by all the new edges are planar pentagons, and you've got a dodecahedron.
So now you know the vertices for a dodecahedron whose radius is the distance from the origin to any vertex of the original cube, which is $\sqrt{3}$. From this information, you should be able to work out what you need.
Hint for finding $a$ and $b$: the vertices at the ends of the ridge need to be the same distance from the origin as are the vertices at the corners of the cube, i.e., \sqrt{3}. And the distance of the edge along the ridge line is $2a$, and must be the same as the distance from the ridge-end $(a, 0, b)$ to the vertex $(1,1,1)$. Those two constraints will tell you $a$ and $b$.
(I see that I've duplicated the idea of the reference that @Blue points to: Cleverest construction of a dodecahedron / icosahedron?; he and I share a fondness for this construction, apparently.)
