How do I find or derive circumspherical radii, surface area, and volume for these 5 non-traditional concave polyhedra? How do I find or derive circumspherical radii, surface area, and volume for these 5 non-traditional concave polyhedra?
As much of a mathematics enthusiast I am, I'm stuck on a huge roadblock in further developing my code for the paper that shows the value in different geometries at the nanoscale for nanomedicine: There are five shapes that I am having trouble deriving the circumspherical radii, and thus, the equations of surface area and volume, to solve my equations:

*

*Medial Rhombic Triacontahedron

*Dodecadodecahedron Medial Triambic

*Icosahedron Ditrigonal

*Dodecadodecahedron Excavated

*Dodecahedron

I've scoured the internet to find some approximations but the only one I could find was based off of one of those solids so far and it's not a modifiable equation as I cannot denote where the side length comes into play (they are just constants based off of a unit edge length): http://dmccooey.com/polyhedra/MedialRhombicTriacontahedron.html
I was wondering if anyone knew how to derivate the shape's equations or where I can find the correct resource? I'd derivate it myself but it's a bit above my scope of knowledge. I need a physical solid to cut and manipulate to try to find basic ratios at the very least.
 A: You can start from the “known” polyhedra. For example, you know that medial rhombic triacontahedron (MRT) is a stellation of a rhombic triacontahedron (RT).
There are 3 types of vertices in MRT. By the symmetry, you can see that the outer vertice $A$ is a scaled vertice of a RT by some coefficient $\alpha$. Medial vertice $B$ is a scaled face center by $\beta$ and the inner vertice is actually a vertice of RT.
By claiming that $A$ and $B$ belong to the plane of the red face, you can find $\alpha=(\sqrt5+1)/2$ (golden ratio) and $\beta=\sqrt5-1$.

After you have found the coordinates of all vertices, it's not hard to find the metrics you are interested in.
As for other polygons, you can do similar things:

*

*Dodecadodecahedron Medial Triambic — not sure what it is

*Ditrigonal icosahedron is as far as I understand is the same as excavated dodecahedron — you start with dodecahedron and downscale face centers until they form regular triangle

*Dodecadodecahedron Excavated — not sure what it is

*Dodecahedron — if it's a regular one, I don't really understand what is the problem with finding anything about one of the platonic solids
A: If the polyhedra are such that any line through "the center" of the polyhedron intersects the surface exactly twice, each face is a simple planar polygon, and you have the 3D coordinates for each face polygon, you can calculate the surface area as the sum area of the face polygons, and the volume as the sum of the volumes of the pyramids with their faces at their bases.
That center does not have to be any specific center (like center of mass, or centroid); it suffices that such a point exists, and it is at origin (wrt. the vertex coordinates), for the following treatment.
Let's assume the polyhedron is centered at origin. For each planar face described by its polygon vertices $\vec{v}_i = (x_i, y_i, z_i)$ counterclockwise or clockwise around its face unit normal vector $\hat{n}$, $i = 1 \dots n$, and point $\vec{o}$ on the face plane, inside the face polygon,
$$\begin{aligned}
\vec{n} &= \left( \vec{v}_{i-1} - \vec{o} \right) \times \left( \vec{v}_i - \vec{o} \right) \\
\hat{n} &= \frac{\vec{n}}{\left\lVert\vec{n}\right\rVert} \\
\end{aligned}$$
where $\vec{v}_0 = \vec{v}_n$.  (Although $\lVert\vec{n}\rVert$ may vary for $i = 1 \dots n$, $\hat{n}$ will always be the same, regardless of which pair of vertices (non-collinear with $\vec{o}$) are chosen; otherwise, the face is not planar.)
The surface area of this face is then calculable via the Shoelace formula, e.g.
$$A = \frac{1}{2} \left\lVert \sum_{i=1}^{n} (\vec{v}_{i-1} - \vec{o}) \times (\vec{v}_i - \vec{o}) \right\rVert$$
and the volume of the pyramid corresponding to this face is
$$V = \frac{A}{3}\left\lvert \vec{o} \cdot \hat{n} \right\rvert$$
The dot product is positive if the normal points outwards, negative if the normal points inwards; its magnitude is the height of the pyramid, measured perpendicular to the base.
Due to symmetries, you can typically calculate the total volume and surface area in exact terms (given exact vertex coordinates), since these polyhedra have many symmetries (similar faces); I suggest using a Computer Algebra System like Maple, Maxima (wxMaxima), or Sagemath.
As far as I know, the mentioned polyhedra are sufficiently symmetric that the circumradius is equal to the maximum distance between a vertex and the center of the polyhedron; with each one of these having at least four vertices at such distances.  For dodecahedron and all other platonic solids, all vertices are at the same distance from the center, as all their vertices pass through the circumsphere.

In purely geometric terms, the above attempts to explain how to compute the surface area of these polyhedra by computing the area of each unique face – these being highly symmetric, there are rarely more than two or three different faces – and the volume as the sum of volumes of the pyramids corresponding to those faces, with the pyramid apex at the center of symmetry.  Compare to fruits and berries that are composed of clusters of individual wedges.
Let's consider medial rhombic triacontahedron (Wikipedia), DU36  (dual uniform polyhedron 36), as an example.  (I recommend you find the exact index of the polyhedra, since I could not determine for sure exactly which polyhedra three of the five listed refer to.)
It consists of 12 "spikes", each with five equal faces, so essentially has 60 identical face polygons.  Each face polygon is a quadrilateral, with two short and two long sides, each pair consecutive (and can be decomposed to two unequal triangles).
If we call the surface area of each face polygon $A_f$, the surface area of the entire medial rhombic triacontahedron is $A = 60 A_f$.
If we call the distance from each face plane from the center of the polyhedron $h$ – due to the symmetries this is the same for all of them –, the volume of the entire medial rhombic triacontahedron is $V = 60 \cdot \frac{A_f}{3} h = 20 A_f h$.
Note how the height $h$ of the face pyramid is measured in the perpendicular direction from the face.  This height is the same as the distance from the face plane – the plane extending the face to infinity – to the center of symmetry of the entire polyhedron.
The circumspherical radius $R$ is the same as the distances from the tips of the "spikes" to the center of symmetry of the polyhedron.
