This is a very interesting question, and I think I understand what ipmcc has in mind. There's a lot of Mathematica code involved in generating these images, so I'll put that at the end.
The first thing to understand is "boards are halfway between being coplanar with the 2 faces on either side". If I understand things correctly, the placement of a board with respect to the dodecahedron should look something like this: 
With all of the boards, we should have one new face for every old edge. "What polyhedron has a face for every edge of an original one?" has been asked and answered in Is there such a thing as the "edge-face dual" of a polyhedron, and is the "edge-face dual" of a cube a rhombic dodecahedron?: It's the dual of the rectified polyhedron. A rectified dodecahedron is an icosidodecahedron, and the dual of that is a rhombic triacontahedron (with 30 golden rhombus faces). If we draw a triacontahedron on the outside of our dodecahedron, we get something like the following: 
The problem is that that doesn't really look anything like "conceptually close" to the link @TonyK posted in a comment. We don't want the maximum possible piece of each board (a golden rhombus), we just want a sliver of each rhombus around the edges of the original dodecahedron. I think the intent was for something that looks like this (possibly with less wide slivers): 
However, the hexagons in that image are infinitely thin, whereas the question referred to "woodworking" and "beveling"; we need to thicken them up. We can't just thicken them naively because of the dihedral angles. They need to be frusta: beveled so they're part of a hexagonal pyramid with "top vertex" at the center of the figure. It should look something like this: 
By symmetry, you only need to build 30 copies of the same frustum. Fixing a length for the long diagonal, a frustum depends on two parameters. Let t be a parameter ranging from 0 to 1 where 0 means you have a whole golden rhombus face and 1 means you have just a line segment that is an edge of the dodecahedron (interpolating linearly). Let s be a parameter ranging from 0 to 1 where 0 means you have the whole pyramid and 1 means you have just the hexagon (interpolating linearly). Then the image with the flat hexagons had t=0.85 and s=1, and the image with the solid frusta had t=s=0.9. To see the effects of changing t and s qualitatively, here is a Youtube animation of the whole dodecahedron as t and s are changed, and here is a Youtube animation of just one frustum.
In sloppy Mathematica notation, where phi
is the golden ratio $\varphi=\frac{1+\sqrt{5}}{2}$, and s
and t
are the parameters just described, the following is a list of faces of a single frustum (of long face diagonal length $\frac{2}{\varphi}$), where each face is a list of vertices (in a right-hand-rule order around the face), and each vertex is is a triple of x, y, z coordinates:
{ {{t - phi t, phi, 1 - t}, {(-1 + phi) t, phi, 1 - t}, {-1 + phi, phi, 0}, {(-1 + phi) t, phi, -1 + t}, {t - phi t, phi, -1 + t}, {1 - phi, phi, 0}}, {{s - phi s, phi s, 0}, {-(-1 + phi) s t, phi s, s (-1 + t)}, {(-1 + phi) s t, phi s, s (-1 + t)}, {(-1 + phi) s, phi s, 0}, {(-1 + phi) s t, phi s, s - s t}, {-(-1 + phi) s t, phi s, s - s t}}, {{s - phi s, phi s, 0}, {1 - phi, phi, 0}, {t - phi t, phi, -1 + t}, {-(-1 + phi) s t, phi s, s (-1 + t)}}, {{-(-1 + phi) s t, phi s, s (-1 + t)}, {t - phi t, phi, -1 + t}, {(-1 + phi) t, phi, -1 + t}, {(-1 + phi) s t, phi s, s (-1 + t)}}, {{(-1 + phi) s t, phi s, s (-1 + t)}, {(-1 + phi) t, phi, -1 + t}, {-1 + phi, phi, 0}, {(-1 + phi) s, phi s, 0}}, {{(-1 + phi) s, phi s, 0}, {-1 + phi, phi, 0}, {(-1 + phi) t, phi, 1 - t}, {(-1 + phi) s t, phi s, s - s t}}, {{(-1 + phi) s t, phi s, s - s t}, {(-1 + phi) t, phi, 1 - t}, {t - phi t, phi, 1 - t}, {-(-1 + phi) s t, phi s, s - s t}}, {{-(-1 + phi) s t, phi s, s - s t}, {t - phi t, phi, 1 - t}, {1 - phi, phi, 0}, {s - phi s, phi s, 0}} }
The angles of the outer hexagon are $\tan^{-1}2\approx 63.43^{\circ}$ and $720^{\circ}-2\tan^{-1}2\approx 148.28^{\circ}$. The dihedral angle between the larger hexagon and one of the four smaller (assuming t near 1) trapezoidal faces is exactly $72^{\circ}$. However, the dihedral angle between the larger hexagon and one of the two bigger (assuming t near 1) trapezoidal faces depends on t. It's $\approx86.07^{\circ}$ for t=0.9, and in general is given by $\cos^{-1}\left(\frac{\left(3-\sqrt{5}\right)\left(1+\sqrt{5}\right)\left(1-t\right)}{4*\sqrt{t^2+\frac{1}{2}\left(3-\sqrt{5}\right)\left(1-t\right)^{2}}}\right)$
If you happen to have Mathematica 7 or above (the code might also work in 6), here is most of the code (with some scaffolding removed and some comments added) for generating the images and formulae referenced above. I make no promises about elegance or efficiency.
(*The Golden Ratio Phi is important*) p = (1 + Sqrt[5])/2; (*Coordinates and grouping from the Wikipedia article for \ dodecahedron.*) groupedvs = {{{1, 1, -1}, {1, 1, 1}, {1, -1, 1}, {1, -1, -1}, {-1, 1, -1}, {-1, 1, 1}, {-1, -1, 1}, {-1, -1, -1}}, {{0, 1/p, p}, {0, -1/p, p}, {0, 1/p, -p}, {0, -1/p, -p}}, {{1/p, p, 0}, {1/p, -p, 0}, {-1/p, p, 0}, {-1/p, -p, 0}}, {{p, 0, 1/p}, {p, 0, -1/p}, {-p, 0, 1/p}, {-p, 0, -1/p}}}; (*Normal vectors to two adjacent faces.*) n2 = -Cross[{1/p, p, 0} - {1, 1, 1}, {0, 1/p, p} - {1, 1, 1}]; n3 = Cross[{-1/p, p, 0} - {1/p, p, 0}, {1, 1, -1} - {1/p, p, 0}]; (*A flat list of vertices of the dodecahedron which should be easier to work with.*) vs = Flatten[groupedvs, 1]; (*A list of lists of vertex indices to represent the 12 faces of the \ dodecahedron.*) faces = {{12, 4, 14, 16, 8}, {18, 4, 12, 11, 1}, {12, 8, 20, 5, 11}, {8, 16, 7, 19, 20}, {20, 19, 6, 15, 5}, {19, 7, 10, 9, 6}, {15, 6, 9, 2, 13}, {5, 15, 13, 1, 11}, {13, 2, 17, 18, 1}, {2, 9, 10, 3, 17}, {17, 3, 14, 4, 18}, {3, 10, 7, 16, 14}}; (*Show the "board" and the vertices of the dodecahedron and the faces \ of the dodecahedron.*)Show[(*Plot the vertices.*) ListPointPlot3D[groupedvs,(*Color the vertices like on Wikipedia.*) PlotStyle -> {Directive[Orange, PointSize[0.05]], Directive[Green, PointSize[0.05]], Directive[Blue, PointSize[0.05]], Directive[Red, PointSize[0.05]]}, AxesLabel -> {x, y, z}],(*Plot the board using the fact that the sum of the two \ normals for the adjacent faces should give a normal right in between.*) ContourPlot3D[ Dot[n3 + n2, {x - 1/p, y - p, z}] == 0, {x, -1, 1}, {y, 0, 2 p}, {z, -p, p}, ContourStyle -> Opacity[0.5], Mesh -> False], Graphics3D[(*A sequence of polygons*) Polygon[(*where for every element of "faces"*) MapThread[(*we make a list of all five corresponding vertices.*) Table[vs[[(#1[[j]])]], {j, 1, 5}] &, {faces}]]](*Finally, make sure Mathematica doesn't \ squish the image.*), BoxRatios -> Automatic] (*For later images, we'll need indices for the vertices that make up \ each edge of the dodecahedron.*) edges = {{5, 11}, {11, 1}, {1, 13}, {13, 15}, {15, 5}, {1, 18}, {18, 4}, {4, 12}, {12, 11}, {13, 2}, {2, 17}, {17, 18}, {15, 6}, {6, 9}, {9, 2}, {9, 10}, {10, 3}, {3, 17}, {6, 19}, {19, 7}, {7, 10}, {19, 20}, {20, 8}, {8, 16}, {16, 7}, {20, 5}, {8, 12}, {14, 16}, {14, 4}, {14, 3}}; (*We also could use a raw list of faces for the rhombic \ tricontahedron where each face is a list of vertices in some cyclic \ order.*)purerhombic = {{{0, 1/2 (1 + Sqrt[5]), -1}, {0, 1/2 (-1 + Sqrt[5]), 1/2 (-1 - Sqrt[5])}, {-1, 0, 1/2 (-1 - Sqrt[5])}, {-1, 1, -1}}, {{1, 1, -1}, {0, 1/2 (1 + Sqrt[5]), -1}, {0, 1/2 (-1 + Sqrt[5]), 1/2 (-1 - Sqrt[5])}, {1, 0, 1/2 (-1 - Sqrt[5])}}, {{1, 1, -1}, {0, 1/2 (1 + Sqrt[5]), -1}, {1/2 (-1 + Sqrt[5]), 1/2 (1 + Sqrt[5]), 0}, {1/2 (1 + Sqrt[5]), 1, 0}}, {{1/2 (1 - Sqrt[5]), 1/2 (1 + Sqrt[5]), 0}, {0, 1/2 (1 + Sqrt[5]), -1}, {1/2 (-1 + Sqrt[5]), 1/2 (1 + Sqrt[5]), 0}, {0, 1/2 (1 + Sqrt[5]), 1}}, {{1/2 (1 - Sqrt[5]), 1/2 (1 + Sqrt[5]), 0}, {0, 1/2 (1 + Sqrt[5]), -1}, {-1, 1, -1}, {1/2 (-1 - Sqrt[5]), 1, 0}}, {{1, 1, -1}, {1/2 (1 + Sqrt[5]), 1, 0}, {1/2 (1 + Sqrt[5]), 0, 1/2 (1 - Sqrt[5])}, {1, 0, 1/2 (-1 - Sqrt[5])}}, {{1/2 (1 + Sqrt[5]), 0, 1/2 (1 - Sqrt[5])}, {1/2 (1 + Sqrt[5]), -1, 0}, {1, -1, -1}, {1, 0, 1/2 (-1 - Sqrt[5])}}, {{1, -1, -1}, {0, 1/2 (-1 - Sqrt[5]), -1}, {0, 1/2 (1 - Sqrt[5]), 1/2 (-1 - Sqrt[5])}, {1, 0, 1/2 (-1 - Sqrt[5])}}, {{1, 0, 1/2 (-1 - Sqrt[5])}, {0, 1/2 (-1 + Sqrt[5]), 1/2 (-1 - Sqrt[5])}, {-1, 0, 1/2 (-1 - Sqrt[5])}, {0, 1/2 (1 - Sqrt[5]), 1/2 (-1 - Sqrt[5])}}, {{1/2 (-1 + Sqrt[5]), 1/2 (1 + Sqrt[5]), 0}, {1/2 (1 + Sqrt[5]), 1, 0}, {1, 1, 1}, {0, 1/2 (1 + Sqrt[5]), 1}}, {{1/2 (1 + Sqrt[5]), 1, 0}, {1, 1, 1}, {1, 0, 1/2 (1 + Sqrt[5])}, {1/2 (1 + Sqrt[5]), 0, 1/2 (-1 + Sqrt[5])}}, {{1/2 (1 + Sqrt[5]), 0, 1/2 (1 - Sqrt[5])}, {1/2 (1 + Sqrt[5]), 1, 0}, {1/2 (1 + Sqrt[5]), 0, 1/2 (-1 + Sqrt[5])}, {1/2 (1 + Sqrt[5]), -1, 0}}, {{1/2 (1 - Sqrt[5]), 1/2 (1 + Sqrt[5]), 0}, {0, 1/2 (1 + Sqrt[5]), 1}, {-1, 1, 1}, {1/2 (-1 - Sqrt[5]), 1, 0}}, {{0, 1/2 (1 + Sqrt[5]), 1}, {0, 1/2 (-1 + Sqrt[5]), 1/2 (1 + Sqrt[5])}, {-1, 0, 1/2 (1 + Sqrt[5])}, {-1, 1, 1}}, {{1, 1, 1}, {0, 1/2 (1 + Sqrt[5]), 1}, {0, 1/2 (-1 + Sqrt[5]), 1/2 (1 + Sqrt[5])}, {1, 0, 1/2 (1 + Sqrt[5])}}, {{0, 1/2 (1 - Sqrt[5]), 1/2 (1 + Sqrt[5])}, {-1, 0, 1/2 (1 + Sqrt[5])}, {0, 1/2 (-1 + Sqrt[5]), 1/2 (1 + Sqrt[5])}, {1, 0, 1/2 (1 + Sqrt[5])}}, {{0, 1/2 (1 - Sqrt[5]), 1/2 (1 + Sqrt[5])}, {1, 0, 1/2 (1 + Sqrt[5])}, {1, -1, 1}, {0, 1/2 (-1 - Sqrt[5]), 1}}, {{1/2 (1 + Sqrt[5]), 0, 1/2 (-1 + Sqrt[5])}, {1/2 (1 + Sqrt[5]), -1, 0}, {1, -1, 1}, {1, 0, 1/2 (1 + Sqrt[5])}}, {{-1, 0, 1/2 (1 + Sqrt[5])}, {-1, 1, 1}, {1/2 (-1 - Sqrt[5]), 1, 0}, {1/2 (-1 - Sqrt[5]), 0, 1/2 (-1 + Sqrt[5])}}, {{-1, -1, 1}, {-1, 0, 1/2 (1 + Sqrt[5])}, {1/2 (-1 - Sqrt[5]), 0, 1/2 (-1 + Sqrt[5])}, {1/2 (-1 - Sqrt[5]), -1, 0}}, {{0, 1/2 (1 - Sqrt[5]), 1/2 (1 + Sqrt[5])}, {-1, 0, 1/2 (1 + Sqrt[5])}, {-1, -1, 1}, {0, 1/2 (-1 - Sqrt[5]), 1}}, {{1/2 (-1 - Sqrt[5]), -1, 0}, {1/2 (-1 - Sqrt[5]), 0, 1/2 (1 - Sqrt[5])}, {1/2 (-1 - Sqrt[5]), 1, 0}, {1/2 (-1 - Sqrt[5]), 0, 1/2 (-1 + Sqrt[5])}}, {{1/2 (-1 - Sqrt[5]), -1, 0}, {1/2 (-1 - Sqrt[5]), 0, 1/2 (1 - Sqrt[5])}, {-1, 0, 1/2 (-1 - Sqrt[5])}, {-1, -1, -1}}, {{0, 1/2 (-1 - Sqrt[5]), -1}, {1/2 (1 - Sqrt[5]), 1/2 (-1 - Sqrt[5]), 0}, {1/2 (-1 - Sqrt[5]), -1, 0}, {-1, -1, -1}}, {{1/2 (1 - Sqrt[5]), 1/2 (-1 - Sqrt[5]), 0}, {1/2 (-1 - Sqrt[5]), -1, 0}, {-1, -1, 1}, {0, 1/2 (-1 - Sqrt[5]), 1}}, {{-1, 1, -1}, {-1, 0, 1/2 (-1 - Sqrt[5])}, {1/2 (-1 - Sqrt[5]), 0, 1/2 (1 - Sqrt[5])}, {1/2 (-1 - Sqrt[5]), 1, 0}}, {{0, 1/2 (-1 - Sqrt[5]), -1}, {0, 1/2 (1 - Sqrt[5]), 1/2 (-1 - Sqrt[5])}, {-1, 0, 1/2 (-1 - Sqrt[5])}, {-1, -1, -1}}, {{0, 1/2 (-1 - Sqrt[5]), -1}, {1/2 (1 - Sqrt[5]), 1/2 (-1 - Sqrt[5]), 0}, {0, 1/2 (-1 - Sqrt[5]), 1}, {1/2 (-1 + Sqrt[5]), 1/2 (-1 - Sqrt[5]), 0}}, {{1, -1, -1}, {1/2 (1 + Sqrt[5]), -1, 0}, {1/2 (-1 + Sqrt[5]), 1/2 (-1 - Sqrt[5]), 0}, {0, 1/2 (-1 - Sqrt[5]), -1}}, {{1/2 (1 + Sqrt[5]), -1, 0}, {1, -1, 1}, {0, 1/2 (-1 - Sqrt[5]), 1}, {1/2 (-1 + Sqrt[5]), 1/2 (-1 - Sqrt[5]), 0}}}; (*If we want to plot this, we need to tell Mathematica these faces \ are polygons command throughout.*) rhombic = MapThread[Polygon[#1] &, {purerhombic}]; Show[ListPointPlot3D[groupedvs, PlotStyle -> {Directive[Orange, PointSize[0.05]], Directive[Green, PointSize[0.05]], Directive[Cyan, PointSize[0.05]], Directive[Pink, PointSize[0.05]]}, AxesLabel -> {x, y, z}], ContourPlot3D[ Dot[n3 + n2, {x - 1/p, y - p, z}] == 0, {x, -1, 1}, {y, 0, 2 p}, {z, -p, p}, ContourStyle -> Opacity[0.5], Mesh -> False], Graphics3D[{Purple, Thickness[ 0.015],(*This time we have lines consisting of pairs of vertex \ indices, not polygons consisting of quintuples of indices.*) Line[MapThread[Table[vs[[(#1[[j]])]], {j, 1, 2}] &, {edges}]]}], Graphics3D[rhombic], BoxRatios -> Automatic] (*To make parametrized hexagons, we need to make sure the lists of \ four vertices for each rhombus begin consistently with a large angle \ vertex. The length of a short diagonal is 2/phi which is about 1.2, \ and the length of a long diagonal is 2. Therefore, the following will \ reorder things appropriately.*) rhombicorderedfaces = MapThread[ If[Norm[N[#1[[1]] - #1[[3]]]] > 1.5, {#1[[2]], #1[[3]], #1[[4]], #1[[1]]}, #1] &, \ {purerhombic}]; (*With a bit of geometry, we can see how to build parametrized \ hexagons now.*) parametrizedrhombic[t_] := MapThread[{#1[[1]], t*#1[[1]] + (1 - t)*#1[[2]], t*#1[[3]] + (1 - t)*#1[[2]], #1[[3]], t*#1[[3]] + (1 - t)*#1[[4]], t*#1[[1]] + (1 - t)*#1[[4]]} &, {rhombicorderedfaces}] (*To have something to plot, we can use the polygon command again.*) pararhombic[t_] := MapThread[Polygon[#1] &, {parametrizedrhombic[t]}]; (*t = .85 seems like a fine start.*) Show[ Graphics3D[pararhombic[85/100], Axes -> True], BoxRatios -> Automatic] (*Next we make solid frusta.*) listofsolidhexes[t_, s_] := MapThread[(*We need to flatten because I don't want to type out all \ six faces around.*) Flatten[{(*The extra braces on the top and bottom faces ensure that \ flatten yields the right result.*){#1}, {s*#1}, Table[{s*#1[[i]], #1[[i]], #1[[6 - Mod[-(i + 1), 6]]], s*#1[[6 - Mod[-(i + 1), 6]]]}, {i, 1, 6}]}, 1] &, {parametrizedrhombic[t]}] (*For the purposes of plotting, it doesn't matter which polygonal \ face belongs to which frustum.*) listofpolysforsolidhexes[t_, s_] := Flatten[listofsolidhexes[t, s], 1] (*For the purposes of plotting, we need polygon commands.*) polylistofpolys[t_, s_] := MapThread[Polygon[#1] &, {listofpolysforsolidhexes[t, s]}]; (*To really see what's going on, we use Manipulate to let us modify t \ and s easily. For t=0.5 and s=1, we essentially have a pentatruncated \ rhombic triacontahedron, which is similar to, but distinct from, a \ European football shape.*) Manipulate[ Graphics3D[polylistofpolys[t, s], Boxed -> False], {{t, .9}, 0, 1}, {{s, .9}, 0, 1}] (*If we only need to look at one piece:*) Manipulate[ Show[Graphics3D[ MapThread[ Polygon[#1] &, {listofsolidhexes[t, s][[(*4 is the index that gives the frustum attached to the \ board in the early images.*)4]]}]], BoxRatios -> Automatic], {{t, .9}, 0, 1}, {{s, .95}, 0, 1}] (*The raw data for that single frustum:*) listofsolidhexes[t, s][[4]]