Description: Regular Dodecahedron A regular dodecahedron or pentagonal dodecahedron is a dodecahedron that is regular, which is composed of twelve regular pentagonal faces, three meeting at each vertex. It is one of the five Platonic solids. It has 12 faces, 20 vertices, 30 edges, and 160 diagonals. It is represented by the Schläfli symbol {5,3}. (WikipediaFaces by sides): 12Dihedral angle: 116.56505° = arccos(−1/√5) Face configuration: V3.3.3.3.35.5.5 (Vertex figure): Regular icosahedron; (dual polyhedron) Rotation group: I,+, (532) Properties: regular, convex Symmetry: Ih, H3,, (*532) Mathematical equations of a Dodecahedron: (http://mathworld.wolfram.com/RegularDodecahedron.html) The polyhedron vertices of a dodecahedron can be given in a simple form for a dodecahedron of side length a=sqrt(5)-1 by (0, +/-phi^(-1), +/-phi), (+/-phi, 0, +/-phi^(-1)), (+/-phi^(-1), +/-phi, 0), and (+/-1, +/-1, +/-1). For a dodecahedron of unit edge length a=1, the circumradius R^’ and inradius r^’ of a pentagonal face are R^’ = 1/(10)sqrt(50+10sqrt(5)) (3) r^’ = 1/(10)sqrt(25+10sqrt(5)). (4) The sagitta x is then given by x=R^’-r^’=1/(10)sqrt(125-10sqrt(5)). (5) Now consider the following figure. DodecahedronTrig Using the Pythagorean theorem on the figure then gives z_1^2+m^2 = (R^’+r^’)^2 (6) z_2^2+(m-x)^2 = 1 (7) ((z_1+z_2)/2)^2+R^(‘2) = ((z_1-z_2)/2)^2+(m+r^’)^2. (8) Equation (8) can be written z_1z_2+r^2=(m+r^’)^2. (9) Solving (6), (7), and (9) simultaneously gives m = r^’=1/(10)sqrt(25+10sqrt(5)) (10) z_1 = 2r^’=1/5sqrt(25+10sqrt(5)) (11) z_2 = R^’=1/(10)sqrt(50+10sqrt(5)). (12) The inradius of the dodecahedron is then given by r=1/2(z_1+z_2), (13) so r^2=1/(40)(25+11sqrt(5)), (14) and solving for r gives r=1/(20)sqrt(250+110sqrt(5))=1.11351…. (15) Now, R^2=R^(‘2)+r^2=3/8(3+sqrt(5)), (16) so the circumradius is R=1/4(sqrt(15)+sqrt(3))=1.40125…. (17) The midradius is given by rho^2=r^(‘2)+r^2=1/8(7+3sqrt(5)), (18) so rho=1/4(3+sqrt(5))=1.30901…. (19) The dihedral angle is alpha=cos^(-1)(-1/5sqrt(5)) approx 116.57 degrees. (20) The area of a single face is the area of a pentagon of unit edge length A=1/4sqrt(25+10sqrt(5)), (21) so the surface area is 12 times this value, namely S=3sqrt(25+10sqrt(5)). (22) The volume of the dodecahedron can be computed by summing the volume of the 12 constituent pentagonal pyramids, V=12(1/3Ar)=1/4(15+7sqrt(5)). (23) Apollonius showed that for an icosahedron and a dodecahedron with the same inradius, (V_(icosahedron))/(V_(dodecahedron))=(A_(icosahedron))/(A_(dodecahedron)), (24) where V is the volume and A the surface area, with the actual ratio being (V_(icosahedron))/(V_(dodecahedron))=(A_(icosahedron))/(A_(dodecahedron))=sqrt(3/(10)(5-sqrt(5))). .