Analysis of the 5 Regular Polyhedra
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In analyzing polyhedra I will do so without the use of trigonometry, unless calculating central or surface angles.
I will use the pyramid method for calculating volumes.
The volume of any n-sided pyramid = 1/3 * area of base * pyramid height.

All of the 5 regular solids have equal sides and equal angles.
Tetrahedron
Octahedron
Cube
Icosahedron
Dodecahedron

Now that we have learned something about these solids, lets summarize our knowledge:

 Centroid to: Centriod to: Centroid to: V(s = 1)    s³ V(r = 1)    r³ Surface Area (s=1)   s² Surface Area (r=1)  r ² Central < Dihedral < Surface < Vertex Mid-edge Mid-face side /radius Tetrahedron 0.11785113 0.513200239 1.73205 4.618802154 109.4712206° 70.52877936° 60° 0.612372436 0.3535534 0.204124 1.632993162 Octahedron 0.47140452 1.333.... 3.46410 6.92820323 90° 109.4712206° 60° 0.707106781 0.5 0.408248 1.414213562 Cube 1.0 1.539600716 6.0 8.0 70.52877936° 90° 90° 0.866025404 0.707106781 0.5 1.154700538 Icosahedron 2.181694991 2.53615071 8.660254 9.57454138 63.4349488° 138.1896852° 60° 0.951056517 0.809016995 0.755761314 1.05146222 Dodecahedron 7.663118963 2.785163863 20.64573 10.51462224 41.81031488° 116.5650512° 108° 1.401258539 1.309016995 1.113616365 0.713644179

Here is the same table without all of the repeating decimals:

Note: Ø = 1.618033989, or (\/¯5 + 1) / 2, the Golden Section.

 Centroid to: Centroid to: Centroid to: V(s = 1) s³ V(r = 1)    r³ Surface Area (s=1)   s² Surface Area (r=1)  r ² Central < Dihedral < Surface < Vertex Mid - Edge Mid - Face side / radius Tetrahedron 1 / 6\/¯2 8 / 9\/¯3 \/¯3 8 / \/¯3 109.4712206° 70.52877936° 60° \/¯3 / (2\/¯2) 1 / (2\/¯2) 1 / (2\/¯6) 2\/¯2/  \/¯3 Octahedron 2 / 3\/¯2 4 / 3 2 \/¯3 4 \/¯3 90° 109.4712206° 60° 1 /  \/¯2 1/2 1 /  \/¯6 \/¯2 Cube 1.0 8 / 3\/¯3 6.0 8.0 70.52877936° 90° 90° \/¯3 / 2 1  / \/¯2 1 / 2 2 / \/¯3 Icosahedron 5Ø² / 6 20Ø² / 3(Ø²+1)^3/2 5 \/¯3 20\/¯3 /  (Ø²+1) 63.4349488° 138.1896852° 60° \/¯(Ø²+1)/  2 Ø / 2 Ø² /    (2 \/¯3) 2 / \/¯(Ø²+1) Dodecahedron 5Ø^5 / 2(Ø²+1) 20Ø² / (3\/¯3)(Ø²+1) 15Ø² / \/¯(Ø²+1) 20 / \/¯(Ø²+1) 41.81031488° 116.5650512° 108° (\/¯3)Ø /  2 Ø² / 2 Ø³ /  2\/¯(Ø²+1) 2 / (\/¯3)Ø

From this chart we can see a number of things:
1)  I have included  2 different ways to measure these solids, both of them helpful: the first keeping length of each side equal, and the second considering each solid inscribed in a sphere of equal radius. The side/radius column describes  the relationship between the Volumes and the Surface Areas for each solid.  For Volume, use (s/r)³ when converting between V(s) and V(r).
For Surface Area, use (s/r)².  For s = k or r = k, the Volume will always increase k times  as fast as the Surface Area.
2) The tetrahedron has the least volume with the most surface area, the dodecahedron has the most volume with the least surface area. The ratio of the volume to surface area goes up as we go down the chart.
3) The central angles decrease as we go down the chart. This makes sense as there are more vertices, so less space between the vertices.
4) The ratio of side to radius also decreases as the volume increases. This also makes sense as in order to get more volume, the radius has to increase. As the radius increases, in the solids with more vertices there is less surface area on the sphere over which to spread out.
5) the central < of the tetra = dihedral < of octa, the central < of the octa = the dihedral < of the cube. So the octa is created from the tetra by using the same equilateral triangles as faces, and applying the tetra central angle to it. And so for the cube. This is consonant with the geometry of these solids, which is  \/¯2 and  \/¯3 geometry.
This pattern breaks down when we get to the icosahedron and the dodecahedron, which are  \/¯5 geometry.

Special characters:
\/¯ ­ ° ¹ ² ³ × ½ ¼ Ø  \/¯(Ø² + 1)