How to Compute Volume of a Regular

12-faces Polyhedron (dodecahedron),
Given Lengths of its Edges.

Some description of a regular 12-faces polyhedron gives a page of ‘Literka’ How to build... (See also Plato's polyhedrons). Let us mention here that there is a similar page of literka, where the volume of a polyhedron built of 30 congruent rhombuses is computed.

Let us find two edges of a regular dodecahedron, which are on opposite sides of this polyhedron and are parallel. Assume that these edges have the lengths L. Cross section by a plane passing through these edges looks like this.

Let us mention here that this polygon is used for a construction of 42-faces polyhedron, which has faces congruent to this polygon (see Example of a polyhedron as a cross section of a polytope).

The sides BC and FE of this polygon – these are edges of regular dodecahedron and have the lengths L. The sides AB, CD, DE, FA have the same lengths and have the same position on faces of a polyhedron. For example side AB looks like this on a face (which is a regular pentagon):

Denote by d the length of AB, CD, DE, FA. Little computation shows that

d=b*L,

where

b=sin 36 + cos 18.

(Here sin 36 means sine of an angle of 36 degrees. All arguments of trigonometric functions are measured in degrees here).

Notice that the lengths BF and CE and AD are equal. In fact, points A and D belong to a similar pair of edges, parallel and on opposite sides of polyhedron (These edges are perpendicular to a cross section). Denote these lengths by x.

The following picture (colored triangle is right triangle of sides x/2, (x-L)/2, d)

shows that

Solving this equation we receive:

Denote by H the center of AD and let K be the projection of H on AB

Then the length of KH is the radius of the inscribed sphere into a polyhedron. Denote this radius by r.

One height of triangle AHB is x/2 and base x/2, another height – r and base b*L. Comparing areas of the triangle AHB we see that

b*L*r=x²/4.

Hence,

r=x²/(4*b*L).

Now, we can divide dodecahedron into 12 regular pyramids. Bases of these pyramids are the faces of dodecahedron and one common vertex of these pyramids is point H – the center of dodecahedron.

Each of these pyramids has the height equal to r.

Since the bases of pyramids are regular pentagons of sides L, the area of each base is

S=5*L²/(4* tan 36).

Hence the volume of each pyramid is

S*r/3.

The volume of polyhedron is 12 times larger, hence

V=4*S*r.

Finally, substituting values we receive the formula