For simplicity, let us assume that a polyhedron W built of 30 congruent rhombuses has all edges of
the length 1. Denote by S the area of each of its faces. S is the half of the product of diagonals of
a rhombus - face of W. Seeing the description of a face of W shown on the page of 'Literka'
(See Few remarks on polyhedrons built of rhombuses).
We can easily deduce (Pitagorean theorem) that these diagonals have lengths
Let us notice that S=h, where h is the height of a face of W. It's because S=a*h, where a - length of a side of rhombus.
Hence,
Substituting this value to the previous formula and doing simple calculation we see that
.
Let us repaint the polyhedron W
Let us take a cross section of this polyhedron by a plane passing through
the centers of blue faces. This plane will also pass through the center of W.
Cross section is a regular polygon with 10 sides. Lengths of sides of this
polygon are equal to h:
Point O is the center of W, A and B - vertices of cross section are points of edges of W.
Angle AOB has 36 degrees. Denote by r the height of AOB. Then
Since
and
Polyhedron W consists of 30 pyramids with vertices O and bases which are faces of W.
Heights of these pyramids are equal r. Hence the volume of W is
V=30*S*r/3=10*S*r=10*h*r.
Hence
This is a final formula for a volume of a polyhedron built of congruent 30 rombuses with
the length of edges equal to L. This formula was used by 'Literka' to find a volume formula
of a regular 120-cells polytope.
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