How to Compute Volume of a Regular 20-faces Polyhedron (Icosahedron).
Above is a picture of a regular icosahedron.
Some basic description of this polyhedron gives a page of 'Literka'
ICOSAHEDRON.
Now we'll show the way that volume formula for a regular icosahedron
can be found.
Let W be a regular icosahedron and let O be the center of W, and assume that lengths of edges of W are equal L>0. Let us cut a
vertex of W by a plane passing through the centers of 10 edges of the following
part of W:
This a polyhedron, which was called 'drum' on the page of 'Literka'
ICOSAHEDRON. (See this page for better description).
In fact, it is a pentagonal antiprism. A plane will pass also through the center O.
Cross section of W by this plane is a regular polygon with 10 sides and each side has
the length L/2.
Let A and B be two vertices of a cross section as in the picture, q - height OE.
The angle AOB has 36 degrees, hence
q=(L*cot18)/4
Let us draw the picture of a face of W, which contains points A,B and E.
Let P be the center of this face. Interval OP is the radius of the inscribed sphere into W - denote it by r.
Point E divides the height (as in the picture) in the half, point P divides this height
in 1/3. Hence EP is equal to 1/6 of length of the height of a triangle, which is a face of W. Hence
The angle OPE is the right angle. Pythagorean theorem for the triangle OPE gives us
Polyhedron W is a union of 20 pyramids with vertices O and bases being faces of W. Each pyramid has a
volume r*S/3, where S is an area of face of W. We have
and volume of W is 20 times larger than a volume of each pyramid.
Hence, from previous equalities it follows that (V denotes a volume of W)
Simplifying it we receive final
formula for a volume of a regular icosahedron with edges of lengths L is:
