Above there are pictures of
hexagonal antiprism, heptagonal antiprism and octagonal antiprism.
Let us mention that there is a page of 'Literka' Polyhedron with 16 faces built of equilateral triangles with a picture of square
antiprism and a page ICOSAHEDRON with pentagonal antiprism. However, these pages are not
scientific and that is why these antiprisms are called 'drums' there (shape
of antiprism reminds a drum).
Antiprisms (that this page is about) are polyhedrons built of 2 congruent regular polygons (top and a bottom) with n sides
and 2n equilateral triangles joining one with another. There are more general definitions, but we are
not interested in them now.
Let W be an antiprism. Assume that a base of W is a regular polygon with n sides, each of length L>0. Projection of W on a plane of base looks like this (case n=7):
Yellow polygon is a base of an antiprism, red triangles are projections of faces of W, which are
equilateral triangles. Red triangles are isosceles triangles.
Two equal angles of this triangle have 90/n degrees. Hence the height d of this triangle is
d=(L/2)*tan(90/n).
Let us denote
Height d is a projection of a height p of a face of W, which is an equilateral triangle.
Let h be the height of W (distance between the top of W and the bottom).
Pythagorean Theorem used for a right triangle with sides p,d and h gives us:
A well known formula for a height of an equilateral triangle gives us:
Now we can find that
Let A be one of vertices of a bottom regular polygon of W and D be the center of this polygon.
Let J be a triangular face of W with a base belonging to the top of W such that one of vertices of J is A (only one vertex A of J
belongs to the bottom face of W). Denote by E the center of J. Using Pythagorean Theorem we find that
But AE=(2/3)*p, OD=h/2 and we can easily compute that
Let us denote by r=EO. Then r is a distance from the center of W to the face J. Combining the previous equalities we obtain
But r is a height of a pyramid with J as a base and vertex O. For each triangular face of W we can create such
pyramid. Volume of all 2*n of them therefore is (here S - area of J)
Now, what is left it is to find volume of 2 pyramids with bases, which are top and bottom of W (regular polygons with 2n sides) and vertices at point O.
Heights of these pyramids are h/2=OD. Area of bases is equal
Hence volume of these 2 pyramids is
Summing the computed volumes we receive final formula for volume of W:
