Unizor - Geometry3D - Prisms - Polygonal Prism
Channel: Zor Shekhtman on UNIZOR Education
Date: 03/08/2015

Unizor - Creative Minds through Art of Mathematics - Math4Teens

A prism with any polygon as its directrix is called a polygonal prism.
Triangular prisms and parallelepipeds are particular cases of polygonal prisms. So are pentagonal (pentagon as a directrix) and hexagonal (hexagon as a directrix) prisms.

According to a mini-theorem proven about all prisms, the upper and lower bases of a polygonal prism are congruent polygons and the side faces are parallelograms.

Obviously, we can talk about right polygonal prism if all the side edges are perpendicular to bases.

Of particular interest is the volume of a polygonal prism. To approach this problem, we will perform the following its transformation.

Assume we have a pentagonal prism with a lower base ABCDE and an upper base A'B'C'D'E' (any other polygonal prism can be addressed similarly). Let's break it into a combination of three triangular prisms by cutting it vertically with planes ACC'A' and ADD'A'. The three triangular prisms are: ABCA'B'C',
ACDA'C'D' and
ADEA'D'E'.
All three triangular prisms have the same altitude H - the distance between the two base planes. The volume Vol of each of them equals to a product of the area of a base S and an altitude:
VolABC = SABC·H
VolACD = SACD·H
VolADE = SADE·H

Summarizing the three equations above, we obtain the volume of an entire prism:
Vol = (SABC+SACD+SADE)·H
But the expression in parenthesis constitutes the area of a polygon of the base of this prism. Therefore, our formula for the volume of any polygonal prism is still the same - product of the area of the base and the altitude:
Vol = S·H

Let's do some counting now.
Assume we are dealing with a polygonal prism with an N-sided polygon as the directrix.
The number of vertices is:
V=N+N=2N.
The number of edges is:
E=N+N+N=3N.
The number of faces is:
F=N+2
Let's check the Euler's formula V+F-E=2 that we mentioned in one of the introductory lectures about polyhedrons:
2N+N+2−3N=2
So, the formula holds, as expected.