﻿ Simple, Elegant yet Neglected Theorems Concerning Areas and Volumes That Includes Circles and Spheres as Limits
San José State University

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Thayer Watkins
Silicon Valley
USA

Simple, Elegant yet Neglected
Theorems Concerning Areas
and Volumes That Includes
Circles and Spheres as Limits

## Symmetric Polygons

Consider a symmetric polygon of n sides. Its area is found by decomposing it into n equilateral triangles of height h and base length b. Its area A is then given by

#### A = n(½hb) = ½nhb =½h(nb) but nb is the length B of its boundary and thus A = ½hB

The distance h is the minimum distance from the center of the polygon to its boundary edge. Thus the theorem is that for a symmetric polygon its area A is equal to ½hB, where B is the length of its boundary and h is the minimum distance from the center of the polygon to its boundary edge.

For a circle h is its radius r and B is its circumference 2πr. Thus

## Symmetric Polyhedra

A symmetric polyhedron of n sides can be decomposed into n pyramids of height h and base area b. Each of these has volume ⅓hb. The volume V of the polyhedron is then given by

#### V = n(⅓hb) = ⅓h(nb).

The area B of the surface of the polyhedron is nb. The quantity h is again the minimum distance from the center of the polyhedron to its boundary surface. The theorem is thus that the volume V of a symmetric polyhedron is

#### V = ⅓hB

For a sphere, h is its radius r and it bounary surface is 4πr². Thus

## Line Segments

Now consider a line segment of length L. Its boundary is the two end points so B=2. The value h for it is L/2. So an analogous formula applies to the one dimensional case; i.e.,

## General Formula for Symmetric Polytopes

Let d be the dimensionality of the symmetric polytope; 1 for line segments, 2 for polygons and 3 for polyhedra. The formula that applies to the three cases is

#### W = (1/d)hU

where W is the d-dimensional volume and U is the (d−1)-dimensional boundary surface.

## Extension to Star-type Polytopes

Consider a Star-of-David. The area of the central hexagon is as above ½h1(6b). The area of the points is 6(h2b). Thus the total area is ½(h1+h2)(6b)=½hB, where B is the border length of the central hexagon and h=h1+h2) is in the nature of a maximum distance from the center to outer border of the figure.

Likewise the volume of a pointed polyhedron is

#### V = ⅓hB

with B being the area of the central polyhedron.

A star-type of a line segment is just a longer line segment so the extension of the theorem is trivial.

## Extension to Sheaves of Polytopes

Symmetric figures can also be created by by sucessively rotating a polygon about an axis of symmetry by one n-th of a full circle. This creates a sheaf of polygons. Similarly a symmetric sheaf of line segments can be created.