# How does the area change of a polyhedra as the number of faces go up but circumference stays the same?

Asked by timetogo (27 ) January 29th, 2011

So imagne we have a very simple polyhedra, I’m not too sure if they would concider an edge from one vertex to another and back again a polyhedra, if not start off with a triangle.

The circumference of this equalatral trangle is 1.
Increase the number of sides keeping all sides the same length and at the same angle.
So next a square then a pentagon and so on.

Is there a graph to show me how the area changes with each face?

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Hint: Imagine a tetrahedron encased in a sphere. Now imagine a cube in a sphere of the same size. Now a dodecahedron. Notice that as the number of faces goes up, the closer the polyhedron approximates a sphere. (It may be easier to imagine a polygon in a circle, but the same principle applies).

submariner (4145 )

Also—the parameters you set out in your question are impossible as you’ve worded them, and triangles do not have circumferences. Framing the problem correctly is half the battle.

submariner (4145 )

First a clarification. Since you are talking about 2 dimensional objects, the correct term is polygon. Polyhedra are 3 dimensional, like cubes and tetrahedrons.

I don’t know how much schooling you have had, but if you are familiar with trigonometry, there is a simple formula for the area. Just connect the center with segments to each of the vertices. Take one of the triangles and drop a perpendicular from the top vertex at the center of the triangle. The lengths of the sides are r sin theta and r cosine theta where theta is 360/2n and n is the number of sides. That gives the area as 2n *½ * r ^2 cos theta sin theta =
n cosine theta sine theta.

In the limit, sin theta = sin 2 pi/ 2n, approximately equal to pi/n and cos theta approaches 1, so the limit would be pi r^2 , the area of the circle. You can see intuitively that as the number of sides increases, the sum of the lengths of the triangles approaches the circumference of the circle and the altitude of the triangles approaches r, so the sum of the areas of the triangles again appraches pi r^2,

This isn’t quite what you asked, since the circumference changes, but almost.

finkelitis (1900 )

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