Polygon Angles

Date: 02/14/97 at 11:20:27
From: Jon, Ryan, and Tony
Subject: Geometry

Yesterday we looked in the archives and found the names of the 
polygons with sides 3-50. We would like to know what is the sum of the 
measure of the angles in those polygons, or do you have a formula for 
figuring that out?

Date: 02/14/97 at 11:36:56
From: Doctor Tom
Subject: Re: Geometry

The formula is this:

For an n-sided polygon, the number of degrees for the sum of the 
internal angles is 180(n-2).

For n=3 (triangle), it's 180.
For n=4 (quadrilateral), it's 360.
And so on.

But rather than memorize a formula, it's easier to think about it 
this way. A triangle has 180 degrees. If you have a 4-sided figure, 
connect 2 vertices forming 2 triangles, and so you have 2*(180) 
degrees.  If you cut up a pentagon, you'll get 3 triangles, and 
so on.

Draw some pictures and see that each time you add a side, you'll 
need one more triangle.  Try it with oddball-shaped figures like 
stars and stuff, and see that it still works.

-Doctor Tom,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/