Polygon AnglesDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/