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Topic: Trying to find perimeter of a regular pentagon
Replies: 36   Last Post: May 31, 2013 9:26 PM

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 Barry Schwarz Posts: 177 Registered: 3/13/08
Re: Trying to find perimeter of a regular pentagon
Posted: May 28, 2013 2:34 PM

On Tue, 28 May 2013 08:15:05 -0700 (PDT), 130bcd
<carl.sundquist@gmail.com> wrote:

>Hello,
>
>I have unsuccessfully tried to find the perimeter of a pentagon. The "radius" to the vertices is 130, which I tried to convert into 5 triangles but I can't find a formula for determining the length of the 3rd side of an isosceles triangle.
>
>The two equal length sides would, of course be 130 and the angle between them would be 72.

Try the law of cosines: c^2 = a^2 + b^2 - 2*a*b*cos(C)

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Date Subject Author
5/28/13 130bcd
5/28/13 JT
5/28/13 JT
5/28/13 Barry Schwarz
5/28/13 JT
5/28/13 JT
5/28/13 JT
5/28/13 JT
5/28/13 Ken.Pledger@vuw.ac.nz
5/28/13 JT
5/28/13 Ken.Pledger@vuw.ac.nz
5/28/13 David Bernier
5/28/13 Brian Q. Hutchings
5/28/13 JT
5/30/13 RGVickson@shaw.ca
5/31/13 Brian Q. Hutchings
5/31/13 JT
5/31/13 JT
5/31/13 JT
5/31/13 JT
5/31/13 JT
5/31/13 JT
5/31/13 Brian Q. Hutchings
5/28/13 JT
5/28/13 JT
5/28/13 Brian Q. Hutchings
5/28/13 JT
5/28/13 Brian Q. Hutchings
5/28/13 JT
5/29/13 Richard Tobin
5/29/13 JT
5/29/13 Richard Tobin
5/29/13 Brian Q. Hutchings
5/29/13 JT
5/29/13 Brian Q. Hutchings
5/30/13 Graham Cooper
5/30/13 RGVickson@shaw.ca