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

 Messages: [ Previous | Next ]
 Ken.Pledger@vuw.ac.nz Posts: 1,412 Registered: 12/3/04
Re: Trying to find perimeter of a regular pentagon
Posted: May 28, 2013 5:24 PM

130bcd <carl.sundquist@gmail.com> wrote:

> .... 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....

If a regular n-gon is inscribed in a circle of radiu r, then each
side of the n-gon has length 2r.sin(pi/n). In your case that means
260.sin(pi/5) or 260.sin(36 degrees).

That can actually be expressed in terms of square roots if you like,
since sin(36 degrees) = (1/4)sqrt(10 - 2.sqrt(5)). Such expressions
involving sqrt(5) underlie the construction of the regular pentagon by
straightedge and compasses (Euclid IV.11, or a neater construction by
Ptolemy).

Ken Pledger.

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