
Re: Trying to find perimeter of a regular pentagon
Posted:
May 28, 2013 5:24 PM


In article <425e7c8f10e6428fbcda44e1fbf8b850@googlegroups.com>, 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 ngon is inscribed in a circle of radiu r, then each side of the ngon 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.

