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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 ]
 JT Posts: 1,448 Registered: 4/7/12
Re: Trying to find perimeter of a regular pentagon
Posted: May 28, 2013 3:19 PM

On 28 Maj, 21:16, JT <jonas.thornv...@gmail.com> wrote:
> On 28 Maj, 20:34, Barry Schwarz <schwa...@dqel.com> wrote:
>

> > On Tue, 28 May 2013 08:15:05 -0700 (PDT), 130bcd
>
> > <carl.sundqu...@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)
>
> > --
> > Remove del for email

>
> The vertice to side ratio seem to be 1.03076923077/1 for a pentagon,
> and for a hexagon it is of course 1/1  and then it all goes downhill
> so to speech. But is it possible to formulate a formula from the
> serie? So one could calculate the perimeter without first doing trig?

Of course for multiples and splits of hexagons it is really easy, but
a general one for any polygon?

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