JT
Posts:
1,386
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?

