JT
Posts:
1,448
Registered:
4/7/12


Re: Trying to find perimeter of a regular pentagon
Posted:
May 28, 2013 3:36 PM


On 28 Maj, 21:19, JT <jonas.thornv...@gmail.com> wrote: > 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? Vertice line/side ratio 12 sides 1/0.5 24 sides 1/0.25 48 sides 1/0.125 ... And the other polygon multiples of six ratisos is not that hard to work out either using fractions. But of course this is not the complete set of ratios for any polygon, just ratios for the polygons based in six. But it seem pretty easy to work out the ratios for polygons based in 3. But of course there is alot of other polygon needed to be integrated to find a general formula?
So could i find a general formula that would calculate the vertice line to side ratio, i am pretty sure i could.

