JT
Posts:
1,254
Registered:
4/7/12


Re: Trying to find perimeter of a regular pentagon
Posted:
May 28, 2013 6:24 PM


On 28 Maj, 23:24, Ken Pledger <ken.pled...@vuw.ac.nz> wrote: > In article <425e7c8f10e6428fbcda44e1fbf8b850@googlegroups.com>, > > 130bcd <carl.sundqu...@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.
I see your formula use Pi so i guess your formula can only calculate the perimeter to the precision of the given Pi and same would go for the area? Would it not be beneficial finding a formula using fractions, that could calculate the perimeter as well as area exact, without using a couple of billions of decimalpoints on Pi?
Except from being accurate it sure would put an ease to the calculation machinwise or humanwise.

