JT
Posts:
1,200
Registered:
4/7/12


Re: Trying to find perimeter of a regular pentagon
Posted:
May 28, 2013 7:41 PM


On 29 Maj, 00:45, JT <jonas.thornv...@gmail.com> wrote: > 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. > > http://www.wolframalpha.com/input/?i=260+sin+%28pi%2F5%29++ > Is 152 correct for the side length? > > Maybe i am lost here but if the side is 130, we can divide the > isoceles into two right angled triangles each with 31 degree angle > correct? > and the hypotenuse of each is 130 and the angle 31. I just plugget it > in herehttp://www.cleavebooks.co.uk/scol/calrtri.htm > And then i see that the height for each must be 67 * 2 = 134 * 5 what > did i do wrong?
I know usally 72/2 = 36 ;D

