JT
Posts:
1,434
Registered:
4/7/12


Re: Trying to find perimeter of a regular pentagon
Posted:
May 28, 2013 6:45 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.
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 here http://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?

