Polygon problem

So basically we have a decagon inscribed in a unit circle, and the problem asks to find the perimeter.

Splitting the decagon up into 10 triangles and observing one of them, you will find that it is an isosceles triangle with vertex angle 360/10 = 36 degrees. Using the law of cosines, the side length of the decagon, s:

\(s^2=1^2+1^2-2*1*1\cos{36^o}\), if you know some trig, you will quickly find that cos(36 degrees) = (sqrt(5) + 1)/4, which can be substituted in to get \(s^2=2-{\sqrt{5}+1\over{2}}={3-\sqrt{5}\over{2}}\). 

Remember to square root both sides and simplify the radical to obtain s, and then multiply by 10 to get the perimeter: I don't want to directly spoil the answer for you Vxrtate- happy solving !

Continuing with proyaop's answer

s^2   = (3 -sqrt 5)   / 2  

So  Pn =  sqrt [ (3 -sqrt 5) / 2) ]  = s

But  P1 * P2     is  =  s * s  =   s^2

So....the given sum should just be  10(s*s) = 10s^2   =   10  (3 - sqrt 5)  /  2 =    5 ( 3 - sqrt 5)  = 15 - 5sqrt5