I'm trying to answer the following question:
"Given a regular pentagon with side length equal to $r$ and a circle of radius $r$ that intersects the pentagon in two of its consecutive vertices and has its centre outside the pentagon, calculate the area of the intersection of the two figures as a function of $r$."
I don't even know where to start, any tip is really really much appreciated! Thanks in advice!
Okay, I think I got the formula:
Basically I added the area of the pentagon, the area of the circle and subtracted the circular sector relative to the angle of 60 degrees.
$r^2\phi+r^2\pi-(\frac{\pi r^2}{6}-\frac{\sqrt3r^2}{4})$
then
$r^2(\phi+(\pi-\frac{\pi}{6}\ ) + \frac{\sqrt3}{4})$
and finally
$r^2(\frac{5}{6}\pi+\phi+\frac{\sqrt3}{4}) \approx r^2(4,7709)$
Hope this is right, and that maybe one day it can be useful to someone!