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!

  • $\begingroup$ What have you tried? Usually a good starting point is to draw the diagram and try to visualise the problem. $\endgroup$
    – sudeep5221
    Commented Dec 23, 2019 at 20:31
  • $\begingroup$ @sudeep5221 I tried to make a graph using geogebra, but I can't quite figure out where to start. I'm looking for possible formulas to get the area of particular sections similar to this one of a circumference, but I don't think I found any useful. P.S. Thanks for the corrections earlier! $\endgroup$
    – GPU'njoyer
    Commented Dec 23, 2019 at 20:36
  • $\begingroup$ Update: i think I'm onto something, using the circular sector formula. $\endgroup$
    – GPU'njoyer
    Commented Dec 23, 2019 at 20:45
  • $\begingroup$ Great, that sounds like the right direction to me. Let me know if you still have some issues with it. $\endgroup$
    – sudeep5221
    Commented Dec 23, 2019 at 20:46

1 Answer 1


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.

$r^2\phi+r^2\pi-(\frac{\pi r^2}{6}-\frac{\sqrt3r^2}{4})$


$r^2(\phi+(\pi-\frac{\pi}{6}\ ) + \frac{\sqrt3}{4})$

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!

  • $\begingroup$ The area of a pentagon of side $r$ is $\frac54\cot\left(\frac{\pi}{5}\right) r^2 = \frac14\sqrt{5(5+2\sqrt{5})} r^2$, not $\varphi r^2$. Furthermore, you have computed the area of the union of the two figures instead of the area of their intersection. $\endgroup$ Commented Dec 24, 2019 at 0:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .