# Calculate the area of a particular pentagon-circumference intersection.

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!

• What have you tried? Usually a good starting point is to draw the diagram and try to visualise the problem. Commented Dec 23, 2019 at 20:31
• @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! Commented Dec 23, 2019 at 20:36
• Update: i think I'm onto something, using the circular sector formula. Commented Dec 23, 2019 at 20:45
• Great, that sounds like the right direction to me. Let me know if you still have some issues with it. Commented Dec 23, 2019 at 20:46

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!

• 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. Commented Dec 24, 2019 at 0:52