Question

If $I_n$ is the area of n sided regular polygon inscribed in a circle of unit radius and $O_n$ be the area of the polygon circumscribing the given circle, prove that
$I_n=\frac{O_n}{2}(1+\sqrt{1-{\left(\frac{2I_n}{n}\right)}^2})$

Answer 1

Area of n-sided regular Polygon $=\frac{l^{2}n}{2}\sin \left(\frac{2\pi }{n}\right)=k^{2}n\tan \left(\frac{\pi }{n}\right)$,

where $l$ is the length of the half of it's diagonal,

k is the length of the half of the perpendicular bisector from one side to it's opposite side $(k=l\cos \left(\frac{\pi }{n}\right))$

and $n$ is the no of sides of the polygon.

Here, $l_{I_n}=k_{O_n}=$ radius of circle$=1$

So, $I_n=\frac{n}{2}\sin \left(\frac{2\pi }{n}\right)$

And, $O_n=n\tan \left(\frac{\pi }{n}\right)$

ie, $I_n=\frac{O_n}{2}2{\cos }^2\left(\frac{\pi }{n}\right)=\frac{O_n}{2}(1+\cos \left(\frac{2\pi }{n}\right))=\frac{O_n}{2}(1+\sqrt{1-{\sin }^2\left(\frac{2\pi }{n}\right)})$

$\Rightarrow I_n=\frac{O_n}{2}(1+\sqrt{1-{\left(\frac{2I_n}{n}\right)}^2})$

Q.E.D