Question

$I_n$ 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
$I_n=\frac{O_n}{2}(1+\sqrt{1-(\frac{2I_n}{n}{)}^2}|bigg)$

Answer 1

We know that, $I_n=\frac{n}{2}{r}^{2}sin\frac{2\pi }{n}$
[since, $I_n$ is area of regular polygon]
$\Rightarrow$ $\frac{2I_n}{n}=sin\frac{2\pi }{n}$ $[\because r=1]$ ...(i)
and $O_n=n{r}^{2}tan\frac{\pi }{2}$
[since, $O_n$ is area of circumscribing polygon]
$\frac{O_n}{n}=tan\frac{\pi }{n}$ ... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{2I_n}{O_n}=\frac{sin\frac{2\pi }{n}}{tan\frac{\pi }{n}}$
$\Rightarrow \frac{I_n}{O_n}=co{s}^2\frac{\pi }{n}=\frac{1+cos\frac{2\pi }{n}}{2}$
$\therefore \frac{I_n}{O_n}=\frac{1+\sqrt{1-(2I_n/n{)}^2}}{2}$ [ from Eq. (i)]
$\Rightarrow I_n=\frac{O_n}{2}(1+\sqrt{(1-2I_n/n{)}^2)}$