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 polygon circumscribing the given circle, then the value of $\frac{O_n(1+\sqrt{1-{\left(\frac{2I_n}{n}\right)}^2})}{I_n}=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$A\left(\Delta AOB\right)=\frac{1}{2}1\times 1\times \sin \left(\frac{2\pi }{n}\right)=\frac{1}{2}\sin \left(\frac{2\pi }{n}\right)$
Area of $n$ sided polygon $=I_n=\frac{n}{2}\sin \frac{2\pi }{n}\cdots \left(i\right)$


In $\Delta O^{\prime }{B}^{\prime }{A}^{\prime },\cos \frac{\pi }{n}=\frac{1}{O^{\prime }{B}^{\prime }}$
$\Rightarrow O^{\prime }{B}^{\prime }=\sec \frac{\pi }{n}$
Area $\left(O^{\prime }{B}^{\prime }{A}^{\prime }\right)=\frac{1}{2}(O^{\prime }{B}^{\prime }{)}^2\sin \left(\frac{2\pi }{n}\right)$
$=\frac{1}{2}{\sec }^2\left(\frac{\pi }{n}\right)\sin \left(\frac{2\pi }{n}\right)$
Therefore, the area of $n$-sided polygon is given by
$\left(O_n\right)=\frac{n}{2}{\sec }^2\frac{\pi }{n}\sin \frac{2\pi }{n}\cdots \left(ii\right)$
$\frac{I_n}{O_n}=\frac{\frac{n}{2}\sin \frac{2\pi }{n}}{\frac{n}{2}{\sec }^2\frac{\pi }{n}sin\frac{2\pi }{n}}=\frac{1}{{\sec }^2\frac{\pi }{n}}$
$\Rightarrow I_n=\left(co{s}^2\frac{\pi }{n}\right)O_n$
$=\frac{O_n}{2}[1+cos\left(\frac{2\pi }{2}\right)]$
$=\frac{O_n}{2}[1+\sqrt{1-{\sin }^2\frac{2\pi }{n}}]$
$\Rightarrow I_n=\frac{O_n}{2}[1+\sqrt{1-{\left(\frac{2I_n}{n}\right)}^2}]$
$\Rightarrow \frac{O_n(1+\sqrt{1-{\left(\frac{2I_n}{n}\right)}^2})}{I_n}=2$