Properties of Triangles 3 Question 16
16. $I _n$ 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
$$ I _n=\frac{O _n}{2} 1+\sqrt{1-\frac{2 I _n^{2}}{n}} $$
$(2003,5 M)$
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Answer:
Correct Answer: 16. 3
Solution:
- We know that, $I _n=\frac{n}{2} r^{2} \sin \frac{2 \pi}{n}$
[since, $I _n$ is area of regular polygon]
$$ \begin{array}{ll} \Rightarrow & \frac{2 I _n}{n}=\sin \frac{2 \pi}{n} \\ \text { and } & O _n=n r^{2} \tan \frac{\pi}{2} \end{array} $$
[since, $O _n$ is area of circumscribing polygon]
$$ \frac{O _n}{n}=\tan \frac{\pi}{n} $$
On dividing Eq. (i) by Eq. (ii), we get
$$ \begin{array}{rlrl} & & \frac{2 I _n}{O _n} & =\frac{\sin \frac{2 \pi}{n}}{\tan \frac{\pi}{n}} \\ \Rightarrow \quad & \frac{I _n}{O _n} & =\cos ^{2} \frac{\pi}{n}=\frac{1+\cos \frac{2 \pi}{n}}{2} \\ \therefore \quad & \frac{I _n}{O _n} & =\frac{1+\sqrt{1-\left(2 I _n / n\right)^{2}}}{2} \quad \text { [from Eq. (i)] } \\ \Rightarrow \quad & I _n & =\frac{O _n}{2}\left(1+\sqrt{\left(1-\left(2 I _n / n\right)^{2}\right)}\right. \end{array} $$
