Question

The area of the circle and the area of a regular polygon inscribed the circle of $n$ sides and of perimeter equal to that of the circle are in the ratio of

• A. $\tan \left(\frac{\pi }{n}\right):\frac{\pi }{n}$
• B. $\cos \left(\frac{\pi }{n}\right):\frac{\pi }{n}$
• C. $\sin \left(\frac{\pi }{n}\right):\frac{\pi }{n}$
• D. $\cot \left(\frac{\pi }{n}\right):\frac{\pi }{n}$

Answer 1

Answer: (A)

$\tan \left(\frac{\pi }{n}\right):\frac{\pi }{n}$

Let $r$ be the radius of the circle and $A$ be its area
$A=\pi r^2$
Let length of side of polygon is ${}^{\prime }{a}^{\prime }$ according to the question
$2\pi r=na$
$\Rightarrow a=\frac{2\pi r}{n}$ ............$\left(1\right)$
If $A_1$ be the area of polygon, then
$A_1=\frac{1}{4}n{a}^2\cot \left(\frac{\pi }{n}\right)$
$=\frac{1}{4}n.\frac{4{\pi }^2{r}^2}{n^2}\cot \left(\frac{\pi }{n}\right)$
$=\frac{{\pi }^2{r}^2}{n}\cot \left(\frac{\pi }{n}\right)$
$\because \frac{A}{A_1}=\frac{\pi r^2}{\frac{{\pi }^2{r}^2}{n}\cot \left(\frac{\pi }{n}\right)}$
$\therefore A:A_1=\tan \left(\frac{\pi }{n}\right):\frac{\pi }{n}$