Question

The area of the circle and the area of an regular polygon of sides of perimeter equal to that of the circle are in the ratio of

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

Answer 1

The correct option is A $\tan \left(\frac{\pi }{n}\right):\frac{\pi }{n}$

Let $r$ be the radius of the circle and a be the side of polygon.

The perimeter of circle $=2\pi r$

The perimeter of a polygon with n sides $=na$

So $2\pi r=na$

So $a=\frac{2\pi r}{n}$

Area of the circle, $A_1=\pi r^2$

Area of a regular polygon, $A_2=(\frac{1}{4})n{a}^2\cot (\frac{\pi }{n})$

$=\frac{1}{4}n(\frac{4{\pi }^2{r}^2}{n^2})\cot (\frac{\pi }{n})$

$=(\frac{{\pi }^2{r}^2}{n})\cot (\frac{\pi }{n})$

$\Rightarrow \frac{A_1}{A_2}=\frac{\pi r^2}{\frac{{\pi }^2{r}^2}{n}\cot (\frac{\pi }{n})}$

$=\frac{\tan (\frac{\pi }{n})}{\frac{\pi }{n}}$

So the ratio is $\tan (\frac{\pi }{n}):\frac{\pi }{n}$