Question

The area of the circle and the area of the regular polygon of $n$ sides of perimeter equal to that of circle are in the ratio of $a\tan \left(\frac{\pi }{n}\right):\frac{b\pi }{n}$, then $a+b=$

Answer 1

Let $r$ be the radius of the circle.

$\therefore$ its area $A_1=\pi r^2$.

Length of one side of a regular polygon of $n$ sides $=($ perimeter of the circle $)/n$

$=\frac{2\pi r}{n}=k$(say)

$\therefore$ Area of the polygon

$=A_2=\frac{\pi k^2}{4}\cot \left(\frac{\pi }{n}\right)=\frac{r^2{\pi }^2}{n}\cot \left(\frac{\pi }{n}\right)$

$\therefore A_1:A_2=\pi r^2:\frac{r^2{\pi }^2}{n}\cot \left(\frac{\pi }{n}\right)=\tan \left(\frac{\pi }{n}\right):\frac{\pi }{n}$

Hence $a+b=2$