Area of a circle inscribed in a polygon If a circle is inscribed in a polygon, show that,   
$$\dfrac{\text{(Area of inscribed circle)}}{\text{(Perimeter of inscribed circle)}} = \dfrac{\text{(Area of Polygon)}}{\text{(Perimeter of Polygon)}}$$

For a regular polygon with $n$ sides with side length $l$. The ends of each side when connected to the centre of the polygon forms a triangle with an angle of $\frac{2\pi}{n}$ at the centre. There will be $n$ such triangles. The altitude of each triangle starting from the centre of the polygon has a length of $\frac{a}{2\tan\frac{\pi}{n}}$ with the opposite side (base of the triangle) of length $a$. This altitude height will also be the radius of the circle inscribed in it.
So, 
$\text{Area of incribed circle} = \pi \left(\frac{a}{2\tan\frac{\pi}{n}}\right)^2$
$\text{Perimeter of incribed circle} = 2\pi \left(\frac{a}{2\tan\frac{\pi}{n}}\right)$
$\text{Area of polygon} = n \cdot \frac{1}{2}a\left(\frac{a}{2\tan\frac{\pi}{n}}\right)$
$\text{Perimeter of polygon} = n \cdot a$
