# Area of a Regular Polygon 2

Area of a regular polygon of n sides when the radius r of the inscribed circle is given

Since $\angle AOB = \frac{{{{360}^ \circ }}}{n}$
$\therefore$ the area of the polygon $= n \times$area of the $\Delta AOB = n \times \frac{{AB \times OD}}{2}$

But $OD = r$, $\frac{{AD}}{{OD}} = \tan \frac{{{{180}^ \circ }}}{n}$
$\therefore$ $AD = OD\tan \frac{{{{180}^ \circ }}}{n}$ and $AD = 2r\tan \frac{{{{180}^ \circ }}}{n}$

Hence, the area of the regular polygon $= n \times \frac{{2r\tan \frac{{{{180}^ \circ }}}{n} \times r}}{2} = n{r^2}\tan \frac{{{{180}^ \circ }}}{n}$

Also, the perimeter of the polygon $= nAB = n{\text{ }}2r\tan \frac{{{{180}^ \circ }}}{n} = 2nr\tan \frac{{{{180}^ \circ }}}{n}$
(Using $AB$ from above)

$\begin{gathered} A = n{r^2}\tan \frac{{{{180}^ \circ }}}{n} \\ P = 2{\text{ }}nr\tan \frac{{{{180}^ \circ }}}{n} \\ \end{gathered}$

Example:

A regular octagon circumscribed a circle of $2m$ radius. Find the area of the octagon.

Solution:

Here $n = 8$, $r = 2m$

$\therefore$ the area of the polygon $= n{r^2}\tan \frac{{{{180}^ \circ }}}{n} = 8 \times {(2)^2}\tan \frac{{{{180}^ \circ }}}{8} = 8 \times 4 \times \tan {22.5^ \circ }$

$\therefore$ the area of the polygon$= 32 \times 0.4142 = 13.3 = 13.2544 =$ square meters, approximately.