# Area of a Regular Polygon 1

Let $AB = a$, be the length of a side of a regular polygon of $n$ sides and $O$ be the center of the polygon.

As there are $n$ sides, $n$ similar triangles are formed.

Let $\Delta AOB$ be one of the $n$ triangles, then
$\angle AOB = \frac{{{{360}^ \circ }}}{n}$ and $\angle AOD = \frac{{{{180}^ \circ }}}{n}$
$\therefore$ Area of the regular polygon $= n \times$area of $\Delta AOB$
Area of the regular polygon $= n \times \frac{{OD \times AB}}{2}$
But $AB = a$, $\frac{{OD}}{{AD}} = \cot \frac{{{{180}^ \circ }}}{n}$
Or $OD = AD\cot \frac{{{{180}^ \circ }}}{n}$
Or $OD = \frac{a}{2}\cot \frac{{{{180}^ \circ }}}{n}$
$\therefore$ Area of the regular polygon $= n \times \frac{{\frac{a}{2}\cot \frac{{{{180}^ \circ }}}{n}}}{2} \times a = \frac{{n{a^2}}}{4}\cot \frac{{{{180}^ \circ }}}{n}$
Also, perimeter of the polygon $= na$
$\begin{gathered} A = \frac{{n{a^2}}}{4}\cot \frac{{{{180}^ \circ }}}{n} \\ P = na \\ \end{gathered}$

Example:

Find the cost of carpeting an octant floor with sides measuring $16m$, if the carpet costs Rs. $2$ per square meter.

Solution:

Here $n = 8$, $a = 16m$
$\therefore$ Area $= \frac{{n{a^2}}}{4}\cot \frac{{{{180}^ \circ }}}{4} = \frac{{8{{(16)}^2}}}{4}\cot \frac{{{{180}^ \circ }}}{8}$
$\therefore$ Area $= \frac{{8 \times 256}}{4}\cot {22.5^ \circ } = 2 \times 256 \times 2.4142 = 512 \times 2.4142 = 1236$ Square meter

$\therefore$ Cost of carpeting $= 1236 \times 2 = 2472$ Rupees