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.


polygon-01

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 $$\Delta AOB$$

The 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 $$OD = \frac{a}{2}\cot \frac{{{{180}^ \circ }}}{n}$$

$$\therefore $$ the 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, the 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 octagonal 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 meters.

 

$$\therefore $$ the cost of carpeting $$ = 1236 \times 2 = 2472$$ Rupees.