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}


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


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