Area of a Regular Polygon 3

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


Let $$OA = R$$ be the radius of the circumscribed circle.

$$\therefore $$ the area of the polygon $$ = n \times $$ area of $$\Delta AOB$$

But the area of $$\Delta AOB = \frac{n}{2} \times OB \times OA \times \sin \frac{{{{360}^ \circ }}}{n}$$
\[\begin{gathered} \Delta AOB = \frac{n}{2} \times OA \times OB \times \sin \frac{{{{360}^ \circ }}}{n} \\ \Delta AOB = \frac{n}{2} \times R \times R \times \sin \frac{{{{360}^ \circ }}}{n} \\ \Delta AOB = \frac{{n{R^2}}}{2}\sin \frac{{{{360}^ \circ }}}{n} \\ \end{gathered} \]


Also, the perimeter of the polygon $$ = n{\text{ }}AB$$

But $$\frac{{AD}}{{OA}} = \sin \frac{{{{180}^ \circ }}}{n}$$
$$\therefore $$ $$AD = OA\sin \frac{{{{180}^ \circ }}}{n}$$

Also $$AD = \frac{{AB}}{2}$$
$$\therefore $$ $$AB = 2AD$$
$$\therefore $$ $$AB = 2AD = 2OA\sin \frac{{{{180}^ \circ }}}{n}$$
$$\therefore $$ Perimeter $$ = nAB$$

Or $$P = n \times 2OA\sin \frac{{{{180}^ \circ }}}{n} = 2nR\sin \frac{{{{180}^ \circ }}}{n}$$ (As$$OA = R$$)
\[\begin{gathered} A = \frac{{n{R^2}}}{2}\sin \frac{{{{360}^ \circ }}}{n} \\ P = 2nR\sin \frac{{{{180}^ \circ }}}{n} \\ \end{gathered} \]



A regular decagon is inscribed in a circle, the radius of which is$$10cm$$. Find the area of the decagon.



Here $$n = 10$$, $$R10cm$$

The area of the decagon $$ = \frac{{n{R^2}}}{2}\sin \frac{{{{360}^ \circ }}}{n} = \frac{{10{{(10)}^2}}}{2}\sin \frac{{{{360}^ \circ }}}{2}$$

The area of the decagon $$ = 500\sin {36^ \circ } = 500(0.5878) = 293.9$$ square cm.