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} (AsOA = 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 is10cm. 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.