question_answer

In Fig. 7, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M. If \[OP=PQ=10\text{ }cm\] show that area of shaded region is 25\[\left( \sqrt{3}-\frac{\pi }{6} \right)c{{m}^{2}}\].

Answer:

Given, \[OP=PQ=10\text{ }cm\]

Since, OP and OQ are radius of circle with centre O.

\[\therefore \,\,\Delta \text{ }OPQ\] is equilateral.

\[\Rightarrow \,\angle POQ=60{}^\circ \]

Now, Area of segment PAQM

= (Area of sector OPAQO\[-\]Area of\[\Delta \text{ }POQ\])

\[=\frac{\pi {{r}^{2}}\theta }{360{}^\circ }-\frac{1}{2}{{r}^{2}}\sin \,i\]

\[=\frac{\pi \times {{(10)}^{2}}\times 60{}^\circ }{360{}^\circ }-\frac{1}{2}{{(10)}^{2}}\sin \,60{}^\circ \]

\[=\left( \frac{100\pi }{6}-\frac{100\sqrt{3}}{4} \right)c{{m}^{2}}\]

and, area of semicircle \[PBQ=\frac{\pi {{r}^{2}}}{2}=\frac{\pi }{2}{{(5)}^{2}}=\frac{25}{2}\pi \,c{{m}^{2}}\]

\[\therefore \] Area of shaded region = Area of semicircle\[-\]Area of segment PAQM

\[=\frac{25}{2}\pi -\left( \frac{100\pi }{6}-\frac{100\sqrt{3}}{4} \right)\]

\[=\frac{25}{2}\pi -\frac{50\pi }{3}+25\sqrt{3}\]

\[=\frac{75\pi -100\pi }{6}+25\sqrt{3}\]

\[=\frac{-25\pi }{6}+25\sqrt{3}\]

\[=25\left( \sqrt{3}-\frac{\pi }{6} \right)c{{m}^{2}}\]      Hence Proved.