Question

The diagram shows two arcs, A and B. Arc A is part of the circle with centre O and radius OP. Arc B is part of the circle with centre M and radius PM, where M is the mid - point of PQ. Show that the area enclosed by the two arcs is equal to 25$(\sqrt{3}-\frac{\pi }{6})c{m}^2$

Answer 1

We have,
Area enclosed by arc B and chord PQ $=$ Area of semi-circle of radius 5 cm

$=\frac{1}{2}\times \pi \times 5^{2}c{m}^2=\frac{25\pi }{2}c{m}^2$

Let $\angle$ MOQ = $\angle$ MOP = $\theta$

In $\Delta$ OMP, we have

sin $\theta$ = $\frac{PM}{OP}=\frac{5}{10}=\frac{1}{2}$

$\Rightarrow \theta ={30}^{\circ }$

$\Rightarrow \angle POQ=2\theta ={60}^{\circ }$

$\therefore$ Area enclosed by arc A and chord PQ

= Area of segment of circle of radius 10 cm and sector containing angle ${60}^{\circ }$

$=\{\frac{\pi \times 60}{360}-sin{30}^{\circ }\times cos{30}^{\circ }\}\times {10}^{2}c{m}^2$ $[\because A=\{\frac{\pi \theta }{360}-sin\frac{\theta }{2}cos\frac{\theta }{2}\}{r}^2]$

$=\{\frac{50\pi }{3}-25\sqrt{3}\}c{m}^2$

Hence,
Required area = $\{\frac{25\pi }{2}-(\frac{50\pi }{3}-25\sqrt{3})\}c{m}^2$

$\Rightarrow$ Required area = $\{25\sqrt{3}-\frac{25\pi }{6}\}c{m}^2=25\{\sqrt{3}-\frac{\pi }{6}\}c{m}^2$