Question

The adjacent figure shows two arcs P and Q. Arc P is part of the circle with centre O and radius OA. Arc Q is a part of the circle with centre M and radius AM, where M is the mid-point of AB. Area enclosed by the arcs P and Q is

• A. $(\sqrt{3}-25\frac{\pi }{4})c{m}^2$
• B. $(25\sqrt{3}-\frac{\pi }{4})c{m}^2$
• C. $25(\sqrt{3}-\frac{\pi }{6})c{m}^2$
• D. None of these

Answer 1

The correct option is C $25(\sqrt{3}-\frac{\pi }{6})c{m}^2$

In $\Delta OMA$, we have
$sin\theta =\frac{AM}{QA}=\frac{5}{10}=\frac{1}{2}\Rightarrow \theta ={30}^{\circ }\therefore \angle AOB=2\times \theta ={60}^{\circ }$
$\therefore OM=5\sqrt{3}cm$

Area of $△AOB=\frac{1}{2}\times AB\times OM=25\sqrt{3}$

Area enclosed by arc Q and chord AB $=\frac{\pi \times 5^2}{2}=\frac{25\pi }{2}$

Area enclosed between arcs P and Q is given by,
= Area enclosed by semicircle AQB - (Area of sector OAPB - Area of $△AOB$)
$=\frac{\pi \times 5^2}{2}-\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \left({10}^2\right)+25\sqrt{3}$
$=\frac{25\pi }{2}-\frac{50\pi }{3}+25\sqrt{3}$
$=25\sqrt{3}-\frac{25\pi }{6}$
$=25[\sqrt{3}-\frac{\pi }{6}]$