Question

Two circles, each of unit radius, are so drawn that the centre of each lies on the circumference of the other. The area of the region, common to both the circles, is

• A. $\left(\frac{4\pi -3\sqrt{3}}{12}\right)$
• B. $\left(\frac{4\pi -6\sqrt{3}}{12}\right)$
• C. $\left(\frac{4\pi -3\sqrt{3}}{6}\right)$
• D. $\left(\frac{4\pi -6\sqrt{3}}{6}\right)$

Answer 1

Answer: (C)

$\left(\frac{4\pi -3\sqrt{3}}{6}\right)$

Consider the $△aob$,
$ob=0.5$ units
$ab=1$ unit
$\Rightarrow oa=\sqrt{1^2-{0.5}^2}=\frac{\sqrt{3}}{2}$ units
Total area of triangles aob and boc $=\frac{\sqrt{3}}{4}$
Area of sector ab $=\frac{1}{2}{r}^2\angle abc=\frac{\pi }{3}$
Area of intersection $=2$(Area of sector $-$ Area of triangles)
$=2(\frac{\pi }{3}-\frac{\sqrt{3}}{4})$
$=\frac{4\pi -3\sqrt{3}}{6}$ sq. units