Question

Two intersecting circles have their radii $1$ and $\sqrt{3}$ meters. The distance between their centres is $2$ meters. Then the overlapping area in sq. meters is

• A. $\frac{19\pi +6\sqrt{3}}{6}$
• B. $\frac{5\pi +6\sqrt{3}}{6}$
• C. $\frac{\pi }{6}$
• D. $\frac{5\pi -6\sqrt{3}}{6}$

Answer 1

Answer: (D)

$\frac{5\pi -6\sqrt{3}}{6}$

Let $r_1=1$ m; $r_2=\sqrt{3}$ m; $d=2$ m

$\sin \frac{\theta }{2}=\frac{\sqrt{3}}{2}$

$⟹\frac{\theta }{2}={\sin }^{-1}\frac{\sqrt{3}}{2}$

$⟹\frac{\theta }{2}=\frac{\pi }{3}$

$⟹$$\theta =\frac{2\pi }{3}$

Now, $\sin \frac{\alpha }{2}=\frac{1}{2}$

$⟹$$\frac{\alpha }{2}={\sin }^{-1}\frac{1}{2}$

$⟹$$\frac{\alpha }{2}=\frac{\pi }{6}$

$⟹$$\alpha =\frac{\pi }{3}$

Area of sector = $\frac{1}{2}{r}^2\theta$

$A_1=\frac{1}{2}{r}_1^2\theta$

$=\frac{1}{2}\times \frac{2\pi }{3}$

$=\frac{\pi }{3}$

And area of triangle, ${\Delta }_1=\frac{1}{2}\times \frac{\sqrt{3}}{2}\times \frac{1}{2}=\frac{\sqrt{3}}{4}$

$A_2$ = $\frac{1}{2}{r}_2^2\alpha$

$=\frac{1}{2}\times 3\times \frac{\pi }{3}$

$=\frac{\pi }{2}$

${\Delta }_2=\frac{1}{2}\times \sqrt{3}\times \frac{3}{2}=\frac{3\sqrt{3}}{4}$

Then the area of intersection of two circles $=(A_1-{\Delta }_1)+(A_2-{\Delta }_2$)

Therefore Area= $\frac{5\pi -6\sqrt{3}}{6}$

Answer is (D)