question_answer

Two intersecting circles have their radii and 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)  \[\frac{5\pi -6\sqrt{3}}{6}\]

Correct Answer: D

Solution :

\[\cos {{30}^{o}}=\frac{AD}{AC}\] \[\Rightarrow \]               \[AD=AC\cos {{30}^{o}}\] \[=\frac{\sqrt{3}}{2}+\sqrt{3}=\frac{3}{2}\]  \[\therefore \]\[DB=1-\frac{3}{2}=\frac{1}{2}\] \[\sin {{30}^{o}}=\frac{CD}{AC}\]   \[\therefore \] \[CD=AC\times \sin {{30}^{o}}\] \[=\sqrt{3}\times \frac{1}{2}=\frac{\sqrt{3}}{2}\] \[\therefore \]  \[DE=\sqrt{3}/2\] Required area = area of sector \[ACE-\]area of \[\Delta ACE+\]area of sector \[BCE-\]area of \[\Delta \Beta CE\] \[=\frac{1}{2}\times {{(\sqrt{3})}^{2}}\times \frac{\pi }{3}-\frac{\sqrt{3}}{2}.\frac{3}{2}+\frac{1}{2}\times 1\times \frac{2\pi }{3}\] \[-\frac{\sqrt{3}}{2}.\frac{1}{2}=\frac{5\pi -6\sqrt{3}}{6}\]