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 \[\Delta \Beta CE\] \[=\frac{1}{2}\times {{(\sqrt{3})}^{2}}\times \frac{\pi }{3}-\frac{\sqrt{3}}{2}.\frac{3}{2}+\frac{1}{2}\times 1\times \frac{2\pi }{3}\] \[-\frac{\sqrt{3}}{2}.\frac{1}{2}=\frac{5\pi -6\sqrt{3}}{6}\]You need to login to perform this action.
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