A) \[6\sqrt{3}\]
B) \[2\pi -6\sqrt{3}\]
C) \[2\pi -3\sqrt{3}\]
D) \[3\sqrt{3}\]
Correct Answer: C
Solution :
Join DP and CP. Now, 3 sectors are formed each subtending an angle of \[{{60}^{o}}\] at P. Area of each sector \[=\frac{\theta }{{{360}^{o}}}\times \pi {{r}^{2}}\] \[=\frac{{{60}^{o}}}{{{360}^{o}}}\times \pi \times {{(2)}^{2}}=\frac{4\pi }{6}=\frac{2\pi }{3}c{{m}^{2}}\] And area of \[\Delta \,DCP=\frac{\sqrt{3}}{4}\,{{(PC)}^{2}}\] \[=\frac{\sqrt{3}}{4}\times 4=\sqrt{3}\,c{{m}^{2}}\] Therefore, area of shaded portion \[=\left( \frac{2\pi }{3}-\sqrt{3} \right)\times 3=2\pi -3\sqrt{3}\]You need to login to perform this action.
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