A) \[\left( 3\sqrt{3}-\frac{\pi }{2} \right)\]\[c{{m}^{2}}\]
B) \[\left( \sqrt{3}-\frac{3\pi }{2} \right)\]\[c{{m}^{2}}\]
C) \[4\left( \sqrt{3}-\frac{\pi }{2} \right)\]\[c{{m}^{2}}\]
D) \[\left( \frac{\pi }{2}-\sqrt{3} \right)\]\[c{{m}^{2}}\]
Correct Answer: C
Solution :
(c): Each angle of the triangle = \[{{60}^{{}^\circ }}\] Required area of the three sectors \[=3\times \frac{60}{360}\times \pi {{(2)}^{2}}=2\pi c{{m}^{2}}\] Area of triangle \[=\frac{\sqrt{3}}{4}\times 16=4\sqrt{3}c{{m}^{2}}\] \[\therefore \]Required Area \[=(4\sqrt{3}-2\pi )c{{m}^{2}}\]You need to login to perform this action.
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