A) \[\left( \frac{\sqrt{3}+\pi }{2} \right)\,\,{{a}^{2}}\,\,sq.cm\]
B) \[\left( \frac{6\sqrt{3-\pi }}{2} \right)\,\,{{a}^{2}}\,\,sq.cm\]
C) \[(\sqrt{3}-\pi )\,\,{{a}^{2}}\,\,sq.cm\]
D) \[\left( \frac{2\sqrt{3-\pi }}{2} \right)\,\,{{a}^{2}}\,\,sq.cm\]
Correct Answer: D
Solution :
\[AB=BC=CA=2a\,\,cm.\] \[\angle BAC=\angle ACB=\angle ABC=60{}^\circ \] Area of \[\Delta ABC=\frac{\sqrt{3}}{4}\times {{(side)}^{2}}\] \[=\frac{\sqrt{3}}{4}\times 4{{a}^{2}}=\sqrt{3}{{a}^{2}}\,\,sq.cm.\] Area of three sectors \[=3\times \frac{60}{360}\times \pi \times {{a}^{2}}\] \[=\frac{\pi {{a}^{2}}}{2}\,\,sq.cm.\] Area of the shaded region \[=\sqrt{3}{{a}^{2}}-\frac{\pi }{2}{{a}^{2}}=\left( \frac{2\sqrt{3}-\pi }{2} \right)\,\,{{a}^{2}}\,\,sq.cm.\]
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