A) \[\frac{\pi {{S}^{2}}}{9}\]
B) \[\frac{3{{S}^{2}}}{\pi }\]
C) \[\frac{3S}{\pi }\]
D) \[\frac{3\sqrt{3}S}{\pi }\]
Correct Answer: D
Solution :
Let a be the side of equilateral triangle, then its area \[\frac{\sqrt{3}}{4}{{a}^{2}}=S\Rightarrow {{a}^{2}}=\frac{4S}{\sqrt{3}}\Rightarrow a=\frac{2\sqrt{S}}{{{3}^{1/4}}}\] \[\therefore \] The perimeter of the triangle \[=3a=2\sqrt{S}{{.3}^{3/4}}\] This perimeter must be equal to the perimeter of the circle formed by the triangle. If r denotes the radius of the circle, then \[2\pi r=2\sqrt{S}{{.3}^{\frac{3}{4}}}\] \[\therefore \] \[r=\frac{1}{\pi }\sqrt{S}{{3}^{3/4}}\] Hence, the area of the circle \[=\pi {{r}^{2}}=\pi \times \frac{1}{{{\pi }^{2}}}.S{{.3}^{3/2}}=\frac{3\sqrt{3}\,S}{\pi }\]
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