A) \[364.5\]
B) \[693.5\]
C) \[346.5\]
D) \[639.5\]
Correct Answer: C
Solution :
Area of the equilateral triangle\[=\frac{\sqrt{3}}{4}\text{sid}{{\text{e}}^{2}}\] \[\Rightarrow \]\[121\sqrt{3}=\frac{\sqrt{3}}{4}\times \text{sid}{{\text{e}}^{2}}\] \[\therefore \]\[\text{Sid}{{\text{e}}^{\text{2}}}=\frac{121\sqrt{3}\times 4}{\sqrt{3}}=121\times 4\] \[\therefore \]Side\[=\sqrt{121\times 4}\] \[=11\times 2=22\,\,\text{cm}\] \[\therefore \]Total length of wire\[=3\times 22=66\,\,\text{cm}\] If the radius of the circle be \[r\]cm, then \[2\pi r=66\] \[\Rightarrow \] \[\frac{2\times 22}{7}\times r=66\] \[\Rightarrow \] \[r=\frac{66\times 7}{2\times 22}=\frac{21}{2}\text{cm}\] \[\therefore \]Area of the circle\[=\pi {{r}^{2}}\] \[=\frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}=346.5\,\,\text{c}{{\text{m}}^{\text{2}}}\]
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