A) \[4\sqrt{6}\pi \,\,\text{c}{{\text{m}}^{2}}\]
B) \[5\pi \,\,\text{c}{{\text{m}}^{2}}\]
C) \[5\frac{1}{3}\pi \,\,\text{c}{{\text{m}}^{2}}\]
D) \[6\pi \,\,\text{c}{{\text{m}}^{2}}\]
Correct Answer: C
Solution :
Area of the equilateral triangle\[=\frac{\sqrt{3}}{4}\times \text{Sid}{{\text{e}}^{\text{2}}}\] |
\[\Rightarrow \] \[4\sqrt{3}=\frac{\sqrt{3}}{4}\times \text{Sid}{{\text{e}}^{\text{2}}}\] |
\[\therefore \] Side = 4 cm |
\[\therefore \]Height of the equilateral triangle\[=\frac{\sqrt{3}}{2}\times 4\] |
\[=2\sqrt{3}\,\,\text{cm}\] |
Radius of the circumcircle \[=\frac{2}{3}\times 2\sqrt{3}=\frac{4}{\sqrt{3}}\,\,\text{cm}\] |
\[\therefore \] Area of the circle\[=\pi \times \frac{4}{\sqrt{3}}\times \frac{4}{\sqrt{3}}=\frac{16}{3}\pi \] |
\[=5\frac{1}{3}\pi \,\,\text{sq}\,\,\text{cm}\] |
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