question_answer

The area of an equilateral triangle, inscribed in a circle is\[4\sqrt{3}\,\,\text{c}{{\text{m}}^{2}}.\]The area of the circle in terms of \[\pi \] will be

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\[=\pi \times \frac{4}{\sqrt{3}}\times \frac{4}{\sqrt{3}}=\frac{16}{3}\pi \]

\[=5\frac{1}{3}\pi \,\,\text{sq}\,\,\text{cm}\]