The perimeters of a circle, a square and an equilateral triangle are same and their areas are C, S and T respectively. Which of the following statement is true?

Answer: Option B

Let radius of circle = R
Side of square = a
Side of equilateral Δ = b
According to question,
$$\begin{array}{rl}& 2\pi R=4a=3b\\ & \therefore a=\frac{\pi R}{2}\\ & b=\frac{2}{3}\pi R\end{array}$$
Ratio of their areas
\begin{array}{*{20}{c}} {\pi {R^2}}&:&{{a^2}}&:&{\frac{{\sqrt 3 }}{4}{b^2}} \\ {\pi {R^2}}&:&{{{\left( {\frac{{\pi R}}{2}} \right)}^2}}&:&{\frac{{\sqrt 3 }}{4}{{\left( {\frac{2}{3}\pi R} \right)}^2}} \\ 1&:&{\frac{\pi }{4}}&:&{\frac{{\sqrt 3 }}{9}\pi } \\ {\text{C}}&:&{\text{S}}&:&{\text{T}} \end{array}
Here, we can see that C > S > T
Quicker Approach: When perimeter of two or more figures are same then the figure which has more vertex is greater in the area. Since, here, circle has infinite vertex.
Therefore, C > S > T