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?

Solution (By Examveda Team)

Let radius of circle = R
Side of square = a
Side of equilateral Δ = b
According to question,
$$\eqalign{ & 2\pi R = 4a = 3b \cr & \therefore a = \frac{{\pi R}}{2} \cr & b = \frac{2}{3}\pi R \cr} $$
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