SSC CGL 26 Oct 2014 Shift 2 Paper

Radius of circle = x cm.
Side of square = y cm.
Side of equilateral triangle = z cm.
Circumference of circle = Perimeter of square = Perimeter of equilateral triangle
⇒ 2 p x = 4 y = 3 z
⇒ x = $\frac{4\mathit{y}}{2\mathit{\pi }}$ = $\frac{2\mathit{y}}{\mathit{\pi }}$
z = $\frac{4\mathit{y}}{3}$
Area of circle ‘C’ = π ${\mathit{x}}^2$
= π × $\frac{4}{{\mathit{\pi }}^2}{\mathit{y}}^2$ = $\frac{4}{\mathit{\pi }}{\mathit{y}}^2$ > ${\mathit{y}}^2$
Area of square ‘S’ = ${\mathit{y}}^2$
Area of triangle ‘T’ = $\frac{\sqrt{3}}{4}{\mathit{z}}^2$
= $\frac{\sqrt{3}}{4}$ × $\frac{4\times 4}{3\times 3}{\mathit{y}}^2$
= $\frac{4}{3\sqrt{3}}{\mathit{y}}^2$ < ${\mathit{y}}^2$
∴ T < S < C