If an equilateral triangle of area X and a square of area Y have the same perimeter, then X is :

Answer: Option C

Let the side of the triangle be a cm and each side of the square be b cm
Then,
$$\begin{array}{rl}& X=\frac{\sqrt{3}}{4}{a}^2\text{ and }Y=b^2\\ & \text{Where }3a=4b,i,e.,b=\frac{3a}{4}\end{array}$$
$$\therefore X=\frac{\sqrt{3}{a}^2}{4}\text{ and }Y=\frac{9a^2}{16}$$     $$\left[\because b=\frac{3a}{4}\right]$$
$$\begin{array}{rl}& \text{Now,}\\ & \frac{\sqrt{3}{a}^2}{4}=\frac{1.732a^2}{4}=0.433a^2\\ & \text{And }\frac{9a^2}{16}=0.5625a^2\\ & \therefore X<Y\end{array}$$