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

Solution(By Examveda Team)

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