If an equilateral triangle of area X and a sq | vedclass

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(C) Let the side of the equilateral triangle be $a$ and the side of the square be $b$.The area of the triangle is $X = \frac{\sqrt{3}}{4} a^2$ and its

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less than $Y$

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(C) Let the side of the equilateral triangle be $a$ and the side of the square be $b$.The area of the triangle is $X = \frac{\sqrt{3}}{4} a^2$ and its perimeter is $P = 3a$.The area of the square is $Y = b^2$ and its perimeter is $P = 4b$.Since both have the same perimeter,$3a = 4b$,which implies $b = \frac{3a}{4}$.Now,$Y = b^2 = (\frac{3a}{4})^2 = \frac{9a^2}{16}$.From the area of the triangle,we have $a^2 = \frac{4X}{\sqrt{3}}$.Substituting this into the expression for $Y$,we get $Y = \frac{9}{16} 	imes \frac{4X}{\sqrt{3}} = \frac{9X}{4\sqrt{3}} = \frac{3\sqrt{3}X}{4}$.Since $\sqrt{3} \approx 1.732$,we have $Y \approx \frac{3 	imes 1.732}{4} X = 1.299X$.Therefore,$Y > X$,which means $X < Y$.

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

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