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The sum of the areas of two squares is $52$ cm$^2$ and difference of their perimeters is $8$ cm. Find the lengths of the sides of the two squares.
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Let the lengths of the sides of two squares be 'x' cm and 'y' cm such that $x > y$.
ATQ
$x^2 + y^2 = 52$ ----- (1)
$4x - 4y = 8$ or $x - y = 2$ ----- (2)
From (1) and (2), we have
$y^2 + 2y - 24 = 0$
$\Rightarrow (y + 6) (y - 4) = 0$
$\therefore y = 4$
So, $x = 2 + 4 = 6$
$\therefore$ Lengths of the sides of two squares are $6$ cm and $4$ cm respectively.
ATQ
$x^2 + y^2 = 52$ ----- (1)
$4x - 4y = 8$ or $x - y = 2$ ----- (2)
From (1) and (2), we have
$y^2 + 2y - 24 = 0$
$\Rightarrow (y + 6) (y - 4) = 0$
$\therefore y = 4$
So, $x = 2 + 4 = 6$
$\therefore$ Lengths of the sides of two squares are $6$ cm and $4$ cm respectively.