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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Step 1: Let the lengths of the sides of two squares be $x$ cm and $y$ cm such that $x > y$.
$x^2 + y^2 = 52$ ... (1)
$4x - 4y = 8$ or $x - y = 2$ ... (2)
Step 2: From (1) and (2), we have
$x = y + 2$
$(y+2)^2 + y^2 = 52$
$y^2 + 4y + 4 + y^2 = 52$
$2y^2 + 4y - 48 = 0$
Step 3: $y^2 + 2y - 24 = 0$
$\Rightarrow (y + 6)(y - 4) = 0$
Step 4: $	herefore y = 4$ (rejecting $y = -6$ as length cannot be negative)
Step 5: So $x = 4 + 2 = 6$.
$	herefore$ Lengths of the sides of two squares are 6 cm and 4 cm respectively.

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