Mathematics — Standard (041) — CBSE Class 10 board question

Let the sides of the two squares be $x$ cm and $y$ cm, where $x > y$.

Setting up equations:

Sum of areas: $x^2 + y^2 = 52$ … (1)

Difference of perimeters: $4x - 4y = 8 \Rightarrow x - y = 2 \Rightarrow x = y + 2$ … (2)

Substituting (2) into (1):

$(y + 2)^2 + y^2 = 52$

$y^2 + 4y + 4 + y^2 = 52$

$2y^2 + 4y - 48 = 0$

$y^2 + 2y - 24 = 0$

Factorising:

$y^2 + 6y - 4y - 24 = 0$

$y(y + 6) - 4(y + 6) = 0$

$(y - 4)(y + 6) = 0$

$y = 4$ or $y = -6$

Since side length cannot be negative, $y = 4$ cm.

Therefore, $x = y + 2 = 6$ cm.

The sides of the two squares are 6 cm and 4 cm.

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• Examiners expect you to clearly define variables, form two equations, substitute to get a quadratic, and solve by factorisation (preferred method in Chapter 4).
• Rejecting the negative root with a reason earns a mark — never skip this step.
• State the final answer clearly; many students lose the concluding mark by not writing it explicitly.