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

Let the side of the first square be $x$ m and the second square be $y$ m, where $x > y$.

Setting up equations:

Sum of areas: $x^2 + y^2 = 640$ … (1)

Difference of perimeters: $4x - 4y = 64 \Rightarrow x - y = 16 \Rightarrow x = y + 16$ … (2)

Substituting (2) in (1):

$(y + 16)^2 + y^2 = 640$

$y^2 + 32y + 256 + y^2 = 640$

$2y^2 + 32y - 384 = 0$

$y^2 + 16y - 192 = 0$

Factorising:

$y^2 + 24y - 8y - 192 = 0$

$y(y + 24) - 8(y + 24) = 0$

$(y - 8)(y + 24) = 0$

$y = 8$ or $y = -24$

Since side cannot be negative, $y = 8$ m.

Therefore, $x = 8 + 16 = 24$ m.

The sides of the two squares are 24 m and 8 m.

Source: Chapter 4, Quadratic Equations

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• Examiners award marks at each step: forming both equations (1 mark), substituting and simplifying to a standard quadratic (1–2 marks), correct factorisation (1 mark), and final answer with rejection of the negative root (1 mark).
• Always state why the negative value is rejected — this is a common mark-earning step.
• The key skill tested is translating word conditions into algebraic equations and solving by factorisation.