The sum of the areas of two squares is 640 m$^2$. If the difference in their perimeters is 64 m, find the sides of the two squares.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
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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Explanation
- 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.