Sum of the areas of two squares is $468\text{ m}^2$. If the difference of their perimeters is $24\text{ m}$, find the sides of the two squares.
Step-by-Step Solution
Key Concept: Let sides of two squares be $x$ m and $y$ m ($x > y$). Difference of perimeters $4x - 4y = 24 \Rightarrow x - y = 6$. Sum of areas $x^2 + y^2 = 468$.
$4x - 4y = 24 \Rightarrow x - y = 6 \Rightarrow x = y + 6$. [0.5 Mark]
Substitute $x = y + 6$ into $x^2 + y^2 = 468$: $(y + 6)^2 + y^2 = 468 \Rightarrow y^2 + 12y + 36 + y^2 = 468$. [1.0 Mark]
$2y^2 + 12y - 432 = 0 \Rightarrow y^2 + 6y - 216 = 0 \Rightarrow (y + 18)(y - 12) = 0$. [1.0 Mark]
Since side cannot be negative, $y = 12$ m. Then $x = 12 + 6 = 18$ m. Sides of squares are $18\text{ m}$ and $12\text{ m}$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Expressing $x = y + 6$ from perimeter difference: 0.5 Mark
Substituting into area equation to get $y^2 + 6y - 216 = 0$: 1.0 Mark
Solving quadratic for $y = 12$: 1.0 Mark
Stating sides 18 m and 12 m: 0.5 Mark
Correct Answer: