(B) The sum of the areas of two squares is 640 m². If the difference in their perimeters is 64 m, find the sides of the two squares.

Quadratic Equations

Word problem on areas of squares

Official Answer

The sides of the two squares are $24\text{ m}$ and $8\text{ m}$ .

quadratic equationarea of squareperimeter of squaresum of areas 640difference of perimeters 64sides 24 m and 8 msubstitution methodreject negative root

Marking Scheme

11 mark: assuming sides as a and b and correctly writing the perimeter condition $4a - 4b = 64 \Rightarrow a - b = 16$ .
21 mark: writing the area condition $a^2 + b^2 = 640$ and substituting $a = b + 16$ .
31 mark: reducing to the quadratic $b^2 + 16b - 192 = 0$ (after dividing by 2).
41 mark: solving to get $b = 8$ (rejecting $b = -24$ with reason).
51 mark: stating both sides — 24 m and 8 m (verification acceptable for the final mark).

Hint

From the perimeter condition, $4a - 4b = 64$ gives $a - b = 16$ ; substitute $a = b + 16$ into $a^2 + b^2 = 640$ and solve the resulting quadratic.

Quick Oral Answer

Let the sides be a and b; the perimeter difference gives $a - b = 16$ and the area sum gives $a^2 + b^2 = 640$ , so substituting $a = b + 16$ yields $b^2 + 16b - 192 = 0$ , giving $b = 8$ and $a = 24$ , i.e. sides of 24 m and 8 m.

Analysis & Explanation

A mensuration + quadratic-equation problem linking the areas and perimeters of two squares.

Concept

Perimeter of a square = 4 × side, so the linear condition $4a - 4b = 64$ simplifies quickly to $a - b = 16$ .
Substituting $a = b + 16$ into the area condition $a^2 + b^2 = 640$ gives $2b^2 + 32b - 384 = 0$ , which reduces (÷2) to $b^2 + 16b - 192 = 0$ , factorising to $(b-8)(b+24) = 0$ .

Common Mistakes

Confusing "difference of perimeters" with "difference of sides" and writing a − b = 64 instead of $4a - 4b = 64$ .
Skipping the division by the common factor 2, which makes the arithmetic heavier and error-prone.
Forgetting to reject the negative root $b = -24$ since a side length must be positive.

Real-World Relevance

Mirrors design situations where a fixed total material area and a fixed boundary-length difference constrain the dimensions of two panels.
Verification: $24^2 + 8^2 = 576 + 64 = 640\text{ m}^2$ and $4(24) - 4(8) = 64\text{ m}$ , confirming both conditions.

Common Mistakes

1Writing $a - b = 64$ (difference of sides) instead of $4a - 4b = 64$ (difference of perimeters), which uses the wrong linear relation and gives incorrect sides.
2Not dividing $2b^2 + 32b - 384 = 0$ by the common factor 2, leading to heavier numbers and factorisation errors.
3Accepting the negative root $b = -24$ or forgetting to reject it with the reason that a side length cannot be negative.

Interesting Facts

The identity being tested here, $a^2 + b^2$ with a linear constraint, is the same structure used in the Pythagorean theorem — many area-based quadratic problems reduce to a sum of squares equal to a constant.

Because perimeter of a square scales linearly with side while area scales with the square of the side, a difference of only 16 m in side length produces a large 512 m² difference in area (576 − 64) — a vivid illustration of why area grows quadratically.

Problems combining 'sum of areas' and 'difference of perimeters' appear regularly in CBSE because they force students to use one condition to eliminate a variable and the other to form the quadratic — testing two skills at once.

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Frequently Asked Questions

Why do we use the perimeter condition to eliminate a variable rather than the area condition?

The perimeter condition $4a - 4b = 64$ is linear, so it easily rearranges to $a = b + 16$ , letting us express one side in terms of the other. Substituting this into the area (which is quadratic) then leaves a single quadratic in one variable. Starting with the area equation, which has two squared terms, would be far harder to reduce.

How do I check my answer is correct?

Substitute back into both original conditions. Areas: $24^2 + 8^2 = 576 + 64 = 640\text{ m}^2$ ✓. Perimeter difference: $4 \times 24 - 4 \times 8 = 96 - 32 = 64\text{ m}$ ✓. Since both conditions hold, the sides 24 m and 8 m are correct.

Is it necessary to divide the quadratic by 2 before factorising?

It is not strictly necessary, but strongly recommended. Dividing $2b^2 + 32b - 384 = 0$ by 2 gives $b^2 + 16b - 192 = 0$ with smaller coefficients that factorise cleanly into $(b-8)(b+24)$ . Smaller numbers reduce the chance of arithmetic mistakes.