Question 385 marksLong AnswerSection D

  1. (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 m24\text{ m} and 8 m8\text{ m}.

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

Marking scheme

  1. 1.1 mark: assuming sides as a and b and correctly writing the perimeter condition 4a−4b=64⇒a−b=164a - 4b = 64 \Rightarrow a - b = 16.
  2. 2.1 mark: writing the area condition a2+b2=640a^2 + b^2 = 640 and substituting a=b+16a = b + 16.
  3. 3.1 mark: reducing to the quadratic b2+16b−192=0b^2 + 16b - 192 = 0 (after dividing by 2).
  4. 4.1 mark: solving to get b=8b = 8 (rejecting b=−24b = -24 with reason).
  5. 5.1 mark: stating both sides — 24 m and 8 m (verification acceptable for the final mark).

Hint

From the perimeter condition, 4a−4b=644a - 4b = 64 gives a−b=16a - b = 16; substitute a=b+16a = b + 16 into a2+b2=640a^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=16a - b = 16 and the area sum gives a2+b2=640a^2 + b^2 = 640, so substituting a=b+16a = b + 16 yields b2+16b−192=0b^2 + 16b - 192 = 0, giving b=8b = 8 and a=24a = 24, i.e. sides of 24 m and 8 m.

Common mistakes

  1. 1.Writing a−b=64a - b = 64 (difference of sides) instead of 4a−4b=644a - 4b = 64 (difference of perimeters), which uses the wrong linear relation and gives incorrect sides.
  2. 2.Not dividing 2b2+32b−384=02b^2 + 32b - 384 = 0 by the common factor 2, leading to heavier numbers and factorisation errors.
  3. 3.Accepting the negative root b=−24b = -24 or forgetting to reject it with the reason that a side length cannot be negative.

Analysis and 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=644a - 4b = 64 simplifies quickly to a−b=16a - b = 16.
  • Substituting a=b+16a = b + 16 into the area condition a2+b2=640a^2 + b^2 = 640 gives 2b2+32b−384=02b^2 + 32b - 384 = 0, which reduces (÷2) to b2+16b−192=0b^2 + 16b - 192 = 0, factorising to (b−8)(b+24)=0(b-8)(b+24) = 0.

Common Mistakes

  • Confusing "difference of perimeters" with "difference of sides" and writing a − b = 64 instead of 4a−4b=644a - 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=−24b = -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: 242+82=576+64=640 m224^2 + 8^2 = 576 + 64 = 640\text{ m}^2 and 4(24)−4(8)=64 m4(24) - 4(8) = 64\text{ m}, confirming both conditions.

Interesting facts

The identity being tested here, a2+b2a^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.

Question details

Q38 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section D
Marks
5
Type
Long Answer
Chapter
Quadratic Equations
Topic
Word problem on areas of squares
Bloom's level
apply
Syllabus topic
Unit 2: Algebra — Quadratic Equations: Application (word problems involving areas and perimeters)
Author
CBSE
Reviewed by
Padhantu Academic Team

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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=644a - 4b = 64 is linear, so it easily rearranges to a=b+16a = 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: 242+82=576+64=640 m224^2 + 8^2 = 576 + 64 = 640\text{ m}^2 ✓. Perimeter difference: 4×24−4×8=96−32=64 m4 \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 2b2+32b−384=02b^2 + 32b - 384 = 0 by 2 gives b2+16b−192=0b^2 + 16b - 192 = 0 with smaller coefficients that factorise cleanly into (b−8)(b+24)(b-8)(b+24). Smaller numbers reduce the chance of arithmetic mistakes.