Question 385 marksLong AnswerSection D

OR

The sum of the areas of two squares is 640 m2640 \text{ m}^2. If the difference in their perimeters is 64 m, find the sides of the two squares.

Quadratic Equations
Areas and Perimeters of Squares (Quadratic Equations)

Official answer

The sides of the two squares are 24 m and 8 m, found by substituting x=y+16x = y + 16 into x2+y2=640x^2 + y^2 = 640 and rejecting the negative root.

Answer keywords:sum of areas of squaresdifference of perimetersquadratic equationside of a squaresubstitution methodrejecting negative rootmensuration word problem

Marking scheme

  1. 1.1 mark: correctly setting up x2+y2=640x^2 + y^2 = 640 from the area condition.
  2. 2.1 mark: correctly setting up x−y=16x - y = 16 from the perimeter condition (or equivalent 4x−4y=644x - 4y = 64).
  3. 3.1 mark: substituting x=y+16x = y + 16 into the area equation and simplifying to y2+16y−192=0y^2 + 16y - 192 = 0.
  4. 4.1 mark: solving the quadratic correctly to get y=8y = 8 (rejecting y=−24y = -24 with reasoning).
  5. 5.1 mark: correct final sides with units (24 m and 8 m), verification earns full credit.

Hint

Let sides be x (larger) and y (smaller); use x−y=16x - y = 16 from the perimeter condition to substitute into x2+y2=640x^2 + y^2 = 640.

Quick oral answer

Taking the sides as x and y, the perimeter condition gives x−y=16x - y = 16, and substituting x=y+16x = y + 16 into x2+y2=640x^2 + y^2 = 640 gives the quadratic y2+16y−192=0y^2 + 16y - 192 = 0, solving which y=8y = 8 and x=24x = 24, so the sides are 24 m and 8 m.

Common mistakes

  1. 1.Writing the perimeter formula incorrectly (using 2×side2\times \text{side} instead of 4×side4\times \text{side} for a square's perimeter).
  2. 2.Not rejecting the negative root y=−24y = -24, or presenting both roots as valid answers.
  3. 3.Sign errors while expanding (y+16)2(y + 16)^2, commonly missing the middle term 32y32y.

Analysis and explanation

A dual-condition mensuration problem combining Area=side2\text{Area} = \text{side}^2 and Perimeter=4×side\text{Perimeter} = 4 \times \text{side} with quadratic solving.


Concept & method

  • Use the linear condition (perimeter difference) to express one side in terms of the other, then substitute into the quadratic area condition — far simpler than solving two quadratics together.
  • Perimeter difference: x−y=16x - y = 16; Area sum: x2+y2=640x^2 + y^2 = 640.
  • Discriminant 1024=3221024 = 32^2 is engineered to be a perfect square, giving whole-number roots.

Common mistakes

  • Trying to solve both equations as simultaneous quadratics instead of substituting the simpler linear relation first.
  • Not rejecting the negative root y=−24y = -24, since a side length cannot be negative — examiners specifically check for this.

Real-world connection

  • Mirrors real design constraints, e.g., an architect fixing a combined plot area and a fixed size difference between two square plots.

Interesting facts

This 'sum of areas / difference of perimeters' problem format is a recurring CBSE favourite because squaring the linear perimeter relation always yields a perfect-square discriminant when the given numbers are chosen carefully, as here (1024=3221024 = 32^2).

The two squares in this problem, with sides 24 m and 8 m, are in the ratio 3:1 — a clean ratio that is often used deliberately in textbook problems to make verification straightforward.

Quadratic modelling of square plots is directly used in real estate and urban planning when subdividing land into square plots of specific combined area under given perimeter/fencing constraints.

Question details

Q38 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section D
Marks
5
Type
Long Answer
Chapter
Quadratic Equations
Topic
Areas and Perimeters of Squares (Quadratic Equations)
Bloom's level
apply
Syllabus topic
Unit 2: Algebra — Quadratic Equations: Application to word problems (mensuration)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why substitute x = y + 16 instead of solving both equations as quadratics simultaneously?

The perimeter condition is linear, so expressing x in terms of y and substituting into the quadratic area equation reduces the problem to a single-variable quadratic, which is far simpler than solving two quadratics together.

Why is y = −24 rejected?

y represents the side length of a square, which must be a positive real number; a negative value has no geometric meaning and is therefore discarded.

How can the answer be quickly verified?

Substitute x=24x = 24 and y=8y = 8 back into both original conditions: 242+82=64024^2 + 8^2 = 640 (area condition) and 4×24−4×8=644\times 24 - 4\times 8 = 64 (perimeter condition), both of which check out.