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.

MathsCBSE Class 10 · Standard · Board exam 2026, set 30/1/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2026: question 33(B)

Set in the CBSE Class 10 board exam of 2026, paper 30/1/1, section D, question 33(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(B) · 5 marks · Long answerQuadratic Equations

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.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Call the sides $x$ and $y$ . Write one equation for the areas and one for the perimeters.

Hint 2

The perimeter equation simplifies to $x - y = 16$ . Substitute $y = x - 16$ into $x^{2} + y^{2} = 640$ .

Hint 3

You get the quadratic $x^{2} - 16 x - 192 = 0$ . Factorise it and reject the negative root.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Let the sides of the two squares be $x$ m and $y$ m, with $x > y$ .
The sum of the areas is 640 m², so
$x^{2} + y^{2} = 640 ...(1)$
1 mark
The perimeters are $4 x$ and $4 y$ , and their difference is 64 m:
$4 x - 4 y = 64 \Rightarrow x - y = 16 \Rightarrow y = x - 16 ...(2)$
1 mark
Substitute (2) in (1):
$x^{2} + (x - 16)^{2} = 640$
$2 x^{2} - 32 x + 256 - 640 = 0$
$x^{2} - 16 x - 192 = 0$
1 mark
Split the middle term ( $- 24 \times 8 = - 192$ and $- 24 + 8 = - 16$ ):
$x^{2} - 24 x + 8 x - 192 = 0 \Rightarrow (x - 24) (x + 8) = 0$
1 mark
So $x = 24$ or $x = - 8$ .
A side cannot be negative, so $x = - 8$ is rejected. Hence $x = 24$ .
½ mark
Then $y = 24 - 16 = 8$ .
Check: $2 4^{2} + 8^{2} = 576 + 64 = 640$ ✓, and $4 (24) - 4 (8) = 64$ ✓.
The sides of the two squares are 24 m and 8 m.
½ mark

Answer: Sides are 24 m and 8 m

Where the marks are

Writes the area equation $x^{2} + y^{2} = 640$ (1 mark)
Uses the perimeter condition to get $y = x - 16$ (1 mark)
Forms the quadratic $x^{2} - 16 x - 192 = 0$ (1 mark)
Factorises as $(x - 24) (x + 8) = 0$ (1 mark)
Takes $x = 24$ and rejects $- 8$ (½ mark)
Finds $y = 8$ and states both sides (½ mark)

Revise the chapter

Chapter: Quadratic Equations (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Quadratic Equations, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 30/1/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2026: question 33(B)

Hints

Step-by-step solution

Revise the chapter

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