The sum of the areas of two squares is 52 cm² and difference of their perimeters is 8 cm. Find the lengths of the sides of the two squares.

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/2/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2025: question 33(a)

Set in the CBSE Class 10 board exam of 2025, paper 30/2/1, section D, question 33(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(a) · 5 marks · Long answerQuadratic Equations

The sum of the areas of two squares is $52$ cm $^{2}$ and difference of their perimeters is $8$ cm. Find the lengths of the sides of the two squares.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/2/1.

Hints

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

Hint 1

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

Hint 2

The perimeter of a square is $4 \times$ its side, so the second equation simplifies to $x - y = 2$ .

Hint 3

Put $x = y + 2$ into $x^{2} + y^{2} = 52$ and solve the quadratic in $y$ . Reject the negative value.

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 $x$ cm and $y$ cm, with $x > y$ .
Sum of areas: $x^{2} + y^{2} = 52$ ...(1)
1 mark
Difference of perimeters: $4 x - 4 y = 8 \Rightarrow x - y = 2 \Rightarrow x = y + 2$ ...(2)
1 mark
Substitute (2) in (1): $(y + 2)^{2} + y^{2} = 52$
$\Rightarrow 2 y^{2} + 4 y + 4 = 52 \Rightarrow 2 y^{2} + 4 y - 48 = 0 \Rightarrow y^{2} + 2 y - 24 = 0$
1 mark
Factorise by splitting the middle term: $y^{2} + 6 y - 4 y - 24 = 0 \Rightarrow (y + 6) (y - 4) = 0$
So $y = - 6$ or $y = 4$ .
1 mark
A side length cannot be negative, so $y = - 6$ is rejected. Hence $y = 4$ .
½ mark
From (2), $x = 4 + 2 = 6$ .
Check: $36 + 16 = 52$ ✓ and $24 - 16 = 8$ ✓
The sides of the two squares are 6 cm and 4 cm.
½ mark

Answer: Sides are 6 cm and 4 cm

Where the marks are

Writes the area equation $x^{2} + y^{2} = 52$ (1 mark)
Writes the perimeter equation $4 x - 4 y = 8$ , i.e. $x - y = 2$ (1 mark)
Forms the quadratic $y^{2} + 2 y - 24 = 0$ (1 mark)
Factorises it as $(y + 6) (y - 4) = 0$ (1 mark)
Rejects the negative root and gets $y = 4$ (½ mark)
Finds $x = 6$ and states both sides (½ mark)

Revise the chapter

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

Practise this chapterBoard-style questions from Quadratic Equations, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 30/2/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2025: question 33(a)

Hints

Step-by-step solution

Revise the chapter

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