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MathsCBSE Class 10 · Standard · Board exam 2025, set 30/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2025: question 32(A)

Set in the CBSE Class 10 board exam of 2025, paper 30/4/1, section D, question 32(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(A) · 5 marks · Long answerQuadratic Equations

The sides of a right triangle are such that the longest side is 4 m more than the shortest side and the third side is 2 m less than the longest side. Find the length of each side of the triangle. Also, find the difference between the numerical values of the area and the perimeter of the given triangle.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/4/1.

Hints

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

Hint 1

Let the shortest side be and write the other two sides in terms of .

Hint 2

The longest side is the hypotenuse; apply Pythagoras and solve the quadratic.

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.

  1. Let the shortest side be m. Then longest side (hypotenuse) m and third side m.

    1 mark
  2. By Pythagoras theorem:

    1 mark
  3. (since is rejected, a length cannot be negative)

    1 mark
  4. Sides: 6 m, 8 m and 10 m.

    ½ mark
  5. Area m²

    ½ mark
  6. Perimeter m

    ½ mark
  7. Difference between numerical values

    ½ mark

Answer: Sides 6 m, 8 m, 10 m; difference = 0

Where the marks are

  • Expresses sides as , , (1 mark)
  • Pythagoras equation leading to (1 mark)
  • Solves to get (1 mark)
  • Sides 6, 8, 10 m (½ mark)
  • Area 24 m² (½ mark)
  • Perimeter 24 m (½ mark)
  • Difference 0 (½ mark)

Revise the chapter

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.