MathsCBSE Class 10 · Basic · Board exam 2025, set 430/5/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2025: question 33(b)

Set in the CBSE Class 10 board exam of 2025, paper 430/5/1, 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 answerTriangles

It is given that sides AB and AC and median AD of $△ A B C$ are respectively proportional to sides PQ and PR and median PM of another $△ P QR$ . Show that $△ A B C \sim △ P QR$ .

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/5/1.

Hints

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

Hint 1

Try doubling each median: extend AD to E with DE = AD (and PM to N with MN = PM).

Hint 2

Show a pair of congruent triangles to move side AB to CE, then use SSS similarity on △CAE and △RPN to get equal angles at A and P.

Hint 3

Once ∠BAC = ∠QPR, use the SAS similarity criterion with AB/PQ = AC/PR.

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.

Given: $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ , where AD and PM are medians. To prove: △ABC ~ △PQR.
Construction (figure): produce AD to E so that DE = AD, and join CE; produce PM to N so that MN = PM, and join RN. Mark $∠ C A D = ∠1$ , $∠ R P M = ∠2$ , $∠ B A D = ∠3$ , $∠ QP M = ∠4$ .
1 mark
In △ADB and △EDC: AD = ED (construction), BD = CD (D is mid-point of BC), $∠ A D B = ∠ E D C$ (vertically opposite angles). So △ADB ≅ △EDC [SAS]. Similarly △PMQ ≅ △NMR.
1 mark
∴ AB = CE and PQ = NR [CPCT]
½ mark
From the given ratios, $\frac{C E}{N R} = \frac{A B}{P Q} = \frac{A D}{P M} = \frac{2 A D}{2 P M} = \frac{A E}{P N}$ and also $= \frac{A C}{P R}$ . So $\frac{C E}{N R} = \frac{A E}{P N} = \frac{A C}{P R}$ .
1 mark
∴ △CAE ~ △RPN [SSS similarity criterion] ⇒ $∠1 = ∠2$ (i.e. $∠ C A D = ∠ R P M$ ).
½ mark
Similarly, joining BE and QN (△ADC ≅ △EDB gives BE = AC, and PR = QN), we get △BAE ~ △QPN, so $∠3 = ∠4$ . Adding: $∠1 + ∠3 = ∠2 + ∠4$ , i.e. $∠ B A C = ∠ QP R$ .
½ mark
Now in △ABC and △PQR: $\frac{A B}{P Q} = \frac{A C}{P R}$ and included $∠ B A C = ∠ QP R$ . Hence △ABC ~ △PQR [SAS similarity criterion].
½ mark

Answer: △ABC ~ △PQR (proved via △CAE ~ △RPN by SSS, giving ∠BAC = ∠QPR, then SAS similarity).

Where the marks are

Correct figure with construction AD = DE, PM = MN (1 mark)
Proves △DAB ≅ △DEC and similarly △MPQ ≅ △MNR (1 mark)
Gets AB = CE, PQ = NR by CPCT (½ mark)
Establishes $\frac{C E}{N R} = \frac{A E}{P N} = \frac{A C}{P R}$ (1 mark)
△CAE ~ △RPN by SSS (½ mark)
Obtains ∠1 = ∠2, ∠3 = ∠4 and adds to get ∠BAC = ∠QPR (½ mark)
Concludes △ABC ~ △PQR by SAS similarity (½ mark)

Revise the chapter

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

Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 430/5/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2025: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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