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

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

Set in the CBSE Class 10 board exam of 2026, paper 30/5/1, 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

Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that:
(i) AQ = QR
(ii) AP = 2PQ
(iii) PR = 2AP

Figure in the paper: Parallelogram ABCD with A bottom left, B bottom right, C upper right, D upper left. Q is the midpoint of side DC (tick marks). Diagonal BD is drawn, and line AR passes through Q and meets BD at P; R lies on the extension of BC beyond C. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/5/1.

Hints

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

Hint 1

Look for pairs of similar triangles formed because DC ∥ AB.

Hint 2

Use QC = DQ = ½AB since Q is the mid-point of DC and DC = AB.

Hint 3

For (iii), write AQ and QR in terms of AP, PQ and PR, then use result (ii).

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.

(i) $D C ∥ A B$ [opposite sides of parallelogram], so $QC ∥ A B$ . In $△ R QC$ and $△ R A B$ : $∠ R$ is common and $∠ R QC = ∠ R A B$ [corresponding angles]. So $△ R QC \sim △ R A B$ (AA) and $\frac{QR}{A R} = \frac{QC}{A B} = \frac{\frac{1}{2} D C}{D C} = \frac{1}{2}$ [Q mid-point of DC, AB = DC].
1 mark
$A R = 2 QR \Rightarrow A Q + QR = 2 QR \Rightarrow A Q = QR$ . Hence proved.
1 mark
(ii) In $△ P Q D$ and $△ P A B$ : $∠ QP D = ∠ A P B$ [vertically opposite] and $∠ P D Q = ∠ P B A$ [alternate angles, $D Q ∥ A B$ ]. So $△ P Q D \sim △ P A B$ (AA) and $\frac{QP}{A P} = \frac{D Q}{B A} = \frac{\frac{1}{2} A B}{A B} = \frac{1}{2}$ .
1 mark
Hence $A P = 2 P Q$ , i.e. $P Q = \frac{1}{2} A P$ .
½ mark
(iii) From (i), $A Q = QR \Rightarrow A P + P Q = P R - P Q$ .
½ mark
Substituting $P Q = \frac{1}{2} A P$ : $A P + \frac{1}{2} A P = P R - \frac{1}{2} A P \Rightarrow P R = 2 A P$ . Hence proved.
1 mark

Answer: (i) AQ = QR; (ii) AP = 2PQ; (iii) PR = 2AP — all proved using similar triangles

Where the marks are

(i) △RQC ~ △RAB giving $\frac{QR}{A R} = \frac{1}{2}$ (1 mark)
Concludes AQ = QR (1 mark)
(ii) △PQD ~ △PAB giving $\frac{QP}{A P} = \frac{1}{2}$ (1 mark)
AP = 2PQ (½ mark)
(iii) Writes $A P + P Q = P R - P Q$ (½ mark)
Substitutes and gets PR = 2AP (1 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 paper2026 board paper, set 30/5/1 (Standard): every question, by section.

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.

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

Hints

Step-by-step solution

Revise the chapter

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