MathsCBSE Class 10 · Standard · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2025-26: question 33

From the official CBSE sample question paper for 2025-26, section D, question 33, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33 · 5 marks · Long answerTriangles

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Hence in $Δ$ PQR, prove that a line $ℓ$ intersects the sides PQ and PR of a $Δ$ PQR at L and M respectively such that LM II QR. If PL = 5.7cm, PQ=15.2cm and MR=5.5cm, then find the length of PM (in cm)

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

Join BE and CD, and drop perpendiculars to compare areas of triangles with the same height.

Hint 2

Triangles BDE and DEC are on the same base and between the same parallels.

Hint 3

For the numerical part, LQ = 15.2 − 5.7. Use PL/LQ = PM/MR, or the equivalent PL/PQ = PM/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: In $△ A B C$ , a line DE ∥ BC meets AB at D and AC at E.
To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Figure: Draw triangle ABC with D on AB and E on AC, with DE parallel to BC.
Construction: Join BE and CD. Draw $E N ⊥ A B$ and $D M ⊥ A C$ .
2 marks
Proof: $ar (A D E) = \frac{1}{2} A D \cdot E N$ and $ar (B D E) = \frac{1}{2} D B \cdot E N$ , so $\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}$ … (1).
Also $ar (A D E) = \frac{1}{2} A E \cdot D M$ and $ar (D E C) = \frac{1}{2} E C \cdot D M$ , so $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ … (2).
1 mark
$△ B D E$ and $△ D E C$ have the same base DE and lie between the same parallels DE and BC, so $ar (B D E) = ar (D E C)$ … (3).
From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Hence proved.
1 mark
Application: In $△ P QR$ , LM ∥ QR. Let $P M = x$ cm, so $P R = x + 5.5$ cm. By the theorem (in the equivalent form), $\frac{P L}{P Q} = \frac{P M}{P R}$ , which gives $\frac{5.7}{15.2} = \frac{x}{x + 5.5}$ .
½ mark
$5.7 x + 31.35 = 15.2 x$ , so $9.5 x = 31.35$ and $x = 3.3$ .
(Check: $L Q = 15.2 - 5.7 = 9.5$ , and $\frac{5.7}{9.5} = \frac{3.3}{5.5} = 0.6$ ✓.) So $P M = 3.3$ cm.
½ mark

Answer: Theorem (BPT) proved; PM = 3.3 cm

Where the marks are

Writes the given, to prove, figure and construction correctly (2 marks)
Proves the result using area ratios and equal areas of triangles BDE and DEC (2 marks)
Sets up $\frac{5.7}{15.2} = \frac{x}{x + 5.5}$ (½ mark)
Finds PM = 3.3 cm (½ mark)

Revise the chapter

Chapter: Triangles (notes, formulas, where marks are lost and every past question)
Topic: Basic Proportionality Theorem (Thales Theorem)

Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2025-26: question 33

Hints

Step-by-step solution

Revise the chapter

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