MathsCBSE Class 10 · Basic · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic sample paper 2023-24: question 33

From the official CBSE sample question paper for 2023-24, 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.

In $Δ$ PQR, S and T are points on PQ and PR respectively. $\frac{P S}{S Q} = \frac{P T}{T R}$ and $∠ P S T = ∠ P R Q$ . Prove that PQR is an isosceles triangle.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

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

Hint 1

For the theorem, join BE and CD and compare areas of triangles that share the same height.

Hint 2

Triangles DBE and DEC have the same base DE and lie between the same parallels, so their areas are equal.

Hint 3

For the second part, the equal ratios give $S T ∥ QR$ (converse of BPT). Compare the corresponding angles with the given angle.

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 $△ A B C$ , $D E ∥ B C$ , with D on AB and E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: Join BE and CD. Draw $E N ⊥ A B$ and $D M ⊥ A C$ .
½ mark
$ar (A D E) = \frac{1}{2} \times A D \times E N$ and $ar (D B E) = \frac{1}{2} \times D B \times E N$
$\Rightarrow \frac{ar ( A D E )}{ar ( D B E )} = \frac{A D}{D B}$ ...(1)
1 mark
$ar (A D E) = \frac{1}{2} \times A E \times D M$ and $ar (D E C) = \frac{1}{2} \times E C \times D M$
$\Rightarrow \frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ ...(2)
1 mark
$△ D B E$ and $△ D E C$ lie on the same base DE and between the same parallels DE and BC, so $ar (D B E) = ar (D E C)$ .
Hence, from (1) and (2), $\frac{A D}{D B} = \frac{A E}{E C}$ . Proved.
½ mark
Second part: In $△ P QR$ , $\frac{P S}{S Q} = \frac{P T}{T R}$ , so $S T ∥ QR$ [converse of BPT].
Therefore $∠ P S T = ∠ P QR$ [corresponding angles].
1 mark
Also $∠ P S T = ∠ P R Q$ [given], so $∠ P QR = ∠ P R Q$ .
$\Rightarrow P R = P Q$ [sides opposite equal angles are equal]. Hence $△ P QR$ is isosceles.
1 mark

Answer: BPT proved; $P Q = P R$ , so $△ P QR$ is isosceles

Where the marks are

Given, to prove, construction and figure (1/2 mark)
Ratio of areas gives $\frac{A D}{D B}$ (1 mark)
Ratio of areas gives $\frac{A E}{E C}$ (1 mark)
Equal areas of $△ D B E$ and $△ D E C$ used to conclude (1/2 mark)
$S T ∥ QR$ by the converse of BPT, and $∠ P S T = ∠ P QR$ (1 mark)
$∠ P QR = ∠ P R Q$ , hence $P Q = P R$ and the triangle is isosceles (1 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 2023-24 (Basic): 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 Basic sample paper 2023-24: question 33

Hints

Step-by-step solution

Revise the chapter

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