MathsCBSE Class 10 · Basic · Board exam 2023, set 430/6/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2023: question 35(a)

Set in the CBSE Class 10 board exam of 2023, paper 430/6/1, section D, question 35(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(a) · 5 marks · Long answerTriangles

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

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/6/1.

Hints

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

Hint 1

Try comparing the areas of triangles that share the same height.

Hint 2

Join BE and CD. Express $a r (A D E)$ as a ratio with $a r (B D E)$ , and then with $a r (C D E)$ .

Hint 3

Triangles BDE and CDE stand on the same base DE and lie between the same parallels. What does that say about their areas?

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.

Figure: Draw $△ A B C$ with a line DE parallel to BC, meeting AB at D and AC at E. Join BE and CD. Draw $E N ⊥ A B$ (N on AB) and $D M ⊥ A C$ (M on AC).
1 mark
Given: In $△ A B C$ , $D E ∥ B C$ , and DE meets AB at D and AC at E.
½ mark
To prove: $\frac{A D}{D B} = \frac{A E}{E C}$
½ mark
Construction: Join BE and CD. Draw $E N ⊥ A B$ and $D M ⊥ A C$ .
½ mark
Proof: $△ A D E$ and $△ B D E$ have the same height EN on bases AD and DB:
$\frac{a r ( △ A D E )}{a r ( △ B D E )} = \frac{\frac{1}{2} \times A D \times E N}{\frac{1}{2} \times D B \times E N} = \frac{A D}{D B}$ ...(i)
1 mark
In the same way, $△ A D E$ and $△ C D E$ have the same height DM on bases AE and EC:
$\frac{a r ( △ A D E )}{a r ( △ C D E )} = \frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times E C \times D M} = \frac{A E}{E C}$ ...(ii)
½ mark
$△ B D E$ and $△ C D E$ lie on the same base DE and between the same parallels DE and BC. So $a r (△ B D E) = a r (△ C D E)$ ...(iii)
½ mark
From (i), (ii) and (iii), the left-hand sides of (i) and (ii) are equal. Hence $\frac{A D}{D B} = \frac{A E}{E C}$ . Proved.
½ mark

Answer: Proved: $\frac{A D}{D B} = \frac{A E}{E C}$ (Basic Proportionality Theorem).

Where the marks are

Correct labelled figure with the construction lines (1 mark)
Given, To prove and Construction stated (½ + ½ + ½ marks)
Shows $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ (1 mark)
Shows $\frac{a r ( A D E )}{a r ( C D E )} = \frac{A E}{E C}$ (½ mark)
States $a r (B D E) = a r (C D E)$ , giving the reason: same base, same parallels (½ mark)
Concludes $\frac{A D}{D B} = \frac{A E}{E C}$ (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 paper2023 board paper, set 430/6/1 (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 2023: question 35(a)

Hints

Step-by-step solution

Revise the chapter

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