Q1. [1] § 6.4 Criteria for Similarity of Triangles
Assertion (A) : ABCD is a trapezium with $DC \parallel AB$. E and F are points on AD and BC respectively, such that $EF \parallel AB$. Then $\dfrac{AE}{ED} = \dfrac{BF}{FC}$.
Reason (R) : Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally.
Select the correct answer from the codes (A), (B), (C) and (D) given below.
- A Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- B Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- C Assertion (A) is true, but Reason (R) is false.
- D Assertion (A) is false, but Reason (R) is true.
Previously asked in CBSE board exam
2024 30/5/1 Q19
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
Option (A)
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation
The Reason directly states the theorem (Basic Proportionality Theorem applied to trapeziums, proved in Example 2 of the textbook) that justifies the Assertion. Since EF ∥ AB ∥ DC, EF divides non-parallel sides AD and BC proportionally, giving AE/ED = BF/FC. So R correctly and completely explains A.
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