Q1. [1] § 6.4 Criteria for Similarity of Triangles
If $\triangle ABC$ and $\triangle DEF$ are similar such that $2\,AB = DE$ and $BC = 8$ cm, then EF is equal to :
- (a) 4 cm
- (b) 8 cm
- (c) 12 cm
- (d) 16 cm
Previously asked in CBSE board exam
2026 30/1/1 Q7
Generated by claude-sonnet-4-6 · 2026-06-15 10:25 · grounding rag
Model Answer
(d) 16 cm
Since △ABC ~ △DEF, $\dfrac{AB}{DE} = \dfrac{BC}{EF}$. Given $2AB = DE$, so $\dfrac{AB}{DE} = \dfrac{1}{2}$. Thus $\dfrac{BC}{EF} = \dfrac{1}{2}$, giving $EF = 2 \times BC = 2 \times 8 = 16$ cm.
Source: Chapter 6, Section 6.3 / 6.4
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Explanation
- The key property of similar triangles is that corresponding sides are in the same ratio.
- Since $2AB = DE \Rightarrow \dfrac{AB}{DE} = \dfrac{1}{2}$, the scale factor from △ABC to △DEF is 2.
- Therefore every side of △DEF is twice the corresponding side of △ABC, so $EF = 2 \times BC = 16$ cm.
- Make sure to identify which sides correspond: BC corresponds to EF (both are between the second and third vertices in the similarity statement △ABC ~ △DEF).
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