1. If $\triangle ABC$ and $\triangle DEF$ are similar such that $2AB = DE$ and $BC = 8\text{ cm}$, then EF is equal to : (a) 4 cm (b) 8 cm (c) 12 cm (d) 16 cm

Since $\triangle ABC \sim \triangle DEF$ and $2\cdot AB = DE$, every side of △DEF is exactly double the matching side of △ABC — so $EF = 2 \times BC = 16\text{ cm}$.

Concept

• In similar triangles, corresponding sides are proportional: $\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}$.
• $2AB = DE \Rightarrow \frac{DE}{AB} = 2$, so $\triangle DEF$ is the enlarged triangle (scale factor 2), not the smaller one.

Key steps

• BC corresponds to EF (both are the "middle" sides in the same vertex order).
• $EF = 2 \times BC = 2 \times 8 = 16\text{ cm}$ → option (d).

Common mistakes

• (a) 4 cm: wrongly halves BC, treating ABC as the larger triangle.
• (b) 8 cm: assumes congruence (ratio 1), ignoring the factor 2.
• (c) 12 cm: a distractor with no valid derivation.

Real-world

• This is exactly the logic used in scale models and map enlargements — doubling every length keeps the shape identical while scaling the size.