Answer: C
In triangle \( \triangle ABC \), PQ is parallel to BC and \( AB = 3 \times PB \), and \( BC = 18 \) cm. Because PQ is parallel to BC, ∴ \( \frac{AP}{AB} = \frac{PQ}{BC} \) ∴ \( \frac{AB - PB}{3PB} = \frac{PQ}{18} \) Or, \( \frac{3PB - PB}{3PB} = \frac{PQ}{18} \) Or, \( \frac{PQ}{18} = \frac{2}{3} \) So, \( PQ = \frac{2}{3} \times 18 \) Therefore, \( PQ = 12 \) cm
In triangle \( \triangle ABC \), PQ is parallel to BC and \( AB = 3 \times PB \), and \( BC = 18 \) cm. Because PQ is parallel to BC, ∴ \( \frac{AP}{AB} = \frac{PQ}{BC} \) ∴ \( \frac{AB - PB}{3PB} = \frac{PQ}{18} \) Or, \( \frac{3PB - PB}{3PB} = \frac{PQ}{18} \) Or, \( \frac{PQ}{18} = \frac{2}{3} \) So, \( PQ = \frac{2}{3} \times 18 \) Therefore, \( PQ = 12 \) cm