Answer: D
In triangle \( \triangle ABC \), PQ is parallel to BC ∴ \( \frac{AP}{PB} = \frac{AQ}{QC} \) ∴ \( \frac{AP}{PB} = \frac{2PB}{QC} \) [because \( AQ = 2PB \)] i.e., \( 2PB^2 = AP \times QC \) ⇒ \( 2PB^2 = 18 \times 9 \) ⇒ \( PB^2 = 81 \) ∴ \( PB = \sqrt{81} = 9 \) ∴ \( PB = 9 \) cm
In triangle \( \triangle ABC \), PQ is parallel to BC ∴ \( \frac{AP}{PB} = \frac{AQ}{QC} \) ∴ \( \frac{AP}{PB} = \frac{2PB}{QC} \) [because \( AQ = 2PB \)] i.e., \( 2PB^2 = AP \times QC \) ⇒ \( 2PB^2 = 18 \times 9 \) ⇒ \( PB^2 = 81 \) ∴ \( PB = \sqrt{81} = 9 \) ∴ \( PB = 9 \) cm