Answer: A
In triangle ∆ABC, PQ is parallel to BC. ∴ \(\frac{AP}{PB} = \frac{AQ}{QC}\) ∴ \(\frac{AP}{PB} = \frac{PB}{QC}\) [because PB = AQ] Therefore, \(PB^2 = AP \times QC\) ⇒ \(PB^2 = 9 \times 16\) ⇒ \(PB = \sqrt{144} = 12\) ∴ PB = 12 cm
In triangle ∆ABC, PQ is parallel to BC. ∴ \(\frac{AP}{PB} = \frac{AQ}{QC}\) ∴ \(\frac{AP}{PB} = \frac{PB}{QC}\) [because PB = AQ] Therefore, \(PB^2 = AP \times QC\) ⇒ \(PB^2 = 9 \times 16\) ⇒ \(PB = \sqrt{144} = 12\) ∴ PB = 12 cm