Answer: A
In triangle \( \triangle ABC \), PQ is parallel to BC and \( AP : PB = 2 : 1 \), with \( AC = 18 \) cm Because PQ is parallel to BC, ∴ \( \frac{AP}{PB} = \frac{AQ}{QC} \) ∴ \( \frac{2}{1} = \frac{AQ}{18 - AQ} \) i.e., \( \frac{AQ}{18 - AQ} = \frac{2}{1} \) ⇒ \( AQ = 36 - 2AQ \) ⇒ \( AQ + 2AQ = 36 \) ⇒ \( 3AQ = 36 \) ⇒ \( AQ = 12 \)
In triangle \( \triangle ABC \), PQ is parallel to BC and \( AP : PB = 2 : 1 \), with \( AC = 18 \) cm Because PQ is parallel to BC, ∴ \( \frac{AP}{PB} = \frac{AQ}{QC} \) ∴ \( \frac{2}{1} = \frac{AQ}{18 - AQ} \) i.e., \( \frac{AQ}{18 - AQ} = \frac{2}{1} \) ⇒ \( AQ = 36 - 2AQ \) ⇒ \( AQ + 2AQ = 36 \) ⇒ \( 3AQ = 36 \) ⇒ \( AQ = 12 \)