Answer: A
Let \( AP = QC = x \) cm In triangle \( \triangle ABC \), PQ is parallel to BC ∴ \( \frac{AP}{PB} = \frac{AQ}{QC} \) ∴ \( \frac{AP}{AB - AP} = \frac{AQ}{QC} \) i.e., \( \frac{x}{12 - x} = \frac{2}{x} \) ⇒ \( x^2 = 24 - 2x \) ⇒ \( x^2 + 2x - 24 = 0 \) ⇒ \( (x + 6)(x - 4) = 0 \) ∴ \( x = 4 \) [because \( x ≠ -6 \)]
Let \( AP = QC = x \) cm In triangle \( \triangle ABC \), PQ is parallel to BC ∴ \( \frac{AP}{PB} = \frac{AQ}{QC} \) ∴ \( \frac{AP}{AB - AP} = \frac{AQ}{QC} \) i.e., \( \frac{x}{12 - x} = \frac{2}{x} \) ⇒ \( x^2 = 24 - 2x \) ⇒ \( x^2 + 2x - 24 = 0 \) ⇒ \( (x + 6)(x - 4) = 0 \) ∴ \( x = 4 \) [because \( x ≠ -6 \)]