Points P and Q lies on side AB and AC of triangle ABC respectively such that segment PQ is parallel to side BC. If the ratio of areas of triangle APQ and triangle ABC is 9:16, then what is the ratio of AP:PB?

In $$\triangle$$ APQ and $$\triangle$$ ABC

$$\angle$$ PAQ = $$\angle$$ BAC  (common)

$$\angle$$ APQ = $$\angle$$ ABC   (Alternate interior angles)

$$\angle$$ AQP = $$\angle$$ ACB    (Alternate interior angles)

=> $$\triangle$$ APQ $$\sim$$ $$\triangle$$ ABC

=> Ratio of Area of ΔAPQ : Area of ΔABC = Ratio of square of corresponding sides = $$(AP)^2$$ : $$(AB)^2$$

=> $$\frac{9}{16} = \frac{(AP)^2}{(AB)^2}$$

=> $$\frac{AP}{AB} = \sqrt{\frac{9}{16}} = \frac{3}{4}$$

Also, PB = AB - AP

=> $$\frac{AP}{PB} = \frac{3}{4-3} = \frac{3}{1}$$

=> Ans - (D)