Points P and Q lie on side AB and AC of triangle ABC respectively such that segment PQ is parallel to side BC. If the ratio of AP:PB is 2:3, and area of Δ APQ is 8 sq cm, what is the area of trapezium PQCB?

It is given that AP : PB = 2 : 3

Let AP = 2 cm and PB = 3 cm

Let area of trapezium PQCB = $$x$$ sq cm

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 $$\triangle$$ APQ : Area of $$\triangle$$ ABC = Ratio of square of corresponding sides = $$(AP)^2$$ : $$(AB)^2$$

= $$\frac{(2)^2}{(2 + 3)^2} = \frac{8}{(8 + x)}$$

=> $$\frac{8}{8 + x} = \frac{4}{25}$$

=> $$8 + x = \frac{25 \times 8}{4} = 50$$

=> $$x = 50 - 8 = 42 cm^2$$

=> Ans - (D)