In △ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find :

(i) area △APO : area △ABC

(ii) area △APO : area △CQO.

(i) Given,

$\dfrac{AP}{PB} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{AP}{AB - AP} = \dfrac{2}{3} \\[1em] \Rightarrow 3AP = 2(AB - AP) \\[1em] \Rightarrow 3AP = 2AB - 2AP \\[1em] \Rightarrow 3AP + 2AP = 2AB \\[1em] \Rightarrow 5AP = 2AB \\[1em] \Rightarrow \dfrac{AP}{AB} = \dfrac{2}{5}.$

Considering △APO and △ABC,

∠ A = ∠ A (Common angles)
∠ APO = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △APO ~ △ABC.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △APO}}{\text{Area of △ABC}} = \dfrac{AP^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △APO}}{\text{Area of △ABC}} = \dfrac{2^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △APO}}{\text{Area of △ABC}} = \dfrac{4}{25}.$

Hence, the ratio of the area of △APO : area of △ABC = 4 : 25.

(ii) In parallelogram PBCQ opposite sides are equal,

so, PB = QC. Hence, $\dfrac{AP}{QC} = \dfrac{AP}{PB} = \dfrac{2}{3}$.

Considering △APO and △CQO,

∠ AOP = ∠ QOC (Vertically opposite angles)
∠ OAP = ∠ OCQ (Alternate angles are equal)

Hence, by AA axiom △APO ~ △CQO.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △APO}}{\text{Area of △CQO}} = \dfrac{AP^2}{QC^2} \\[1em] \Rightarrow \dfrac{\text{Area of △APO}}{\text{Area of △CQO}} = \Big(\dfrac{AP}{QC}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of △APO}}{\text{Area of △CQO}} = \Big(\dfrac{2}{3}\Big)^2 = \dfrac{4}{9}.$

Hence, the ratio of the area of △APO : area of △CQO = 4 : 9.