In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q.

If area of △CPQ = 20 cm², find

(i) area of △BPQ.

(ii) area of △CDP.

(iii) area of ||gm ABCD.

(i) Draw QN ⊥ CB as shown in the figure below:

$\Rightarrow \dfrac{\text{Area of △BPQ}}{\text{Area of △CPQ}} = \dfrac{\dfrac{1}{2} BP \times QN}{\dfrac{1}{2} PC \times QN} \\[1em] \Rightarrow \dfrac{\text{Area of △BPQ}}{\text{Area of △CPQ}} = \dfrac{BP}{PC} \\[1em] \Rightarrow \dfrac{\text{Area of △BPQ}}{\text{Area of △CPQ}} = \dfrac{1}{2} \\[1em] \therefore \text{Area of △BPQ} = \dfrac{1}{2}\text{Area of △CPQ} = \dfrac{1}{2} \times 20 = 10 \text{cm}^2.$

Hence, the area of △BPQ = 10 cm².

(ii) Considering △CDP and △BQP,

∠CPD = ∠QPB (Vertically opposite angles are equal)
∠PDC = ∠PQB (Alternate angles are equal)

Hence, by AA axiom △CDP ~ △BQP.

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

$\therefore \dfrac{\text{Area of △CDP}}{\text{Area of △BQP}} = \dfrac{PC^2}{BP^2} \\[1em] \Rightarrow \dfrac{\text{Area of △CDP}}{\text{Area of △BQP}} = \dfrac{2^2}{1^2} \\[1em] \Rightarrow \dfrac{\text{Area of △CDP}}{\text{Area of △BQP}} = \dfrac{4}{1}$

∴ Area of △CDP = 4 × Area of △BQP = 4 × 10 = 40 cm².

Hence, the area of △CDP = 40 cm².

(iii) Area of ||gm ABCD = 2 Area of △DCQ (As △DCQ and ||gm ABCD have same base and are between same parallels)

$= 2(\text{Area of △CDP + Area of △CPQ}) \\[1em] = 2(40 + 20) \\[1em] = 2 \times 60 \\[1em] = 120 \text{ cm}^2.$

Hence, the area of ||gm = 120 cm².