In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given ar (ΔCPQ) = 20 cm². Calculate :

(i) ar (ΔCDP)

(ii) ar (∥ gm ABCD)

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

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

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.

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

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

Hence, the area of △CDP = 40 cm².

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

= 2(Area of △CDP + Area of △CPQ)

= 2(40 + 20)

= 2 × 60

= 120 cm².

Hence, the area of ||gm ABCD = 120 cm².