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 ∆CDP.
(iii) area of parallelogram ABCD.

Solution:-

From the question it is given that, ABCD is a parallelogram.

BP: PC = 1: 2

area of ∆CPQ = 20 cm²

Construction: draw QN perpendicular CB and Join BN.

Then, area of ∆BPQ/area of ∆CPQ = ((½BP) × QN)/((½PC) × QN)

= BP/PC = ½

(i) So, area ∆BPQ = ½ area of ∆CPQ

= ½ × 20

Therefore, area of ∆BPQ = 10 cm²

(ii) Now we have to find area of ∆CDP,

Consider the ∆CDP and ∆BQP,

Then, ∠CPD = ∠QPD … [because vertically opposite angles are equal]

∠PDC = ∠PQB … [because alternate angles are equal]

Therefore, ∆CDP ~ ∆BQP … [AA axiom]

area of ∆CDP/area of ∆BQP = PC²/BP²

area of ∆CDP/area of ∆BQP = 2²/1²

area of ∆CDP/area of ∆BQP = 4/1

area of ∆CDP = 4 × area ∆BQP

Therefore, area of ∆CDP = 4 × 10

= 40 cm²

(iii) We have to find the area of parallelogram ABCD,

Area of parallelogram ABCD = 2 area of ∆DCQ

= 2 area (∆DCP + ∆CPQ)

= 2 (40 + 20) cm²

= 2 × 60 cm²

= 120 cm²

Therefore, the area of parallelogram ABCD is 120 cm².