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.