Solution:-

From the question it is given that,

PB = 2: 3

PO is parallel to BC and is extended to Q so that CQ is parallel to BA.

(i) we have to find the area ∆APO: area ∆ABC,

Then,

∠A = ∠A … [common angles for both triangles]

∠APO = ∠ABC … [because corresponding angles are equal]

Then, ∆APO ~ ∆ABC … [AA axiom]

We know that, area of ∆APO/area of ∆ABC = AP²/AB²

= AP²/(AP + PB)²

= 2²/(2 + 3)²

= 4/5²

= 4/25

Therefore, area ∆APO: area ∆ABC is 4: 25

(ii) we have to find the area ∆APO: area ∆CQO

Then, ∠AOP = ∠COQ … [because vertically opposite angles are equal]

∠APQ = ∠OQC … [because alternate angles are equal]

Therefore, area of ∆APO/area of ∆CQO = AP²/CQ²

area of ∆APO/area of ∆CQO = AP²/PB²

area of ∆APO/area of ∆CQO = 2²/3²

area of ∆APO/area of ∆CQO = 4/9

Therefore, area ∆APO: area ∆CQO is 4: 9