Solution:-

From the question it is given that,

DC is parallel to AB

AB = 9 cm, DC = 6 cm and BB = 12 cm

(i) Consider the ∆APB and ∆CPD

∠APB = ∠CPD … [because vertically opposite angles are equal]

∠PAB = ∠PCD … [because alternate angles are equal]

So, ∆APB ~ ∆CPD

Then, BP/PD = AB/CD

BP/(12 – BP) = 9/6

6BP = 108 – 9BP

6BP + 9BP = 108

15BP = 108

BP = 108/15

Therefore, BP = 7.2 cm

(ii) We know that, area of ∆APB/area of ∆CPD = AB²/CD²

area of ∆APB/area of ∆CPD = 9²/6²

area of ∆APB/area of ∆CPD = 81/36

By dividing both numerator and denominator by 9, we get,

area of ∆APB/area of ∆CPD = 9/4

Therefore, the ratio of areas of ∆APB and ∆DPC is 9: 4