This is the basic and conceptual question from similarity chapter in which we are to find

(i) length of BP

(ii) the ratio of areas of △APB and △DPC.

And we have given a figure of trapezium in which it’s parallel sides are AB and DC

ML Aggarwal Avichal Publication Class 10, Chapter 13.3, question no 14a

## From the question it is given that,

DC is parallel to AB

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

(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=7.2cm

(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.