Question

In the
given figure , AB C D is a trapezium
in which AB II CD and AD I AB. If to AB= 10 cm,
CD=6cm and B C 5 cm find the
area oi trapezium
​

Answer 1

Answer:

Answer

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=AB

2

/CD

2

area of △APB/area of △CPD=9

2

/6

2

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.

Answer 2

Step-by-step explanation:

Here is ur answer mate