Question

In the above figure, $ABCD$ is a parallelogram. $P$ is a point on $BC$ such that $BP:PC=1:2.$ $DP$ produced meets $AB$ produced at $Q$. Given $ar\left(△BPQ\right)=20c{m}^2$. Calculate
$\left(i\right)ar\left(△CDP\right)$

• A. 80
• B. 120
• C. 160
• D. 200

Answer 1

The correct option is A 80

Given $BP:PC=1:2$

$\angle DPC=\angle BPQ\angle PQB=\angle PDC\angle QBP=\angle DCB$

So $AAA$

The triangles are similar

$\frac{Ar\left(CPD\right)}{Ar\left(BPQ\right)}={\left(\frac{PC}{BP}\right)}^2\frac{Ar\left(CPD\right)}{20}=\frac{4}{1}Ar\left(CPD\right)=4\times 20=80c{m}^2$