Question

In the adjoining figure, $ABCD$ is a parallelogram. $P$ is a point on $BC$ such that $BP:PC=1:2$ and $DP$ produced meets $AB$ produced at $Q$. If area of $△CPQ=20c{m}^2$, find
(i) area of $△BPQ$.
(ii) area $△CDP$.
(iii) area of parallelogram $ABCD$.

Answer 1

From the question it is given that, $ABCD$ is a parallelogram.
$BP:PC=1:2$
area of $△CPQ=20c{m}^2$
Construction: draw $QN$ perpendicular $CB$ and Join $BN$.
Then, area of $△BPQ$/area of $△CPQ=\left(\right(\frac{1}{2}BP)\times QN)/\left(\right(\frac{1}{2}PC)\times QN)$
$=BP/PC=\frac{1}{2}$
(i) So, area $△BPQ=\frac{1}{2}$ area of $△CPQ$
$=\frac{1}{2}\times 20$
Therefore, area of $△BPQ=10c{m}^2$
(ii) Now we have to find area of $△CDP$,
Consider the $△CDP$ and $△BQP$,
Then, $\angle CPD=\angle QPD$ … [because vertically opposite angles are equal]
$\angle PDC=\angle PQB$ … [because alternate angles are equal]
Therefore, $△CDP\sim △BQP$ … [$AA$ axiom]
area of $△CDP$/area of $△BQP=P{C}^2/B{P}^2$
area of $△CDP$/area of $△BQP=2^2/1^2$
area of $△CDP$/area of $△BQP=4/1$
area of $△CDP=4\times area△BQP$
Therefore, area of $△CDP=4\times 10$
$=40c{m}^2$
(iii) We have to find the area of parallelogram $ABCD$,
Area of parallelogram $ABCD=2$ area of $△DCQ$
$=2area(△DCP+△CPQ)$
$=2(40+20)c{m}^2$
$=2\times 60c{m}^2$
$=120c{m}^2$
Therefore, the area of parallelogram $ABCD$ is $120c{m}^2$.