Question

In the above figure, $ABCD$ is a parallelogram. $P$ is a point on $DC$ such that $ar\left(△APD\right)=25c{m}^2$ and $ar\left(△BPC\right)=15c{m}^2$. Calculate
$\left(i\right)ar(\parallel gmABCD)\left(ii\right)DP:PC$

• A. $\left(i\right)80c{m}^2\left(ii\right)5:3$
• B. $\left(i\right)50c{m}^2\left(ii\right)8:3$
• C. $\left(i\right)30c{m}^2\left(ii\right)5:2$
• D. $\left(i\right)100c{m}^2\left(ii\right)10:7$

Answer 1

The correct option is A $\left(i\right)80c{m}^2\left(ii\right)5:3$

Given-

$ABCD$ is a parallelogram.

P is a point on DC such that $ar\left(△APD\right)=25c{m}^2$ and $ar\left(△BPC\right)=15c{m}^2$.

To calculate-
$\left(i\right)ar(\parallel gmABCD)\left(ii\right)DP:PC$

Solution-

We draw $PQ\parallel AD\left(orBC\right).$

PQ meets AB at Q.

Here

$PQ\parallel AD\left(orBC\right)\&AB\parallel DC$

So both ADPQ & BCPQ are parallelograms with AP & BP as diagonals

respectively.

$\therefore ar.\Delta ADP=ar.\Delta AQP=25{cm}^2\&ar.\Delta BCP=ar.\Delta BPQ=15{cm}^2.\therefore ar.ABCD=ar.\Delta ADP+ar.\Delta AQP+ar.\Delta BCP+ar.\Delta BPQ=25{cm}^2+25{cm}^2+15{cm}^2+15{cm}^2=80{cm}^2.$

Again let us take the height of ABCD=h.

Then arADPQ=$ar.\Delta ADP+ar.\Delta AQP=25{cm}^2+25{cm}^2=50{cm}^2=DP\times h....\left(i\right).$

Also ar.BCPQ=$ar.\Delta BCP+ar.\Delta BPQ=15{cm}^2+15{cm}^2={30cm}^2=PC\times h........\left(ii\right).$

$\therefore \frac{DP\times h}{PC\times h}=\frac{50}{30}⟹\frac{DP}{PC}=\frac{5}{3}⟹DP:PC=5:3$ and

ar.ABCD=$80{cm}^2.$

Ans- Option A.