Question

In $△ABC,AP:PB=2:3.PO$ is parallel to $BC$ and is extended to $Q$ so that $CQ$ is parallel to $BA$. Find:
(i) area $△APO:$ area $△ABC$.
(ii) area $△APO:$ area $△CQO$.

Answer 1

From the question it is given that,
$PB=2:3$
$PO$ is parallel to $BC$ and is extended to $Q$ so that $CQ$ is parallel to $BA$.
(i) we have to find the area $△APO:$ area $△ABC$,
Then,
$\angle A=\angle A$ … [common angles for both triangles]
$\angle APO=\angle ABC$ … [because corresponding angles are equal]
Then, $△APO\sim △ABC$ … [$AA$ axiom]
We know that, area of $△APO$/area of $△ABC=A{P}^2/A{B}^2$
$=A{P}^2/(AP+PB{)}^2$
$=22/(2+3{)}^2$
$=4/5^2$
$=4/25$
Therefore, area $△APO:$ area $△ABC$ is $4:25$
(ii) we have to find the area $△APO$ : area $△CQO$
Then, $\angle AOP=\angle COQ$ … [because vertically opposite angles are equal]
$\angle APQ=\angle OQC$ … [because alternate angles are equal]
Therefore, area of $△APO$/area of $△CQO=A{P}^2/C{Q}^2$
area of $△APO$/area of $△CQO=A{P}^2/P{B}^2$
area of $△APO$/area of $△CQO=2^2/3^2$
area of $△APO$/area of $△CQO=4/9$
Therefore, area $△APO$ : area $△CQO$ is $4:9$.