Question

In below figure $PQ\parallel BC$ and $AP:PB=1:2$. Then the ratio of area $\Delta APQ$ and $\Delta ABC$ will be:

• A. $1:2$
• B. $1:4$
• C. $1:9$
• D. $4:1$

Answer 1

The correct option is C $1:9$

Here $BC||PQ$ and $\frac{AP}{PB}=\frac{1}{2}$, So

$⟹\frac{AP}{AB}=\frac{AP}{AP+PB}=\frac{1}{1+2}=\frac{1}{3}$

In $△ABC$ and $△APQ$

$\angle BAC=\angle PAQ$---same angle

$\angle ABC=\angle APQ$---Corresponding angles as $BC||PQ$ and $AB$ as transversal line

So,

$△ABC\sim △APQ$ ---AAA Postulate

We know the ratio of area of similar triangles :

$\frac{Areaof△APQ}{Areaof△ABC}=\frac{(correspondingside{)}^2}{(correspondingside{)}^2}$

$⟹\frac{Areaof△APQ}{Areaof△ABC}=\frac{(AP{)}^2}{(AB{)}^2}$

$⟹\frac{Areaof△APQ}{Areaof△ABC}=\frac{(1{)}^2}{(3{)}^2}$

$⟹\frac{Areaof△APQ}{Areaof△ABC}=\frac{1}{9}$

$⟹Areaof△APQ:Areaof△ABC=1:9$ Option C