Question

In a triangle $ABC$, a line $PQ$ is drawn parallel to $BC$, points P, Q being on $AB$ and $AC$ respectively. If $AB=3AP$, then what is the ratio of the area of triangle $APQ$ to the area of triangle $ABC$?

• A. $1:3$
• B. $1:5$
• C. $1:7$
• D. $1:9$

Answer 1

The correct option is D $1:9$

R.E.F image

Given, $AB=3AP...\left(1\right)$

we can prove the two triangles

are similar using

1) AA (Angle-Angle)

2) SSS (side-side-side)

3) SAS (side-Angle-side)

In this case.

$\angle A$ is common to both the

$\Delta ABC$ and $\Delta APQ$

Also, $\angle B\cong \angle P$ (corresponding angles)

$\therefore$ using AA theorem it is proved

that $\Delta ABC\sim \Delta APQ$

[Note :- Similarity sign is $\sim$]

Therm : The two similar $\Delta s$ , the ratio of

their area is the square of the

ratio of their sides.

Hence.

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

$={\left(\frac{AB}{AP}\right)}^2$

from (1)

$\frac{Areaof\Delta APQ}{Areaof\Delta ABC}={\left(\frac{AP}{AB}\right)}^2={\left(\frac{1}{3}\right)}^2$

$=\frac{1}{9}$