Question

In the adjacent figure, P and Q are points on the sides AB and AC respectively of a triangle ABC. PQ is parallel to BC and divides the triangle ABC into 2 parts, equal in area. The ratio of $PB:AB=$

• A. $1:1$
• B. $(\sqrt{2}-1):\sqrt{2}$
• C. $1:\sqrt{2}$
• D. $(\sqrt{2}-1):1$

Answer 1

The correct option is A $1:1$

As in $\Delta ABC$ and $\Delta APQ$:

$\angle A=\angle A$

$\angle APQ=\angle ABC$

$\angle AQP=\angle ACB$

$\therefore$ by $AAA$ criteria of Similarity:

$\Delta APQ\sim \Delta ABC$

Now,

we know that in similar triangles,

Ratio of area of triangle is equal to ratio of square of corresponding sides:

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

$\frac{Areaof\Delta APQ+AreaofPBCQ}{AreaofAPQ}={\left(\frac{AB}{AP}\right)}^2$

$\Rightarrow \frac{Areaof\Delta APQ+Areaof\Delta APQ}{AreaofAPQ}$ $={\left(\frac{AB}{AP}\right)}^2$ ........ [As Area of $\Delta APQ$ $=$ Area of $PBCQ$]

$\Rightarrow \frac{2\times Areaof\Delta APQ}{Areaof\Delta APQ}={\left(\frac{AB}{AP}\right)}^2$

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

$\Rightarrow \frac{AB}{AP}=\sqrt{2}$

$\Rightarrow \frac{AP}{AB}=\frac{1}{\sqrt{2}}\to ⟨1⟩$

We have to find $\frac{PB}{AB}$:

Since, $AP+PB=AB$

$\Rightarrow \frac{AB}{\sqrt{2}}+PB=AB$ $\left[Using⟨1⟩\right]$

$\Rightarrow PB=AB-\frac{AB}{\sqrt{2}}$

$\Rightarrow \frac{PB}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}$

Hence, Option (B) is correct.