Question

P and Q are points on the sides AB and AC respectively of $△$ABC such that PQ || BC and divides $△$ABC into two parts, equal in area. Find PB : AB.

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{(\sqrt{3}-1)}{\sqrt{3}}$
• C. $\frac{(\sqrt{2}-1)}{\sqrt{2}}$
• D. $\frac{(\sqrt{3}-1)}{\sqrt{2}}$

Answer 1

The correct option is C $\frac{(\sqrt{2}-1)}{\sqrt{2}}$

Given : Area ($\Delta$APQ) = Area (Trap. PBCQ)
We can see that, area ($\Delta$ APQ) = [Area ($\Delta$ABC) - Area (Trap. PBCQ)]
$\Rightarrow 2$ Area ($\Delta$APQ) = Area ($\Delta$ABC)
$\Rightarrow \frac{\text{Area}\left(\Delta APQ\right)}{\text{Area}\left(\Delta ABC\right)}=\frac{1}{2}$ ....(1)
Now, in $\Delta$APQ and $\Delta$ABC, we have $\angle PAQ=\angle BAC$ [Common]
$\angle APQ=\angle ABC$ [since PQ || BC, corresponding angles are equal]
$\therefore \Delta APQ\sim \Delta ABC$ [by AA similarity criterion]
We know that the areas of similar triangles are proportinal to the squares of their corresponding sides.
$\therefore \frac{Ar\left(\Delta APQ\right)}{Ar\left(\Delta ABC\right)}=\frac{A{P}^2}{A{B}^2}$
$\Rightarrow \frac{A{P}^2}{A{B}^2}=\frac{1}{2}$ [Using (i)]
$\to \frac{AP}{AB}=\frac{1}{\sqrt{2}}$
$\Rightarrow AB=\sqrt{2}$ (AB - PB)
$\Rightarrow \sqrt{2}PB=(\sqrt{2}-1)$ AB
$\frac{PB}{AB}=\frac{(\sqrt{2}-1)}{\sqrt{2}}$