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

Answer 1

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

Given: Area ($△$APQ) = Area (Trap. PBCQ)

We can see that,
area ($△$APQ) = [Area ($△$ABC) - Area (Trap. PBCQ)]

$\Rightarrow$ 2 Area ($△$APQ) = Area ($△$ABC)

$\Rightarrow$ $\frac{Area\left(△APQ\right)}{Area\left(△ABC\right)}$ = $\frac{1}{2}$ ......(i)

Now, in $△$APQ and $△$ABC , we have

$\angle$PAQ = $\angle$BAC [Common]

$\angle$APQ = $\angle$ABC [since PQ|| BC, corresponding angles are equal ]

$\therefore$ $△$APQ $\sim$ $△$ABC [ by AA similarity criterion ]

We know that the areas of similar triangles are proportional to the squares of their corresponding sides.

$\therefore$ $\frac{Ar\left(△APQ\right)}{Ar\left(△ABC\right)}=\frac{A{P}^2}{A{B}^2}$

$\Rightarrow$ $\frac{A{P}^2}{A{B}^2}=\frac{1}{2}$ [ Using (i) ]

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

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

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

$\therefore$ $\frac{PB}{AB}$ =$\frac{(\sqrt{2}-1)}{\sqrt{2}}$