Question

The line segment XY is parallel to side AC of Δ ABC and it divides the triangle into two parts of equal areas. Find the ratio $\frac{BX}{AB}$.

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{4}{\sqrt{1}}$
• D. $\frac{\sqrt{2}}{1}$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}$

Given: XY || AC
$Ar\left(△ABC\right)=2Ar\left(△XBY\right)$

Here, $\angle$ BXY = $\angle$ A and $\angle$ BYX = $\angle$ C (Corresponding angles)

$\Rightarrow$ $△$ ABC $\sim$ $△$ XBY (AA similarity criterion)

$\Rightarrow$ $\frac{Ar\left(△ABC\right)}{Ar\left(△XBY\right)}={\left(\frac{AB}{XB}\right)}^2$ ​

$\because$ $Ar\left(△ABC\right)=2Ar\left(△XBY\right)$

$\frac{Ar\left(△ABC\right)}{Ar\left(△XBY\right)}=\frac{2}{1}$

$\Rightarrow$ ${\left(\frac{AB}{XB}\right)}^2$= $\frac{2}{1}$
$\Rightarrow$ $\frac{AB}{XB}$ = $\frac{\sqrt{2}}{1}$

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

Rationalising the denominator we get,

$\frac{XB}{AB}=\frac{\sqrt{2}}{2}$