question_answer

In \[\Delta ABC,\] X and Y are points on sides AB and BC respectively, such that \[XY||AC\] and XY divides triangular region ABC into two parts equal in area. Then, \[\frac{AX}{AB}\] is equal to

A) \[\frac{2+\sqrt{2}}{2}\]

B) \[\frac{\sqrt{2}+3}{2}\]

C) \[\frac{\sqrt{2}-1}{\sqrt{2}}\]

D) \[\frac{3-\sqrt{2}}{2}\]

Correct Answer: C

Solution :

According to the question,

\[\frac{\text{Area}\,\text{of}\,\Delta ABC}{\text{Area}\,\text{of}\,\Delta BXY}=\frac{2}{1}\]

\[\Rightarrow \]   \[\frac{AB}{BX}=\frac{\sqrt{2}}{1}\]

\[\Rightarrow \]   \[\frac{AB-BX}{BX}=\frac{\sqrt{2}-1}{1}\]

\[\frac{AX}{BX}=\frac{\sqrt{2}-1}{1}\]

\[\Rightarrow \]   \[\frac{AX}{AX+BX}=\frac{\sqrt{2}-1}{\sqrt{2}-1+1}\]

\[\Rightarrow \]   \[\frac{AX}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}\]