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                                  [SSC (10+2) 2013]

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

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

C) \[\frac{2-\sqrt{2}}{2}\]  

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

Correct Answer: C

Solution :

According to the question,

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

\[\Rightarrow \]   \[\frac{A{{B}^{2}}}{B{{X}^{2}}}=\frac{2}{1}\]\[\Rightarrow \]\[\frac{AB}{BX}=\frac{\sqrt{2}}{1}\]

[\[\because \]ratio of area of similar triangles \[={{(ratio\,\,corresponding\,\,side)}^{2}}\]]

Subtracting 1 on both sides, we get

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

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

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

On adding 1 in both sides, we get

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

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

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

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