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) \[\frac{2-\sqrt{2}}{2}\]
D) \[\frac{3-\sqrt{2}}{2}\]
Correct Answer: C
Solution :
According to the question, |
\[\frac{\text{Area}\,\,\text{of}\Delta ABC}{\text{Area}\,\,\text{of}\,\,\Delta BXY}=\frac{2}{1}\] |
\[\Rightarrow \] \[\frac{A{{B}^{2}}}{B{{X}^{2}}}=\frac{2}{1}\]\[\Rightarrow \]\[\frac{AB}{BX}=\frac{\sqrt{2}}{1}\] |
\[\begin{align} & \text{ }\!\![\!\!\text{ }\because \text{ratio of area of similar triangles} \\ & \text{= (ratio corresponding side}{{)}^{\text{2}}}\text{ }\!\!]\!\!\text{ } \\ \end{align}\] |
On subtracting 1 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 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] |
\[\Rightarrow \] \[\frac{AX}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\] |
\[\Rightarrow \] \[\frac{AX}{AB}=\frac{2-\sqrt{2}}{2}\] |
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