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, |
\[\frac{\text{Area}\,\text{of}\,\Delta ABC}{\text{Area}\,\text{of}\,\Delta BXY}=\frac{2}{1}\] |
\[\Rightarrow \] \[\frac{AB}{BX}=\frac{\sqrt{2}}{1}\] |
\[\Rightarrow \] \[\frac{AB-BX}{BX}=\frac{\sqrt{2}-1}{1}\] |
\[\frac{AX}{BX}=\frac{\sqrt{2}-1}{1}\] |
\[\Rightarrow \] \[\frac{AX}{AX+BX}=\frac{\sqrt{2}-1}{\sqrt{2}-1+1}\] |
\[\Rightarrow \] \[\frac{AX}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}\] |
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