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
$ \Rightarrow $ $ \frac{AB^{2}}{BX^{2}}=\frac{2}{1} $
$ \Rightarrow $ $ \frac{AB}{BX}=\frac{\sqrt{2}}{1} $
$ \begin{aligned} & [\because \text{ratio of area of similar triangles} \\ & \text{= (ratio corresponding side}{{)}^{2}}] \\ \end{aligned} $
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} $
BETA
