(i) \[AX:AB\] (ii) \[\frac{AC}{XY}\]
A)
(i) (ii) \[\left( 2+\sqrt{2} \right):2\] \[\sqrt{2}-2\]
B)
(i) (ii) \[\left( 2-\sqrt{2} \right):2\] \[\sqrt{2}-1\]
C)
(i) (ii) \[\left( 2-\sqrt{3} \right):3\] \[3\]
D)
(i) (ii) \[\left( 2+\sqrt{2} \right):3\] \[\sqrt{2}-3\]
Correct Answer: B
Solution :
(i) Since, \[XY\text{ }\!\!|\!\!\text{ }\!\!|\!\!\text{ }AC,\]we have \[\angle A=\angle BXY\] and \[\angle C=\angle BYX\] [Corresponding angles] ; \[\therefore \] \[\Delta MBC\tilde{\ }\Delta XBY\] [By AA similarity]; \[\Rightarrow \] \[\frac{ar\,(\Delta ABC)}{ar\,(\Delta XBY)}=\frac{A{{B}^{2}}}{X{{B}^{2}}}\] But, \[ar(\Delta ABC)=2\times ar\,(\Delta XBY)\] [Given] \[\therefore \] \[\frac{ar(\Delta \,ABC)}{ar(\Delta \,XBY)}=2\] ?.(ii) From (i) and (ii), we get \[\frac{A{{B}^{2}}}{X{{B}^{2}}}=2\,\,\Rightarrow \,\,{{\left( \frac{AB}{XB} \right)}^{2}}=2\] \[\Rightarrow \] \[\frac{AB}{XB}=\sqrt{2}\,\,\Rightarrow \,\,AB=\sqrt{2}(XB)\] \[\Rightarrow \]\[AB=\sqrt{2}(AB-AX)\Rightarrow (\sqrt{2}AX)=(\sqrt{2}-1)AB\] \[\Rightarrow \]\[\frac{AX}{AB}=\frac{(\sqrt{2}-1)}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{(2-\sqrt{2})}{2}\] Hence, \[AX:AB=(2-\sqrt{2}):2\] (ii) Since, \[\Delta ABC\tilde{\ }\Delta XBY\] \[\therefore \] \[\frac{BY}{YC}=\frac{BX}{XA}=\frac{XY}{AC}\Rightarrow \frac{XY}{AC}=\frac{BX}{XA}\Rightarrow \frac{AC}{XY}=\frac{XA}{BX}\] \[\Rightarrow \] \[\frac{AC}{XY}=\frac{AB-BX}{BX}\Rightarrow \frac{AC}{XY}=\frac{AB}{BX}-1\] \[\Rightarrow \] \[\frac{AC}{XY}=\frac{\sqrt{2}-1}{1}\]
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