question_answer

            In a triangle, a line XY is drawn parallel to BC meeting AB in X and AC in Y. The area of \[\Delta ABC\]is 2 times the area of \[\Delta AXY.\] In what ratio X divides AB?                                    

A)  \[1:\sqrt{2}\]                

B)  \[\sqrt{2}:1\]                       

C)  \[(\sqrt{2}-1):1\]           

D)  \[1:(\sqrt{2}-1)\]

Correct Answer: D

Solution :

\[\frac{\text{Area}\,\,(\Delta ABC)}{\text{Area}\,\,(\Delta AXY)}=\frac{A{{B}^{2}}}{A{{X}^{2}}}\] \[\Rightarrow \]   \[\frac{2\,\text{Area}\,\,(\Delta AXY)}{\text{Area}\,\,(\Delta AXY)}=\frac{A{{B}^{2}}}{A{{X}^{2}}}\] \[\Rightarrow \]   \[\frac{2}{1}=\frac{A{{B}^{2}}}{A{{X}^{2}}},\frac{AB}{AX}=\sqrt{2}\] \[\Rightarrow \]   \[AB=\sqrt{2}AX\] \[\Rightarrow \]   \[AX+BX=\sqrt{2}AX\] \[\therefore \]      \[BX=AX(\sqrt{2}-1)\Rightarrow \frac{AX}{BX}=\frac{1}{\sqrt{2}-1}\] \[\therefore \]\[X\]divides AB in 1 : \[(\sqrt{2}-1).\]