question_answer

A student moves \[\sqrt{2}\,\,x\,\,km\] East from his residence and then moves \[x\,\,km\]North. He then goes x km North-East and finally he takes a turn of \[90{}^\circ \] towards right and moves a distance \[x\,\,km\] and reaches his school. What is the shortest distance of the school from his residence?

A)  \[(2\times \sqrt{2}+1)x\,\,km\]   

B) \[3x\,\,km\]

C)  \[2\sqrt{2}x\,\,km\]                   

D) \[3\sqrt{2}x\,\,km\]

Correct Answer: B

Solution :

In\[\Delta BCD,\]                         \[B{{D}^{2}}=B{{C}^{2}}+C{{D}^{2}}={{x}^{2}}+{{x}^{2}}\]             \[\Rightarrow \]   \[BD=\sqrt{2}x\]             \[\Rightarrow \]   \[BD=AE=\sqrt{2}x\]             \[\therefore \]      \[OE=OA+AE=\sqrt{2}\cdot x+\sqrt{2}\cdot x=2\sqrt{2}x\]             \[\because \]       \[BA=DE=x\] \[\therefore \]In \[\Delta ODE,\] \[O{{D}^{2}}=O{{E}^{2}}+D{{E}^{2}}\] \[\therefore \]Minimum distance \[\therefore \]\[OD=\sqrt{{{(2\sqrt{2}\cdot x)}^{2}}+{{x}^{2}}}=\sqrt{8{{x}^{2}}+{{x}^{2}}}=3x\,\,km\]