A) 50 m
B) 100 m
C) 40 m
D) 260 m
Correct Answer: B
Solution :
The movements of the child from A to E are as shown in figure. Clearly, the child meets his father at E. Now, \[AF=(AB-FB)\] \[=(AB-DC)=(90-30)m=60m\] \[EF=(DE-DF)=(DE-BC)\] \[=(100-20)m=80\,m\] \[\therefore \] Required distance \[=AE=\sqrt{A{{F}^{2}}+E{{F}^{2}}}=\sqrt{{{(60)}^{2}}+{{(80)}^{2}}}\] \[=\sqrt{3600+6400}=\sqrt{10000}=100\,m\]You need to login to perform this action.
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