A) 80 m
B) 100 m
C) 140 m
D) 260 m
Correct Answer: B
Solution :
Clearly, the child moves form A, 90 m eastwards upto B then turns right and moves 20 m upto C, then turns right am moves 30 m upto D. Finally, he turns right and moves 100 m upto E. Clearly, \[AB=90\,\,m,\,\,BF=CD=30\,\,m\] So \[AF=AB-BF=60\,\,m\] Also, \[DE=100\,\,m,\,\,DF=BC=20\,\,m\] So, \[EF=DE-DF=80\,\,m\] \[\therefore \]His distance from starting point A \[=AE=\sqrt{A{{F}^{2}}+E{{F}^{2}}}\] \[=\sqrt{{{(60)}^{2}}+{{(80)}^{2}}}=\sqrt{3600+6400}\] \[=\sqrt{10000}=100\,\,m\]You need to login to perform this action.
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