A) \[300\sqrt{67}\,km\]
B) \[400\sqrt{61}\,km\]
C) \[200\sqrt{61}\,km\]
D) \[300\sqrt{61}\,km\]
Correct Answer: D
Solution :
Let the point A represent the airport. Let the plane-l fly towards North. \[\therefore \] Distance of the plane-l from the airport after \[1\frac{1}{2}\] hours = Speed \[\times \]Time \[=1000\times 1\frac{1}{2}\,km=1500\,km\] Let the plane-II fly towards West. \[\therefore \] Distance of the plane-II from the airport after \[1\frac{1}{2}\] hours \[=1200\times 1\frac{1}{2}km=1800\,km\] Now, in right \[\Delta ABC,\] using Pythagoras theorem, we have \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\] \[\Rightarrow \] \[B{{C}^{2}}={{(1800)}^{2}}+{{(1500)}^{2}}=5490000\] \[\Rightarrow \] \[B{{C}^{2}}=\sqrt{5490000}=300\sqrt{61}km\] Thus, after \[1\frac{1}{2}\]hours, the two planes will be \[300\sqrt{61}km,\] apart from each other.You need to login to perform this action.
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