An airplane goes 14 km due east and then 48 km due north. How far is it from its initial position?
Answer & explanation
Correct answer: option 3
⇒ Using pythagoras theorem,
⇒ \( {14 }^{2 } \) + \( {48 }^{2 } \) = 196 + 2304 = 2500
⇒ \(\sqrt { 2500}\) = 50 Km.
Therefore, plane is 50 Km far from the initial point.