An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after \({1} {\dfrac{1}{2}}\) hours?

Given,
Speed of first aeroplane = 1000 km/hr

Distance covered by first aeroplane flying due north in

\({1} {\dfrac{1}{2}}\) hours = 100 × \(\frac{3}{2} \) km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance covered by second aeroplane flying due west in

\({1} {\dfrac{1}{2}}\) hours (OB) = 1200 × \(\frac{3}{2} \) km = 1800 km



In right angle \(\triangle\) AOB, by Pythagoras Theorem,
\(\Rightarrow AB^2 = AO^2 + OB^2\)
\(\Rightarrow AB^2 = (1500)^2 + (1800)^2\)
\(\Rightarrow AB = \sqrt {(2250000 + 3240000)}\)
\( \Rightarrow AB = \sqrt{5490000}\)
\(\Rightarrow \) AB = \(300 \sqrt{61}\) km