An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after \(3\over 2\) hours.

Solution :

Let the first plane starts from O and goes upto A towards north.triangle

Where OA = (\(1000 \times {3\over 2}\)) km = 1500 km

Let the second plane starts from O at the same time and goes upto B towards west, where OB = (\(1200 \times {3\over 2}\)) km = 1800 km

According to the question, the required distance = BA.

In right triangle ABC, by Pythagoras theorem, we have :

\({AB}^2\) = \({OA}^2\) + \({OB}^2\)

= \((1500)^2\) + \((1800)^2\)

= 2250000 + 3240000

= 5490000

AB = \(3\times 100\sqrt{61}\) = \(300\sqrt{61}\).

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