An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after $1\frac{1}{2}$ hours?

AB is the distance travelled by aeroplane travelling towards north

AB = 1000 km/hr × $1\frac{1}{2}$hr

AB = 1000 × 3/2 km

AB = 1500 km

BC is the distance travelled by another aeroplane travelling towards the west.

BC = 1200 km/hr × $1\frac{1}{2}$hr

BC = 1200 × 3/2 km

BC = 1800 km

Now, In ΔABC , ∠ABC = 90°

AC² = AB² + BC²    (By Pythagoras theorem)

= (1500)² + (1800)²

= 2250000 + 3240000

AC² = 5490000

AC = √5490000

= 300√61 km

The distance between two planes after $1\frac{1}{2}$hr = 300√61 km.