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

(N/A) Distance travelled by the plane flying towards north in $1 \frac{1}{2}$ hours $= 1000 \times 1.5 = 1500\, km$.
Distance travelled by the plane flying towards west in $1 \frac{1}{2}$ hours $= 1200 \times 1.5 = 1800\, km$.
Let the airport be at the origin $O$. The plane flying north reaches point $A$ and the plane flying west reaches point $B$.
Since north and west directions are perpendicular,$\triangle AOB$ is a right-angled triangle with $\angle AOB = 90^\circ$.
By Pythagoras theorem,the distance between the two planes is the hypotenuse $AB = \sqrt{OA^2 + OB^2}$.
$AB = \sqrt{1500^2 + 1800^2} = \sqrt{2250000 + 3240000} = \sqrt{5490000}$.
$AB = \sqrt{90000 \times 61} = 300\sqrt{61}\, km$.
Thus,the distance between the two planes after $1 \frac{1}{2}$ hours is $300\sqrt{61}\, km$.