# 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(1/2) hours?

**Solution:**

We have to find the distance travelled by aeroplanes, we need to use

Distance = speed × time

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. [Pythagoras theorem]

AB is the distance travelled by aeroplane travelling towards north

AB = 1000 km/hr × 1(1/2) hr

= 1000 × 3/2 km

AB = 1500 km

BC is the distance travelled by another aeroplane travelling towards south

BC = 1200 km/hr × 1(1/2) hr

= 1200 × 3/2 hr

BC = 1800 km

Now, In ΔABC , ∠ABC = 90°

AC^{2} = AB^{2} + BC^{2} (Pythagoras theorem)

= (1500)^{2} + (1800)^{2}

= 2250000 + 3240000

AC² = 5490000

AC = √549000

= 300√61 km

The distance between two planes after 1(1/2) hr = 300√61 km

**Video Solution:**

## 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(1/2) hours?

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.5 Question 11:

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(1/2) hours?

The distance between two planes after 1(1/2) hr = 300√61 km.