# 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 \(1{\Large\frac{1}{2}}\) hours?

**Solution:**

We know that,

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.

AB is the distance travelled by aeroplane travelling towards north

AB = 1000 km/hr × \(1{\Large\frac{1}{2}}\) h

= 1000 × 3/2 km

AB = 1500 km

BC is the distance travelled by another aeroplane travelling towards west

BC = 1200 km/hr × \(1{\Large\frac{1}{2}}\) h

= 1200 × 3/2 h

BC = 1800 km

Now, In ΔABC , ∠ABC = 90°

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

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

= 2250000 + 3240000

AC^{2} = 5490000

AC = √549000

= 300√61 km

The distance between two planes after \(1{\Large\frac{1}{2}}\) hr = 300√61 km

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 6

**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 \(1{\Large\frac{1}{2}}\) hours?

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 11

**Summary:**

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. The distance between two planes after \(1{\Large\frac{1}{2}}\) hr will be 300√61 km.

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