 Hello and welcome to the session. Let us understand the following question today. An aeroplane leaves an airport and flies due north at a speed of 10 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 will be the two planes after one and a half hours? First aeroplane flies due north at a speed of 1000 km per hour. And second aeroplane flies due west at a speed of 1200 km per hour. Now, let us write the solution. Let us see the diagram. Let the first aeroplane starts from O and goes up to A towards north where distance covered away is equal to speed into time. That is, away is equal to 1000 multiplied by 3 by 2 which is given to us. So, this gets cancelled by 500 so away is equal to 1500 km. Now, let us consider the second aeroplane which starts from O at the same time but goes up to B towards west where distance covered OB is equal to speed into time. That is, OB is equal to 1200 multiplied by 3 by 2. Here 1200 get cancelled by 2 so we get here 600 therefore OB is equal to 1800 km. Here we can see OB is equal to 1800 km. Now, according to the problem AB is the required distance. So, now in right triangle OAB by Pythagoras theorem AB square is equal to OA square plus OB square. AB square is equal to 1500 square plus 1800 square which implies AB square is equal to 225000 plus 324000 which implies AB square is equal to 549000 which implies AB is equal to square root of 549000 which implies AB is equal to 361 hence distance between two planes after one and a half hours is equal to 361 km. Hence, the required answer is 361 km. I hope you understood the question. Bye and have a nice day.