Question

An aeroplane leaves an airport and files due north at a speed of 1000Km per hour. at the same time, another aeroplane leaves the same airport and files due west at a speed of 1200 km per hour . How far apart will be two planes after 1½ hours​

Answer 1

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.

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?

AB is the distance travelled by aeroplane travelling towards north

AB = 1000 km/hr × 112 h

= 1000 × 3/2 km

AB = 1500 km

BC is the distance travelled by another aeroplane travelling towards west

BC = 1200 km/hr × 112 h

= 1200 × 3/2 h

BC = 1800 km

Now, In ΔABC , ∠ABC = 90°

AC2 = AB2 + BC2 (Pythagoras theorem)

= (1500)2 + (1800)2

= 2250000 + 3240000

AC2 = 5490000

AC = √549000

= 300√61 km

• The distance between two planes after 112 hr = 300√61 km

Answer 2

Answer:

The first aeroplane leaves an airport and flies due north at a speed of 1000 km per hr.

∴ Distance travelled in 1.5 hrs.

= 1000 × 1.5 = 1500 km.

Similarly, Distance traveled by second

aeroplane,

= 1200 × 1.5 = 1800 km.

In △ABC,

BC is distance travelled by first aeroplane and BA is the distance traveled by second aeroplane.

Hence, applying Pythagoras Theorem,

AC² = AB² + BC².

∴ AC² = 1500² + 1800².

∴ AC = 300√61 km.

Hence, two planes are 300√61 km apart from each other.

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