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°
AC2 = AB2 + BC2 (Pythagoras theorem)
= (1500)2 + (1800)2
= 2250000 + 3240000
AC2 = 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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