Chapter 6 Class 10 Triangles

Class 10
Important Questions for Exam - Class 10

This video is only available for Teachoo black users

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month

### Transcript

Ex 6.5, 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 11/2 hours? Given : Speed of north flying aeroplane = 1000 km/hr. Speed of west-flying aeroplane = 1200 km/hr. To find: Distance between the two planes after 1.5 hours , i.e. , BC Solution: We know that Speed = (π·ππ π‘ππππ )/ππππ Distance = speed Γ time Now, we have to find distance BC Since North and West are perpendicular, β  BAC = 90Β° So, Ξ ABC is a right triangle Using Pythagoras theorem in right angle triangle ACB (Hypotenuse)2 = (Height)2 + (Base)2 (BC)2 = (AB)2 + (AC)2 (BC)2 = (1500)2 + (1800)2 (BC)2 = 2250000 + 3240000 (BC) = β5490000 = β(3Γ3Γ61Γ100Γ100) = β((3)^2Γ(100)^2 )Γβ61 = 3Γ100β61 = 300 β61 km Hence, two planes would be 300β61 km from each other.