Check sibling questions

Ex 6.5, 11 - An aeroplane leaves an airport and flies north - Ex 6.5

Ex 6.5, 11 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.5, 11 - Chapter 6 Class 10 Triangles - Part 3
Ex 6.5, 11 - Chapter 6 Class 10 Triangles - Part 4

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


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 = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 )/𝑇𝑖𝑚𝑒 Distance = speed × time Now, we have to find distance BC Since North and West are perpendicular, ∠ BAC = 90° So, Δ 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) = √5490000 = √(3×3×61×100×100) = √((3)^2×(100)^2 )×√61 = 3×100√61 = 300 √61 km Hence, two planes would be 300√61 km from each other.

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.