Last updated at May 29, 2018 by Teachoo

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 = (๐ท๐๐ ๐ก๐๐๐๐ )/๐๐๐๐ Distance = speed ร time Now, we have to find distance BC Since North and West are perpendicular, โ BAC = 90ยฐ So, ฮ 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) = โ5490000 = โ(3ร3ร61ร100ร100) = โ((3)^2ร(100)^2 )รโ61 = 3ร100โ61 = 300 โ61 km Hence, two planes would be 300โ61 km from each other.

Chapter 6 Class 10 Triangles (Term 1)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.