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

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  1. Chapter 6 Class 10 Triangles
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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 = (๐ท๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ )/๐‘‡๐‘–๐‘š๐‘’ Distance = speed ร— time Now, we have to find distance BC Since North and West are perpendicular, โˆ  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) = โˆš5490000 = โˆš(3ร—3ร—61ร—100ร—100) = โˆš((3)^2ร—(100)^2 )ร—โˆš61 = 3ร—100โˆš61 = 300 โˆš61 km Hence, two planes would be 300โˆš61 km from each other.

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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.