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Ex 6.5

Ex 6.5, 1
Deleted for CBSE Board 2023 Exams

Ex 6.5, 2 Important Deleted for CBSE Board 2023 Exams

Ex 6.5, 3 Important Deleted for CBSE Board 2023 Exams

Ex 6.5, 4 Deleted for CBSE Board 2023 Exams

Ex 6.5, 5 Deleted for CBSE Board 2023 Exams

Ex 6.5, 6 Deleted for CBSE Board 2023 Exams

Ex 6.5, 7 Important Deleted for CBSE Board 2023 Exams

Ex 6.5, 8 Important Deleted for CBSE Board 2023 Exams

Ex 6.5, 9 Deleted for CBSE Board 2023 Exams

Ex 6.5, 10 Deleted for CBSE Board 2023 Exams

Ex 6.5, 11 Important Deleted for CBSE Board 2023 Exams You are here

Ex 6.5, 12 Important Deleted for CBSE Board 2023 Exams

Ex 6.5, 13 Deleted for CBSE Board 2023 Exams

Ex 6.5, 14 Deleted for CBSE Board 2023 Exams

Ex 6.5, 15 Important Deleted for CBSE Board 2023 Exams

Ex 6.5, 16 Deleted for CBSE Board 2023 Exams

Ex 6.5, 17 (MCQ) Deleted for CBSE Board 2023 Exams

Chapter 6 Class 10 Triangles

Serial order wise

Last updated at May 29, 2018 by Teachoo

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·ππ π‘ππππ )/ππππ 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) = β5490000 = β(3Γ3Γ61Γ100Γ100) = β((3)^2Γ(100)^2 )Γβ61 = 3Γ100β61 = 300 β61 km Hence, two planes would be 300β61 km from each other.