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Ex 6.5, 12 - Two poles of heights 6 m and 11 m stand on - Pythagoras Theoram - Finding value

Ex 6.5, 12 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.5, 12 - Chapter 6 Class 10 Triangles - Part 3

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Ex 6.5, 12 Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops. Given: Height of first pole = AB = 6 m Height of second pole = CD = 11 m Distance b/w feet of poles = AC = 12 m To Find :- Distance between the tops of the pole ,i.e., BD Solution :- Let we draw a line BE perpendicular to DC i.e. BE DC Since AC is also perpendicular to DC as pole is vertical to ground, So, BE = AC = 12 m Similarly , AB = EC = 6 m Now, DE = DC EC DE = 11 6 DE = 5 m Since BED = 90 as BE DC BED is right triangle Using Pythagoras theorem in right angle triangle AEB (Hypotenuse)2 = (Height)2 + (Base)2 (BD)2 = (DE)2 + (BE)2 (BD)2 = (5)2 + (12)2 (BD)2 = 25 + 144 (BD)2 = 169 BD = 169 BD = (13 13) BD = ((13)2) BD = 13 Hence, the distance between tops of the pole = 13 metre .

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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.