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  1. Chapter 6 Class 7 Triangle and its Properties
  2. Serial order wise
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Ex 6.5, 5 A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree. Given that tree breaks down Let the part after breaking down be AB Distance of its top from base of tree be BC Since Tree is vertical ∠ B = 90° and AB = 5 m BC = 12 m In Δ ABC By Pythagoras Theorem, 〖(𝐴𝐶)〗^2 = 〖(𝐴𝐵)〗^2 + 〖(𝐵𝐶)〗^2 〖(𝐴𝐶)〗^2 = 〖(12)〗^2 + 〖(5)〗^2 〖(𝐴𝐶)〗^2 = 144 + 25 〖(𝐴𝐶)〗^2 = 169 〖(𝐴𝐶)〗^2 = 〖(13)〗^2 Cancelling square AC = 13 m Original height of tree = AB + AC = 5 + 13 = 18 m ∴ Original height of tree is 18 m

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.