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  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Theorem 6.9: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. Given: A triangle ABC in which 〖𝐴𝐢〗^2=〖𝐴𝐡〗^2+〖𝐡𝐢〗^2 To Prove: ∠B=90Β° Construction: Draw Ξ” PQR right angled at Q, such that PQ = AB and QR = BC. Proof: In βˆ†PQR ∠Q=90Β° By Pythagoras theorem, 〖𝑃𝑅〗^2=〖𝑃𝑄〗^2+〖𝑄𝑅〗^2 Since PQ = AB and QR = BC 〖𝑃𝑅〗^2=〖𝐴𝐡〗^2+〖𝐡𝐢〗^2 Also, given that 〖𝐴𝐢〗^2=〖𝐴𝐡〗^2+〖𝐡𝐢〗^2 From (1) & (2) 〖𝑃𝑅〗^2=〖𝐴𝐢〗^2 PR = AC In Ξ” ABC & Ξ” PQR AC = PR AB = PQ BC = QR ∴ Ξ” ABC β‰… Ξ” PQR β‡’ ∠ B = ∠ Q Since ∠ Q = 90Β° ∴ ∠ B = 90Β° Hence Proved.

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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.