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  1. Chapter 6 Class 7 Triangle and its Properties
  2. Serial order wise
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Ex 6.5, 1 PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.In βˆ†PQR, right angled at P, PR = 24 cm, PQ = 10 cm, QR = ? By Pythagoras theorem, 〖𝑄𝑅〗^2 = 〖𝑃𝑄〗^2 + 〖𝑃𝑅〗^2 〖𝑄𝑅〗^2 = γ€–(10)γ€—^2 + γ€–(24)γ€—^2 〖𝑄𝑅〗^2 = 100 + 576 〖𝑄𝑅〗^2 = 676 〖𝑄𝑅〗^2 = γ€–(26)γ€—^2 Cancelling squares 𝑄𝑅 = 26 ∴ 𝑸𝑹 = 26 Rough 676 = 2 Γ— 2 Γ— 13 Γ— 13 = (2 Γ— 13) Γ— (2 Γ— 13) = 26 Γ— 26 = 262

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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.