Theorem 9.2 - Class 9 - Two triangles on same base between parallels

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Theorem 9.2 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

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Theorem 9.2 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3 Theorem 9.2 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 4

  1. Chapter 9 Class 9 - Areas of Parallelograms and Triangles (Deleted)
  2. Serial order wise

Transcript

Theorem 9.2 (Method 1) Two triangles on the same base (or equal bases) and between the same parallels are equal in area. Given : βˆ†ADB and βˆ†ABC are triangles on same base AB and between the same parallels AB and DC To Prove : π‘Žr (βˆ†ADB) = π‘Žr (βˆ†ABC) Proof: We know that Area of triangle = 1/2Γ—π΅π‘Žπ‘ π‘’Γ—π»π‘’π‘–π‘”β„Žπ‘‘ Here AB is base We draw DE βŠ₯ AB So, DE is the height of βˆ† ABD ∴ Area βˆ† ABD = 1/2Γ—π΅π‘Žπ‘ π‘’Γ—π»π‘’π‘–π‘”β„Žπ‘‘ = 1/2×𝐴𝐡×𝐷𝐸 Here AB, is base We draw CF βŠ₯ AB So, CF is the height of βˆ† ABC ∴ Area βˆ† ABC = 1/2Γ—π΅π‘Žπ‘ π‘’Γ—π»π‘’π‘–π‘”β„Žπ‘‘ = 1/2×𝐴𝐡×𝐢𝐹 DE & CF are perpendicular between the same parallel line ∴ DE = CF So, from (1) Area βˆ† ABD = 1/2×𝐴𝐡×𝐷𝐸 = 1/2×𝐴𝐡×𝐢𝐹 = Area βˆ† ABC ∴ Area βˆ† ABD = Area of βˆ† ABC Hence proved. Theorem 9.2 Two triangles on the same base (or equal bases) and between the same parallels are equal in area. Given : βˆ†ADB and βˆ†ABC are triangles on same base AB and between the same parallels AB and DC To Prove : π‘Žr (βˆ†ADB) = π‘Žr (βˆ†ABC) Construction : Construct a line through A parallel to be meeting DC at E i.e. AE βˆ₯ BC & Construct a line through B parallel to AD meeting DC at F i.e. BF βˆ₯ AD Proof : From (1) π‘Žr (ABFD) = π‘Žr (ABCE) 2 π‘Žr (βˆ†ABD) = 2 π‘Žr (βˆ†ABC) π‘Žr (βˆ†ABD) = π‘Žr (βˆ†ABC) 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 10 years. He provides courses for Maths and Science at Teachoo.