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  1. Chapter 9 Class 9 Areas of Parallelograms and Triangles
  2. Serial order wise
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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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.