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  1. Chapter 6 Class 10 Triangles
  2. Serial order wise


Theorem 6.6: The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides. Given: βˆ†ABC ~ βˆ†PQR To Prove: (π‘Žπ‘Ÿ (𝐴𝐡𝐢))/(π‘Žπ‘Ÿ (𝑃𝑄𝑅)) = (𝐴𝐡/𝑃𝑄)^2 = (𝐡𝐢/𝑄𝑅)^2 = (𝐴𝐢/𝑃𝑅)^2 Construction: Draw AM βŠ₯ BC and PN βŠ₯ QR. Proof: ar (ABC) = 1/2 Γ— Base Γ— Height = 1/2 Γ— BC Γ— AM ar (PQR) = 1/2 Γ— Base Γ— Height = 1/2 Γ— QR Γ— PN ar (PQR) = 1/2 Γ— Base Γ— Height = 1/2 Γ— QR Γ— PN In βˆ†ABM and βˆ†PQN ∠B = ∠Q ∠M = ∠N βˆ†ABM ∼ βˆ†PQN ∴ 𝐴𝐡/𝑃𝑄=𝐴𝑀/𝑃𝑁From (A) (π‘Žπ‘Ÿ (𝐴𝐡𝐢))/(π‘Žπ‘Ÿ (𝑃𝑄𝑅))=(𝐡𝐢\ Γ— 𝐴𝑀)/(𝑄𝑅 Γ— 𝑃𝑁) (π‘Žπ‘Ÿ (𝐴𝐡𝐢))/(π‘Žπ‘Ÿ (𝑃𝑄𝑅))=𝐡𝐢/𝑄𝑅 Γ— 𝐴𝐡/𝑃𝑄 Now, Given βˆ†ABC ~ βˆ†PQR β‡’ 𝐴𝐡/𝑃𝑄=𝐡𝐢/𝑄𝑅=𝐴𝐢/𝑃𝑅 Putting in (C) β‡’ (π‘Žπ‘Ÿ (𝐴𝐡𝐢))/(π‘Žπ‘Ÿ (𝑃𝑄𝑅)) = 𝐴𝐡/𝑃𝑄 Γ— 𝐴𝐡/𝑃𝑄 = (𝐴𝐡/𝑃𝑄)^2 Now, again using 𝐴𝐡/𝑃𝑄=𝐡𝐢/𝑄𝑅=𝐴𝐢/𝑃𝑅 β‡’ (π‘Žπ‘Ÿ (𝐴𝐡𝐢))/(π‘Žπ‘Ÿ (𝑃𝑄𝑅)) = (𝐴𝐡/𝑃𝑄)^2 = (𝐡𝐢/𝑄𝑅)^2 = (𝐴𝐢/𝑃𝑅)^2 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 9 years. He provides courses for Maths and Science at Teachoo.