Theorem 6.6 - Chapter 6 Class 10 Triangles

Last updated at Dec. 16, 2024 by

Theorem 6.6 Class 10 - Prove that Ratio of Areas of 2 Similar Triangle

Theorem 6.6 - Chapter 6 Class 10 Triangles - Part 2
Theorem 6.6 - Chapter 6 Class 10 Triangles - Part 3
Theorem 6.6 - Chapter 6 Class 10 Triangles - Part 4

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo