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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Transcript

Theorem 6.6: The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides. Given: ∆ABC ~ ∆PQR To Prove: (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅)) = (𝐴𝐵/𝑃𝑄)^2 = (𝐵𝐶/𝑄𝑅)^2 = (𝐴𝐶/𝑃𝑅)^2 Construction: Draw AM ⊥ BC and PN ⊥ QR. Proof: ar (ABC) = 1/2 × Base × Height = 1/2 × BC × AM ar (PQR) = 1/2 × Base × Height = 1/2 × QR × PN ar (PQR) = 1/2 × Base × Height = 1/2 × QR × PN In ∆ABM and ∆PQN ∠B = ∠Q ∠M = ∠N ∆ABM ∼ ∆PQN ∴ 𝐴𝐵/𝑃𝑄=𝐴𝑀/𝑃𝑁From (A) (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅))=(𝐵𝐶\ × 𝐴𝑀)/(𝑄𝑅 × 𝑃𝑁) (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅))=𝐵𝐶/𝑄𝑅 × 𝐴𝐵/𝑃𝑄 Now, Given ∆ABC ~ ∆PQR ⇒ 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅=𝐴𝐶/𝑃𝑅 Putting in (C) ⇒ (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅)) = 𝐴𝐵/𝑃𝑄 × 𝐴𝐵/𝑃𝑄 = (𝐴𝐵/𝑃𝑄)^2 Now, again using 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅=𝐴𝐶/𝑃𝑅 ⇒ (𝑎𝑟 (𝐴𝐵𝐶))/(𝑎𝑟 (𝑃𝑄𝑅)) = (𝐴𝐵/𝑃𝑄)^2 = (𝐵𝐶/𝑄𝑅)^2 = (𝐴𝐶/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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.