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Theorems
Theorem 6.2 - Converse of Basic Proportionality Theorem
Theorem 6.3
AA Similarity Criteria
Theorem 6.4 Important
Theorem 6.5
Theorem 6.6 Important Deleted for CBSE Board 2024 Exams You are here
Theorem 6.7 Important Deleted for CBSE Board 2024 Exams
Theorem 6.8 Important Deleted for CBSE Board 2024 Exams
Theorem 6.9 Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
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.