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Theorems

Theorem 6.1 - Basic Proportionality Theorem (BPT)
Important

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 2023 Exams You are here

Theorem 6.7 Important Deleted for CBSE Board 2023 Exams

Theorem 6.8 Important Deleted for CBSE Board 2023 Exams

Theorem 6.9

Chapter 6 Class 10 Triangles

Serial order wise

Last updated at March 23, 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.