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 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 April 16, 2024 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.