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 2025 Exams You are here

Theorem 6.7 Important Deleted for CBSE Board 2025 Exams

Theorem 6.8 Important Deleted for CBSE Board 2025 Exams

Theorem 6.9 Deleted for CBSE Board 2025 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: βABC ~ βPQR To Prove: (ππ (π΄π΅πΆ))/(ππ (πππ )) = (π΄π΅/ππ)^2 = (π΅πΆ/ππ )^2 = (π΄πΆ/ππ )^2 Construction: Draw AM β₯ BC and PN β₯ 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 βABM and βPQN β B = β Q β M = β N βABM βΌ βPQN β΄ π΄π΅/ππ=π΄π/ππFrom (A) (ππ (π΄π΅πΆ))/(ππ (πππ ))=(π΅πΆ\ Γ π΄π)/(ππ Γ ππ) (ππ (π΄π΅πΆ))/(ππ (πππ ))=π΅πΆ/ππ Γ π΄π΅/ππ Now, Given βABC ~ βPQR β π΄π΅/ππ=π΅πΆ/ππ =π΄πΆ/ππ Putting in (C) β (ππ (π΄π΅πΆ))/(ππ (πππ )) = π΄π΅/ππ Γ π΄π΅/ππ = (π΄π΅/ππ)^2 Now, again using π΄π΅/ππ=π΅πΆ/ππ =π΄πΆ/ππ β (ππ (π΄π΅πΆ))/(ππ (πππ )) = (π΄π΅/ππ)^2 = (π΅πΆ/ππ )^2 = (π΄πΆ/ππ )^2 Hence Proved.