Theorems

Chapter 6 Class 10 Triangles
Serial order wise

### 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: βABC ~ βPQR To Prove: (ππ (π΄π΅πΆ))/(ππ (πππ)) = (π΄π΅/ππ)^2 = (π΅πΆ/ππ)^2 = (π΄πΆ/ππ)^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 βABM and βPQN β B = β Q β M = β N βABM βΌ βPQN β΄ π΄π΅/ππ=π΄π/ππFrom (A) (ππ (π΄π΅πΆ))/(ππ (πππ))=(π΅πΆ\ Γ π΄π)/(ππ Γ ππ) (ππ (π΄π΅πΆ))/(ππ (πππ))=π΅πΆ/ππ Γ π΄π΅/ππ Now, Given βABC ~ βPQR β π΄π΅/ππ=π΅πΆ/ππ=π΄πΆ/ππ Putting in (C) β (ππ (π΄π΅πΆ))/(ππ (πππ)) = π΄π΅/ππ Γ π΄π΅/ππ = (π΄π΅/ππ)^2 Now, again using π΄π΅/ππ=π΅πΆ/ππ=π΄πΆ/ππ β (ππ (π΄π΅πΆ))/(ππ (πππ)) = (π΄π΅/ππ)^2 = (π΅πΆ/ππ)^2 = (π΄πΆ/ππ)^2 Hence Proved.