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Last updated at June 1, 2020 by Teachoo

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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.

Area of similar triangles

Theorem 6.6
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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.