Prove that, 'the ratio of areas of two triangles is equal to the square of the ratio of their corresponding sides
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Prove that, 'the ratio of areas of two triangles is equal to the square of the ratio of their corresponding sides
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Verified answer
Theorem :
Given :
To prove :
Construction :
Proof :
1. Area of ΔABC/Area of ΔDEF :
∵ Area of Δ = 1/2 × Base × Height
∴ Area of ΔABC/Area of ΔDEF :
2. In ΔALB and ΔDME, we have :
∵ Each angle to 90°.
∵ ΔABC ~ ΔDEF ⇒ ∠B = ∠E
∵ AA-axiom for similarity of Δs.
∵ Corresponding sides of similar Δs are proportional.
3. ΔABC ~ ΔDEF, we have :
∵ Corresponding sides of similar Δs are proportional.
4. AL/DM = BC/EF
∵ From (i) and (iii).
5. Substituting AL/DM = BC/EF in (i), we get ;
⇒ Area of ΔABC/Area of ΔDEF = BC²/EF² .. (iv)
6. Combining (iii) and (iv), we get :
⇒ Area of ΔABC/Area of ΔDEF :
Hence,