Question

Prove that, 'the ratio of areas of two triangles is equal to the square of the ratio of their corresponding sides ​

Answer 1

Theorem :

• The areas of two similar triangles are proportional to the squares on their corresponding sides.

Given :

• ΔABC ~ ΔDEF.

To prove :

• $\bold{\dfrac{Area\;of\;\triangle ABC}{Area\;of\;\triangle DE{F}} =\dfrac{AB^{2}}{DE^{2}}=\dfrac{BC^{2}}{EF^{2}}=\dfrac{AC^{2}}{DF^{2}} }$.

Construction :

• Draw AL ⊥ BC and DM ⊥ EF.

Proof :

1. Area of ΔABC/Area of ΔDEF :

• [1/2 × BC × AL]/[1/2 × EF × DM]

∵ Area of Δ = 1/2 × Base × Height

∴ Area of ΔABC/Area of ΔDEF :

• BC/EF × AL/DM .. (i)

2. In ΔALB and ΔDME, we have :

• ∠ALB = ∠DME

∵ Each angle to 90°.

• ∠ABL = ∠DEM

∵ ΔABC ~ ΔDEF ⇒ ∠B = ∠E

• ΔALB ~ ΔDME

∵ AA-axiom for similarity of Δs.

• AL/DM = AB/DE .. (ii)

∵ Corresponding sides of similar Δs are proportional.

3. ΔABC ~ ΔDEF, we have :

• AB/DE = BC/EF = AC/DF .. (iii)

∵ 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 :

• AB²/DE² = BC²/EF² = AC²/DF²

Hence,  

• $\bold{\dfrac{Area\;of\;\triangle ABC}{Area\;of\;\triangle DE{F}} =\dfrac{AB^{2}}{DE^{2}}=\dfrac{BC^{2}}{EF^{2}}=\dfrac{AC^{2}}{DF^{2}} }$.