Question

if the areas of two similar triangles are equal,prove that they are congruent.

Answer 1

Let △ABC and △DEF are two similar triangles having same area.

Therefore, we can write ar(△ABC)ar(△DEF)=1     (1)

 

But, we also have ar(△ABC)ar(△DEF)=AB2DE2=BC2EF2=AC2DF2               (2)

{The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides}

 

From (1) and (2), we can say that

AB2DE2=BC2EF2=AC2DF2=1

⇒AB2=DE2,BC2=EF2,AC2=DF2

⇒AB=DE,BC=EF,AC=DF

 

Therefore, by SSS, we, can say that △ABC≅△DEF

 

Hence Proved

Answer 2

Use the theorem that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides , then prove that they are congruent.
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Solution:

[Fig is in the attachment]

Given: ΔABC ~ ΔPQR. &

ar ΔABC =ar ΔPQR

To Prove: ΔABC ≅ ΔPQR

Proof: Since, ΔABC ~ ΔPQR
ar ΔABC =ar ΔPQR. (given)

ΔABC / ar ΔPQR = 1

⇒ AB²/PQ² = BC²/QR² = CA²/PR² = 1

[ USING THEOREM OF AREA OF SIMILAR TRIANGLES]

⇒ AB= PQ , BC= QR & CA= PR

Thus, ΔABC ≅ ΔPQR
[BY SSS criterion of congruence]

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Hope this will help you.....