Math, asked by manaalpreearf, 1 year ago

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

Answers

Answered by nikitasingh79
1634
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.....
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Answered by Anonymous
501
▶ Answer :-

▶ Step-by-step explanation :-

➡ Given :-

→ ∆ABC ~ ∆DEF such that ar(∆ABC) = ar( ∆DEF) .

➡ To prove :-

→ ∆ABC ≅ ∆DEF .

➡ Proof :-

→ ∆ABC ~ ∆DEF . ( Given ) .

\tiny \sf \implies \frac{ar( \triangle ABC )}{ ar( \triangle D EF )} = \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2} = \frac{BC^2}{EF^2} .........(1) . \\

Now, ar(∆ABC) = ar( ∆DEF ) [ given ] .

\sf \implies \frac{ar( \triangle ABC )}{ ar( \triangle D EF )} = 1..........(2). \\ <br />

▶ From equation (1) and (2), we get

\sf \implies \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2} = \frac{BC^2}{EF^2} = 1 . \\ <br />

⇒ AB² = DE² , AC² = DF² , and BC² = EF² .

[ Taking square root both sides, we get ] .

⇒ AB = DE , AC = DF and BC = EF .

 \large\pink{ \boxed{ \tt \therefore \triangle ABC \cong \triangle D EF .}}

[ by SSS-congruency ] .

Hence, it is proved.
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