if the areas of two similar triangles are equal prove that they are congruent
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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.
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)
Hope this will help you.....
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)
Hope this will help you.....
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abhik6:
,thanks brother
Answered by
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Step-by-step explanation:
Given :-
→ ∆ABC ~ ∆DEF such that ar(∆ABC) = ar( ∆DEF) .
➡ To prove :-
→ ∆ABC ≅ ∆DEF .
➡ Proof :-
→ ∆ABC ~ ∆DEF . ( Given ) .
Now, ar(∆ABC) = ar( ∆DEF ) [ given ] .
▶ From equation (1) and (2), we get
⇒ AB² = DE² , AC² = DF² , and BC² = EF² .
[ Taking square root both sides, we get ] .
⇒ AB = DE , AC = DF and BC = EF .
[ by SSS-congruency ] ..... . .... ......
Hence, it is proved.
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