if the area of two similar triangles are equal prove that they are congruent
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16
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.________________________________________________________
Solution:
[Fig is in the attachment]
Given: ΔABC ~ ΔPQR. &
ar ΔABC =ar ΔPQR
To Prove: ΔABC ≅ ΔPQR
Proof: Since, ΔABC ~ ΔPQRar 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]
Solution:
[Fig is in the attachment]
Given: ΔABC ~ ΔPQR. &
ar ΔABC =ar ΔPQR
To Prove: ΔABC ≅ ΔPQR
Proof: Since, ΔABC ~ ΔPQRar 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]
manish410:
thanks
Answered by
6
ar(∆ABC)=ar(∆pqr). - 1
√ar(∆abc)\ar(pqr)=ab\pq
from both,
√1=ab\pq
1=ab\pq
ab=pq
therefore, ∆abc=~∆pqr
√ar(∆abc)\ar(pqr)=ab\pq
from both,
√1=ab\pq
1=ab\pq
ab=pq
therefore, ∆abc=~∆pqr
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