Question

If triangle ABC triangle PQR and area of triangle ABC = area of triangle PQR, then prove that triangle ABC is congruent to triangle PQľ ​

Answer 1

Step-by-step explanation:

Given:ΔABC~ΔPQR

Ar(ABC)=Ar(PQR)

To prove :ΔABC is congruent to ΔPQR

Proof :ΔABC~ΔPQR(given)

Ar(ABC)/Ar(PQR)=AB²/PQ²=BC²/QR²=AC²/PR²

Ar(ABC)/Ar(ABC)=AB²/PQ²=BC²/QR²=AC²/PR²

1=AB²/PQ²=BC²/QR²=AC²/PR²

1=AB²/PQ², 1=BC²/QR², 1=AC²/PR²

AB²=PQ², BC²=QR², AC²=PR²

AB=PQ, BC=QR,AC=PR

Now in ΔABC and ΔPQR

AB=PQ

BC=QR

AC=PR

so by SSS congruency ΔABC is congruent to ΔPQR

Hence, proved

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Answer 2

ΔABC∼ΔPQR

ar(ΔABC)=4ar(ΔPQR)

area ΔPQRarea ΔABC=14

area(ΔPQR)area(ΔABC)=QR2BC2 

14=QR2BC2

⇒(12)2=(QR12)2

12=QR12

⇒QR=212×1=6

QR=6cm