Question

# In the given figure, △ABC∼△PQR. Then, area of △ ABCarea of △ PQR equals

A

AB2PQ2

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B

BC2QR2

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C

AC2PR2

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D
All of the above.
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Solution

## The correct option is D All of the above.We are given two triangles ABC and PQR such that △ABC∼△PQR. For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles ABC and PQR respectively, as shown below. Now, area of △ABC=12×BC×AM and area of △PQR=12×QR×PN So, area of △ABCarea of △PQR=12×BC×AM12×QR×PN=BC×AMQR×PN⋯(1) Now, in △ABM and △PQN, ∠B=∠Q (As △ABC∼△PQR) and ∠AMB=∠PNQ=90∘. So, △ABM∼△PQN ( By AA similarity criterion) ∴AMPN=ABPQ ⋯(2) Also, △ABC∼△PQR (Given) So, ABPQ=ACPR=BCQR ⋯(3) From (1) and (3), we get, area of (ABC)area of (PQR)=AB×AMPQ×PN =AB×ABPQ×PQ =AB2PQ2 Now using (3), we get, area of △ABCarea of △PQR=AB2PQ2=BC2QR2=AC2PR2

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