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Question

Prove that the ratio of the area of two similar triangles is equal to the ratio of squares on the corresponding sides.

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Solution

Given:ABCPQR

To prove :ar(ABC)ar(PQR)=(ABPQ)2=(BCQR)2=(ACPR)2

Proof:-
arABC=12×base×height=12×BC×AM(i)

ar(PQR)=12×base×hight=12×QP×PN(ii)

ar(ABC)ar(PQR)=BC×AMQR×PN(A)

In ABM and PQN,<B=<Q,<M=<N

ABMPQN

ABPQ=AMPN(B)

From (A)

ar(ABC)ar(PQR)=BC×AMQR×PN=BCQR×ABPQ(C)

Now, given ABCPQRABPQ=BCQP=ACPR

Putting in (C)ar(ABC)ar(PQR)=(ABPQ)2

Similarly
ar(ABC)ar(PQR)=(ABPQ)2(BCQR)2=(ACPR)2
Hence proved


1378457_1146691_ans_783dcd07848a422a9d8b0da03f7f2b0e.png

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