Question

prove that the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides ​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

Proof : We are given 2 triangles ABC and POR such that     ABC ~     PQR.  

We need to prove that (Ar (ABC))/(Ar(POR))=(AB/PQ)^2=(BC/QR)^2= (CA/RP)^2  

For finding the area of the 2 triangles, we draw altitudes AM and PN of the Triangles.

Now, ar (ABC)= 1/2BC*AM

And    ar (PQR) = 1/2QR*PN

(ar (ABC))/(ar (PQR))=(1/2 BC*AM)/(1/2 QR*PN)=(BC*AM)/(QR*PN)

Now, in     ABM and      PQN

      <B=<Q   (As      ABC ~     PQR)

and                              <M=<N   (Each is of 90°)

So,                       ABM    ~      PQN  (AA similarity criterion)

Therefore,  AM/PN=AB/PQ

Also,                     ABC    ~        PQR  (Given)

So,          AB/PQ=BC/QR=CA/RP

Therefore,         (Ar(ABC))/(Ar (PQR))=AB/PQ*AM/PN

                         (Ar(ABC))/(Ar (PQR))=AB/PQ*AB/PQ

                      (Ar(ABC))/(Ar (PQR))=(AB/PQ)^2

(Ar(ABC))/(Ar (PQR))=(AB/PQ)^2=(BC/QR)^2=(CA/RP)^2

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