Question

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

Answer 1

Let the two triangles be:

ΔABC and ΔPQR

Area of ΔABC=21×BC×AM……………..(1)

Area of ΔPQR=21×QR×PN……………………..(2)

Dividing (1) by (2)

ar(PQR)ar(ABC)=21×QR×PN21×BC×AM

ar(PQR)ar(ABC)=QR×PNBC×AM…………………..(1)

In ΔABM and ΔPQN

∠B=∠Q (Angles of similar triangles

M=∠N (Both 90∘)

Therefore, ΔABM∼ΔPQN

So, AMAB=PNPQ…………………….(2)

From 1 and 2

ar(PQR)ar(ABC)=QRBC×PNAM

⇒ar(PQR)ar(ABC)=QRBC×PQAB…………………..(3)

PQAB=QRBC=PRAC………….(ΔABC∼ΔPQR)

Putting in ( 3 )ar(PQR)ar(ABC)=PQAB×PQAB=(PQAB)2

⇒ar(PQR)ar(ABC)=(PQAB)2=(QRBC)2=(PRAC)2

Answer 2

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