Areas of Similar Triangles

Class X

Theorem

"The ratios of areas of two triangles is equal to the square of the ratio of their corresponding sides".

Solution:-

Given

△ABC ~ △PQR

Therefore,

AB/PQ = BC/QR = CA/RP

To prove that

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

Construction

Draw AM ⫡ BC and PN ⫡ QR

Proof

ar(ABC) = BCxAM/2 ---------------- (i)

ar(PQR) = QRxPN/2 ----------------- (ii)

Now, from (i) and (ii)

ar(ABC)/ar(PQR) = BCxAM/QRxPN ------------ (iii)

Now, consider in △ABM and △PQN

<B = <Q     (△ABC ~ △PQR)

<M = <N    ( Each 90⁰)

Hence,

△ABM ~ △PQN  (AA similarity)

Therefore,

AM/PN = AB/PQ

Hence,

AM/PN = AB/PQ = BC/QR = CA/RP

Now, from (iii)

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)² Proved.