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

Answer:

Given : Δ ABC ~ Δ PQR

To Prove: (ΔABC) / (ΔPQR) = (AB/PQ)²=(BC/QR)²=(AC/PR)²

Draw AM ⊥ BC, PN ⊥ QR

(ΔABC) / (ΔPQR) = (½ × BC × AM) / (½ × QR × PN)

= BC/QR × AM/PN........................................... [I]

In Δ ABM and Δ PQN,

∠B = ∠Q (Δ ABC ~ Δ PQR)

∠M = ∠N (each 90°)

So, Δ ABM ~ Δ PQN  

AM/PN = AB/PQ ... ………………. [ii]

AB/PQ = BC/QR = CA/RP (Δ ABC ~ Δ PQR)..................... [iii]

Therefore Equation (i)

(ΔABC) / (ΔPQR) = BC/QR × AM/PN

= AB/PQ × AB/PQ [From Equation (ii) and Equation (iii)]

Using Equation (iii)

(ABC) / (PQR) =>  (AB/PQ)²=(BC/QR)²=(AC/PR)²

Step-by-step explanation: