Question

prove that the ratios of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Answer 1

it is given that Δabc≈Δdef

Answer 2

Step 1:

Given Data: Δ ABC ~ Δ PQR

To Prove: (ΔABC) / (ΔPQR) = $(\mathrm{AB} / \mathrm{PQ})^{2}=(\mathrm{BC} / \mathrm{QR})^{2}=(\mathrm{CA} / \mathrm{RP})^{2}$

Step 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,

Step 3:

∠B = ∠Q (Δ ABC ~ Δ PQR)

∠M = ∠N (each 90°)

Step 4:

So, Δ ABM ~ Δ PQN  

AM/PN = AB/PQ ... ………………. [ii]

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

Step 5:

Therefore Equation (i)

(ΔABC) / (ΔPQR) = BC/QR × AM/PN

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

$=(\mathrm{AB} / \mathrm{PQ})^{2}$

Step 6:

Using Equation (iii)

(ABC) / (PQR) => $(\mathrm{AB} / \mathrm{PQ})^{2}=(\mathrm{BC} / \mathrm{QR})^{2}=(\mathrm{CA} / \mathrm{RP})^{2}$