Question

Prove that ratio of areas of two similar triangle is equal to the square of the ratio of their corresponding sides.

Answer 1

Answer:

Step-by-step explanation:Given: Δ ABC ~ Δ PQR

To Prove: ar(ΔABC) / ar(ΔPQR) = (AB/PQ)2 = (BC/QR)2 = (CA/RP)2

Construction: Draw AM ⊥ BC, PN ⊥ QR

ar(ΔABC) / ar(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 (AA similarity criterion)

Therefore, AM/PN = AB/PQ ... [ii]

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

Hence, from (i)

ar(ΔABC) / ar(ΔPQR) = BC/QR × AM/PN

= AB/PQ × AB/PQ [From (ii) and (iii)]

= (AB/PQ)2

Using (iii)

ar(ΔABC) / ar(ΔPQR) = (AB/PQ)2 = (BC/QR)2 = (CA/RP)2

Answer 2

Answer:

Step-by-step explanation:

Let similar triangles be ABC and PQR

Since, triangles are similar

Therefore, AB/PQ=BC/QR=AC/PR

Area of ABC /Area of PQR=1/2.BC.AD/1/2.QR.PM

Triangle ABD is similar to triangle PQM

Therefore, AB/PQ=BD/QM=AD/PM

from above

Area of ABC/Area of PQR=BC2/QR2=AB2/PQ2=AC2/PR2

Hence, proved