Prove that the ratio of the areas of two similar triangles is equal to

1.	Prove that the ratio of the areas of two similar triangles is equal to the square of the ratioof their corresponding sides.29.Using the above, do the followingIn a trapezium ABCD, AC and BD intersecting at O, AB II CD, ABAAOB :-84 cm2, find the area of ÎCOD.2CD. If area of
Answer» 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 (ΔABC ~ 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

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratioof their corresponding sides.29.Using the above, do the followingIn a trapezium ABCD, AC and BD intersecting at O, AB II CD, ABAAOB :-84 cm2, find the area of ÎCOD.2CD. If area of

Answer»

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 (ΔABC ~ 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