Ratio of area of two smiller triangle is equal to the ratio of square of two crosponding side prove
Math class 10th
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HERE IS YOUR ANSWER##
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Area Theorem: The ratio of areas of two similar triangles is equal to the squares of the ratio of their corresponding sides.
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
HOPE IT HELP YOU
THANK YOU!!!
HERE IS YOUR ANSWER##
PLEASE REFER TO THE ATTACHMENT =>
Area Theorem: The ratio of areas of two similar triangles is equal to the squares of the ratio of their corresponding sides.
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
HOPE IT HELP YOU
THANK YOU!!!
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