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Areas of Similar Triangles

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theorem

Theorem1: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: ∆ABC ∼ ∆PQR
Construction: Draw AM ⊥ BC and PN ⊥ QR


To prove: `"ar∆ABC"/"ar∆PQR" = ("AB"/"PQ")^2= ("BC"/"QR")^2= ("AC"/"PR")^2`


Proof: ∆ABC ∼ ∆PQR (Given)


`"AB"/"PQ"= "BC"/"QR" = "AC"/"PR"` (Corresponding parts of Congruent triangles) ...eq1


and ∠B= ∠Q (Corresponding parts of Congruent triangles)


Now, `"ar(∆ABC)"= 1/2 × "BC" × "AM"`


          `"ar(∆PQR)"= 1/2 × "QR" × "PN"`

 

`"ar(∆ABC)"/"ar(∆PQR)"= (1/2 xx "BC" xx "AM")/(1/2 xx "QR" xx "PN")`


`"ar(∆ABC)"/"ar(∆PQR)"`= `"BC"/"QR"×"AM"/"PN"` .......eq2


Again, In ∆ABM and ∆PQN
∠AMB = ∠PNQ = 90°
∠B= ∠Q (Corresponding parts of Congruent triangles)
∆ABM ∼ ∆PQN


So, `"AB"/"PQ"= "AM"/"PN"` (Corresponding parts of Congruent triangles) .....eq3


From eq2 and eq3


`"ar(∆ABC)"/"ar(∆PQR)"= "BC"/"QR"×"AB"/"PQ"` ......eq4


From eq1 and eq4


`"ar(∆ABC)"/"ar(∆PQR)"= "BC"/"QR" × "BC"/"QR"`


`"ar(∆ABC)"/"ar(∆PQR)"= ("BC"/"QR")^2` .........eq5


From eq1 and eq5


`"ar(∆ABC)"/"ar(∆PQR)"= ("AB"/"PQ")^2 = ("BC"/"QR")^2 = ("AC"/"PR")^2`


Hence proved. 

Shaalaa.com | Triangles part 37 (Example Ratio of Area of triangle)

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Triangles part 37 (Example Ratio of Area of triangle) [00:10:05]
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