Question

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

Answer 1

I think it helps you to solve your question

Answer 2

Step-by-step explanation:

Given :∆ABC ~ ∆PQR

RTP:

$\frac{Area\triangle ABC}{Area\triangle PQR}=\left(\frac{AB}{PQ}\right)^{2}=\left(\frac{BC}{QR}\right)^{2}=\left(\frac{CA}{RP}\right)^{2}$

Construction:

Draw AM perpendicular to BC and PN perpendicular to QR.

Proof:

$\frac{Area\triangle ABC}{Area\triangle PQR}\\=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times QR\times PN}\\=\frac{BC\times AM}{QR\times PN}\:---(1)$

$In \: \triangle \:and \: \triangle PQN$

$\angle B = \angle Q\\(Since\: \triangle ABC = \triangle PQR )$

$\angle M = \angle N=90\degree$

∆ABM~∆PQN ( By AA similarity )

$\frac{AM}{PN}=\frac{AB}{PQ}\:--(2)$

Also ∆ABC ~ ∆PQR (given)

$\boxed {\frac{AB}{PQ}=\frac{BC}{QR}}=\frac{AC}{PR}\:---(3)$

$\frac{Area\triangle ABC}{Area\triangle PQR}=\frac{AB}{PQ}\times \frac{AB}{PQ}$

\* From (1),(2) and (3)

$= \left(\frac{AB}{PQ}\right)^{2}.$

Now, by using (3) , we get

$\frac{Area\triangle ABC}{Area\triangle PQR}=\left(\frac{AB}{PQ}\right)^{2}=\left(\frac{BC}{QR}\right)^{2}=\left(\frac{CA}{RP}\right)^{2}$

Hence proved .

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