Question

theorem 2.6
the ratio of the areas of two similar triangle is equal to square of the ratio of their corresponding sides
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Answer 1

Appropriate Question :

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

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Step-by-step Explanation :

Given :

ΔABC ∼ ΔPQR

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RTP :

$\green\star \: \displaystyle \tt \frac{ar(ΔABC )}{ar(ΔPQR)} = \bigg( \frac{AB}{PQ} \bigg)^{2} = \bigg( \frac{BC}{QR} \bigg)^{2} = \bigg( \frac{CA}{RP} \bigg)^{2}$

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Construction :

Draw AM ⊥ BC and PN ⊥ QR.

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Proof :

$\red \star \: \displaystyle \tt \frac{ar(ΔABC )}{ar(ΔPQR)} = \frac{ \frac{1}{2} \times BC \times AM}{ \frac{1}{2} \times QR×PN } = \frac{BC \times AM}{ QR×PN} \: - - - (1)$

In ΔABM & ΔPQN

• ∠B = ∠Q ( ∵ ΔABM ∼ ΔPQN )
• ∠M = ∠N = 90°

∴ ΔABM ∼ ΔPQN (by AA similarly)

$\red \star \: \displaystyle \tt \frac{AM}{PN} = \frac{AB}{PQ} \: - - - (2)$

Also ΔABC ∼ ΔPQR (given)

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

$∴ \: \displaystyle \tt \frac{ar(ΔABC )}{ar(ΔPQR)} = \frac{AB}{PQ} \times \frac{AB}{PQ} \: \: \: \: \: \: \: from \: (1), \: (2) \: and(3) \\ \tt = \bigg(\frac{AB}{PQ} \bigg)^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now by using (3), We get

$\purple \star\: \displaystyle \tt \frac{ar(ΔABC )}{ar(ΔPQR)} = \bigg( \frac{AB}{PQ} \bigg)^{2} = \bigg( \frac{BC}{QR} \bigg)^{2} = \bigg( \frac{CA}{RP} \bigg)^{2}$

Hence proved.

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