Question

I the given figure, DE||BC and if DE:BC =3:5,Find ar(ADE)/ar(bced). QUES.NO. 3​

Answer 1

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$given \: that$

DE : BC

__3 : 5__

DE || BC.

∴ ∠ADE = ∠ABC 

»  ∠AED = ∠ACB

Applying AA similarity theorem, we can conclude that ∆ ADE ~ ∆ABC.

» $\frac{ar(∆ ABC)}{ar(∆ ADE)} - 1 = \frac{ { 5}^{2} }{ {3}^{2} } - 1$

» $\frac{ar(∆ ABC) - ar(∆ ADE)}{ar(∆ ADE)} = \frac{25 - 9}{9}$

» $\frac{ar(BCED)}{ar(∆ ADE)} = \frac{16}{ 9}$

» $or \\ \frac{ar(∆ADE)}{ar(BCED)} = \frac{9}{16}$

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