In ∆ABC and ∆DEF, we have:


In $\triangle A B C$ and $\Delta D E F$, we have: $\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}=\frac{5}{7}$, then

ar(∆ABC) : ar(∆DEF) = ?

(a) 5 : 7
(b) 25 : 49
(c) 49 : 25
(d) 125 : 343



(b) 25 : 49

- In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$, we have :

$\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}=\frac{5}{7}$

Therefore, by SSS criterion, we conclude that $\triangle A B C \sim \triangle D E F$.

$\Rightarrow \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D E F)}=\frac{A B^{2}}{D E^{2}}=\left(\frac{5}{7}\right)^{2}=\frac{25}{49}=25: 49$


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