Theo đề bài, ta có: ∆A'B'C' ~ ∆ABC
=> $\frac{A'B'}{AB}$=$\frac{B'C'}{BC}$=$\frac{A'C'}{AC}$ <=> $\frac{A'B'}{3}$=$\frac{B'C'}{7}$=$\frac{A'C'}{5}$=$\frac{A'B'+B'C'+A'C'}{3+5+7}$
Mà chu vi ∆A'B'C' bằng 45cm
=> A'B'+B'C'+A'C'=45
=> $\frac{A'B'+B'C'+A'C'}{3+5+7}$ =$\frac{45}{15}$=3 (cm)
=> $\frac{A'B'}{3}$=3 => A'B'=9
$\frac{A'C'}{5}$=3 =>A'C'=15(cm)
$\frac{B'C'}{7}$ =3 =>B'C'=21(cm)
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