If ΔABC ~ ΔDEF, AB = 4 cm, DE = 6, EF = 9 cm

Given AB = 4cm, DE = 6cm and EF = 9cm and FD = 12 cm

Also,  $\triangle A B C \sim \triangle D E F$

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

$\Rightarrow$ $\frac{4}{6}=\frac{B C}{9}=\frac{A C}{12}$

On taking first two terms, we get

$\frac{4}{6}=\frac{B C}{9}$

$\Rightarrow$ $B C=\frac{4 \times 9}{6}=6 \mathrm{~cm}$

$=A C=\frac{6 \times 12}{9}=8 \mathrm{~cm}$

Now, $\quad$ perimeter of $\triangle A B C=A B+B C+A C$ 

$=4+6+8=18 \mathrm{~cm}$