If triangles ABC and DEF are similar and AB = 4 cm, DE = 6 cm, EF = 9 cm, FD = 12 cm, then the perimeter of triangle ABC is:

As we know, two triangles are similar if

1. their corresponding angles are equal and
2. their corresponding sides are in the same ratio.

$\because △\mathrm{ABC}~△\mathrm{DEF}$   [Given]

$\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$

$\Rightarrow \frac{4}{6}=\frac{\mathrm{BC}}{9}=\frac{\mathrm{AC}}{12}$

So, $\mathrm{BC}=\frac{4\times 9}{6}=6 \mathrm{cm}$

$\mathrm{AC}=\frac{4\times 12}{6}=8 \mathrm{cm}$

Therefore, the perimeter of the triangle ABC = AB + BC + CA

= 4 + 6 + 8

= 18 cm