In ∆ABC, it is given that AB = 9 cm, BC = 6 cm and CA = 7.5 cm.

(d) 30 cm

Perimeter of $\triangle A B C=A B+B C+C A=9+6+7.5=22.5 \mathrm{~cm}$

$\because \triangle D E F \sim \triangle A B C$

$\therefore \frac{\text { Perimeter } \triangle(A B C)}{\text { Perimeter }(\triangle D E F)}=\frac{B C}{E F}$

$\Rightarrow \frac{22.5}{\text { Perimeter }(\triangle D E F)}=\frac{6}{8}$

$\Rightarrow$ Perimeter $(\triangle D E F)=\frac{22.5 \times 8}{6}=30 \mathrm{~cm}$