∆ABC ∼ ∆DEF such that ar(∆ABC) = 64 cm2 and ar(∆DEF) = 169 cm2.

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

$\therefore \frac{a r(\triangle A B C)}{a r(\triangle D E F)}=\frac{B C^{2}}{E F^{2}}$

$\Rightarrow \frac{64}{169}=\frac{4^{2}}{E F^{2}}$

$\Rightarrow E F^{2}=\frac{16 \times 169}{64}$

$\Rightarrow E F=\frac{4 \times 13}{8}=6.5 \mathrm{~cm}$