∆ABC ∼ ∆PQR and ar(∆ABC) = 4 ar(∆PQR).

Given : $\operatorname{ar}(\triangle A B C)=4 \operatorname{ar}(\Delta P Q R)$

$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{4}{1}$

$\because \Delta A B C-\Delta P Q R$

$\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{B C^{2}}{Q R^{2}}$

$\therefore \frac{B C^{2}}{Q R^{2}}=\frac{4}{1}$

$\Rightarrow Q R^{2}=\frac{12^{2}}{4}$

$\Rightarrow Q R^{2}=36$

$\Rightarrow Q R=6 \mathrm{~cm}$

Hence, QR = 6 cm