The areas of two similar triangles ABC and PQR are in the ratio

It is given that $\triangle A B C \sim \triangle P Q R$.

Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.

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

$\Rightarrow \frac{9}{16}=\frac{4.5^{2}}{Q R^{2}}$

$\Rightarrow Q R^{2}=\frac{4.5 \times 4.5 \times 16}{9}$

$\Rightarrow Q R=\sqrt{\frac{4.5 \times 4.5 \times 16}{9}}$

$=\frac{4.5 \times 4}{3}$

$=6 \mathrm{~cm}$

Hence, QR = 6 cm