Given that : $\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$, If $\frac{\text { area }(\Delta \mathrm{PQR})}{\text { area }(\Delta \mathrm{ABC})}$ $=\frac{256}{441}$ and $\mathrm{PR}=12 \mathrm{~cm}$, then $\mathrm{AC}$ is equal to

The ratio of the areas of two similar triangles is equal to the ratio of squares of any two corresponding sides.

$\therefore \frac{\text { Area of } \Delta \mathrm{PQR}}{\text { Area of } \Delta \mathrm{ABC}}=\frac{\mathrm{PR}^{2}}{\mathrm{AC}^{2}}$

$\Rightarrow \frac{\mathrm{PR}^{2}}{\mathrm{AC}^{2}}=\frac{256}{441} \Rightarrow \quad \frac{12^{2}}{\mathrm{AC}^{2}}=\frac{256}{441}$

Taking square roots of both sides, $\frac{12}{\mathrm{AC}}=\frac{16}{21}$

$\Rightarrow 16 \times \mathrm{AC}=12 \times 21$

$\Rightarrow \mathrm{AC}=\frac{12 \times 21}{16}=\frac{63}{4}=15.75 \mathrm{~cm} .$