Let $△ABC\sim △PQR$ and $\frac{\text{ar}(△ABC)}{\text{ar}(△QPR)}=\frac{25}{169}$. If $AB=15\text{cm}$, $BC=7\text{cm}$ and $AC=8\text{cm}$, then $PQ$ (in $\text{cm}$) is equal to:

Since $△ABC\sim △PQR$, the triangles are similar, and the ratio of their areas is equal to the square of the ratio of their corresponding sides:
$$\frac{\text{ar}(△ABC)}{\text{ar}(△PQR)}={\left(\frac{AB}{PQ}\right)}^2$$
Taking the square root of the given area ratio:
$$\frac{AB}{PQ}=\sqrt{\frac{25}{169}}=\frac{5}{13}$$
Substitute the given value $AB=15\text{cm}$:
$$\frac{15}{PQ}=\frac{5}{13}$$
$$5\times PQ=15\times 13$$
$$PQ=3\times 13=39\text{cm}$$