Let $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ and $\frac{ar(\triangle A B C)}{ar(\triangle Q P R)}=\frac{144}{49}$. If AB = 12 cm, BC = 7 cm and AC = 9 cm, then PR(in cm) is equal to:

$\frac{21}{4}$

\(\frac{area\;of\;ABC}{area\;of\;PQR}\) = \(\frac{AC}{PR}\)^2

= \(\frac{144}{49}\) = \(\frac{AC}{PR}\)^2

= \(\frac{AC}{PR}\) = \(\sqrt {144/49 }\)= \(\frac{12}{7}\)

= \(\frac{9}{PR}\) = \(\frac{12}{7}\)

= PR = \(\frac{9\;×\;7}{12}\) = \(\frac{21}{4}\) cm.

Therefore, PR = \(\frac{21}{4}\) cm.