If ΔABC ∼ ΔQRP, $\frac{ar(Δ ABC)}{ar (Δ QRP)} = \frac{9}{4}$, Ab = 18 cm, BC = 15 cm, then the length of PR is _________.

10 cm

By using the property of similar triagle

ΔABC and ΔQRP is similar triangle. Then ,

\(\frac{Area\;of\; ΔABC }{Area \;of \;ΔQRP }\) = \(\frac{( Side \;of \;ΔABC)² }{( Side\; of\; ΔQRP)² }\)

\(\frac{Area\;of\; ΔABC }{Area \;of \;ΔQRP }\) = \(\frac{(BC)² }{(PR)² }\)

\(\frac{9 }{4 }\) = \(\frac{(15)² }{(PR)² }\)

PR² = 100

PR = 10 cm