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

10 cm

We know that the area of the triangle is the square of the side .

Given,

$ΔABC ∼ Δ QRP, \frac{ar(ΔABC)}{ar(ΔQRP)}=\frac{9}{4}$

= \(\frac{BC}{PR}\) = \(\frac{15}{PR}\)  

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

= \(\frac{15}{PR}\)  = $\frac{3}{2}$

PR = 10 cm