Let $$\triangle$$ABC $$\sim$$ $$\triangle$$QPR  and  $$\frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{4}{25}$$. If AB = 12 cm, BC = 8 cm and AC = 9 cm, then PR is equal to:

Let AD is the altitude of $$\triangle$$ABC and QT is the altitude of $$\triangle$$QPR as shown in figure

Given, $$\triangle$$ABC $$\sim$$ $$\triangle$$QPR

$$=$$>  $$\frac{AB}{QP}=\frac{BC}{PR}=\frac{AC}{QR}=\frac{AD}{QT}$$

Let $$\frac{AB}{QP}=\frac{BC}{PR}=\frac{AC}{QR}=\frac{AD}{QT}=t$$

Given,  $$\frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{4}{25}$$

$$=$$>  $$\frac{\frac{1}{2}\times AD\times BC}{\frac{1}{2}\times QT\times PR}=\frac{4}{25}$$

$$=$$>  $$\frac{tQT\times tPR}{QT\times PR}=\frac{4}{25}$$

$$=$$>  $$t^2=\frac{4}{25}$$

$$=$$>  $$t=\frac{2}{5}$$

$$\therefore\ $$ $$\frac{BC}{PR}=t=\frac{2}{5}$$

$$=$$>  $$\frac{8}{PR}=\frac{2}{5}$$

$$=$$>  PR = 20 cm

Hence, the correct answer is Option D