​$$If \ \triangle ABC \sim \triangle QRP , \frac{ar(\triangle ABC)}{ar(QRP)}=\frac{9}{4}, AB =18 cm, BC = 15cm, \text{then the length of PR is:}$$​​

Given: 

​$$\triangle ABC \sim \triangle QRP$$ 

​$$\frac{ar(\triangle ABC)}{ar(QRP)}=\frac{9}{4}, \\ AB =18 cm,\\ BC = 15cm,$$​​

Concept Used: 

If two triangles are similar, then

$$\frac{\text{Area of triangle1}}{\text{Area of triangle2}} = \left(\frac{\text{Side1}}{\text{Side2}}\right)^2$$​  

Solution: 

As we know, 

​$$\Delta ABC \sim \Delta QRP \\ \ \\ \text{thus,} \\ \ \\ \frac{\text{area of } \Delta ABC}{\text{area of } \Delta QRP} = \left(\frac{BC}{PR}\right)^2 \\ \ \\ \Rightarrow \sqrt{\frac{9}{4}} = \frac{15}{PR} \\ \ \\\frac{15}{PR} = \frac{3}{2}\\ \ \\ \implies \bf PR = 10 \, \text{cm}$$​​​​​