if triangle ABC is similar to the triangle QRP, ar(ABC)/ar(PQR)= 9/4, AB=18 cm BC=15 cm, then PR= ?

Given :

$\frac{Area of ∆ ABC}{Area of ∆QRP} = \frac{9}{4}$


AB  = 18  cm  , BC  = 15  cm  So  PR =  ?

We know when two triangles are similar then " The areas of two similar triangles are proportional to the squares of their corresponding sides.

$\frac{Area of ∆ ABC}{Area of ∆ QRP} = \frac{A{B}^2}{Q{R}^2} = \frac{B{C}^2}{P{R}^2} = \frac{A{C}^2}{Q{P}^2}$
So , we take 
$\frac{Area of ∆ ABC}{Area of ∆ QRP} = \frac{B{C}^2}{P{R}^2}$

Now substitute all given values and get

$\frac{9}{4} = \frac{{15}^2}{P{R}^2}$

Taking square root on both hand side , we get

$\frac{3}{2} = \frac{15}{PR}$

PR  = 10    cm