Question

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

Answer 1

Given :

$\frac{Areaof∆ABC}{Areaof∆QRP}=\frac{9}{4}$


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.

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

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