∆ABC is similar to ∆DEF and the ratio of the area of triangle ABC to triangle DEF is 25 ∶ 144. If AB = p cm,AC = q cm and BC = r cm,then DE = ?

Correct option is D

Given:
∆ABC ~ ∆ DEF
Ratio of  $$\frac{(ar(ABC))}{(ar(DEF))} = \frac{25}{144}$$​​
AB=p cm, BC= q cm, AC= r cm
Formula Used:
​$$\frac{ar(ABC)}{ar(DEF)}= \frac{(AB)^2}{(DE)^2}= \frac{(BC)^2}{(EF)^2} = \frac{(AC)^2}{(DF)^2}$$​​
Solution:
​$$\frac{ar(ABC)}{(ar(DEF)} = \frac{(AB)^2}{(DE)^2} \\\ \\\frac{(25)}{(144)} = \frac{(AB^2)}{(DE)^2} \\\ \\\sqrt{ \frac{25}{144}} =\frac{AB}{DE} \\\ \\\frac{5}{12} = \frac{AB}{DE }\\\ \\DE= \frac{12}{5} AB (AB=p) \\\ \\DE = \frac{12}{5} p$$​​