Which of the following is correct?

1. If D is a point on side AB of ΔABC such that AD : DB = 5 : 2 and E is a point on BC such that DE ∥ AC, then ar(ΔABC) : ar(ΔDBE) = 9 : 4.

2. If the areas of two similar triangles are in the ratio 25 : 64, then their perimeters are in the ratio 5 : 8.

3.

In the adjoining figure if AB ∥ CD, then ΔAOB ∼ ΔCOD.

4.

In the adjoining figure, if D and E are the mid-points of AB and AC respectively, then ar(ΔADE) = $\dfrac{1}{2}$ × ar(ΔABC).

Solving option 2,

Since the triangles are similar,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides (or perimeters).

$\Rightarrow \dfrac{Area$1}{Area2} = \Big(\dfrac{\text{Perimeter}1}{\text{Perimeter}2}\Big)^2 \\[1em] \Rightarrow \dfrac{25}{64} = \Big(\dfrac{\text{Perimeter}1}{\text{Perimeter}2}\Big)^2 \\[1em] \Rightarrow \sqrt{\dfrac{25}{64}} = \dfrac{\text{Perimeter}1}{\text{Perimeter}2} \\[1em] \Rightarrow \dfrac{\text{Perimeter}1}{\text{Perimeter}2} =\dfrac{5}{8} \\[1em]

Hence proved.

Hence, option 2 is the correct option.