In the adjoining figure, DE || CA and D is a point on BD such that BD : DC = 2 : 1. The ratio of area of △ABC to area of △BDE is

1. 4 : 1

2. 9 : 1

3. 9 : 4

4. 3 : 2

Given DE || CA.

Considering △BDE and △BCA,

∠B = ∠B (Common angles)
∠BDE = ∠BCA (Corresponding angles are equal)

Hence, by AA axiom △BDE ~ △BCA.

Since triangles are similar. We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △BDE}} = \dfrac{BC^2}{BD^2}$ …..(i)

Given,

$\Rightarrow \dfrac{BD}{DC} = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{BD}{BC - BD} = \dfrac{2}{1} \\[1em] \Rightarrow BD = 2(BC - BD) \\[1em] \Rightarrow BD = 2BC - 2BD \\[1em] \Rightarrow BD + 2BD = 2BC \\[1em] \Rightarrow 3BD = 2BC \\[1em] \Rightarrow \dfrac{BC}{BD} = \dfrac{3}{2}.$

Putting this value in (i) we get,

$\Rightarrow \dfrac{\text{Area of △ABC}}{\text{Area of △BDE}} = \dfrac{BC^2}{BD^2} \\[1em] = \dfrac{3^2}{2^2} \\[1em] = \dfrac{9}{4} \\[1em] = 9 : 4.$

Hence, Option 3 is the correct option.