In the given figure, AD : DB = 2 : 5, then area of △ ODE : area of △ OCB is :

1. 4 : 49

2. 49 : 4

3. 4 : 25

4. 25 : 4

Given,

AD : DB = 2 : 5

Let AD = 2x and DB = 5x

From figure,

AB = AD + DB = 2x + 5x = 7x.

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∠ADE = ∠ABC (Corresponding angles are equal)

∴ △ ADE ~ △ ABC (By A.A. postulate)

We know that,

Corresponding sides of similar triangles are in proportion.

$\therefore \dfrac{DE}{BC} = \dfrac{AD}{AB} \\[1em] \Rightarrow \dfrac{DE}{BC} = \dfrac{2x}{7x} \\[1em] \Rightarrow \dfrac{DE}{BC} = \dfrac{2}{7}.$

In △ ODE and △ OCB,

⇒ ∠DOE = ∠BOC (Vertically opposite angle are equal)

⇒ ∠ODE = ∠OCB (Alternate angles are equal)

∴ △ ODE ~ △ OCB (By A.A. postulate)

We know that,

The areas of two similar triangles are proportional to the squares on their corresponding sides.

$\Rightarrow \dfrac{\text{Area of △ ODE}}{\text{Area of △ OCB}} = \dfrac{DE^2}{BC^2} \\[1em] = \dfrac{2^2}{7^2} \\[1em] = \dfrac{4}{49} \\[1em] = 4 : 49.$

Hence, Option 1 is the correct option.