In the given figure, area of △ ADE : area of trapezium BCED = 25 : 39, then AD : BD is :

1. 5 : 8

2. 8 : 5

3. 3 : 5

4. 5 : 3

In △ ADE and △ ABC,

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

⇒ ∠DAE = ∠BAC (Common angle)

∴ △ ADE ~ △ ABC (By A.A. postulate)

Given,

Area of △ ADE : Area of trapezium BCED = 25 : 39

Let area of △ ADE = 25x and area of trapezium BCED = 39x.

From figure,

Area of △ ABC = Area of △ ADE + Area of trapezium BCED = 25x + 39x = 64x.

We know that,

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

$\therefore \dfrac{\text{Area of △ ADE}}{\text{Area of △ ABC}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{25x}{64x} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{25}{64} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{AD}{AB} = \sqrt{\dfrac{25}{64}} = \dfrac{5}{8}.$

Let AD = 5y and AB = 8y.

From figure,

⇒ BD = AB - AD = 8y - 5y = 3y.

⇒ AD : BD = 5y : 3y = 5 : 3.

Hence, Option 4 is the correct option.