If AD = 5 cm and BD = 2 cm, then area of △ ADE : area of trapezium DBCE is equal to :

1. 5 : 2

2. 2 : 5

3. 24 : 25

4. 25 : 24

From figure,

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC (Common angle)

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

∴ △ ADE ~ △ ABC (By A.A. postulate)

From figure,

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

We know that,

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

$\therefore \dfrac{\text{Area of △ ADE}}{\text{Area of △ ABC}} = \dfrac{AD^2}{AB^2} \\[1em] = \dfrac{5^2}{7^2} \\[1em] = \dfrac{25}{49} \\[1em] = 25 : 49.$

Let area of △ ADE = 25x and area of △ ABC = 49x.

From figure,

Area of trapezium DBCE = Area of △ ABC - Area of △ ADE = 49x - 25x = 24x.

$\therefore \dfrac{\text{Area of △ ADE}}{\text{Area of trapezium DBCE}} = \dfrac{25x}{24x} = \dfrac{25}{24}$

⇒ Area of △ ADE : Area of trapezium DBCE = 25 : 24.

Hence, Option 4 is the correct option.