Question

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

Options

• 5 : 8

• 8 : 5

• 3 : 5

• 5 : 3

Answer 1

In the given figure, area of △ADE : area of trapezium BCED = 25 : 39, then AD : BD is 5 : 3. 

Explanation: 

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.

∴ `("Area of"△ ADE)/("Area of" △ ABC) ​= (AD^2)/(AB^2)​`

⇒ `(25x)/(64x) ​= (AD^2)/(AB^2)`

​⇒ `25/64 = (AD^2)/(AB^2)​`

⇒ `(AD)/(AB)​ = sqrt(25/64) = 5/8​`

Let AD = 5y and AB = 8y.

From figure,

⇒ BD = AB − AD

= 8y − 5y

= 3y

⇒ AD : BD

= 5y : 3y

= 5 : 3