Question

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

Options

• 4 : 49

• 49 : 4

• 4 : 25

• 25 : 4

Answer 1

In the given figure, AD : DB = 2 : 5, then area of ∆ODE : area of ∆OCB is 4 : 49.

Explanation:

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.

∴ `(DE​)/(BC) = (AD)/(AB​)`

⇒ `(DE​)/(BC) = (2x​)/(7x)`

⇒ `(DE​)/(BC)​ = 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.

⇒ `("Area of" △ ODE)/("Area of" △ OCB) ​= (DE^2)/(BC^2)`

= `2^2/7^2`

​= `4/94​`

= 4 : 49