Question

In the given figure, ∠BAC = 90°, AD is perpendicular to BC, BC = 13 cm and AC = 5 cm, then area of △ ADC : area of △ DBA is ______.

Options

• 5 : 13

• 13 : 5

• 25 : 144

• 144 : 25

Answer 1

In the given figure, ∠BAC = 90°, AD is perpendicular to BC, BC = 13 cm and AC = 5 cm, then area of △ ADC : area of △ DBA is 25 : 144.

Explanation:

From figure,

In Δ BAC and Δ ADC,

⇒ ∠BAC = ∠ADC (Both equal to 90°)

⇒ ∠ACB = ∠ACD (Common angle)

∴ Δ BAC ~ Δ ADC (By A.A. postulate)

We know that,

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

∴ `("Area of" Δ BAC)/("Area of" Δ ADC) ​= (BC^2)/(AC^2)`

⇒ `("Area of" Δ BAC)/("Area of" Δ ADC) ​= 13^2/5^2`

​⇒ `("Area of" Δ BAC)/("Area of" Δ ADC)​ = 169/25​`

Let, area of Δ BAC = 169x and area of Δ ADC = 25x.

From figure,

Area of Δ DBA = Area of Δ BAC − Area of Δ ADC = 169x − 25x = 144x.

Substituting values we get:

area of △ ADC : area of △ DBA = 25x : 144x = 25 : 144