In $$\triangle$$ABC, D is a point on side AB such that BD = 2 cm and DA = 3 cm. E is a point on BC such that DE $$\parallel$$ AC, and AC = 4 cm. Then (Area of $$\triangle$$BDE) : (Area of trapezium ACED) is:

we have :

Now DE||AC

: $$\triangle\ $$BDE is similar to $$\triangle\ $$ABC
The ratio of sides of two similar triangles is equal to the ratio of the square root of the areas of two triangles.

so, $$\dfrac{A(\triangle BDE)}{A(\triangle ABC)}\ =\ \dfrac{2^2}{5^2}$$

$$=\ \dfrac{4}{25}$$

Area of trapezium ACED= Area of triangle ABC - Area of triangle BDE= 25 - 4 = 21

The required ratio : (Area of $$\triangle$$BDE) : (Area of trapezium ACED)

i.e 4 : 21 

Hence, Option D is correct.