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 A 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, $$\frac{\left(Area.\ \triangle\ BDE\right)}{\left(Area.\ \triangle\ ABC\right)}\ =\ \frac{2^2}{5^2}$$

$$=\ \frac{4}{25}$$

$$ \text{Area of trapezium}\ ACED= \text{Area of } \triangle\ ABC- \text{Area of }\triangle\ BDE=25-4\ =\ 21$$

The required ratio : (Area of A BDE) : (Area of trapezium ACED)

i.e 4 : 21 

Hence, Option D is correct.