The area of $$\triangle$$ ABC is 44 $$cm^2$$. If D is the midpoint of BC and E is the mid point of AB,then the area (in $$cm^2$$) of $$\triangle$$BDE is:

D is the midpoint of BC and E is the mid point of AB.

Let the BE be x cm.

AB = 2BE = 2x cm

$$\frac{AB}{BE} = \frac{2x}{x} = \frac{2}{1}$$ 

$$\angle B$$ is common in both triangle.

$$\triangle$$ ABC ~ $$\triangle$$BDE

So,

$$\frac{area of \triangle ABC}{area of \triangle BDE} = (\frac{AB}{BE})^2$$

$$\frac{44}{area of \triangle BDE} = (\frac{2}{1})^2$$

$$\frac{44}{area of \triangle BDE} = (\frac{4}{1})$$

Area of $$\triangle$$ BDE = 11 $$cm^2$$.