In $$\triangle ABC, \angle = 90^\circ, AB = 16 cm$$ and $$AC = 12 cm$$. D is the midpoint of AC and $$DE \perp CB$$ at E. What is the area (in cm$$^2$$) of $$\triangle$$CDE?

from the given question we draw the diagram is given below 

From the diagram ,  Let the $$ \triangle DEC, \angle c= \theta $$

$$ \tan \theta = \frac {16}{12}$$

$$\Rightarrow \tan \theta = \frac{4}{3}$$

  $$\Rightarrow  \theta = 53.13 $$

So,  $$ \sin theta = \frac{DE}{DC} $$

  $$\Rightarrow \sin 53.13 \times 6=  DE $$ 

$$\Rightarrow DE = 4.8 $$

from the Pythagoras' Theorem

EC = 3.6 

then Area of $$ \triangle CDE  = \frac{1}{2} \times 3.6 \times 4.8 $$

$$ \Rightarrow 8.64  cm^2 $$ Ans