Finding the unknown area. I tried several options, but certainly do not know anything about this figure, since I can not solve it. 
a) The length of one side of given regular hexagon is $4cm$. What is the are of the shaded region:

b) In the figure, $ABCD$ is a parallelogram. $|AF|=8cm,|BF|=4cm, |CE|=9cm.$ What is $|EF|.$ (I find DE, but nothing helps me.)

Please help me. Thank you very much.
 A: (a) As we know that the side of a regular hexagon is equal to the radius of circumscribed circle.
and we also know that the hypotenuse is the diameter the circle.
Hence,  The hypotenuse of the given triangle  $H=2\times4=8cm$ .
therefore the perpendicular $P=\sqrt{8^2-4^2}$        ...(Pythagoras theorem)
$P=4\sqrt{3} $
so,
Area of shaded region $=\frac{1}{2}\times4\sqrt{3}\times4$
$=8\sqrt{3}cm^2 $ 
(b) $\angle CDB=\angle DBF$  ...(Alternate interior Angles)
$\angle DCF=\angle CFB$  ...(Alternate interior Angles)
Therefore, $\triangle DEC\sim \triangle BEF$  ...(AA-Similarity)
As you said you had found $|DE|$ you can use it to find $|EF|$
A: For (a):
Since the given regular hexagon each internal angle is $120$ degrees and every side is equal. So by the given data, we have each angle=$120$ and each side= $4cm$.
The height in your black shaded triangle can be easily calculated by applying cosine rule.
$H^2=4^2+4^2-2\times 4\times 4\times \cos120$.
Which gives $H=4\sqrt{3} $. So, area of right triangle $=\frac{1}{2}\times4\times 4\sqrt{3}=8\sqrt{3}$.
For (b):
Notice that $\triangle DEC$ is similar to $\triangle BEF $.    
So, $\frac{DC}{BF}=\frac{EC}{EF}$ $\implies \frac{12}{4}=\frac{9}{EF}\implies EF=3$
