How to get the shaded region of the rectangle? I have this problem:

So my development was:
Denote side of rectangle with: $2a, 2b$. 
So, $4ab= 64, ab = 16$
Denote shaded region with $S$
Denote area of triangle $DGH = A_1$ and triangle $FBE = A_2$.
So, $A_1 + A_2 + S = 64$
$S = 64 - A_1 - A_2$
The triangles $A_1, A_2$ are congruent because $LAL$ congruence criterion.
The area of $A_1$ and $A_2$, is the same and i got it with this way:
Since, the $\angle{GDH} = 90$ and the median from this angle to the base $HG$, that is the altitude of the triangle $DGH$,  will measure the half of the $HG$ side.
And the $HG$ side by Pythagorean theorem, will be $\sqrt{a^2 + b^2}$, that will be the base of the triangle.
And the altitude will be: $\frac{\sqrt{a^2 + b^2}}{2} $,
So the Area of $A_1 = \frac{a^2 + b^2}{4}$
So, $A_1 + A_2 = \frac{a^2 + b^2}{2}$
Then, $64 - (\frac{a^2 + b^2}{2}) = S$
And, $-(a^2 - 8ab + b^2) = 2S$
And I have not been able to continue from here, what should I do? Thanks in advance.
 A: 
If you notice that if you combine two right triangles then they occupy the area of $\dfrac14$ of the total area.
So, the area of the shaded region is $=64-\dfrac14(64)=64-16=48$
A: Hint:



*

*there are eight small red triangles all with the same area 

*six of them are shaded
A: Hint. Because the triangles connecting the midpoints with the vertices are all congruent, we have:
$$\color{grey}{A}=64-2A_{DHG}$$
$$\color{grey}{A}=2A_{DHG}+A_{GHEF}$$
Addinng the two relations:
$$2\color{gray}{A}=64+A_{GHEF}$$
But, $A_{GHEF}=\frac{A_{ABCD}}{2}$ (I'll let you figure out why yourself as an (easy) exercise), and therefore:
$$\color{gray}{A}=\frac{96}{2}=48$$
A: By Midpoint theorem, area of triangle DHG is $\frac{1}{4}$ of triangle ACD. So, $$ar(DHG)=0.25*ar(ACD)=64/(2*4)=8$$
By symmetry, ar(unshaded region)=$8*2=16$
So, ar(Shaded region)=$64-16=48$
A: The area of hatched region of the rectangle  is the area of the rectangle minus the sum of the areas of the right triangles $BEF$ and $DGH$. Now if you glue these triangles along their hypotenuses, you obtain a rectangle  with dimensions half the dimensions of the big rectangle, hence with area equal to ¼ area of the big rectangle.
Hence the required area is ¾ big area, i.e. $\color{red}{48}$.
A: If your sidelengths are $2a$ and $2b$, how do you think you could represent $|\triangle GHD|$ and $|\triangle EFB|$ in terms of $a$ and $b$? (Keeping in mind that $E,G,H,F$ are midpoints.)
