This image has been floating in the net for a while now:
with some approaches to "clarify" it:
but is there really an analytical way of solving it? The proportions and are of course out of scale (per Gimp, magic wand and histogram).
Being new here, please add appropriate tags if needed.
Updated Answer
I've rotated the image (and redrawn it to correct scale) and added some labels & lines.
Drawing in the four red lines divides each quadrilateral in the original shape into isosceles triangles (blue) and a second triangle (white). The four isosceles triangles are identical.
If we call the length of the red line $x$ and lets call the height of each triangle $a$, $b$, $c$ and $d$ respectively. Note that $a+c=x=b+d$.
Lets find the area of each triangle.
$$Area_A=\frac{ax}{2}, Area_B=\frac{bx}{2}, Area_C=\frac{cx}{2}, Area_D=\frac{dx}{2}$$
As $a+c=x=b+d$ then we have:
$$Area_A+Area_C=\frac{ax}{2}+\frac{cx}{2}=\frac{x^2}{2}=\frac{bx}{2}+\frac{dx}{2}=Area_B+Area_D$$
We can add on the area of each blue isosceles triangles to each of $Area_A, Area_B, Area_C, Area_D$ to get: $16+32=20+?$ which gives the unknown area as $28$.
Note the length $x$ is not needed to be known (or even really mentioned unless you are doing long in-depth steps like this updated answer).
Original Quicky Answer
With the inclusion of the red lines you have four triangles with identical base of the red line. The area of a triangle is found by $\frac{1}{2}\times base \times height$. Using this to calculate the sum of the area of two triangles directly opposite each other you will get $\frac{1}{2}$ times the length of the red lines squared. Then you can add the identical green triangles onto this to get the following: $32+16=20+?$ resulting in an answer of $28$.
There's a simply solution. Consider cutting up the shape as such
$a, b, x, y$ are areas of their triangles. We can then construct a system of equations:
$a+x=16$
$a+b=20$
$b+y=32$
and see
$a+x+b+y=48$
$a+x+b+y-a-b=x+y=48-20=28$
Which is our desired result.
