Hint The area of the right triangle is obviously $$\frac{ab}{2}$$ And the areas of the squares are respectively (see: Pythagorean Theorem)$$a^2\qquad b^2\qquad a^2+b^2$$
Now use the fact that the area of a triangle can be expressed as $$\frac{ab\cdot \sin(\gamma)}{2}$$ in order to prove that the remeaning triangles all have the same area, namely $$\frac{ab}{2}$$ You can find several proofs here. I also posted a simple proof here to the first Lemma.
Thus the area of the yellow region is
$$\text{Area$_{\text{yellow}}$}=2ab+a^2+b^2+a^2+b^2=2\cdot(a^2+ab+b^2)$$
Let $A$ denote the area between the sides of lengths $b,\,c$ in the central right-angled triangle, and define $B$ similarly. There is another triangle congruent to this. The area is $$a^2+b^2+c^2+ab+\frac{bc\sin(\pi-A)+ac\sin(\pi-B)}{2}=2(a^2+ab+b^2).$$
We have 3 squares, which are easy.
2 right triangles that are easy.
And two obtuse triangles that are less obvious.
We know the base, what are the heights?
Divide the big square into triangles as in the figure.
What is the heights of the obtuse triangles?
In the drawing below it's obvious that the angle $AXZ = \alpha$. Furthermore, $AW=YZ = a$ as the right triangle $AXW$ is congruent with the one in the center. Then, the area of triangle AYX is same as the area of triangle AYZ as they have the same base $(AY)$ and same height $(YZ)$, which is $ab/2$. Analogously for the other non-right triangle. Thus, the additional area (of the non-right triangles) is $ab$.
Consider the two triangles $ABC$ and $CDF$. If we rotate $CDF$ counterclockwise through 90 degree, so points $A$ and $F$ coincide then we observe that the two triangles have the same base length and height, therefore they have equal area.
This is true for other triangles too, hence all four triangles have equal area.
The area of the hexagol is:
Sum of the areas of three squares $+$ Sum of the areas of four triangles.
$2a^2+2b^2+2ab$
