3 different shapes are inscribed in an isosceles right triangle. Given that the side shorter side of the isosceles right triangle to be 2. Find their areas.
I have found the area of the 1st and 2nd diagram but could not find the area of the 3rd figure. Appreciate any help. 
Reflect the triangle across its legs $AB$ and $BC$. Because $AB = BC = 2$, the resulting figure is a square. Moreover, since the inscribed yellow shape is also a square, it is easy to see that their reflections must form a central square that is congruent, and in fact, the entire figure consists of $9$ congruent squares. From this, it is trivial to compute the desired area. See the figure below.
If we label your figure as follows:
then show that we must have $EF = FC$, and similarly, $AG = GD$. Hence conclude that $GF = AC/3$ and the rest is obvious.