Yellow Figure

Let's break this figure into 5 rectangles as shown below.

The area of the yellow region in the 1st rectangle is exactly half the area of the $4\times 2$ rectangle, so the area is $4\times2 \div 2= 4$ .

Similarly, the area of the yellow region in the 3rd rectangle is exactly half the area of the $2\times 4$ rectangle, so the area is $2\times4 \div 2= 4$ .

We can work out the areas of the yellow region in the 4th and 5th rectangle to $\dfrac32$ and $\dfrac92$ .

And the area of the yellow region in the 3rd rectangle is just $2\times1 = 2$ .

So the total area of the yellow region is the sum of these areas, $4 + 4 + \dfrac32 + \dfrac92 + 2 = \boxed{16}$ .

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Bonus : Anyone wants to solve this question using Pick's theorem ?

This can easily be done by subtracting the remaining area from the main square.The remaining areas are all 4 triangles whose total area is 14 so subtracting it from the square area i.e. 30 we will get our required answer i.e. 16......

The area of the yellow region is equal to the area of the rectangle inscribing the yellow region minus the sum of the areas of the four triangles. We have

$A=6(5)-\dfrac{1}{2}\left[2(4)+1(3)+3(3)+2(4)\right]=16$

Count all the squares, all full squares are square unit and all partial squares have another partial somewhere to complete it. 9 full squares and 14 partials, 9 + 14÷2 = 16. I believe this only works for vertex of polygons on integer coordinates