Simple fractal area

The four largest blue corners have an area which is the result of subtracting the areas of the largest square and the largest circle: $2^2-\pi \cdot1^2=4-\pi$ . The next square in the sequence has an area which is one-half of the largest square since its side is diminished by a factor of $\frac{\sqrt{2}}{2}$ . The area of its four blue corners is then $\left(\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}$ the area of the four largest, and this ratio is the same for every step in the sequence.

The total area is the sum of the infinite terms of a geometric progression of ratio $r=\frac{1}{2}$ and $a_1=4-\pi$ :

$S_{\infty}=\dfrac{a_1}{1-r}=\dfrac{4-\pi}{1-\frac{1}{2}}=8-2\pi$

So the answer is $A=8,B=2$ and $A+B=\boxed{10}$

$\begin{aligned} A_1 &= \text{the blue shaded area under the biggest square} \\ A_2 &= \text{the blue shaded area under the second square} \\ L_1 &=\text{the side length of the biggest square}=2\\ L_2 &=\text{the side length of the second square}=\sqrt{2}\\ A_2 &= A_1 - \text{the area of the largest four blue corners} \\ A_2 &= A_1 - (4 - \pi) \\ \frac{A_1}{A_2} &=\left ( \frac{L_1}{L_2} \right )^2 \\ \frac{A_1}{A_1 - (4 - \pi)} &=\left ( \frac{2}{\sqrt{2}} \right )^2\\ \frac{A_1}{A_1 - (4 - \pi)} &=2\Rightarrow A_1=8-2\pi \end{aligned}$