Total Area with stacking Squares

In the given animation, each square with side parallel to the $x-$ and $y-$ axes , i.e., each square in the sum $A_1$ has a corresponding square of half the area in the squares in the sum $A_1$ .

$\implies A_1 = 2\times A_2$

$\implies \dfrac{A_1}{A_2} = \dfrac{2\times A_2}{A_2}$

$\implies \dfrac{A_1}{A_2} = 2$

A square with sides parallel to the $x-$ and $y-$ axes can be broken up into $8$ congruent triangles so that $4$ of those triangles make up a square with slanted sides.

Therefore, the ratio of the area of one square in $A_1$ to the area of its connected square in $A_2$ is $\frac{8}{4} = 2$ .

Since each square in $A_1$ has one connected square in $A_2$ , this ratio is maintained throughout the whole diagram, so $\frac{A_1}{A_2}$ is also $\boxed{2}$ .

If the value of $a$ is say $2$ , then the length of the inner square $(A_{2})$ will be $\sqrt{2}$

\[\begin{cases} A_{1} = (2)^2 = 4 \\ A_{2} = (\sqrt{2})^2 = 2 \end{cases}

\implies \dfrac{A_{1}}{A_{2}} = \dfrac{4}{2} = \boxed{2}\]