What is the area of the shaded region?

Since $CA\parallel JH\parallel GE$ , the transformation we see in the animation preserves the areas of moving triangles with base $JH$ , thus, the area of the red quadrilateral $HGJC$ equals the area of the square $JIHE$ . But, $\left[ JIHE \right]=\dfrac{1}{2}J{{H}^{2}}$ . Hence, $\left[ HGJC \right]=\dfrac{1}{2}{{\left( 4\sqrt{2} \right)}^{2}}=\boxed{16}$

Place point $D$ at the origin of the $xy-$ plane. Let squares $ABCD, DEFG, EHIJ$ each have side lengths equal to $p, q, 4$ respectively. The red area is bounded by vertices:

$C(0,p); G(0,q); H(q+4,0); J(q,4)$

If point $I(q+4,4)$ lies on diagonal $AC$ , which can be represented as the line $y=-x+q,$ then we have:

$4 = -(p+4) + q \Rightarrow q = p+8$ .

The red area, $CGHJ$ , can be computed as the sum of triangular areas $\triangle{GHJ}$ and $\triangle{CHJ}$ :

$A_{CGHJ} = \frac{1}{2} \cdot | \begin{vmatrix} 1 & 0 & p \\ 1 & p & 4 \\ 1 & p+4 & 0 \end{vmatrix}| + \frac{1}{2} \cdot | \begin{vmatrix} 1 & 0 & p+8 \\ 1 & p & 4 \\ 1 & p+4 & 0 \end{vmatrix}| = |\frac{4p-4p-16}{2}| + |\frac{4p-4p-16+32}{2}| = |-8| + |8| = \boxed{16}.$

Inelegant (but straightforward) solution:

Goal will be to start with the area of triangle $ACD$ and subtract areas of triangles $AHC$ , $GDH$ , and $CJG$ .
We are given that the distance $HJ = 4\sqrt{2}$ . Thus the side length of this square is $4$ . And since I is on the main diagonal of square $ABCD$ , $AH = HI = 4.$

We are not given the side length of the small square $DEFG$ , but that will not matter: let $x = DE$ . Then $AD = 8 + x$ and the area of the square $ABCD$ is $(8 + x)^2 = 64 + 16x + x^2$ . And the area of the triangle $ACD$ is half of this quantity, i.e. $32 + 8x + (1/2)x^2$ .

The area of triangle $ACH$ is $(1/2)(8+x)4 = 2(8+x) 16 + 2x.$

The area of triangle $HDG$ is $(1/2)x(4+x) = 2x + (1/2)x^2.$

The area of triangle $CJG$ is $(1/2)(8)x = 4x$ .

Thus the area of the red shape is $(32 + 8x + (1/2)x^2) - (16 + 2x) - (2x + (1/2)x^2) - (4x) = 16$ . The terms involving $x$ have disappeared.

If the length of the side, $\overline { HJ } = 4\sqrt2$ , then $\square { EHIJ } \rightarrow \overline { IJ } = \overline { HI } \rightarrow \overline { IJ } ^ 2 + \overline { HI } ^ 2 = ( 4\sqrt2 ) ^ 2 \rightarrow 2\overline { HI } = 8 \rightarrow \overline { IJ } = \overline { HI } = 4 \cm$ . Since $\square { CHGJ } = \square { EHIJ }$ , so the area of $\square { CHGJ }$ is $\boxed { 16 \cm ^ 2 }$ .