If the sides of both squares were parallel to the other, then then shaded area would be 1/4 of the total square area.
If the partially tilted square were at a 45 degree incline, then you could see that again it is 1/4 of the total shaded area.
You need to determine an additional intermediate area to state that the area is static and does not ossilate as the squares rotate.
Since you've solved it but haven't explained it
here is my version.
Name the squares 'upper square' and 'lower square'.
Draw the squares so that the sides of the squares are
parallel to each other. The shaded region is
in the lower-right quadrant of the upper square,
and in the higher-left quadrant of the lower square.
The distance from the centerpoint of the upper square
to any side of the upper square is 6 or label it 'h'.
Restated: the distance from the center of the upper
square to the lower or bottom side of the square is
distance h, which is also the distance from the center
point to the right side of the upper square.
That is re-stating what should be obvious.
Cut the shaded region into two triangles.
Cut from the center of the upper-square to
the lower-right corner of the upper square.
You have two shaded triangles.
On lower-left triangle, label the vertical side
as 'h' and label the bottom as 'b'
On the upper-right triangle label the horizonal
distance as 'h' and the vertical distance is the same
as 'c'. (It is the same length as 'b'.)
The area of the lower-left triangle is: $\displaystyle \frac {h \times b}{2} $
The area of the upper-right triangle is: $\displaystyle \frac {h \times c}{2} $
Since b=c the two triangles have the same area which
sums to the area of the shaded square region.
NOW.
Rotate the lower square counter-clockwise through a small angle theta.
We need to calculate the area lost by the lower-left triangle.
The distance along the bottom edge of the upper-square is 'd'
$\displaystyle d = h\times \tan \theta $
The area lost is $\displaystyle \frac {h \times d}{2}$
The area gained by the upper-right triangle:
$\displaystyle d = h\times \tan \theta $
When rotated,
The area lost by the lower-left triangle will always be
equal to the area gained by the upper-right triangle,
thus the area is constant.