Notes
tilted rectangle in a square solution

Tilted Rectangle in a Square

Solution by Symmetry

With the points labelled as in the above diagram, point $O$ is the centre of the rectangle.

As $F$ is the same way along $E G$ as $B$ is along $A C$, the line $F B$ is centrally placed in the square and so its midpoint, which is $O$, is the same as the centre of the square. Therefore the circle centred on $O$ sits symmetrically in the outer square and so the points where it intersects with the square are proportionally the same along each side. This means that $H$ and $D$ are also one third of the way along their respective sides.

An alternative way to see this is to consider the triangles $O A B$ and $O A H$. The sides $O B$ and $O H$ are the same length, and the angles at $H \hat{O} A$ and $A \hat{O} B$ are the same. This shows that $A H$ and $A B$ are the same length.

From the grid, the shaded area covers four of the smaller squares and so the fraction of the square covered by the rectangle is $\frac{4}{9}$.