Notes
two squares and a rectangle in an octagon solution

Solution to the Two Squares and a Rectangle in an Octagon Puzzle

Solution by Lengths in a Square or Pythagoras' Theorem

As the area of the squares is $8$, the side length is $2 \sqrt{2}$. Triangle $H G B$ is an isosceles right-angled triangle so is half a square. Therefore, either by using Pythagoras' theorem or from lengths in a square, the length of $H G$ is $\frac{2 \sqrt{2}}{\sqrt{2}} = 2$. The length of $E H$ is then $4 + 2 \sqrt{2}$.

This is also the length of $B J$, so the length of $C I$ is:

The area of the shaded rectangle is then: