Notes
triangle in a circle in a quarter circle in a square solution

Solution to the Triangle in a Circle in a Quarter Circle in a Square Puzzle

Solution by Angle Between a Radius and Tangent, Lengths in an Isosceles Right-Angled Triangle

With the points labelled as above, let $r$ be the radius of the inner circle and $x$ the side length of the outer square. Then $B G$ has length $x$, and $B D$ is the diagonal of the square so has length $x \sqrt{2}$. Therefore, the length of $D G$ is $x (\sqrt{2} - 1)$. This is also the diagonal of the square of area $4$, so:

and hence $x = 4 + 2 \sqrt(2)$.

Since the angle between a radius and tangent is $90^\circ$, angles $O \hat{F} B$ and $O \hat{E} B$ are $90^\circ$, so $E B F O$ is a rectangle, but since $O F$ and $O E$ are both radii of the circle they are the same length and so $E B F O$ is a square. Hence $O B$ has length $r \sqrt{2}$, so $B G$ has length $r(1 + \sqrt{2})$ and thus:

and hence $r = 2 \sqrt{2}$, and then $O B$ has length $4$.

Since $F E$ is also a diagonal of square $E B F O$ (and so also has length $4$), it bisects $O B$ and so $H O$ has length $2$. Therefore $H G$ has length $2 + 2 \sqrt{2}$, so triangle $E F G$ has area: