Notes
a quarter circle and a semi-circle in a triangle solution

Solution to the A Quarter Circle and a Semi-Circle in a Triangle Puzzle

Solution by Pythagoras' Theorem and Angle Between a Radius and Tangent

Note: the semi-circle seems redundant in this puzzle

With the points labelled as above, let $r$ be the radius of the quarter circle, $a$ the length of line segment $O A$, and $b$ the length of line segment $O B$.

Since the red region is a quarter circle, angle $A \hat{O} B$ is $90^\circ$. Therefore the outer triangle is right-angled. Applying Pythagoras' theorem shows that:

As the angle between a radius and tangent is also $90^\circ$, triangles $O B A$ and $O B C$ are also right-angled. Applying Pythagoras' theorem to those shows that:

Adding these equations together shows that $a^2 + b^2 = 2r^2 + 20$ and so $2 r^2 + 20 = 36$, from which $r^2 = 8$.

Therefore, the area of the quarter circle is:

Solution by Angle Between a Radius and Tangent, Angles in a Triangle, and Similar Triangles

As above, triangles $A B O$ and $C B O$ are right-angled. Angles $A \hat{O} B$ and $B \hat{O} C$ add to $90^\circ$, but then so also do angles $A \hat{O} B$ and $B \hat{A} O$. So angles $B \hat{A} O$ and $B \hat{O} C$ are the same, so triangles $O B C$ and $A B O$ are similar.

Therefore, the ratios $4 : r$ and $r : 2$ are the same, leading to $r^2 = 8$ as before.