Notes
circles resting on triangles solution

Circles Resting on Triangles

Solution by Lengths in an Isosceles Right-Angled Triangle, Angles in a Triangle, and Angles at a Point on a Straight Line

The triangles are identical and $A B C D$ is a straight line, so angles $F \hat{B} A$ and $C \hat{B} F$ are the same and add to $180^\circ$, hence both are right-angles. Then angles $F \hat{C} B$ and $D \hat{C} E$ are equal to the non-right-angles, so since the angles in a triangle add up to $180^\circ$, these must add up to $90^\circ$. This leaves $90^\circ$ for angle $E \hat{C} F$. Since $F C$ and $E C$ are the same length, triangle $F C E$ is therefore an isosceles right-angled triangle. The length of $F E$ is then $\sqrt{2}$ times that of $F C$ and so the length scale factor from the dark blue semi-circle to the cyan one is $\sqrt{2}$. The area scale factor is thus $2$ and so the area of the cyan semi-circle is $28$.