Notes
two squares overlapping a semi-circle solution

Two Squares Overlapping a Semi-Circle

Solution by Angle in a Semi-Circle, Vertically Opposite Angles, and Similar Triangles

With the points labelled as in the above diagram, the direct line from $D$ to $B$ forms a right-angle with the line $A D$ since this is the angle in a semi-circle. Since $D E$ is also at right-angles to $A D$, this means that $D B$ is an extension of $D E$, so $D E B$ is a straight line. Similarly, $A E C$ is a straight line.

Since vertically opposite angles are equal, angles $C \hat{E} D$ and $A \hat{E} B$ are equal. Then $A E$ is a diagonal of a square of which $D E$ is a side, and $B E$ is a diagonal of a square of which $C E$ is a side. Put together, these mean that triangle $A E B$ is similar to triangle $D E B$ with scale factor $\sqrt{2}$. The length of $A B$ is therefore $10 \sqrt{2}$ and so the area of the semi-circle is $25 \pi$.