Notes
three equilateral triangles and a semi-circle solution

Solution to the Three Equilateral Triangles and a Semi-Circle Puzzle

Solution by Lengths in Equilateral Triangles, Angle in a Semi-Circle

With the points labelled as in the diagram, the reflective symmetry of the semi-circle shows that the points $E$ and $A$ must be at the same height, which shows that the overlaps between the two pairs of triangles are equal. This means that $E A C$ is an equilateral triangle with side length $6$.

The point $B$ is where the semi-circle meets $A C$, so angle $O \hat{B} A$ is the angle between a radius and tangent which is $90^\circ$. This means that triangle $O B A$ is right-angled. Since angle $B \hat{A} O$ is the internal angle of an equilateral triangle, it is $60^\circ$, meaning that angle $A \hat{O} B$ is $30^\circ$ and so angle $B \hat{O} D$ is again $60^\circ$. As triangle $D O B$ must be isosceles, this shows that it is in fact equilateral.

Then triangles $O B A$, $B D C$, and $D O E$ are congruent and are half of an equilateral triangle. So the length of $A B$ is half that of $O A$, and so of $B C$, meaning that $A B$ has length $2$. The length of $O B$ is then $\sqrt{3}$ times the length of $A B$, from lengths in an equilateral triangle, so is of length $2\sqrt{3}$.

This is the radius of the semi-circle, so the area of the semi-circle is $6 \pi$.