# Solution to the Three Equilateral Triangles and a Semi-Circle Puzzle +-- {.image} [[ThreeEquilateralTrianglesandaSemiCircle.png:pic]] > The three equilateral triangles have sides of length $12$. What's the area of the semicircle? =-- ## Solution by [[equilateral|Lengths in Equilateral Triangles]], [[Angle in a Semi-Circle]] +-- {.image} [[ThreeEquiTrianglesandaSemiCircleLabelled.png:pic]] =-- 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 triangle|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 [[equilateral|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 $6 \pi$.