# Solution to the [[Subdivided Semi-Circle]] Puzzle +-- {.image} [[SubdividedSemiCircle.jpeg:pic]] > The dots around the circumference of this semicircle are equally spaced. What fraction is shaded? =-- ## Solution by [[Dissection]], Lengths in a [[Regular Hexagon]], and the [[Area of a Triangle]] +-- {.image} [[SubdividedSemiCircleLabelled.jpeg:pic]] =-- Label the points as in the diagram above, where $O$ is the centre of the semi-circle. Consider the region formed by the arc $A C$ and line segments $C E$ and $E A$. As the points are equally spaced around the semi-circle, points $A$, $C$, $E$, and $G$ form part of a [[regular hexagon]]. Therefore, $O E$ is [[parallel]] to $A C$. This means that triangles $A C E$ and $A C O$ have the same area as both have the same "height" above $A C$. The area of the region formed by the arc $A C$ and line segments $C E$ and $E A$ is therefore the same as the area of the sector $A O C$. Consider the region formed by the arc $F G$ and line segments $G A$ and $A F$. The two triangles $O G F$ and $A O F$ have the same length base, as they are both radii of the same circle, and the same height, as it is the height of $F$ above $G A$, so they have the same area. Therefore, the area of triangle $A O F$ is the same as that of triangle $F O E$. So the region formed by the arc $F G$ and line segments $G A$ and $A F$ has the same area as that formed by the arc $F G$ and line segments $G O$, $O E$, and $E F$. Lastly, the area of the segment cut off by line segment $C D$ is the same as that cut off by line segment $E F$, so combining this with the region above produces the same area as sector $E O G$. In total, the shaded regions have the same area as two $60^\circ$ sectors of the semi-circle, so their combined area is $\frac{2}{3}$rds of the total.