# Solution to the [[Striped Semi-Circle]] Puzzle +-- {.image} [[StripedSemiCircle.png:pic]] > The points on the circumference are equally spaced. What’s the total shaded area? =-- ## Solution by [[Congruent Triangles]], Lengths in [[Equilateral Triangles]], and the Area of a [[Semi-Circle]]. +-- {.image} [[StripedSemiCircleLabelledv2.jpg:pic]] =-- Consider the diagram as labelled above, in which $H$ and $I$ are such that angles $O \hat{H} F$ and $O \hat{I} B$ are [[right-angles]]. Then $H F J O$ is a [[rectangle]], so triangles $J F O$ and $H O F$ are [[congruent]]. Triangles $H O F$ and $K O C$ are also [[congruent]], so $J F O$ can be placed over $K O C$ to fill in [[sector]] $D O C$. Similarly, triangle $J B O$ can be placed over $K O E$ to fill in sector $D O E$. Therefore, the shaded region has the same area as four sectors. Its area is therefore $\frac{4}{6}$ths of the area of the semi-circle. To find the area of the semi-circle, we need to find its [[radius]]. Triangle $J F O$ is [[right-angled triangle|right-angled]] and angle $F \hat{O} J$ is $\frac{2}{6}$ths of $180^\circ$, which is $60^\circ$. Therefore, triangle $J F O$ is half of an [[equilateral triangle]] which means that the length of $F J$ is $\frac{\sqrt{3}}{2}$ of the length of $O F$. Therefore, the length of $O F$ is $\frac{3}{2} \div \frac{\sqrt{3}}{2}$ = \sqrt{3}$. The shaded region therefore has area: $$ \frac{4}{6} \times \frac{1}{2} \times \pi \times (\sqrt{3})^2 = \pi $$