# Solution to the Semi-Circle on a Hexagon Puzzle +-- {.image} [[SemiCircleonaHexagon.png:pic]] > One third of this regular hexagon is shaded. What's the area of the semicircle? =-- ## Solution by [[Area of a Trapezium]] and [[Semi-circle]], [[Similarity]], and Properties of a [[Regular Hexagon]] +-- {.image} [[SemiCircleonaHexagonLabelled.png:pic]] =-- In the above diagram, the point labelled $I$ lies on $F C$ below $E$ so that angle $F \hat{I} E$ is a [[right-angle]]. Since $I$ is the [[midpoint]] of $F O$, which has length $4$, $F I$ has length $2$. Let $x$ be the length of $G J$, so $G H$ has length $2 x + 4$, and let $h$ be the length of $E I$. Then as triangles $F I E$ and $G J E$ are [[similar]], $E J$ has length $\frac{x h}{2}$. From the formula for the [[area of a trapezium]], the shaded region has area $$ \frac{1}{2} \times \frac{x h}{2} \times (4 + 2 x + 4) = \frac{x h(4 + x)}{2} $$ Since $h$ is the height of one of the six [[equilateral triangles]] making up the [[hexagon]], the area of the is $6 \times \frac{1}{2} \times 4 h = 12 h$. So since the shaded area is one third of the hexagon: $$ \frac{x h (4 + x)}{2} = \frac{1}{3} \times 12 h $$ which simplifies to $x(4 + x) = 8$. The radius of the semi-circle is $2 + x$ so its area is $$ \frac{1}{2} \times \pi \times (2 + x)^2 = \frac{\pi}{2} (4 + 4x + x^2) $$ From above, $4 x + x^2 = x(4 + x) = 8$, so the area simplifies to: $$ \frac{\pi}{2}(4 + 8) = 6 \pi $$