Geometry Puzzle (finding a fraction of an area)

Geometry Puzzles in Felt Tip: A Compilation of puzzles from 2018 – Catriona Shearer

Band $1$ and $3$ have the same area.

We want to find the area of the shaded segment.

As the dots are equally spaced, the sector’s angle is $\frac{\pi}{2} = (\frac{2\pi}{12}\times 3)$

Remember the area of a segment is $A=\frac{1}{2}r^2(\theta-sin(\theta))$ where the angle measurement is in radians.

(1) $\begin{equation*}A=\frac{1}{2}r^2(\frac{\pi}{2}-sin(\frac{\pi}{2}))=\frac{1}{2}r^2(\frac{\pi}{2}-1))=\frac{\pi r^2}{4}-\frac{r^2}{2}\end{equation*}$

We want to find the area of the shaded segment.

As the dots are equally spaced, the sector’s angle is $\frac{2\pi}{12} = (\frac{\pi}{6})$

(2) $\begin{equation*}A=\frac{1}{2}r^2(\frac{\pi}{6}-sin(\frac{\pi}{6}))=\frac{\pi r^2}{12}-\frac{r^2}{4}\end{equation*}$

The area of band $1$ is equation $1 -$ equation $2$ .

(3) $\begin{equation*}\frac{\pi r^2}{4}-\frac{r^2}{2}-(\frac{\pi r^2}{12}-\frac{r^2}{4})=\frac{\pi r^2}{6}-\frac{r^2}{4}\end{equation*}$

Band 2 consists of two congruent triangles and two congruent sectors.

$\begin{equation*}\theta=\frac{2\pi}{12}\times 5=\frac{5\pi}{6}, \alpha=\frac{\pi}{6}\end{equation}$

(4) $\begin{equation*}A=2(\frac{1}{2}r^2sin(\frac{5\pi}{6}))+2(\frac{1}{2}r^2\frac{\pi}{6})=\frac{r^2}{2}+\frac{r^2 \pi}{6}\end{equation*}$

Hence the shaded area is $2(\frac{\pi r^2}{6}-\frac{r^2}{4})+\frac{r^2}{2}+\frac{r^2 \pi}{6}=\frac{\pi r^2}{2}$

The area of the circle is $\pi r^2$

Hence the fraction of the shaded area is $=\frac{\frac{\pi r^2}{2}}{\pi r^2}=\frac{1}{2}$

Geometry Puzzle (finding a fraction of an area)

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