Monthly Archives: January 2026

January 2, 2026 · 7:08 am

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}

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