Suppose that $A B C D E F$ is a regular hexagon with sides of length 6 . Each interior angle of $A B C D E F$ is equal to $120^{\circ}$.

(a) A circular arc with centre $D$ and radius 6 is drawn from $C$ to $E$, as shown. Determine the area of the shaded sector.

(b) A circular arc with centre $D$ and radius 6 is drawn from $C$ to $E$, as shown. A second arc with centre $A$ and radius 6 is drawn from $B$ to $F$, as shown. These arcs are tangent (that is, touch) at the centre of the hexagon. Line segments $B F$ and $C E$ are also drawn. Determine the total area of the shaded regions.

(c) Along each edge of the hexagon, a semi-circle with diameter 6 is drawn. Determine the total area of the shaded regions; that is, determine the total area of the regions that lie inside exactly two of the semi-circles.

![](https://cdn.mathpix.com/cropped/2024_04_30_aeb4d8fddfe5f1048ea8g-4.jpg?height=303&width=336&top_left_y=862&top_left_x=428)

Diagram for (a)

![](https://cdn.mathpix.com/cropped/2024_04_30_aeb4d8fddfe5f1048ea8g-4.jpg?height=301&width=347&top_left_y=863&top_left_x=867)

Diagram for (b)

![](https://cdn.mathpix.com/cropped/2024_04_30_aeb4d8fddfe5f1048ea8g-4.jpg?height=300&width=344&top_left_y=861&top_left_x=1300)

Diagram for $(\mathrm{c})$
(a) Each interior angle in a regular hexagon measures $120^{\circ}$. (One way to verify this is to use the fact that the sum of the interior angles in a regular polygon with $n$ sides is $(n-2) \cdot 180^{\circ}$. When $n=6$, this sum is $4 \cdot 180^{\circ}=720^{\circ}$. In a regular hexagon, each of these angles has measure $\frac{1}{6} \cdot 720^{\circ}=120^{\circ}$.)

Since $120^{\circ}$ is equal to $\frac{1}{3}$ of $360^{\circ}$, then the area of the shaded sector is $\frac{1}{3}$ of the area of a complete circle of radius 6 .

Therefore, the shaded area is $\frac{1}{3} \cdot \pi\left(6^{2}\right)=12 \pi$.

(b) To determine the area of region between the arc through $C$ and $E$ and the line segment $C E$, we take the area of the sector obtained in (a) and subtract the area of $\triangle C D E$.

$\triangle C D E$ has $D E=D C=6$ and $\angle C D E=120^{\circ}$.

There are many ways to find the area of this triangle.

One way is to consider $E D$ as the base of this triangle and to draw an altitude from $C$ to $E D$ extended, meeting $E D$ extended at $T$.
![](https://cdn.mathpix.com/cropped/2024_04_30_0c682e608943d35a21aag-06.jpg?height=304&width=698&top_left_y=1366&top_left_x=816)

Since $\angle C D E=120^{\circ}$, then $\angle C D T=180^{\circ}-\angle C D E=60^{\circ}$.

This means that $\triangle C D T$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, and so $C T=\frac{\sqrt{3}}{2} C D=\frac{\sqrt{3}}{2} \cdot 6=3 \sqrt{3}$.

This means that the area of $\triangle C D E$ is $\frac{1}{2} \cdot E D \cdot C T=\frac{1}{2} \cdot 6 \cdot 3 \sqrt{3}=9 \sqrt{3}$.

Therefore, the area of the shaded region between the arc through $C$ and $E$ and the line segment $C E$ is $12 \pi-9 \sqrt{3}$.

Since the region between line segment $B F$ and the arc through $B$ and $F$ is constructed in exactly the same way, its area is the same.

Therefore, the total area of the shaded regions is $2(12 \pi-9 \sqrt{3})=24 \pi-18 \sqrt{3}$.
(c) Let $M$ be the midpoint of $D E$ and $N$ be the midpoint of $E F$.

This means that $M$ and $N$ are the centres of two of the semi-circles.

Let $P$ be the point other than $E$ where the semi-circles with centres $M$ and $N$ intersect. Join $E$ to $P$.

![](https://cdn.mathpix.com/cropped/2024_04_30_0c682e608943d35a21aag-07.jpg?height=312&width=352&top_left_y=386&top_left_x=992)

By symmetry, $E P$ divides the shaded region between these two semi-circles into two pieces of equal area.

Let the area of one of these pieces be $a$.

Furthermore, by symmetry in the whole hexagon, each of the six shaded regions between two semi-circles is equal in area.

This means that the entire shaded region is equal to $12 a$.

Therefore, we need to determine the value of $a$.

Consider the region between $E P$ and the arc with centre $M$ through $E$ and $P$.

![](https://cdn.mathpix.com/cropped/2024_04_30_0c682e608943d35a21aag-07.jpg?height=299&width=458&top_left_y=1138&top_left_x=931)

Since $D E=E F=6$, then $D M=M E=E N=N F=3$. Each of the two semi-circles has radius 3 .

Since $P$ is on both semi-circles, then $N P=M P=3$.

Consider $\triangle E M P$. Here, we have $M E=M P=3$.

Also, $\angle M E P=60^{\circ}$ since $\angle D E F=120^{\circ}$ and $\angle M E P=\angle N E P$ by symmetry.

Since $M E=M P$, then $\angle M P E=\angle M E P=60^{\circ}$.

Since $\triangle E M P$ has two $60^{\circ}$ angles, then its third angle also has measure $60^{\circ}$, which means that $\triangle E M P$ is equilateral.

Therefore, $P E=3$ and $\angle E M P=60^{\circ}$.

Now, we can calculate the area $a$.

The area $a$ is equal to the area of the sector of the circle with centre $M$ defined by $E$ and $P$ minus the area of $\triangle E M P$.

Since $\angle E M P=60^{\circ}$, which is $\frac{1}{6}$ of a complete circle, and the radius of the circle from which the sector comes is 3 , then the area of the sector is $\frac{1}{6} \cdot \pi\left(3^{2}\right)=\frac{3}{2} \pi$.

The area of $\triangle E M P$, which is equilateral with side length 3 , can be found in many different ways.

One way to do this is to use the formula that the area of a triangle with two side lengths $x$ and $y$ and an angle of $\theta$ between these two sides is equal to $\frac{1}{2} x y \sin \theta$.

Thus, the area of $\triangle E M P$ is $\frac{1}{2} \cdot 3 \cdot 3 \sin 60^{\circ}=\frac{9 \sqrt{3}}{4}$.

This means that $a=\frac{3}{2} \pi-\frac{9 \sqrt{3}}{4}$.

Finally, the total area of the shaded regions equals $12 a$ which equals $12\left(\frac{3}{2} \pi-\frac{9 \sqrt{3}}{4}\right)$, or $18 \pi-27 \sqrt{3}$.