, 2016

Triangle $ABC$ is equilateral with side length 6, and so $AB=BC=CA=6$.

Since $D$ is the midpoint of $BC$, then $BD=DC=3$.

In $\triangle ADC$, $\angle ADC=90^{\circ}$ and so by the Pythagorean Theorem $AD^2=AC^2-DC^2$.

Therefore, $h^2=6^2-3^2=36-9=27$ and so $h=\sqrt{27}=\sqrt{9\times3}=\sqrt{9}\times\sqrt{3}=3\sqrt{3}$,since $h>0$.
The shaded region lies inside the circle and outside the hexagon and thus its area is determined by subtracting the area of the hexagon from the area of the circle.

First we find the area of the hexagon.

Each vertex of hexagon $EFGHIJ$ lies on the circle.

Since the circle has centre $O$ and radius 6, then $OE=OF=OG=OH=OI=OJ=6$.

Each side length of the hexagon is also 6, and so the hexagon is formed by six congruent equilateral triangles with side length 6. (For example, $\triangle OGH$ is one these 6 triangles.)

Each of these triangles is congruent to $\triangle ABC$ from part (a) and thus has height $h=3\sqrt{3}$.

The area of each of the six congruent triangles is $\frac12(6)(3\sqrt{3})=9\sqrt{3}$.

Therefore, the area of hexagon $EFGHIJ$ is $6\times9\sqrt{3}=54\sqrt{3}$.

The area of the circle with centre $O$ and radius 6 is $\pi(6)^2=36\pi$.

Finally, the area of the shaded region is $36\pi-54\sqrt{3}$.
Let the area of the shaded region that we are required to find be $A$.

Let the area of the shaded region in the diagram to the right be $S$.

[[IMAGE0]]

We may determine $A$ by subtracting $S$ from the area of the semi-circle with centre $P$.

First we determine $S$.

Consider the circle with centre $O$. The shaded region having area $S$ lies inside sector $MON$ of this circle, but outside $\triangle MON$.

That is, $S$ is determined by subtracting the area of $\triangle MON$ from the area of sector $MON$.

In $\triangle MON$, $MN=ON=OM=r$ (since $ON$ and $OM$ are radii), and so the triangle is equilateral. Join $O$ to $P$.

Since $ON=OM$ and $P$ is the midpoint of $MN$, then $OP$ is the altitude (height) of $\triangle MON$ with base $MN$.

In $\triangle OPN$, $\angle OPN=90^{\circ}$ and so by the Pythagorean Theorem $OP^2=ON^2-PN^2$.

Since $PN=\frac12(MN)=\frac12r$, $OP^2=r^2-(\frac12r)^2=r^2-\frac14r^2=\frac34r^2$, and so $OP=\sqrt{\frac34r^2}=\frac{\sqrt{3}}{2}r$.

Therefore, the area of $\triangle MON$ is $\frac12(MN)(OP)=\frac12(r)\left(\frac{\sqrt{3}}{2}r\right)=\frac{\sqrt{3}}{4}r^2$.

Next, we determine the area of sector $MON$.

Since $\triangle MON$ is equilateral, then $\angle MON=60^{\circ}$.

Thus, the area of sector $MON$ is $\frac{60^{\circ}}{360^{\circ}}=\frac16$ of the area of the circle with centre $O$ and radius $r$, or $\frac16\pi r^2$.

Therefore, $S=\frac16\pi r^2-\frac{\sqrt{3}}{4}r^2$.

Finally, one-half of the area of the circle with centre $P$ and radius $PN=\frac12r$ is$\frac12\pi\left(\frac12r \right)^2=\frac18\pi r^2$, and so $A=\frac18\pi r^2-S=\frac18\pi r^2-\left(\frac16\pi r^2-\frac{\sqrt{3}}{4}r^2\right)=\left(\frac18\pi -\frac16\pi +\frac{\sqrt{3}}{4}\right)r^2$.

Simplifying further, the exact area of the shaded region is $\frac{6\sqrt{3}-\pi}{24}r^2$.