In the diagram, OBC is the sector of a circle, centre O, and angle BOC = $60^\circ$. A and D are the midpoints of OB and OC respectively, and AB = DC = 6 cm. AED is a semicircle with AD as diameter.

1. Find the radius of the sector.

Since A is the midpoint of $OB$ and $AB=6$ cm,

$OB=2\times 6=12$ cm.

So the radius of sector $OBC$ is $12$ cm.

2. Calculate the arc length $BC$.

Arc length of $BC$

$=\frac{60}{360}\times 2\pi(12)$

$=4\pi$

1. Find the area of sector $OBC$.

Since $OB=OC=12$ cm and $\angle BOC=60^\circ$,
$\text{Area of sector }OBC=\frac{60}{360}\times \pi(12)^2=24\pi\text{ cm}^2.$

2. Find the length of $AD$, the diameter of the semicircle.

Since $A$ and $D$ are midpoints of $OB$ and $OC$ respectively,
$OA=OD=6\text{ cm}.$
In $\triangle OAD$, $OA=OD=6$ cm and $\angle AOD=60^\circ$, so $\triangle OAD$ is equilateral.
$AD=6\text{ cm}.$

3. Find the total unshaded area to subtract.

The radius of the semicircle is
$\frac{AD}{2}=3\text{ cm}.$
$\text{Area of semicircle}=\frac{1}{2}\pi(3)^2=\frac{9\pi}{2}\text{ cm}^2.$
Also,
$\text{Area of }\triangle OAD=\frac{1}{2}(6)(6)\sin60^\circ=9\sqrt{3}\text{ cm}^2.$

4. Subtract the triangle and semicircle from the sector area.

$\text{Shaded area}=24\pi-\frac{9\pi}{2}-9\sqrt{3}$
$=\frac{39\pi}{2}-9\sqrt{3}\text{ cm}^2.$