Let's go through the problem step by step based on the diagram and the information given in the question:

Part (a): Finding the area of the sector $OAB$

• The sector $OAB$ is part of a circle with radius $6 \, \text{cm}$ and an angle $\theta = \frac{\pi}{3}$ radians.

The formula for the area of a sector is: $A_{\text{sector}} = \frac{1}{2} r^2 \theta$ where $r$ is the radius of the sector and $\theta$ is the angle in radians.

Substitute the values $r = 6 \, \text{cm}$ and $\theta = \frac{\pi}{3}$: $A_{\text{sector}} = \frac{1}{2} \times 6^2 \times \frac{\pi}{3}$ $A_{\text{sector}} = \frac{1}{2} \times 36 \times \frac{\pi}{3}$ $A_{\text{sector}} = 18 \times \frac{\pi}{3} = 6\pi \, \text{cm}^2$

So, the area of the sector $OAB$ is $6\pi \, \text{cm}^2$.

Part (b): Finding the radius of the circle $C$

• The circle $C$ is inscribed in the sector. It touches the two straight sides $OA$ and $OB$, as well as the arc $AB$.

We can use the formula for the radius of a circle inscribed in a sector: $r_{\text{circle}} = \frac{r_{\text{sector}} (\sin \frac{\theta}{2})}{1 + \sin \frac{\theta}{2}}$ where $r_{\text{sector}} = 6 \, \text{cm}$ and $\theta = \frac{\pi}{3}$.

First, calculate $\sin \frac{\theta}{2} = \sin \frac{\pi}{6} = \frac{1}{2}$.

Now, substitute the values into the formula: $r_{\text{circle}} = \frac{6 \times \frac{1}{2}}{1 + \frac{1}{2}}$ $r_{\text{circle}} = \frac{3}{1.5} = 2 \, \text{cm}$

So, the radius of the circle $C$ is $2 \, \text{cm}$.

Part (c): Finding the area of the shaded region

The shaded region is the area of the sector minus the area of the inscribed circle.

1. Area of the sector $OAB$ (already calculated): $A_{\text{sector}} = 6\pi \, \text{cm}^2$

2. Area of the circle $C$: The area of a circle is given by $A = \pi r^2$. $A_{\text{circle}} = \pi \times 2^2 = 4\pi \, \text{cm}^2$

3. Shaded area: $A_{\text{shaded}} = A_{\text{sector}} - A_{\text{circle}} = 6\pi - 4\pi = 2\pi \, \text{cm}^2$

So, the area of the shaded region is $2\pi \, \text{cm}^2$.

Summary of the solutions:

1. The area of the sector $OAB$ is $6\pi \, \text{cm}^2$.
2. The radius of the circle $C$ is $2 \, \text{cm}$.
3. The area of the shaded region is $2\pi \, \text{cm}^2$.

Would you like further details or have any questions? Here are 5 questions that build on this problem:

1. What would the area of the sector be if the angle $\theta$ were doubled?
2. How does the radius of the inscribed circle change if the angle $\theta$ increases?
3. What is the perimeter of the sector $OAB$, including the arc and the two sides $OA$ and $OB$?
4. Can you derive the formula for the radius of an inscribed circle in a sector from first principles?
5. How would the shaded area change if the radius of the sector were increased to 12 cm?

Tip: When working with sectors and circles, remember that angles in radians often simplify formulas for area and arc length, especially when inscribed shapes are involved.