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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.

The shape shown in Figure 1 is a pattern for a pendant. It consists of a sector OAB of a circle center O, of radius 6 cm, and angle AOB = π/3. The circle C, inside the sector, touches the two straight edges, OA and OB, and the arc AB as shown.
Find
(a) the area of the sector OAB,
(b) the radius of the circle C.
The region outside the circle C and inside the sector OAB is shown shaded in Figure 1.
(c) Find the area of the shaded region.

Step-by-step solution for finding the area of a sector and the radius of an inscribed circle in a sector pattern. This problem explores geometric concepts, including circle properties and sector area calculations. Suitable for students in Grades 10-12.

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