To find the area of the shaded region, we need to follow these steps:

1. Determine the radius of the inner circle:

   • The circumference $C$ of the inner circle is given by the formula: $C = 2\pi r_{\text{inner}}$ Given $C = 111.784$ mm, solve for $r_{\text{inner}}$: $r_{\text{inner}} = \frac{C}{2\pi} = \frac{111.784 \text{ mm}}{2\pi} \approx 17.8 \text{ mm}$

2. Determine the radius of the outer circle:

   • We know that the difference between the radii of the outer circle and the inner circle is given as 18.1 mm. So, the radius of the outer circle $r_{\text{outer}}$ is: $r_{\text{outer}} = r_{\text{inner}} + 18.1 \text{ mm} \approx 17.8 \text{ mm} + 18.1 \text{ mm} = 35.9 \text{ mm}$

3. Calculate the area of the inner circle:

   • The area $A_{\text{inner}}$ is given by: $A_{\text{inner}} = \pi r_{\text{inner}}^2 \approx \pi (17.8 \text{ mm})^2 \approx 995.58 \text{ mm}^2$

4. Calculate the area of the outer circle:

   • The area $A_{\text{outer}}$ is given by: $A_{\text{outer}} = \pi r_{\text{outer}}^2 \approx \pi (35.9 \text{ mm})^2 \approx 4047.27 \text{ mm}^2$

5. Calculate the area of the shaded region:

   • The area of the shaded region is the difference between the area of the outer circle and the inner circle: $A_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}} \approx 4047.27 \text{ mm}^2 - 995.58 \text{ mm}^2 \approx 3051.69 \text{ mm}^2$

The area of the shaded region is approximately 3051.69 square millimeters.

Would you like to ask any more questions or need further details? Here are some related questions to consider:

1. How does the circumference of a circle relate to its radius?
2. What would happen to the area of the shaded region if the difference between the radii increased?
3. Can you calculate the circumference of the outer circle based on the given data?
4. What would be the area of the shaded region if the radii difference was 20 mm instead of 18.1 mm?
5. How does the area of a circle change if its radius is doubled?

Tip: Always double-check units when solving geometry problems to ensure the final answer is in the correct form.