Sure! Let's break down the solution step by step.
1. Diameter of the Circle:
- The diameter of the circle given is [tex]\( 12 \sqrt{2} \)[/tex] millimeters.
2. Radius of the Circle:
- The radius of the circle is half of the diameter.
[tex]\[
\text{Radius} = \frac{12 \sqrt{2}}{2} = 6 \sqrt{2} \text{ millimeters}
\][/tex]
3. Side Length of the Inscribed Square:
- In a circle, the diagonal of the inscribed square is equal to the diameter of the circle.
- For a square, the relation between the side length [tex]\( s \)[/tex] and the diagonal [tex]\( d \)[/tex] (using the properties of a 45°-45°-90° triangle) is:
[tex]\[
d = s \sqrt{2}
\][/tex]
- Therefore, we can find the side length [tex]\( s \)[/tex] using:
[tex]\[
s \sqrt{2} = 12 \sqrt{2}
\][/tex]
- Solving for [tex]\( s \)[/tex]:
[tex]\[
s = 12 \, \text{ millimeters}
\][/tex]
4. Area of the Square:
- The area of the square [tex]\( A_{\text{square}} \)[/tex] is given by:
[tex]\[
A_{\text{square}} = s^2 = 12^2 = 144 \, \text{square millimeters}
\][/tex]
5. Area of the Circle:
- The area of the circle [tex]\( A_{\text{circle}} \)[/tex] is given by [tex]\( \pi \)[/tex] times the square of the radius:
[tex]\[
A_{\text{circle}} = \pi (6 \sqrt{2})^2 = \pi (6^2 \cdot 2) = \pi \cdot 72 = 72\pi \, \text{square millimeters}
\][/tex]
6. Area of the Shaded Region:
- The shaded region is the area of the circle minus the area of the square.
[tex]\[
A_{\text{shaded}} = A_{\text{circle}} - A_{\text{square}} = 72\pi - 144 \, \text{square millimeters}
\][/tex]
Thus, the area of the shaded region is:
[tex]\[
72\pi - 144 \, \text{square millimeters}
\][/tex]
The correct option from the given choices is:
[tex]\[
(72\pi - 144) \, \text{mm}^2
\][/tex]
