A room is built in the shape of the region between two semicircles with the same center and parallel diameters. The farthest distance between two points with a clear line of sight is $12$m. What is the area (in $m^2$) of the room?

[img]https://cdn.artofproblemsolving.com/attachments/b/c/d2b2fc9fa9cca27dc9d692a1bb3089ba792063.png[/img]
1. **Identify the given information and the goal:**
   - The farthest distance between two points with a clear line of sight is \(12\) meters.
   - The room is the region between two semicircles with the same center and parallel diameters.
   - We need to find the area of the room in square meters.

2. **Understand the geometry of the problem:**
   - Let \(R\) be the radius of the outer semicircle and \(r\) be the radius of the inner semicircle.
   - The farthest distance between two points with a clear line of sight is the length of a chord of the outer semicircle that is tangent to the inner semicircle.

3. **Use the given maximum distance:**
   - The length of the chord tangent to the inner semicircle is given by \(2\sqrt{R^2 - r^2} = 12\).
   - Solving for \(R^2 - r^2\):
     \[
     2\sqrt{R^2 - r^2} = 12 \implies \sqrt{R^2 - r^2} = 6 \implies R^2 - r^2 = 36
     \]

4. **Calculate the area of the room:**
   - The area of the room is the difference between the areas of the outer and inner semicircles.
   - The area of the outer semicircle is \(\frac{1}{2} \pi R^2\).
   - The area of the inner semicircle is \(\frac{1}{2} \pi r^2\).
   - Therefore, the area of the room is:
     \[
     \frac{1}{2} \pi R^2 - \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (R^2 - r^2)
     \]
   - Substitute \(R^2 - r^2 = 36\):
     \[
     \frac{1}{2} \pi (R^2 - r^2) = \frac{1}{2} \pi \times 36 = 18\pi
     \]

The final answer is \(\boxed{18\pi}\)