A special wedge in the shape of a regular pyramid has a square base $12.0 \mathrm{~mm}$ on a side. The height of the wedge is $40.0 \mathrm{~mm}$. What is the total surface area of the wedge?
\[
A=\square \mathrm{mm}^{2}
\]
(Do not round until the final answer. Then round to the nearest ten as needed.)

Step 1 ：First, we find the area of the square base using the formula \(A = s^2\), where s is the side length. In this case, the side length is 12 mm, so the area of the square base is \(A = 12^2 = 144\) square millimeters.

Step 2 ：Next, we need to find the height of the triangular faces. We can use the Pythagorean theorem, where the height of the wedge (40 mm) is the height of the triangular faces, and half the side length of the square base (6 mm) is the base of the triangular faces. So, \(h^2 = 40^2 - 6^2\), which gives us \(h \approx 39.55\) mm.

Step 3 ：Now that we have the height of the triangular faces, we can calculate the area of each triangular face using the formula \(A = 0.5 * base * height\). In this case, the base is 12 mm and the height is approximately 39.55 mm, so the area of each triangular face is \(A \approx 0.5 * 12 * 39.55 = 237.28\) square millimeters.

Step 4 ：Finally, we find the total surface area of the wedge by adding the area of the square base and the areas of the four triangular faces. The total surface area is approximately \(144 + 4 * 237.28 = 1093.14\) square millimeters.

Step 5 ：\(\boxed{1093.14}\) square millimeters is the total surface area of the wedge.