1. **Problem Statement:** Calculate the surface area of the 3D shape with base 6 cm, height 12 cm, and depth 2 cm, given the net with side lengths 12, 72, 24, 6, 2, and $\sqrt{40}$.\n\n2. **Understanding the shape and net:** The shape is a rectangular prism with an additional triangular face (since $a^2 + 2^2 = 40$ implies a triangle side length $\sqrt{40}$). The net consists of rectangles and triangles representing the faces.\n\n3. **Calculate areas of each face:**\n- Rectangle 1 (base $6 \times 12$): area = $6 \times 12 = 72$ cm$^2$\n- Rectangle 2 (base $12 \times 2$): area = $12 \times 2 = 24$ cm$^2$\n- Rectangle 3 (base $6 \times 2$): area = $6 \times 2 = 12$ cm$^2$\n- Triangle face with sides 6, 2, and $\sqrt{40}$: area = $\frac{1}{2} \times 6 \times 2 = 6$ cm$^2$\n\n4. **Sum all areas:**\n$$\text{Surface Area} = 72 + 24 + 12 + 6 = 114 \text{ cm}^2$$\n\n5. **Answer:** The total surface area of the shape is $114$ cm$^2$.