1. **State the problem:** Calculate the total surface area of the pyramid with a square base BCDE where each side is 12 cm, and the vertex F is directly above the center M of the base with height FM = 9 cm. 2. **Identify the components of the surface area:** The total surface area $A_{total}$ is the sum of the base area $A_{base}$ and the lateral surface area $A_{lateral}$. 3. **Calculate the base area:** Since BCDE is a square with side length 12 cm, $$A_{base} = 12 \times 12 = 144 \text{ cm}^2.$$ 4. **Calculate the slant height of the triangular faces:** The diagonals of the square intersect at M, the center. The length of diagonal BD (or CE) is $$BD = 12\sqrt{2}.$$ Since M is the midpoint of BD, $$BM = \frac{12\sqrt{2}}{2} = 6\sqrt{2}.$$ 5. **Calculate the slant height $s$ of the triangular faces:** The slant height is the length from F to a base vertex (e.g., B). Using the Pythagorean theorem in triangle FBM, $$s = \sqrt{FM^2 + BM^2} = \sqrt{9^2 + (6\sqrt{2})^2} = \sqrt{81 + 72} = \sqrt{153} = 3\sqrt{17}.$$ 6. **Calculate the area of one triangular face:** Each triangular face has base 12 cm and height equal to the slant height $s$, $$A_{triangle} = \frac{1}{2} \times 12 \times 3\sqrt{17} = 6 \times 3\sqrt{17} = 18\sqrt{17}.$$ 7. **Calculate the total lateral surface area:** There are 4 triangular faces, $$A_{lateral} = 4 \times 18\sqrt{17} = 72\sqrt{17}.$$ 8. **Calculate the total surface area:** $$A_{total} = A_{base} + A_{lateral} = 144 + 72\sqrt{17}.$$ 9. **Approximate the numerical value:** Since $\sqrt{17} \approx 4.123$, $$A_{total} \approx 144 + 72 \times 4.123 = 144 + 296.856 = 440.856 \text{ cm}^2.$$ **Final answer:** The total surface area of the pyramid is $$\boxed{144 + 72\sqrt{17} \approx 440.86 \text{ cm}^2}.$$