1. **State the problem:** Calculate the surface area of a rhombicuboctahedron with 18 square faces each of edge length 24 inches, and 8 triangular faces each with height approximately 20.8 inches. 2. **Formula for surface area:** The total surface area $A$ is the sum of the areas of all square faces plus the areas of all triangular faces: $$A = 18 \times A_{square} + 8 \times A_{triangle}$$ 3. **Area of one square face:** For a square with edge length $s$, area is: $$A_{square} = s^2$$ Given $s = 24$ inches, $$A_{square} = 24^2 = 576$$ square inches. 4. **Area of one triangular face:** For a triangle, area is: $$A_{triangle} = \frac{1}{2} \times base \times height$$ We know the height $h = 20.8$ inches, but the base length is the same as the edge length of the square, $b = 24$ inches. Calculate: $$A_{triangle} = \frac{1}{2} \times 24 \times 20.8 = 12 \times 20.8 = 249.6$$ square inches. 5. **Calculate total surface area:** $$A = 18 \times 576 + 8 \times 249.6 = 10368 + 1996.8 = 12364.8$$ square inches. 6. **Final answer:** The surface area of the rhombicuboctahedron is approximately **12364.8 square inches**.