1. **State the problem:** We need to find the surface area of a square-based pyramid where the base is a square with side length 8.6 cm and the triangular faces are identical isosceles triangles with slant edges of 12 cm. 2. **Formula for surface area of a square-based pyramid:** $$\text{Surface Area} = \text{Base Area} + \text{Lateral Area}$$ where $$\text{Base Area} = s^2$$ and $$\text{Lateral Area} = 4 \times \text{Area of one triangular face}$$ 3. **Calculate the base area:** $$s = 8.6 \text{ cm}$$ $$\text{Base Area} = 8.6^2 = 73.96 \text{ cm}^2$$ 4. **Find the height of one triangular face:** Each triangular face is isosceles with two equal sides of 12 cm and base 8.6 cm. The height $h$ of the triangle can be found using the Pythagorean theorem by splitting the base into two halves: $$h = \sqrt{12^2 - \left(\frac{8.6}{2}\right)^2} = \sqrt{144 - 4.3^2} = \sqrt{144 - 18.49} = \sqrt{125.51} \approx 11.2 \text{ cm}$$ 5. **Calculate the area of one triangular face:** $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8.6 \times 11.2 = 48.16 \text{ cm}^2$$ 6. **Calculate the lateral area:** $$4 \times 48.16 = 192.64 \text{ cm}^2$$ 7. **Calculate total surface area:** $$73.96 + 192.64 = 266.6 \text{ cm}^2$$ 8. **Round to nearest cm²:** $$\boxed{267 \text{ cm}^2}$$