1. **State the problem:** Find the surface area of a triangular prism with given side lengths 15 yd, 12 yd, 13 yd, and 9 yd. 2. **Identify the shape and dimensions:** The prism has a triangular base with sides 15 yd, 12 yd, and 13 yd, and the length (height) of the prism is 9 yd. 3. **Formula for surface area of a triangular prism:** $$\text{Surface Area} = 2 \times \text{Area of triangular base} + \text{Perimeter of base} \times \text{Length}$$ 4. **Calculate the area of the triangular base using Heron's formula:** - Semi-perimeter $s = \frac{15 + 12 + 13}{2} = 20$ yd - Area $= \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{20(20 - 15)(20 - 12)(20 - 13)} = \sqrt{20 \times 5 \times 8 \times 7}$ 5. Simplify inside the square root: $$\sqrt{20 \times 5 \times 8 \times 7} = \sqrt{5600}$$ 6. Calculate $\sqrt{5600}$: $$\sqrt{5600} = \sqrt{100 \times 56} = 10 \sqrt{56} = 10 \times 2 \sqrt{14} = 20 \sqrt{14}$$ 7. So, the area of the triangular base is: $$20 \sqrt{14} \text{ yd}^2$$ 8. Calculate the perimeter of the base: $$15 + 12 + 13 = 40 \text{ yd}$$ 9. Calculate the lateral surface area: $$\text{Perimeter} \times \text{Length} = 40 \times 9 = 360 \text{ yd}^2$$ 10. Calculate total surface area: $$2 \times 20 \sqrt{14} + 360 = 40 \sqrt{14} + 360 \text{ yd}^2$$ 11. Approximate $\sqrt{14} \approx 3.7417$: $$40 \times 3.7417 = 149.668$$ 12. Final surface area: $$149.668 + 360 = 509.668 \text{ yd}^2$$ **Answer:** The surface area of the triangular prism is approximately **509.67 yd\textsuperscript{2}**.