1. **State the problem:** We need to find the surface area of a triangular pyramid (tetrahedron) with a base that is an equilateral triangle of side length 15 inches, a height inside the base of 13 inches, and three slant heights from the apex to the base edges each measuring 20 inches. 2. **Identify the base area:** The base is an equilateral triangle with side $s=15$ inches. The formula for the area of an equilateral triangle is: $$\text{Area} = \frac{\sqrt{3}}{4} s^2$$ Calculate the base area: $$\text{Area}_{base} = \frac{\sqrt{3}}{4} \times 15^2 = \frac{\sqrt{3}}{4} \times 225 = 56.25\sqrt{3}$$ 3. **Calculate the lateral faces' areas:** Each lateral face is a triangle with base 15 inches and slant height 20 inches. The area of each lateral face is: $$\text{Area}_{face} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \times 20 = 150$$ Since there are 3 such faces: $$\text{Total lateral area} = 3 \times 150 = 450$$ 4. **Calculate total surface area:** $$\text{Surface area} = \text{Area}_{base} + \text{Total lateral area} = 56.25\sqrt{3} + 450$$ 5. **Approximate the numerical value:** $$\sqrt{3} \approx 1.732$$ $$56.25 \times 1.732 \approx 97.425$$ So, $$\text{Surface area} \approx 97.425 + 450 = 547.425$$ **Final answer:** The surface area of the triangular pyramid is approximately **547.43 square inches**.