1. **State the problem:** Find the surface area of a triangular pyramid (tetrahedron) with a triangular base of sides 8 m, 8 m, and 6.9 m, and slant heights (heights of the triangular faces) of 8 m and 10 m. 2. **Formula for surface area:** The surface area $S$ of a triangular pyramid is the sum of the base area plus the areas of the three triangular faces: $$S = \text{Area}_{base} + \text{Area}_{face1} + \text{Area}_{face2} + \text{Area}_{face3}$$ 3. **Calculate the base area:** The base is a triangle with sides $a=8$, $b=8$, and $c=6.9$ meters. Use Heron's formula: $$s = \frac{a+b+c}{2} = \frac{8+8+6.9}{2} = 11.45$$ $$\text{Area}_{base} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{11.45(11.45-8)(11.45-8)(11.45-6.9)}$$ Calculate inside the root: $$= \sqrt{11.45 \times 3.45 \times 3.45 \times 4.55} = \sqrt{11.45 \times 3.45^2 \times 4.55}$$ $$= \sqrt{11.45 \times 11.9025 \times 4.55} = \sqrt{620.1} \approx 24.9$$ So, $$\text{Area}_{base} \approx 24.9 \text{ m}^2$$ 4. **Calculate the areas of the three triangular faces:** Each face area is $\frac{1}{2} \times \text{base edge} \times \text{slant height}$. - Face 1 with base 8 m and slant height 10 m: $$\text{Area}_{face1} = \frac{1}{2} \times 8 \times 10 = 40$$ - Face 2 with base 8 m and slant height 8 m: $$\text{Area}_{face2} = \frac{1}{2} \times 8 \times 8 = 32$$ - Face 3 with base 6.9 m and slant height 8 m: $$\text{Area}_{face3} = \frac{1}{2} \times 6.9 \times 8 = 27.6$$ 5. **Sum all areas for total surface area:** $$S = 24.9 + 40 + 32 + 27.6 = 124.5$$ **Final answer:** The surface area of the triangular pyramid is approximately **124.5 square meters**.