1. **Problem:** The base of the triangular pyramid is equilateral and has an area of approximately 5.1 cm². Find the total surface area of the triangular pyramid. 2. **Formula and rules:** - The total surface area of a triangular pyramid (tetrahedron) is the sum of the areas of all its faces. - Since the base is equilateral, all sides are equal, and the lateral faces are congruent triangles. - If the base area is $A_b = 5.1$ cm², and the pyramid is regular, the lateral faces are equilateral triangles with the same side length. 3. **Find the side length of the base:** - Area of an equilateral triangle with side length $s$ is $$A = \frac{\sqrt{3}}{4} s^2$$ - Solve for $s$: $$s^2 = \frac{4A}{\sqrt{3}} = \frac{4 \times 5.1}{1.732} \approx 11.77$$ $$s = \sqrt{11.77} \approx 3.43 \text{ cm}$$ 4. **Find the area of one lateral face:** - Each lateral face is an equilateral triangle with side $s = 3.43$ cm. - Area of one lateral face: $$A_{lat} = \frac{\sqrt{3}}{4} s^2 = 5.1 \text{ cm}^2$$ (same as base area since equilateral) 5. **Number of faces:** - A triangular pyramid has 4 faces: 1 base + 3 lateral faces. 6. **Calculate total surface area:** $$\text{Total Surface Area} = A_b + 3 \times A_{lat} = 5.1 + 3 \times 5.1 = 5.1 + 15.3 = 20.4 \text{ cm}^2$$ **Final answer:** $$\boxed{20.4 \text{ cm}^2}$$