1. **State the problem:** We have a regular triangular pyramid (tetrahedron) with an altitude (height) of 11 m and a volume of 45.8 m³. We need to find the lateral area of the pyramid in m². 2. **Recall the formulas:** - Volume of a pyramid: $$V = \frac{1}{3} \times B \times h$$ where $B$ is the area of the base and $h$ is the height (altitude). - Lateral area of a regular triangular pyramid: $$L = \frac{1}{2} \times P \times l$$ where $P$ is the perimeter of the base and $l$ is the slant height. 3. **Find the base area $B$:** Given volume $V = 45.8$ and height $h = 11$, solve for $B$: $$45.8 = \frac{1}{3} \times B \times 11$$ Multiply both sides by 3: $$137.4 = B \times 11$$ Divide both sides by 11: $$B = \frac{137.4}{11} = 12.49 \text{ m}^2$$ 4. **Find the side length $a$ of the equilateral triangle base:** Area of an equilateral triangle: $$B = \frac{\sqrt{3}}{4} a^2$$ Solve for $a$: $$a^2 = \frac{4B}{\sqrt{3}} = \frac{4 \times 12.49}{1.732} = 28.82$$ $$a = \sqrt{28.82} = 5.37 \text{ m}$$ 5. **Find the perimeter $P$ of the base:** $$P = 3a = 3 \times 5.37 = 16.11 \text{ m}$$ 6. **Find the slant height $l$:** The slant height is the height of each triangular face. It can be found using the Pythagorean theorem: - The height of the base triangle (from center to midpoint of a side) is: $$h_b = \frac{\sqrt{3}}{2} a = \frac{1.732}{2} \times 5.37 = 4.65 \text{ m}$$ - The altitude of the pyramid is $h = 11$ m. - The slant height $l$ is: $$l = \sqrt{h^2 + h_b^2} = \sqrt{11^2 + 4.65^2} = \sqrt{121 + 21.62} = \sqrt{142.62} = 11.94 \text{ m}$$ 7. **Calculate the lateral area $L$:** $$L = \frac{1}{2} \times P \times l = 0.5 \times 16.11 \times 11.94 = 96.1 \text{ m}^2$$ **Final answer:** The lateral area of the pyramid is approximately **96.1 m²**.