1. **Problem statement:** We have a regular hexagonal pyramid with a lateral edge of length 6 cm and a base side length of 3 cm. We need to find its volume in cubic centimeters. 2. **Relevant formulas:** - The volume $V$ of a pyramid is given by $$V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$$ - The base is a regular hexagon, so its area $A$ can be calculated by $$A = \frac{3\sqrt{3}}{2} s^2$$ where $s$ is the side length of the hexagon. 3. **Calculate the base area:** Given $s = 3$ cm, $$A = \frac{3\sqrt{3}}{2} \times 3^2 = \frac{3\sqrt{3}}{2} \times 9 = \frac{27\sqrt{3}}{2} \text{ cm}^2$$ 4. **Find the height of the pyramid:** - The lateral edge is the slant edge from the apex to a vertex of the base, length 6 cm. - The height $h$ is the perpendicular distance from the apex to the base plane. - The distance from the center of the hexagon to a vertex (circumradius $R$) is $$R = s = 3 \text{ cm}$$ for a regular hexagon. - The lateral edge, height, and circumradius form a right triangle with the lateral edge as hypotenuse: $$6^2 = h^2 + 3^2$$ $$36 = h^2 + 9$$ $$h^2 = 27$$ $$h = \sqrt{27} = 3\sqrt{3} \text{ cm}$$ 5. **Calculate the volume:** $$V = \frac{1}{3} \times \frac{27\sqrt{3}}{2} \times 3\sqrt{3} = \frac{1}{3} \times \frac{27\sqrt{3}}{2} \times 3\sqrt{3}$$ Simplify inside: $$3\sqrt{3} \times \sqrt{3} = 3 \times 3 = 9$$ So, $$V = \frac{1}{3} \times \frac{27}{2} \times 9 = \frac{1}{3} \times \frac{243}{2} = \frac{243}{6} = 40.5 \text{ cm}^3$$ **Final answer:** The volume of the pyramid is $40.5$ cubic centimeters.