1. **Stating the problem:** We have a truncated regular quadrilateral pyramid (a frustum) with height $h=6$ cm. The base edge length of the larger base is $a = \frac{8}{3} \times h = \frac{8}{3} \times 6 = 16$ cm. The smaller base edge length is $b = 25\%$ of $a = 0.25 \times 16 = 4$ cm. We need to find: a) Volume of the frustum. b) Length of the lateral edge (slant height) of the frustum. c) Measure of the dihedral angle between two opposite lateral faces. d) Distance from a vertex of the larger base to the opposite lateral face. 2. **Formulas and important rules:** - Volume of a frustum of a pyramid with square bases: $$V = \frac{h}{3} (A_1 + A_2 + \sqrt{A_1 A_2})$$ where $A_1 = a^2$ and $A_2 = b^2$ are areas of the larger and smaller bases. - The lateral edge length $l$ can be found using the Pythagorean theorem in the vertical cross-section: $$l = \sqrt{h^2 + \left(\frac{a-b}{2}\right)^2}$$ - The dihedral angle $\theta$ between two opposite lateral faces can be found by considering the angle between their normal vectors or by geometry of the frustum. - Distance from a vertex of the larger base to the opposite lateral face can be found using plane equations or geometric relations. 3. **Calculations:** **a) Volume:** $$A_1 = 16^2 = 256$$ $$A_2 = 4^2 = 16$$ $$V = \frac{6}{3} (256 + 16 + \sqrt{256 \times 16}) = 2 (272 + \sqrt{4096}) = 2 (272 + 64) = 2 \times 336 = 672 \text{ cm}^3$$ **b) Lateral edge length:** $$l = \sqrt{6^2 + \left(\frac{16 - 4}{2}\right)^2} = \sqrt{36 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \approx 8.49 \text{ cm}$$ **c) Dihedral angle between two opposite lateral faces:** The angle $\theta$ satisfies: $$\cos \theta = \frac{h^2 - \left(\frac{a-b}{2}\right)^2}{h^2 + \left(\frac{a-b}{2}\right)^2} = \frac{36 - 36}{36 + 36} = 0$$ So, $$\theta = \arccos(0) = 90^\circ$$ **d) Distance from a vertex of the larger base to the opposite lateral face:** The lateral face is a trapezoid connecting corresponding edges. The distance from a vertex of the larger base to the opposite lateral face equals the height of the frustum times the ratio of the smaller base edge to the larger base edge: $$d = h \times \frac{b}{a} = 6 \times \frac{4}{16} = 6 \times 0.25 = 1.5 \text{ cm}$$ **Final answers:** - Volume $V = 672$ cm$^3$ - Lateral edge length $l = 6\sqrt{2} \approx 8.49$ cm - Dihedral angle $\theta = 90^\circ$ - Distance $d = 1.5$ cm