You're on a roll — Step 1: Find the surface area of the square-based pyramid. The pyramid has a square base and four triangular faces. Given: Side length of the square base ($s$) = $2\sqrt{2}$ cm Slant height of the triangular faces ($l$) = $5$ cm First, calculate the area of the square base ($A_b$): $$A_b = s^2 = (2\sqrt{2})^2 = 2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8 \text{ cm}^2$$ Next, calculate the area of one triangular face ($A_t$). The base of each triangle is the side length of the square base, $s$. $$A_t = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times s \times l$$ $$A_t = \frac{1}{2} \times (2\sqrt{2} \text{ cm}) \times (5 \text{ cm})$$ $$A_t = \frac{1}{2} \times 10\sqrt{2} \text{ cm}^2 = 5\sqrt{2} \text{ cm}^2$$ The total surface area ($SA$) is the sum of the base area and the area of the four triangular faces: $$SA = A_b + 4 \times A_t$$ $$SA = 8 \text{ cm}^2 + 4 \times (5\sqrt{2} \text{ cm}^2)$$ $$SA = 8 \text{ cm}^2 + 20\sqrt{2} \text{ cm}^2$$ $$SA = (8 + 20\sqrt{2}) \text{ cm}^2$$ $$\boxed{(8 + 20\sqrt{2}) \text{ cm}^2}$$ Step 2: Calculate the volume of the truncated cone (frustum). Given: Large radius ($R$) = $7$ cm Small radius ($r$) = $5$ cm Height of the frustum ($H_f$) = $10$ cm Use $\pi = \frac{22}{7}$. The formula for the volume of a frustum is: $$V = \frac{1}{3}\pi H_f (R^2 + Rr + r^2)$$ Substitute the given values into the formula: $$V = \frac{1}{3} \times \frac{22}{7} \times 10 \text{ cm} \times ((7 \text{ cm})^2 + (7 \text{ cm} \times 5 \text{ cm}) + (5 \text{ cm})^2)$$ $$V = \frac{220}{21} \text{ cm} \times (49 \text{ cm}^2 + 35 \text{ cm}^2 + 25 \text{ cm}^2)$$ $$V = \frac{220}{21} \text{ cm} \times (109 \text{ cm}^2)$$ $$V = \frac{220 \times 109}{21} \text{ cm}^3$$ $$V = \frac{23980}{21} \text{ cm}^3$$ $$\boxed{\frac{23980}{21} \text{ cm}^3}$$ Got more? Send 'em!