1. **Stating the problem:** We have two truncated pyramids with given dimensions: base widths, top widths, heights, and depths of the shaded truncated portions. We want to analyze or calculate properties related to these truncated pyramids. 2. **Formula for volume of a truncated pyramid:** The volume $V$ of a truncated pyramid with height $h$, lower base area $B_1$, and upper base area $B_2$ is given by: $$V = \frac{h}{3} (B_1 + B_2 + \sqrt{B_1 B_2})$$ 3. **Important rules:** - The base areas $B_1$ and $B_2$ depend on the shape of the bases. Assuming square bases (since only widths are given), the area is width squared. - Heights and depths are vertical measurements. 4. **Calculate volumes for each truncated pyramid:** **Top truncated pyramid:** - Base width $= 10$ cm, so $B_1 = 10^2 = 100$ cm$^2$ - Top width $= 10$ cm, so $B_2 = 10^2 = 100$ cm$^2$ - Height $h = 16$ cm Calculate volume: $$V_1 = \frac{16}{3} (100 + 100 + \sqrt{100 \times 100}) = \frac{16}{3} (200 + 100) = \frac{16}{3} \times 300 = 1600 \text{ cm}^3$$ **Bottom truncated pyramid:** - Base width $= 8$ cm, so $B_1 = 8^2 = 64$ cm$^2$ - Top width $= 6$ cm, so $B_2 = 6^2 = 36$ cm$^2$ - Height $h = 14$ cm Calculate volume: $$V_2 = \frac{14}{3} (64 + 36 + \sqrt{64 \times 36}) = \frac{14}{3} (100 + 48) = \frac{14}{3} \times 148 = \frac{2072}{3} \approx 690.67 \text{ cm}^3$$ 5. **Explanation:** We used the volume formula for truncated pyramids assuming square bases. The shaded truncated portion's depth is not directly used in volume calculation but may relate to other properties. **Final answers:** - Volume of top truncated pyramid: $1600$ cm$^3$ - Volume of bottom truncated pyramid: approximately $690.67$ cm$^3$