1. **Problem 1: Find the slant height $s$ of the square-based pyramid.** Given: - Height $h = 8.2$ cm - Base side length $6.4$ cm The slant height $s$ is the length from the apex to the midpoint of one side of the base. 2. **Formula:** The slant height $s$ can be found using the Pythagorean theorem in the right triangle formed by the height $h$, half the base side length, and the slant height $s$: $$s = \sqrt{h^2 + \left(\frac{\text{base side}}{2}\right)^2}$$ 3. **Calculate half the base side:** $$\frac{6.4}{2} = 3.2$$ cm 4. **Calculate $s$:** $$s = \sqrt{8.2^2 + 3.2^2} = \sqrt{67.24 + 10.24} = \sqrt{77.48}$$ 5. **Simplify:** $$s \approx 8.8$$ cm (rounded to 1 decimal place) --- 6. **Problem 2: Find the maximum length $t$ of a pencil that can fit entirely inside the open box (net).** Given dimensions of the box: - Width = 3 cm - Height = 9 cm - Length = 7 cm 7. **Explanation:** The maximum length $t$ is the space diagonal of the box, which can be found using the 3D Pythagorean theorem: $$t = \sqrt{\text{length}^2 + \text{width}^2 + \text{height}^2}$$ 8. **Calculate $t$:** $$t = \sqrt{7^2 + 3^2 + 9^2} = \sqrt{49 + 9 + 81} = \sqrt{139}$$ 9. **Simplify:** $$t \approx 11.8$$ cm (rounded to 1 decimal place) --- **Final answers:** - Slant height $s = 8.8$ cm - Maximum pencil length $t = 11.8$ cm