1. **Problem 1: Find the diagonal $x$ of a cuboid with edges 6.3 cm, 5 cm, and 3.1 cm.** The diagonal $x$ inside a cuboid can be found using the 3D Pythagorean theorem: $$x = \sqrt{a^2 + b^2 + c^2}$$ where $a$, $b$, and $c$ are the edges of the cuboid. 2. Substitute the given values: $$x = \sqrt{6.3^2 + 5^2 + 3.1^2} = \sqrt{39.69 + 25 + 9.61} = \sqrt{74.3}$$ 3. Calculate the square root: $$x \approx 8.6$$ So, the diagonal $x$ is approximately 8.6 cm. --- 1. **Problem 2: Find the height $x$ of triangle ABD in a cube corner where edges AB, BC, and BD are each 4.2 cm.** Since the shape is a corner of a cube, triangle ABD is a right triangle with legs AB and BD both equal to 4.2 cm. The height $x$ from point B to line AD is the altitude in this right triangle. 2. First, find the length of AD using Pythagoras: $$AD = \sqrt{AB^2 + BD^2} = \sqrt{4.2^2 + 4.2^2} = \sqrt{17.64 + 17.64} = \sqrt{35.28} \approx 5.9$$ 3. The area of triangle ABD is: $$\text{Area} = \frac{1}{2} \times AB \times BD = \frac{1}{2} \times 4.2 \times 4.2 = 8.82$$ 4. The height $x$ from B to AD satisfies: $$\text{Area} = \frac{1}{2} \times AD \times x \Rightarrow x = \frac{2 \times \text{Area}}{AD} = \frac{2 \times 8.82}{5.9} \approx 3.0$$ So, the height $x$ is approximately 3.0 cm. --- 1. **Problem 3: Find the length $x$ of edge EA in a square-based pyramid with base side 3.2 cm and slant edge EB = 5.1 cm, where E is directly above D.** 2. Since ABCD is a square with side 3.2 cm, the diagonal $AD$ is: $$AD = \sqrt{3.2^2 + 3.2^2} = \sqrt{10.24 + 10.24} = \sqrt{20.48} \approx 4.5$$ 3. Triangle EAD is right-angled at D because E is directly above D. 4. Using Pythagoras in triangle EAD: $$EA = \sqrt{ED^2 + AD^2}$$ We know $EB = 5.1$ cm and since E is above D, $ED = EB = 5.1$ cm. 5. Calculate $x = EA$: $$x = \sqrt{5.1^2 + 4.5^2} = \sqrt{26.01 + 20.25} = \sqrt{46.26} \approx 6.8$$ So, the length $x$ is approximately 6.8 cm.