The problem involves finding the diagonals of a cube:

1. $x$ is the diagonal of one face of the cube.
2. $y$ is the diagonal of the entire cube.

Let’s solve this step by step:

Step 1: Finding $x$ (Diagonal of one face)

The diagonal $x$ of a square face can be found using the Pythagorean Theorem:

$x = \sqrt{a^2 + a^2} = \sqrt{2a^2}$

Here $a = 6$ (edge length of the cube):

$x = \sqrt{2(6^2)} = \sqrt{72} \approx 8.5 \, \text{units.}$

Step 2: Finding $y$ (Diagonal of the cube)

To find the diagonal $y$ of the entire cube, we apply the Pythagorean Theorem again. The diagonal $y$ forms a right triangle with two sides:

• One is the edge of the cube ($a = 6$).
• The other is the diagonal of the face ($x = 8.5$).

$y = \sqrt{x^2 + a^2} = \sqrt{(8.5)^2 + 6^2}$

Calculate step-by-step: $y = \sqrt{72.25 + 36} = \sqrt{108.25} \approx 10.4 \, \text{units.}$

Final Answer:

• $x \approx 8.5$ units
• $y \approx 10.4$ units

Would you like a breakdown of the calculations or further clarification?

Here are 5 related questions to deepen your understanding:

1. How would the calculations change if the cube's edge length was different?
2. What is the formula for the diagonal of a face in terms of edge length?
3. How is the 3D diagonal $y$ generalized for cubes with edge length $a$?
4. Can we solve this problem using vector geometry instead of the Pythagorean theorem?
5. What are real-life applications of calculating 3D diagonals in cubes?

Tip: Always visualize the 3D problem step by step to ensure clarity when applying the Pythagorean theorem in multiple dimensions.