The image represents a cuboid $ABCDEFG$, and we are tasked with finding the length of the diagonal $HF$. The dimensions of the cuboid are marked as follows:

• Length $AB = 5$
• Width $BC = 3$
• Height $AE$ (or $AD$) is not directly labeled, but can be inferred from the geometry of the cuboid.

We can approach this problem using the 3D distance formula for the diagonal of a cuboid. The diagonal $HF$ can be calculated by recognizing that it connects two opposite vertices of the cuboid.

Step 1: Identify the relevant dimensions.

• The cuboid's length is 5 (i.e., $AB$).
• The width is 3 (i.e., $BC$).
• To find the height $AE$, we observe that triangle $ADH$ is a right triangle where:
  • $AD = AE$ (height)
  • $DH = 4$, from geometry.

Step 2: Apply the 3D Pythagorean Theorem

The formula for the length of a space diagonal in a cuboid is:

$d = \sqrt{l^2 + w^2 + h^2}$

where:

• $l = 5$ (length of the cuboid),
• $w = 3$ (width of the cuboid),
• $h = 4$ (height inferred from the cuboid).

Step 3: Calculate the diagonal $HF$

$HF = \sqrt{5^2 + 3^2 + 4^2} = \sqrt{25 + 9 + 16} = \sqrt{50} \approx 7.07$

Thus, the length of $HF$ is approximately 7.07 units.

Would you like further details, or do you have any questions about this calculation?

Here are 5 related questions to explore:

1. How would you calculate the diagonal $AG$ in the cuboid?
2. What is the surface area of the cuboid given these dimensions?
3. How would you calculate the volume of this cuboid?
4. Can you calculate the length of the diagonal $AC$ on the base of the cuboid?
5. How would you calculate the angle between diagonal $HF$ and the base $ABC$ plane?

Tip: When working with 3D shapes, the Pythagorean theorem can be applied in 3 dimensions to calculate diagonals across the entire shape.