To calculate the length of $DE$ in the cuboid using Pythagoras' theorem, we follow a two-step process, as $DE$ is the diagonal across the cuboid.

Step 1: Calculate the diagonal $DA$ of the base

The base of the cuboid is a rectangle with dimensions:

• $AD = 17 \, \text{cm}$
• $AB = 21 \, \text{cm}$

Using Pythagoras' theorem for triangle $DAB$: $DA^2 = AD^2 + AB^2$ $DA = \sqrt{(17^2 + 21^2)} = \sqrt{(289 + 441)} = \sqrt{730}$ $DA \approx 27.02 \, \text{cm}$

Step 2: Use Pythagoras' theorem in triangle $DEA$

Now, we use the height $AE = 8 \, \text{cm}$ and the base diagonal $DA \approx 27.02 \, \text{cm}$ to find $DE$.

Applying Pythagoras' theorem: $DE^2 = DA^2 + AE^2$ $DE = \sqrt{(27.02^2 + 8^2)} = \sqrt{(730 + 64)} = \sqrt{794}$ $DE \approx 28.2 \, \text{cm}$

Thus, the length of $DE$ is approximately $28.2 \, \text{cm}$ to 1 decimal place.

Would you like more details or have any questions?

Here are some related questions to expand on this topic:

1. How does Pythagoras' theorem work in three-dimensional problems?
2. What is the geometric significance of the diagonal in a cuboid?
3. Can you generalize the formula to find the longest diagonal of any cuboid?
4. How would the process change if the cuboid were irregular?
5. What real-world applications require the use of cuboid diagonals?

Tip: In a 3D space, the Pythagorean theorem can be applied repeatedly to solve for distances across different planes of a cuboid.