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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.

Use Pythagoras' theorem to work out the length of DE in the cuboid below. The base dimensions are AD = 17 cm, AB = 21 cm, and the height AE = 8 cm.

Learn how to calculate the diagonal DE in a cuboid using Pythagoras' theorem. This step-by-step problem involves a 3D geometry question where you use two applications of Pythagoras' theorem to find the diagonal of the base and then the full diagonal. The base measures 17 cm by 21 cm, with a height of 8 cm. Suitable for students in Grades 9-10.

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