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To solve the problem of finding the absolute error in the area of the dorm room:

### Problem Setup
The area \( A \) of a rectangular room is calculated as:
\[
A = L \cdot W
\]
where:
- \( L = 6.5 \pm 0.1 \, \text{m} \) (length with uncertainty),
- \( W = 4.2 \pm 0.1 \, \text{m} \) (width with uncertainty).

### Formula for Absolute Error in Area
The absolute error in \( A \) is given by:
\[
\Delta A = \left( \Delta L \cdot W \right) + \left( L \cdot \Delta W \right)
\]
where \( \Delta L = 0.1 \, \text{m} \) and \( \Delta W = 0.1 \, \text{m} \).

### Calculation
1. **Compute the area**:
   \[
   A = L \cdot W = 6.5 \cdot 4.2 = 27.3 \, \text{m}^2
   \]

2. **Compute the absolute error**:
   \[
   \Delta A = (0.1 \cdot 4.2) + (6.5 \cdot 0.1)
   \]
   \[
   \Delta A = 0.42 + 0.65 = 1.07 \, \text{m}^2
   \]

### Final Result
- The area of the dorm room is \( 27.3 \, \text{m}^2 \).
- The absolute error in the area is \( \Delta A = 1.07 \, \text{m}^2 \).

Would you like a more detailed explanation of any step? Let me know!

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Here are 5 related questions for practice:
1. How would the error change if \( \Delta L = 0.2 \, \text{m} \)?
2. What is the percentage error in the area?
3. How is relative error in area calculated for this problem?
4. What would happen to the area if both \( L \) and \( W \) increased by their uncertainties?
5. How can you reduce measurement errors in determining the area of a rectangular space?

**Tip**: Always report uncertainties with your measurements to ensure accuracy and precision in calculations!

You want to calculate the area of your dorm room which is in a rectangular shape. You measure the length and the width of the room as L = 6.5 ± 0.1 m, W = 4.2 ± 0.1 m. What is the absolute error in the area in m² units?

Learn how to calculate the absolute error in the area of a rectangular dorm room with dimensions L = 6.5 ± 0.1 m and W = 4.2 ± 0.1 m. Includes step-by-step error analysis and uncertainty propagation, suitable for high school students.

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