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!

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!