Let me transcribe the mathematical content from the image and solve it:

Problem:

When measuring the dimensions of a rectangular field, an error was introduced in which one side was recorded as being 9% larger than its actual length, while the other side was documented as being 7% shorter than its true length. Determine the percentage of error in the calculated area that arises from these measurements.

Options:

• A. 1.7%
• B. 1.4%
• C. 1.8%
• D. 2.3%

Solution:

1. Let the true dimensions of the rectangle be $L$ and $W$:

   • True area $A_{\text{true}} = L \cdot W$.

2. Dimensions after errors:

   • The length becomes $L_{\text{measured}} = 1.09L$ (9% larger).
   • The width becomes $W_{\text{measured}} = 0.93W$ (7% smaller).

3. Measured area: $A_{\text{measured}} = L_{\text{measured}} \cdot W_{\text{measured}} = (1.09L) \cdot (0.93W) = 1.09 \cdot 0.93 \cdot L \cdot W$

   Simplify: $A_{\text{measured}} = 1.0137 \cdot L \cdot W$

4. Error in area: $\text{Error} = \frac{A_{\text{measured}} - A_{\text{true}}}{A_{\text{true}}} \cdot 100\%$

   Substitute: $\text{Error} = \frac{1.0137 \cdot L \cdot W - L \cdot W}{L \cdot W} \cdot 100\% = (1.0137 - 1) \cdot 100\%$

   Simplify: $\text{Error} = 0.0137 \cdot 100\% = 1.37\%$

5. Closest option:

   • The percentage error is approximately 1.4%, which matches Option B.

Final Answer:

B. 1.4%

Would you like a further breakdown or clarification on any step?

Follow-up Questions:

1. How does each measurement error affect the total area independently?
2. Can this error compound differently for other shapes, like circles or triangles?
3. What are real-world implications of such small percentage errors in large fields?
4. How can one mitigate or correct such measurement errors practically?
5. Could alternative percentage error definitions yield a different result?

Tip:

Always consider the multiplicative nature of errors in products like area, as errors in dimensions are not simply additive.