To calculate the percent error in the area of the rectangle, we first need to find the area using the measured dimensions and then calculate the error based on the possible measurement errors in the dimensions.
The area \( A \) of a rectangle is calculated as:
\[ A = \text{length} \times \text{width} \]
Given:
- Length = 14 yd
- Width = 4 yd
Calculating the area:
\[ A = 14 , \text{yd} \times 4 , \text{yd} = 56 , \text{yd}^2 \]
Next, we need to consider measurement errors. Assuming the simplest case where each dimension has a possible error of ±0.5 yd (a common assumption for dimensions if not specified), we need to calculate the maximum and minimum areas:
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Maximum Length = \( 14.5 , \text{yd} \), Width = \( 4.5 , \text{yd} \): \[ A_{\text{max}} = 14.5 , \text{yd} \times 4.5 , \text{yd} = 65.25 , \text{yd}^2 \]
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Minimum Length = \( 13.5 , \text{yd} \), Width = \( 3.5 , \text{yd} \): \[ A_{\text{min}} = 13.5 , \text{yd} \times 3.5 , \text{yd} = 47.25 , \text{yd}^2 \]
Now we can find the percent error based on the calculated area of \( 56 , \text{yd}^2 \).
To find the maximum percent error:
\[ \text{Error} = A_{\text{max}} - A = 65.25 , \text{yd}^2 - 56 , \text{yd}^2 = 9.25 , \text{yd}^2 \]
Percent error:
\[ \text{Percent Error} = \left( \frac{9.25 , \text{yd}^2}{56 , \text{yd}^2} \right) \times 100 \approx 16.5% \]
Now for the minimum percent error:
\[ \text{Error} = A - A_{\text{min}} = 56 , \text{yd}^2 - 47.25 , \text{yd}^2 = 8.75 , \text{yd}^2 \]
Percent error:
\[ \text{Percent Error} = \left( \frac{8.75 , \text{yd}^2}{56 , \text{yd}^2} \right) \times 100 \approx 15.6% \]
To find the overall percent error in the area, we can average the two percent errors calculated for maximum and minimum:
\[ \text{Average Percent Error} = \frac{16.5% + 15.6%}{2} \approx 16.05% \]
Rounding to the nearest tenth of a percent:
\[ \text{Final Percent Error} \approx 16.1% \]
Therefore, the percent error in the calculated area is 16.1%.
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