To find the relative error in calculating the perimeter and area of a rectangle with the given dimensions, we first need to clarify the dimensions. Since a rectangle has width and length (two dimensions), I'll assume the dimensions provided are as follows:
- Length (L) = 7.78 cm
- Width (W) = 5.6 cm
- Height (H) = 6.767 cm
However, the third dimension seems to be irrelevant for the calculation of the perimeter and area of a rectangle, as it generally refers to a 2D shape. I will calculate based on the width and length.
Step 1: Calculate the Perimeter and Area
- Perimeter (P) of a rectangle is given by: \[ P = 2(L + W) \]
- Area (A) of a rectangle is given by: \[ A = L \times W \]
Using the provided dimensions:
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For the length \(L = 7.78\) cm and width \(W = 5.6\) cm:
Perimeter: \[ P = 2(7.78 + 5.6) = 2(13.38) = 26.76 \text{ cm} \]
Area: \[ A = 7.78 \times 5.6 = 43.648 \text{ cm}^2 \]
Step 2: Calculate the Absolute Errors
Next, we need to find the absolute errors in the measurements:
-
For Length (L): The absolute error can be derived from the measurement precision. We'll assume a general measurement error of 0.01 cm (if no specific error is mentioned).
\[ \text{Absolute Error in Length} = 0.01 \text{ cm} \]
-
For Width (W): Similarly, we'll assume the same measurement error.
\[ \text{Absolute Error in Width} = 0.01 \text{ cm} \]
Relative Errors for Perimeter and Area
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Relative Error in Perimeter:
- The relative error in circumference (perimeter) can be derived as: \[ \text{Relative Error}_{P} = \frac{\text{Absolute Error in Perimeter}}{P} \]
To find the absolute error in perimeter: \[ \text{Absolute Error in Perimeter} = 2(\text{Abs. Error in Length} + \text{Abs. Error in Width}) = 2(0.01 + 0.01) = 0.04 \text{ cm} \]
Thus, \[ \text{Relative Error in Perimeter} = \frac{0.04}{26.76} \approx 0.00149 \text{ or } 0.149% \]
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Relative Error in Area:
- The formula for the relative error in area: \[ \text{Relative Error}_{A} = \frac{\sqrt{\left( \frac{\partial A}{\partial L} \cdot \text{Abs. Error in Length} \right)^2 + \left( \frac{\partial A}{\partial W} \cdot \text{Abs. Error in Width} \right)^2}}{A} \]
The derivatives of the area with respect to length and width: \[ \frac{\partial A}{\partial L} = W \quad \text{and} \quad \frac{\partial A}{\partial W} = L \]
Therefore, the contributions are: \[ \text{Absolute Error in Area} = \sqrt{(W \cdot 0.01)^2 + (L \cdot 0.01)^2} \] \[ = \sqrt{(5.6 \cdot 0.01)^2 + (7.78 \cdot 0.01)^2} \] \[ = \sqrt{(0.056)^2 + (0.0778)^2} \] \[ = \sqrt{(0.003136) + (0.00604884)} \] \[ = \sqrt{0.00918484} \approx 0.0959 \text{ cm}^2 \]
Thus, \[ \text{Relative Error in Area} = \frac{0.0959}{43.648} \approx 0.0022 \text{ or } 0.22% \]
Summary of Relative Errors
- Relative error in Perimeter: approximately 0.149%
- Relative error in Area: approximately 0.22%
These calculations assume uniform absolute errors; actual relative errors would depend on the specifics of the measurement conditions.