Question

A rectangle has the following dimensions 5.6cm,7.78cm and 6.767cm.find the relative error in finding its perimeter and its area.

Answer 1

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:

- 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

1. **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,
\[
\text{Relative Error in Perimeter} = \frac{0.04}{26.76} \approx 0.00149 \text{ or } 0.149\%
\]

2. **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.