Question

A trapezoidal land parcel for a residential plot measures 10.335 m in the south, 13.971 m in the north, 5.047 m in the west and 6.302 m in the east directions. The north and south sides are perpendicular to the western side of the plot, making the eastern side skewed. Given the lengths of the sides are measured with a precision of 1 mm, the precision of the area of the plot calculated from these measurements is __________ m² (Rounded off to two decimal values).

Hint: \( \text{Area of a trapezium} = \left(\dfrac{\text{sum of parallel sides}}{2}\right) \times \text{height} \)

Hint:

Since north and south are perpendicular to west, treat south and north as the parallel sides of the trapezium and west as the height, then apply the general law of propagation of variance to the area formula.

Answer 1

Correct Answer: 0

Step 1: Identify the geometry and the working formula.
The north and south sides are both perpendicular to the west side, so they are parallel to each other, and the west side is the perpendicular distance (height) between them. Taking south \(a = 10.335\ m\), north \(b = 13.971\ m\) and west \(h = 5.047\ m\) as the parallel sides and height of the trapezium, the area is \( A = \left(\dfrac{a+b}{2}\right)h \). The east side does not enter this formula since it is skewed and is not required once south, north and west are known.

Step 2: Write the law of propagation of variance.
Each of \(a\), \(b\) and \(h\) is measured independently with the same precision \( \sigma = 1\ mm = 0.001\ m \). Treating \(A\) as a function \(A(a,b,h)\), the variance of \(A\) is obtained from the general law of propagation of errors: \[ \sigma_A^2 = \left(\frac{\partial A}{\partial a}\right)^2 \sigma_a^2 + \left(\frac{\partial A}{\partial b}\right)^2 \sigma_b^2 + \left(\frac{\partial A}{\partial h}\right)^2 \sigma_h^2 \]

Step 3: Compute the partial derivatives.
\( \dfrac{\partial A}{\partial a} = \dfrac{h}{2} = \dfrac{5.047}{2} = 2.5235\ m \), \( \dfrac{\partial A}{\partial b} = \dfrac{h}{2} = 2.5235\ m \), \( \dfrac{\partial A}{\partial h} = \dfrac{a+b}{2} = \dfrac{10.335+13.971}{2} = 12.153\ m \)

Step 4: Substitute into the propagation formula.
\[ \sigma_A^2 = (2.5235)^2(0.001)^2 + (2.5235)^2(0.001)^2 + (12.153)^2(0.001)^2 \] \[ \sigma_A^2 = (6.3681 + 6.3681 + 147.6954)\times 10^{-6} = 160.4315\times 10^{-6}\ m^4 \]

Step 5: Take the square root and round off.
\[ \sigma_A = \sqrt{160.4315\times 10^{-6}} = 0.01267\ m^2 \] Rounded off to two decimal places, this is \(0.01\ m^2\), which lies in the accepted range of \(0.00\) to \(0.02\ m^2\).

\[ \boxed{\sigma_A \approx 0.01\ m^2} \]