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: \( \text{Area of a trapezium} = \left(\dfrac{\text{sum of parallel sides}}{2}\right) \times \text{height} \)
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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.
Step 1: Set up the area function and the base values.
Let $A(a,b,h) = \left(\dfrac{a+b}{2}\right)h$ with $a = 10.335\ m$ (south), $b = 13.971\ m$ (north) and $h = 5.047\ m$ (west), since north and south are parallel and west is the perpendicular height between them. Base area $A_0 = \left(\dfrac{10.335+13.971}{2}\right)(5.047) = (12.153)(5.047) = 61.336\ m^2$.
Step 2: Perturb each variable by the measurement precision and estimate its individual effect on area.
With a precision of $\delta = 1\ mm = 0.001\ m$ on each side, perturbing $a$ alone changes area by $\Delta A_a = \dfrac{h}{2}\delta = \dfrac{5.047}{2}(0.001) = 0.0025235\ m^2$. By symmetry perturbing $b$ alone gives the same $\Delta A_b = 0.0025235\ m^2$. Perturbing $h$ alone gives $\Delta A_h = \left(\dfrac{a+b}{2}\right)\delta = (12.153)(0.001) = 0.012153\ m^2$.
Step 3: Combine the three independent contributions in quadrature.
Since $a$, $b$ and $h$ are independently measured, the random errors combine as a root-sum-square, not a simple sum: \[ \sigma_A = \sqrt{(\Delta A_a)^2+(\Delta A_b)^2+(\Delta A_h)^2} \]
Step 5: Round off and confirm.
Rounded to two decimal places, $\sigma_A \approx 0.01\ m^2$, confirming the analytical result and lying within the accepted range $0.00$ to $0.02\ m^2$.
