Surveying error propagation: a rectangular plot measuring 60 m × 20 m is measured with a steel tape; if the standard error in each side length is ±0.01 m (±1 cm), what is the standard error of the computed area?

Introduction / Context:
In engineering surveying and construction layout, areas are typically computed from measured lengths. Because each length is subject to random error, the area inherits uncertainty. This problem tests your ability to propagate measurement errors from side lengths to the area using first-order (linear) error propagation.

Given Data / Assumptions:

• Measured sides: L = 60 m and W = 20 m.
• Standard error in each dimension: σ_L = σ_W = 0.01 m (1 cm).
• Area A = L * W for a rectangle.
• Errors in L and W are independent and small (linearization valid).

Concept / Approach:
For a function A(L, W) = L * W, the variance of A is approximated by σ_A^2 = (∂A/∂L)^2 σ_L^2 + (∂A/∂W)^2 σ_W^2 when errors are uncorrelated. Here ∂A/∂L = W and ∂A/∂W = L. Compute σ_A from these partial derivatives and the given standard errors.

Step-by-Step Solution:
Compute partial derivatives: ∂A/∂L = W = 20, ∂A/∂W = L = 60.Form variance: σ_A^2 = (20)^2 (0.01)^2 + (60)^2 (0.01)^2.Evaluate: (400 + 3600) * 0.0001 = 4000 * 0.0001 = 0.4.Standard error: σ_A = sqrt(0.4) ≈ 0.632455… m^2.Rounded to three significant figures: ±0.632 m^2.

Verification / Alternative check:
If one side were error-free, σ_A would be W * σ_L = 20 * 0.01 = 0.2 m^2 (or L * σ_W = 0.6 m^2). Including both errors and the larger sensitivity to L (because L = 60) explains why σ_A exceeds 0.6 m^2 in quadrature, landing at ≈0.632 m^2.

Why Other Options Are Wrong:

• ±0.1414 m^2: corresponds to using only a fraction of the variance terms.
• ±0.566 m^2: underestimates because it ignores the larger sensitivity to L.
• ±0.8484 m^2: overestimates by adding linearly instead of in quadrature.

Common Pitfalls:

• Adding absolute errors instead of combining variances.
• Forgetting independence and covariance assumptions.

Final Answer:
± 0.632 m2