Question Bank • QB_SWCE_EXP_052 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Surveying and Levelling

Subtopic: Computation of areas and volume

Question Statement:

Offsets were taken from a survey line to an irregular boundary at $10\text{ m}$ intervals in the order: $0, 2.5, 4.0, 5.5, 6.0, 4.5, 3.0,$ and $0\text{ m}$. Using Simpson's one-third rule for the first six intervals and the trapezoidal rule for the final seventh interval, the total enclosed area in $\text{m}^2$ is ________ (round off to 1 decimal place).

Answer Key:

258.3

Step-by-Step Solution & Explanation

1. Simpson's rule for first 6 intervals (7 offsets $0, 2.5, 4.0, 5.5, 6.0, 4.5, 3.0$): $A_1 = \frac{d}{3} [ (O_1 + O_7) + 4(O_2 + O_4 + O_6) + 2(O_3 + O_5) ]$ $A_1 = \frac{10}{3} [ (0 + 3.0) + 4(2.5 + 5.5 + 4.5) + 2(4.0 + 6.0) ] = \frac{10}{3} [ 3.0 + 50.0 + 20.0 ] = \frac{730}{3} = 243.33\text{ m}^2$ 2. Trapezoidal rule for the 7th interval: $A_2 = \frac{3.0 + 0}{2} \times 10 = 15.0\text{ m}^2$ 3. Total Area: $A = 243.33 + 15.0 = 258.33\text{ m}^2 \approx 258.3\text{ m}^2$
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