---
title: "Area from eight offsets at 20 m interval"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:area-offsets-20m-interval-eight-offsets"
source: https://www.bctengineer.com/q/semester_3/surveying_i/area-offsets-20m-interval-eight-offsets
site: BCT Engineering
---

# Area from eight offsets at 20 m interval

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2074 Chaitra, 4 marks
- Video: [Area of an Irregular Plot Using Average Ordinate Rule, Trapezoidal and Simpson's Method|Problem](https://www.youtube.com/watch?v=gKs34m0AC0M) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/area-offsets-20m-interval-eight-offsets
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9

## Question

The following offsets were taken at 20 m interval from a survey line to an irregular boundary line: 0.00 m, 1.53 m, 5.37 m, 3.50 m, 4.32 m, 7.25 m, 4.30 m, 6.55 m. Calculate the area enclosed between the survey line and the boundary by (i) Trapezoidal rule (ii) Simpson's 1/3 rule.

## Short answer

Trapezoidal rule: 590.90 m². Simpson’s 1/3 rule applied to the first 6 intervals (7 offsets) and the last strip by trapezoidal gives 593.83 m².

## Answer

**Given:** 8 offsets at a common interval $d = 20$ m, so the length of the chain line = 7 × 20 = 140 m.

#### (i) Trapezoidal rule
$$
\text{Area} = d\left[\frac{o_0 + o_n}{2} + o_1 + o_2 + \dots + o_{n-1}\right]
$$

Sum of the intermediate offsets = 26.27 m; (first + last)/2 = (0.00 + 6.55)/2 = 3.275 m.

$$
\text{Area} = 20\times\left[3.275 + 26.27\right] = 590.900\ \text{m}^2
$$

#### (ii) Simpson's one-third rule
There are 8 offsets, i.e. 7 (an odd number of) intervals, so Simpson's rule cannot be used for the whole length. It is applied to the first 6 intervals (offsets 0 to 6), and the last strip is found by the trapezoidal rule.

$$
\text{Area}_1 = \frac{20}{3}\left[(0.00 + 4.30) + 4(12.28) + 2(9.69)\right] = 485.333\ \text{m}^2
$$

$$
\text{Area}_2 = 20\times\frac{4.30 + 6.55}{2} = 108.500\ \text{m}^2
$$

$$
\text{Total area} = 485.333 + 108.500 = 593.833\ \text{m}^2
$$

**Answer:** trapezoidal rule = **590.90 m²**; Simpson's rule = **593.83 m²**.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. 8 offsets at 20 m interval give how many intervals and what chain length?
   A. 8 intervals, 160 m
   B. 7 intervals, 140 m
   C. 7 intervals, 160 m
   D. 9 intervals, 180 m
   Answer: B. n offsets give n-1 intervals, so 7 x 20 = 140 m. (Key point: Number of intervals)

2. In the trapezoidal rule for offsets at equal interval d, which expression gives the area?
   A. d x [ (first + last)/2 + sum of intermediate offsets ]
   B. d/3 x [ first + last + 4(odd) + 2(even) ]
   C. d x [ first + last + sum of intermediate offsets ]
   D. d/2 x [ sum of all offsets ]
   Answer: A. The trapezoid rule halves only the end offsets and counts every intermediate offset fully, then multiplies by d. (Key point: Trapezoidal formula)

3. Why can Simpson's 1/3 rule not be applied to the whole length here?
   A. Because the first offset is zero
   B. Because the interval is too large
   C. Because the offsets are irregular
   D. Because there is an odd number of intervals (7), but Simpson needs an even number
   Answer: D. Simpson's rule pairs strips, so it needs an even number of intervals (odd number of offsets). Here there are 7 intervals. (Key point: Simpson's condition)

4. How is the area found by Simpson's rule when intervals are odd?
   A. Trapezoidal for all and Simpson for none
   B. Drop the last offset and ignore its strip
   C. Simpson on the first 6 intervals, trapezoidal rule on the last strip
   D. Use Simpson with half the interval
   Answer: C. Simpson's rule is applied to the largest even group of intervals and the remaining strip is added by the trapezoidal rule (108.5 m^2 here). (Key point: Handling the leftover strip)

## Close variants set in other exams

- [Area from twelve offsets at 20 m interval](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aarea-offsets-twelve-offsets-4-19) — Surveying I, chapter 9, asked 2×

## Related in Surveying I

- [Area from offsets, 10 m interval](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aarea-offsets-4-6-7-2-9-6-6-4) — Surveying I, chapter 9
- [Area enclosed by chain line and hedge](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aarea-offsets-hedge-5m-interval-simpson-trapezoidal) — Surveying I, chapter 9
- [Offsets area: trapezoidal, Simpson, coordinate](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aarea-offsets-curved-boundary-trapezoidal-simpson-coordinate) — Surveying I, chapter 9
- [Area between boundary, offsets and base line](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aarea-offsets-chainage-0-12-20-25-34-42-52) — Surveying I, chapter 9
- [Area from perpendicular offsets to a hedge](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aarea-offsets-hedge-5-5-12-7) — Surveying I, chapter 9
- [Area by coordinate method from offsets](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aarea-coordinate-method-chainage-3-75) — Surveying I, chapter 9