---
title: "Area enclosed by chain line and hedge"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:area-offsets-hedge-5m-interval-simpson-trapezoidal"
source: https://www.bctengineer.com/q/semester_3/surveying_i/area-offsets-hedge-5m-interval-simpson-trapezoidal
site: BCT Engineering
---

# Area enclosed by chain line and hedge

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2074 Asoj, 8 marks
- Video: [(LEC-19) Surveying - Numericals on Computing Area and Volume by Trapezoidal and Simpson's Rules](https://www.youtube.com/watch?v=05o6y67YDg8) — Diksha Point
- 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-hedge-5m-interval-simpson-trapezoidal
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9

## Question

The following offsets were taken from a chain line to a hedge.

| Distance (m) | 0 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
|---|---|---|---|---|---|---|---|---|---|
| Offset (m) | 0 | 2.5 | 5 | 7.5 | 8.8 | 7.5 | 6.5 | 3.5 | 0 |

Calculate the area enclosed between the chain line and hedge by (i) Simpson's rule (ii) the trapezoidal rule.

## Short answer

With a common interval of 5 m and 8 strips, Simpson’s rule gives 207.67 m², the trapezoidal rule 206.50 m².

## Answer

**Given:** 9 offsets at a common interval $d = 5$ m, so the length of the chain line = 8 × 5 = 40 m.

#### (i) Simpson's one-third rule
The number of intervals (8) is even, so the rule applies to the whole length:

$$
\text{Area} = \frac{d}{3}\left[(o_0 + o_n) + 4(\text{odd offsets}) + 2(\text{even offsets})\right]
$$

Sum of the odd-numbered offsets = 21.00 m; sum of the even-numbered (intermediate) offsets = 20.30 m.

$$
\text{Area} = \frac{5}{3}\left[(0.00 + 0.00) + 4(21.00) + 2(20.30)\right] = 207.667\ \text{m}^2
$$

#### (ii) 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 = 41.30 m; (first + last)/2 = (0.00 + 0.00)/2 = 0.000 m.

$$
\text{Area} = 5\times\left[0.000 + 41.30\right] = 206.500\ \text{m}^2
$$

**Answer:** Simpson's rule = **207.67 m²**; trapezoidal rule = **206.50 m²**.

## Check yourself

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

1. Nine offsets at 5 m spacing: how many intervals and why does Simpson's rule apply to the whole length?
   A. 9 intervals, odd, so Simpson cannot be used
   B. 8 intervals, even, so Simpson applies to the whole length
   C. 8 intervals, but Simpson needs an odd count
   D. 9 intervals, and Simpson works for any count
   Answer: B. 9 offsets give 8 intervals, an even number, which is what Simpson's 1/3 rule needs. (Key point: Interval count)

2. The first and last offsets are both 0. What does the trapezoidal formula reduce to?
   A. Area = d x (sum of intermediate offsets)
   B. Area = d x (sum of all offsets)/2
   C. Area = d/3 x (sum of intermediate offsets)
   D. Area = 2d x (sum of intermediate offsets)
   Answer: A. With (o0+on)/2 = 0, the area is just d times the sum of the intermediate offsets: 5 x 41.30 = 206.5 m^2. (Key point: Trapezoidal setup)

3. In Simpson's rule, which offsets get multiplier 4?
   A. Even-numbered intermediate offsets
   B. Only the first and last offsets
   C. All intermediate offsets
   D. Odd-numbered offsets (1st, 3rd, 5th, 7th intermediate)
   Answer: D. Odd-numbered offsets are multiplied by 4 and even-numbered intermediate ones by 2; the ends count once. (Key point: Simpson multipliers)

4. Which results does the answer give?
   A. Trapezoidal 207.67 m^2, Simpson 206.50 m^2
   B. Both 206.50 m^2
   C. Trapezoidal 206.50 m^2, Simpson 207.67 m^2
   D. Trapezoidal 413.0 m^2, Simpson 207.67 m^2
   Answer: C. Trapezoidal gives 5 x 41.30 = 206.50 m^2 and Simpson gives (5/3)(84 + 40.6) = 207.67 m^2. (Key point: Comparing results)

## Related in Surveying I

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

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