---
title: "Area from offsets, 10 m interval"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:area-offsets-4-6-7-2-9-6-6-4"
source: https://www.bctengineer.com/q/semester_3/surveying_i/area-offsets-4-6-7-2-9-6-6-4
site: BCT Engineering
---

# Area from offsets, 10 m interval

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2075 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-4-6-7-2-9-6-6-4
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9

## Question

The offsets in meter from a survey line to an irregular boundary line are given below.

| Chainage (m) | 0 | 10 | 20 | 30 |
|---|---|---|---|---|
| Offset (m) | 4.6 | 7.2 | 9.6 | 6.4 |

Calculate the area enclosed by the 1st line, last line, survey line and boundary line using Simpson's rule and trapezoidal rule.

## Short answer

Trapezoidal rule with 10 m interval gives 223.00 m². Simpson's 1/3 rule for the first two intervals plus trapezoidal for the last yields 223.33 m².

## Answer

**Given:** 4 offsets at a common interval $d = 10$ m, so the length of the chain line = 3 × 10 = 30 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 = 16.80 m; (first + last)/2 = (4.60 + 6.40)/2 = 5.500 m.

$$
\text{Area} = 10\times\left[5.500 + 16.80\right] = 223.000\ \text{m}^2
$$

#### (ii) Simpson's one-third rule
There are 4 offsets, i.e. 3 (an odd number of) intervals, so Simpson's rule cannot be used for the whole length. It is applied to the first 2 intervals (offsets 0 to 2), and the last strip is found by the trapezoidal rule.

$$
\text{Area}_1 = \frac{10}{3}\left[(4.60 + 9.60) + 4(7.20)\right] = 143.333\ \text{m}^2
$$

$$
\text{Area}_2 = 10\times\frac{9.60 + 6.40}{2} = 80.000\ \text{m}^2
$$

$$
\text{Total area} = 143.333 + 80.000 = 223.333\ \text{m}^2
$$

**Answer:** trapezoidal rule = **223.00 m²**; Simpson's rule = **223.33 m²**.

*Alternative, Simpson's three-eighths rule (valid for exactly 3 intervals):* Area $= \frac{3d}{8}[o_0 + 3o_1 + 3o_2 + o_3] = \frac{3 \times 10}{8}[4.6 + 3(7.2) + 3(9.6) + 6.4] = 230.250\ \text{m}^2$.

## Check yourself

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

1. What is the trapezoidal area for offsets 4.6, 7.2, 9.6, 6.4 at 10 m?
   A. 230.25 m2
   B. 143.33 m2
   C. 250.00 m2
   D. 223.00 m2
   Answer: D. 10 x [5.5 + 16.8] = 223 m2. (Key point: Trapezoidal area)

2. There are 3 intervals. How is Simpson's rule applied?
   A. Simpson on first 2 intervals, trapezoidal on the last strip
   B. Directly to all three intervals
   C. Trapezoidal on the first, Simpson on the last
   D. It cannot be used at all
   Answer: A. Even intervals only; the remaining strip uses the trapezoidal rule, giving 223.33 m2. (Key point: Simpson's with odd intervals)

3. Which rule is valid for exactly 3 intervals as a check?
   A. Mid-ordinate rule only
   B. Simpson's three-eighth rule
   C. Average ordinate with 2 intervals
   D. Prismoidal rule
   Answer: B. Area = 3d/8 [o0 + 3o1 + 3o2 + o3] = 230.25 m2. (Key point: Simpson's 3/8)

4. What is the length of the chain line?
   A. 40 m
   B. 10 m
   C. 30 m
   D. 20 m
   Answer: C. 3 intervals x 10 m = 30 m. (Key point: Chain length)

## Related in Surveying I

- [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 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×
- [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 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
- [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 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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