---
title: "Area from twelve offsets at 20 m interval"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:area-offsets-twelve-offsets-4-19"
source: https://www.bctengineer.com/q/semester_3/surveying_i/area-offsets-twelve-offsets-4-19
site: BCT Engineering
---

# Area from twelve offsets at 20 m interval

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2068 Baisakh, 8 marks
- Video: [Offset Method: Step-by-step Area Calculation Approach](https://www.youtube.com/watch?v=R3yYgE1BvwE) — Study Force
- 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-twelve-offsets-4-19
- 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, 4.25 m, 7.30 m, 6.25 m and 4.19 m. Calculate the area enclosed between the survey line, the irregular boundary line, and the first and last offsets, by (i) Trapezoidal rule and (ii) Simpson's rule.

## Short answer

Trapezoidal rule: 1054.30 m². Simpson’s 1/3 rule applies to the first 11 offsets (10 strips) and the last strip by trapezoidal, yielding 1086.07 m².

## Answer

There are 12 offsets, so there are 11 equal strips of width $d = 20$ m, and the first offset is $O_1 = 0$ and the last is $O_{12} = 4.19$ m.

#### (i) Trapezoidal rule

$$
A = d\left[\frac{O_1 + O_n}{2} + O_2 + O_3 + \dots + O_{n-1}\right]
$$

Sum of the intermediate offsets ($O_2$ to $O_{11}$):

$$
\sum O_{mid} = 50.62\ \text{m}
$$

$$
A = 20\left[\frac{0.00 + 4.19}{2} + 50.62\right] = 20 \times 52.715 = 1054.30\ \text{m}^2
$$

#### (ii) Simpson's rule

Simpson's one-third rule needs an **odd number of offsets** (an even number of strips). Here there are 12 offsets, so apply Simpson's rule to the first 11 offsets (10 strips) and the trapezoidal rule to the last strip.

$$
A = \frac{d}{3}\left[O_1 + O_{11} + 4(O_2 + O_4 + O_6 + O_8 + O_{10}) + 2(O_3 + O_5 + O_7 + O_9)\right] + d\,\frac{O_{11} + O_{12}}{2}
$$

- Sum of even-numbered offsets: $O_2 + O_4 + O_6 + O_8 + O_{10} = 26.13$ m
- Sum of odd-numbered intermediate offsets: $O_3 + O_5 + O_7 + O_9 = 18.24$ m
- First and last offset of the Simpson part: $O_1 + O_{11} = 6.25$ m

$$
\begin{aligned} A_{1-11} &= \frac{20}{3}\left[6.25 + 4(26.13) + 2(18.24)\right] = \frac{20}{3} \times 147.25 = 981.67\ \text{m}^2 \\ A_{last} &= 20 \times \frac{6.25 + 4.19}{2} = 104.40\ \text{m}^2 \\ A &= 981.67 + 104.40 = 1086.07\ \text{m}^2 \end{aligned}
$$

| Rule | Area (m²) |
|---|---|
| Trapezoidal | 1054.30 |
| Simpson's | 1086.07 |

**Answer: Trapezoidal area = 1054.30 m²; Simpson's area = 1086.07 m².**

## Check yourself

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

1. 12 offsets at 20 m: how many strips?
   A. 12
   B. 11
   C. 10
   D. 13
   Answer: B. 12 offsets give 11 intervals. (Key point: Strips)

2. With first offset 0 and last 4.19, what is the first term inside the trapezoidal bracket?
   A. (0 + 4.19)/2 = 2.095, added to the sum of O2 to O11
   B. 0 + 4.19 = 4.19, added to the sum of O2 to O11
   C. (0 + 4.19)/3
   D. 4 x 4.19
   Answer: A. Only the end offsets are halved; the 10 intermediate offsets sum to 50.62, so area = 20 x 52.715 = 1054.30 m^2. (Key point: Trapezoidal rule)

3. How is Simpson's rule applied here?
   A. To all 12 offsets directly
   B. To the last 11 offsets only, ignoring the first strip
   C. Trapezoidal for the first strip and Simpson with 12 offsets
   D. To the first 11 offsets (10 strips), then trapezoidal on the last strip
   Answer: D. Simpson needs an odd number of offsets, so use the first 11 and add the last strip with the trapezoidal rule. (Key point: Simpson with 12 offsets)

4. What are the final areas?
   A. Trapezoidal 1086.07 m^2, Simpson 1054.30 m^2
   B. Both 1070 m^2
   C. Trapezoidal 1054.30 m^2, Simpson 1086.07 m^2
   D. Trapezoidal 981.67 m^2, Simpson 104.40 m^2
   Answer: C. Simpson part 981.67 plus last strip 104.40 gives 1086.07 m^2; trapezoidal gives 1054.30 m^2. (Key point: Result)

## Close variants set in other exams

- [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×

## 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 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 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 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

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