---
title: "Thiessen average rainfall, four gauges at rectangle corners"
subject: "Engineering Hydrology"
chapter: 2
key: "semester_5/engineering_hydrology:thiessen-rectangle-four-gauges"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/thiessen-rectangle-four-gauges
site: BCT Engineering
---

# Thiessen average rainfall, four gauges at rectangle corners

Engineering Hydrology · Chapter 2: Precipitation

- Asked in: 2066 Magh (old course), 4 marks
- Video: [Numerical on Average Rainfall Data - Hydrology - Water Resource Engineering 1](https://www.youtube.com/watch?v=DNSjoMBH5Cc) — Ekeeda
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/thiessen-rectangle-four-gauges
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2

## Question

There are four rain gauges at four corners of a rectangle. The two sides of the rectangle are 100 km and 150 km. The yearly rainfalls of the four gauges are 1200 mm, 1300 mm, 1500 mm and 1100 mm respectively. Calculate the average rainfall by Thiessen polygon method.

## Short answer

Each corner gauge has equal area (3750 km²) due to symmetry. Mean rainfall = (1200+1300+1500+1100)/4 = 1275 mm.

## Answer

Four gauges at the corners of a rectangle are equally spaced from the centre, so the perpendicular bisectors are the two lines through the centre parallel to the sides. The rectangle divides into four equal rectangles, one for each gauge.

```text
  G1 +--------+--------+ G2
     | A/4    | A/4    |
     +--------+--------+
     | A/4    | A/4    |
  G4 +--------+--------+ G3
```

Total area $= 100 \times 150 = 15\,000$ km²; each gauge area $= 3750$ km² (weight 0.25).

$$
\bar P = \frac{\sum A_i P_i}{\sum A_i} = \frac{3750(1200+1300+1500+1100)}{15\,000} = \frac{5100}{4} = 1275\ \text{mm}
$$

**Answer: Average rainfall by Thiessen polygon method = 1275 mm** (equal to the arithmetic mean).

## Check yourself

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

1. How is the rectangle divided?
   A. Into two triangles
   B. Unequal parts
   C. Into four equal rectangles by lines through the centre
   D. By diagonals
   Answer: C. Corner gauges are symmetric about the centre. (Key point: Polygon division)

2. What is each gauge's weight and area?
   A. 0.5 and 7500
   B. 0.25 and 15000
   C. 0.1 and 1500
   D. 0.25 and 3750 km2
   Answer: D. 15000/4 = 3750 km2. (Key point: Weights)

3. What is the average rainfall?
   A. 1275 mm
   B. 1300 mm
   C. 1200 mm
   D. 5100 mm
   Answer: A. (1200+1300+1500+1100)/4 = 1275 mm, equal to arithmetic mean here. (Key point: Result)

## Related in Engineering Hydrology

- [Thiessen polygon mean rainfall, pentagon catchment](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Athiessen-pentagon-pqrs) — Engineering Hydrology, chapter 2
- [Number of additional rain gauges required](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aoptimum-raingauges-7stn) — Engineering Hydrology, chapter 2
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- [Additional rain gauges for 5% error, seven stations](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aoptimum-raingauges-7stn-130) — Engineering Hydrology, chapter 2, asked 2×
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## Related in other subjects

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- [Arithmetic mean and Thiessen average precipitation](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aarithmetic-thiessen-numerical) — Hydro Power, chapter 2
- [Empirical formula; voltage for 150 MW over 100 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aempirical-formula-150mw-100km) — Transmission and Distribution System Design, chapter 2