---
title: "Thiessen polygon mean rainfall, pentagon catchment"
subject: "Engineering Hydrology"
chapter: 2
key: "semester_5/engineering_hydrology:thiessen-pentagon-pqrs"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/thiessen-pentagon-pqrs
site: BCT Engineering
---

# Thiessen polygon mean rainfall, pentagon catchment

Engineering Hydrology · Chapter 2: Precipitation

- Asked in: 2071 Shrawan, 6 marks
- Video: [Thiessen Polygon Method](https://www.youtube.com/watch?v=YjuOfu_1rhs) — Trevor Ray
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/thiessen-pentagon-pqrs
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2

## Question

The shape of a catchment is in the form of a pentagon ABCDE. There are 4 rain gauge stations P, Q, R and S inside the catchment. The position co-ordinates in km are: A(0,0), B(50,75), C(100,70), D(150,0), E(75,-50), P(50,25), Q(100,25), R(100,-25) and S(50,-25). If rainfalls recorded at P, Q, R and S are 90, 105, 114 and 120 mm respectively, determine the mean rainfall by Thiessen Polygon method.

## Short answer

Perpendicular bisectors x=75 and y=0 divide the pentagon. Using shoelace method, polygon areas sum to 11000 km² and Σ(A_i·P_i)=1,144,218.75. Mean rainfall = 1,144,218.75/11000 = 104.0 mm.

## Answer

**Method:** Draw perpendicular bisectors between neighbouring stations P(50,25), Q(100,25), R(100,-25), S(50,-25). The catchment is divided by the lines $x = 75$ (between P-Q and S-R) and $y = 0$ (between P-S and Q-R). Each polygon is clipped by the catchment boundary A(0,0), B(50,75), C(100,70), D(150,0), E(75,-50).

```text
      B(50,75)  C(100,70)
   A(0,0)  P    |   Q          D(150,0)
   --------+----+---+-----------
           S    |   R
              E(75,-50)
          x=75 divides left/right, y=0 upper/lower
```

**Polygon areas (by coordinates, shoelace formula):**

| Station | Polygon vertices | Area (km²) | Rain (mm) | A×P |
|---|---|---|---|---|
| P | (0,0),(50,75),(75,72.5),(75,0) | 3718.75 | 90 | 334687.5 |
| Q | (75,72.5),(100,70),(150,0),(75,0) | 3531.25 | 105 | 370781.25 |
| R | (150,0),(75,-50),(75,0) | 1875 | 114 | 213750 |
| S | (0,0),(75,0),(75,-50) | 1875 | 120 | 225000 |
| **Total** |  | **11000** |  | **1144218.75** |

Total area check: pentagon area $= 11000$ km².

$$
\bar P = \frac{\sum A_i P_i}{\sum A_i} = \frac{1\,144\,218.75}{11\,000} = 104.02\ \text{mm}
$$

**Answer: Mean rainfall by Thiessen polygon method = 104.0 mm.**

## Check yourself

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

1. Which lines divide the catchment into polygons for P, Q, R, S?
   A. x = 75 and y = 0
   B. x = 50 and y = 25
   C. x = 100 and y = -25
   D. The diagonals of the pentagon
   Answer: A. Stations form a rectangle; its bisectors are x = 75 and y = 0. (Key point: Bisectors for the four stations)

2. What must be done after drawing the bisectors?
   A. Ignore the boundary
   B. Clip each polygon by the catchment boundary
   C. Use only stations on the boundary
   D. Average the polygons
   Answer: B. Polygons are trimmed at the pentagon ABCDE before measuring areas. (Key point: Clipping to boundary)

3. Which formula gives mean rainfall?
   A. Sum of P_i / 4
   B. Sum of A_i / sum of P_i
   C. Sum of A_i P_i / sum of A_i
   D. Max of P_i
   Answer: C. Thiessen mean weights each gauge by polygon area: 1,144,218.75/11000. (Key point: Mean formula)

4. What is the Thiessen mean rainfall?
   A. 107.3 mm
   B. 90 mm
   C. 114 mm
   D. 104.0 mm
   Answer: D. 1,144,218.75 / 11,000 = 104.02 mm, which differs from the simple mean 107.25. (Key point: Result)

## Related in Engineering Hydrology

- [Thiessen weights for semicircular catchment](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Athiessen-weights-semicircle) — Engineering Hydrology, chapter 2
- [Thiessen average rainfall, four gauges at rectangle corners](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Athiessen-rectangle-four-gauges) — Engineering Hydrology, chapter 2
- [Empirical flood formula, 100 sq km catchment](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aempirical-flood-100sqkm) — Engineering Hydrology, chapter 6
- [Optimum rain gauges for 10% error, six stations](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aoptimum-raingauges-6stn-240) — Engineering Hydrology, chapter 2
- [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×
- [Procedure of Thiessen mean rainfall](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Athiessen-procedure) — Engineering Hydrology, chapter 2

## Related in other subjects

- [Thiessen polygon method with rainfall data](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Athiessen-polygon-explain-numerical) — Hydro Power, chapter 2
- [Conductor selection, 100 MW 70 km 132 kV](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aconductor-selection-100mw-70km-132kv) — Transmission and Distribution System Design, chapter 4
- [Vertical curve +1.25% and +4.75%, 1267.70 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-125-475-126770) — Surveying II, chapter 6, asked 3×
- [Received power: 100 MHz, 0.75 km](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-100mhz-0-75km) — Propagation and Antenna, chapter 5