1. Catchment Geometry and Total Area

To simplify calculations, let's establish a coordinate system with station Q at the origin $(0, 0)$. Based on the provided dimensions:

• Station Q: $(0, 0)$
• Station P: $(-2, 0)$
• Station R: $(4, 0)$
• Station S: $(-2, 2)$
• Station U: $(0, 4)$
• Station V: $(4, 4)$
• Point T (on boundary): $(2, 2)$
• Center of the arc C: $(0, 2)$

The catchment area consists of two parts:

• Lower Triangle (QST): Base $ST = 4 \text{ km}$, Height $= 2 \text{ km}$. $$A_{\text{tri}} = \frac{1}{2} \times 4 \times 2 = 4 \text{ km}^2$$
• Upper Semi-circle (SUT): Radius $R = 2 \text{ km}$. $$A_{\text{semi}} = \frac{1}{2} \times \pi \times 2^2 = 2\pi \approx 6.283 \text{ km}^2$$

Total Catchment Area: $$A_{\text{total}} = 4 + 2\pi \approx 10.283 \text{ km}^2$$

2. Determination of Influence Areas ($A_i$)

The Thiessen polygon method assigns an influence area to each gauge based on perpendicular bisectors between adjacent stations. Only stations Q, S, and U have influence within the catchment boundaries:

• Station Q: The bisector of $QR$ is $x = 2$, and the bisector of $QU$ is $y = 2$. Within the lower part, the bisector of $QS$ passes through $C(0,2)$ and $M(-1,1)$. The area closer to Q includes the right triangle $QCT$ (Area = 2) and the left sub-triangle $QMC$ (Area = 1). $$A_Q = 2 + 1 = 3 \text{ km}^2$$
• Station S: It influences the left part of the lower triangle above the bisector (triangle $SMC$, Area = 1) and half of the left quadrant (sector of $45^\circ$, Area = $\pi/2$). $$A_S = 1 + \frac{\pi}{2} \approx 2.571 \text{ km}^2$$
• Station U: It influences the other half of the left quadrant (Area = $\pi/2$) and the entire right quadrant $TCU$ (Area = $\pi$) because all points in $TCU$ are closer to U than to Q, R, or V. $$A_U = \frac{\pi}{2} + \pi = \frac{3\pi}{2} \approx 4.712 \text{ km}^2$$
• Stations P, R, V: Their influence areas lie entirely outside the catchment boundary ($A_P = A_R = A_V = 0$).

3. Mean Rainfall Calculation

The mean rainfall $\bar{P}$ is calculated as the weighted average:

$$ \bar{P} = \frac{\sum (P_i \times A_i)}{A_{\text{total}}} $$

Given rainfall depths: $P_Q = 25 \text{ mm}, P_S = 15 \text{ mm}, P_U = 22 \text{ mm}$.

𝑃 ˉ = 25 ( 3 ) + 15 ( 1 + 𝜋 2 ) + 22 ( 3 𝜋 2 ) 4 + 2 𝜋 P ˉ = 4+2π 25(3)+15(1+ 2 π ​ )+22( 2 3π ​ ) ​ 𝑃 ˉ = 75 + 15 + 7.5 𝜋 + 33 𝜋 4 + 2 𝜋 P ˉ = 4+2π 75+15+7.5π+33π ​ 𝑃 ˉ = 90 + 40.5 𝜋 4 + 2 𝜋 P ˉ = 4+2π 90+40.5π ​

Substituting $\pi \approx 3.14159$:

$$ \bar{P} \approx \frac{90 + 127.2345}{10.2832} \approx \frac{217.2345}{10.2832} \approx 21.125 $$

Final Answer:

The mean rainfall over the catchment is 21.13 mm (rounded to two decimal places).