Calculation of Catchment Area from Triangular Unit Hydrograph:

• Formula: A unit hydrograph represents a total direct runoff volume ($V$) resulting from $1\text{ cm} = 0.01\text{ m}$ of effective rainfall excess over the catchment area ($A$):
  $$V = \frac{1}{2} \times B \times Q_p = A \times R$$
• Given Data:
  Base width $B = 144\text{ h} = 144 \times 3600 = 518,400\text{ s}$,
  Peak discharge $Q_p = 23\text{ m}^3/\text{s}$,
  Runoff depth $R = 1\text{ cm} = 0.01\text{ m}$.
• Total Runoff Volume ($V$):
  $$V = \frac{1}{2} \times 518,400 \times 23 = 5,961,600\text{ m}^3$$
• Catchment Area ($A$):
  $$A = \frac{V}{R} = \frac{5,961,600\text{ m}^3}{0.01\text{ m}} = 596,160,000\text{ m}^2$$
  Converting to $\text{km}^2$ ($1\text{ km}^2 = 10^6\text{ m}^2$):
  $$A = \frac{596,160,000}{10^6} = 596.16\text{ km}^2 \approx 596\text{ km}^2$$