1. **Problem Statement:** Find the area of a triangular fish pen with sides 16 m, 18 m, and 25 m using Heron's formula. 2. **Formula:** Heron's formula for the area $A$ of a triangle with sides $a$, $b$, and $c$ is: $$ A = \sqrt{s(s-a)(s-b)(s-c)} $$ where $s$ is the semi-perimeter: $$ s = \frac{a+b+c}{2} $$ 3. **Calculate the semi-perimeter:** $$ s = \frac{16 + 18 + 25}{2} = \frac{59}{2} = 29.5 $$ 4. **Calculate the area:** $$ A = \sqrt{29.5(29.5-16)(29.5-18)(29.5-25)} = \sqrt{29.5 \times 13.5 \times 11.5 \times 4.5} $$ Calculate the product inside the square root: $$ 29.5 \times 13.5 = 398.25 $$ $$ 11.5 \times 4.5 = 51.75 $$ $$ 398.25 \times 51.75 = 20610.9375 $$ 5. **Final area:** $$ A = \sqrt{20610.9375} \approx 143.56 \text{ m}^2 $$ **Answer:** The area of the triangular fish pen is approximately **143.56 m²**.