1. **State the problem:** We need to find the area of triangle $\triangle ABC$ with sides $AB=44$ km, $BC=48$ km, and $AC=39$ km using Heron's formula. 2. **Heron's formula:** The area $A$ of a triangle with sides $a$, $b$, and $c$ is given by: $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ where $s$ is the semi-perimeter: $$s = \frac{a+b+c}{2}$$ 3. **Calculate the semi-perimeter:** $$s = \frac{44 + 48 + 39}{2} = \frac{131}{2} = 65.5$$ 4. **Apply Heron's formula:** $$A = \sqrt{65.5(65.5 - 44)(65.5 - 48)(65.5 - 39)}$$ $$= \sqrt{65.5 \times 21.5 \times 17.5 \times 26.5}$$ 5. **Calculate the product inside the square root:** $$65.5 \times 21.5 = 1408.25$$ $$17.5 \times 26.5 = 463.75$$ $$1408.25 \times 463.75 = 653388.4375$$ 6. **Find the square root:** $$A = \sqrt{653388.4375} \approx 808.32$$ 7. **Final answer:** The area of $\triangle ABC$ is approximately **808.32** square kilometers.