1. **Problem statement:** Find the volume of the triangular pyramid (tetrahedron) with a right triangle base of legs 9 km and 9 km, and a height of 11 km. 2. **Formula:** The volume $V$ of a pyramid is given by $$V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$$ 3. **Calculate the base area:** The base is a right triangle with legs 9 km and 9 km, so $$\text{Base Area} = \frac{1}{2} \times 9 \times 9 = \frac{1}{2} \times 81 = 40.5 \text{ km}^2$$ 4. **Use the height:** The height $h$ is given as 11 km. 5. **Calculate the volume:** $$V = \frac{1}{3} \times 40.5 \times 11 = \frac{1}{3} \times 445.5 = 148.5 \text{ km}^3$$ 6. **Round to the nearest unit:** $$V \approx 149 \text{ km}^3$$ **Final answer:** The volume of the tetrahedron is approximately **149 km³**.