1. **State the problem:** Find the perimeter of triangle $\triangle ABC$ with vertices $A(-6,4)$, $B(8,-1)$, and $C(0,-9)$. The perimeter is the sum of the lengths of sides $AB$, $BC$, and $CA$. 2. **Formula for distance between two points:** $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ This formula calculates the length of a segment between points $(x_1,y_1)$ and $(x_2,y_2)$. 3. **Calculate each side length:** - $AB = \sqrt{(8 - (-6))^2 + (-1 - 4)^2} = \sqrt{14^2 + (-5)^2} = \sqrt{196 + 25} = \sqrt{221}$ - $BC = \sqrt{(0 - 8)^2 + (-9 - (-1))^2} = \sqrt{(-8)^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}$ - $CA = \sqrt{(-6 - 0)^2 + (4 - (-9))^2} = \sqrt{(-6)^2 + 13^2} = \sqrt{36 + 169} = \sqrt{205}$ 4. **Sum the side lengths to find the perimeter:** $$\text{Perimeter} = \sqrt{221} + 8\sqrt{2} + \sqrt{205}$$ 5. **Approximate the values:** - $\sqrt{221} \approx 14.866$ - $8\sqrt{2} = 8 \times 1.414 = 11.312$ - $\sqrt{205} \approx 14.318$ 6. **Add the approximations:** $$14.866 + 11.312 + 14.318 = 40.496$$ 7. **Round to the nearest tenth:** $$\boxed{40.5}$$ units Thus, the perimeter of $\triangle ABC$ is approximately 40.5 units.