5. Find the perimeter of \( \triangle A B C \) with vertices \( A(2,4), B(8,12) \), and \( C(24,0) \).

To find the perimeter of \( \triangle ABC \), we first need to determine the lengths of the sides \( AB \), \( BC \), and \( CA \) using the distance formula: The distance formula is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] 1. **Calculate \( AB \)**: Points \( A(2, 4) \) and \( B(8, 12) \): \[ AB = \sqrt{(8 - 2)^2 + (12 - 4)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \] 2. **Calculate \( BC \)**: Points \( B(8, 12) \) and \( C(24, 0) \): \[ BC = \sqrt{(24 - 8)^2 + (0 - 12)^2} = \sqrt{16^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \] 3. **Calculate \( CA \)**: Points \( C(24, 0) \) and \( A(2, 4) \): \[ CA = \sqrt{(2 - 24)^2 + (4 - 0)^2} = \sqrt{(-22)^2 + 4^2} = \sqrt{484 + 16} = \sqrt{500} = 10\sqrt{5} \] Now that we have all the sides, we can find the perimeter \( P \): \[ P = AB + BC + CA = 10 + 20 + 10\sqrt{5} \] Thus, the perimeter of \( \triangle ABC \) is: \[ P = 30 + 10\sqrt{5} \] This is approximately \( 30 + 22.36 \) (since \( \sqrt{5} \approx 2.236 \)), equating to around \( 52.36 \). Final answer: \[ \text{The perimeter of } \triangle ABC \text{ is } 30 + 10\sqrt{5} \text{ units.} \]