1. (Ch. 1) Find the perimeter of triangle \( X Y Z \). Round to the nearest hundredth. \( X(-2,-5) \quad Y(4,1) Z(1,3) \)

To find the perimeter of triangle \( XYZ \), we first need to calculate the lengths of each side using the distance formula, which is given by \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). 1. Length of side \( XY \): \[ d_{XY} = \sqrt{(4 - (-2))^2 + (1 - (-5))^2} = \sqrt{(4 + 2)^2 + (1 + 5)^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \approx 8.49 \] 2. Length of side \( YZ \): \[ d_{YZ} = \sqrt{(1 - 4)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.61 \] 3. Length of side \( XZ \): \[ d_{XZ} = \sqrt{(1 - (-2))^2 + (3 - (-5))^2} = \sqrt{(1 + 2)^2 + (3 + 5)^2} = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73} \approx 8.54 \] Now, we can add up the lengths to find the perimeter: \[ \text{Perimeter} = d_{XY} + d_{YZ} + d_{XZ} \approx 8.49 + 3.61 + 8.54 \approx 20.64 \] So, the perimeter of triangle \( XYZ \) is approximately \( 20.64 \) when rounded to the nearest hundredth.