What is the area of rectangle \( A B C D \) with vertices \( A(-6,3), B(2,5), C(3,1) \), and \( D(-5,-1) \) \( A=\square \) units \( ^{2} \)

To find the area of rectangle \( ABCD \), we first need to determine the lengths of its sides. We can do this by calculating the distances between two pairs of adjacent vertices. 1. **Distance between \( A(-6,3) \) and \( B(2,5) \)**: \[ AB = \sqrt{(2 - (-6))^2 + (5 - 3)^2} = \sqrt{(2 + 6)^2 + (5 - 3)^2} = \sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17} \] 2. **Distance between \( B(2,5) \) and \( C(3,1) \)**: \[ BC = \sqrt{(3 - 2)^2 + (1 - 5)^2} = \sqrt{(1)^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \] The area \( A \) of rectangle \( ABCD \) is calculated by multiplying the lengths of two adjacent sides: \[ A = AB \times BC = (2\sqrt{17}) \times (\sqrt{17}) = 2 \times 17 = 34 \] Thus, the area of rectangle \( ABCD \) is \( A = 34 \) square units.