What is the area of a rectangle with vertices at (-3, - 1), (1, 3), (3, 1), and (-1, - 3) ?

Enter your answer in the box. Do not round any side lengths.

16 square units. Let's use the pythagorean theorem to determine the lengths of each side of the rectangle. A = (-3,-1), B = (1,3), C = (3,1), D = (-1,-3) Length AB^2 = (-3 - 1) ^2 + (-1 - 3) ^2 = - 4^2 + - 4^2 = 16+16 = 32 Length BC^2 = (1 - 3) ^2 + (3-1) ^2 = - 2^2 + 2^2 = 4 + 4 = 8 Length CD^2 = (3 - (-1)) ^2 + (1 - (-3)) ^2 = 4^2 + 4^2 = 16+16 = 32 Length AD^2 = (-3 - (-1)) ^2 + (-1 - (-3)) ^2 = - 2^2 + 2^2 = 4 + 4 = 8 And just to make certain I haven't accidentally included the diagonal of the rectangle, I'll check AC and BD. So Length AC^2 = (-3 - 3) ^2 + (-1 - 1) ^2 = - 6^2 + - 2^2 = 36 + 4 = 40 Length BD^2 = (1 - (-1)) ^2 + (3 - (-3)) ^2 = 2^2 + 6^2 = 4 + 36 = 40 So I now know that length of the rectangle is sqrt (32) and the width is sqrt (8). And the area will be the product of those 2 numbers. So sqrt (32) * sqrt (8) = sqrt (256) = 16. So the area of the rectangle is 16 square units.