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.

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