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.

First, we use the formula for the distance between two points to find the width and the length of the rectangle.

[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]

Using the formula above, we get

Width = distance between (-3,-1) and (1,3) = [tex] \sqrt{(1-(-3))^2 + (3-(-1))^2} = \sqrt{32} = 4 \sqrt{2} [/tex]

Length = distance between (1,3) and (-1,-3) =

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Kalahira Kalahira · Answer 2 · 2017-10-28T23:57:47+02:00

16 square units. Using the pythagorean theorem, calculate the length and width of the rectangle. sqrt((-3 - 1)^2 + (-1 - 3)^2) = sqrt(-4^2 + -4^2) = sqrt(16+16) = sqrt(32) = 4sqrt(2) sqrt((1-3)^2 + (3-1)^2) = sqrt(-2^2 + 2^2) = sqrt(4 + 4) = sqrt(8) = 2sqrt(2) You can check using the rest of the points to insure you actually have the length and width of the rectangle, I've done so, but am not going to demonstrate it here. We now have the length and width of the rectangle at 4sqrt(2) and 2sqrt(2). So just multiply them together to get the area, so 4sqrt(2) * 2sqrt(2) = 4 * 2 * sqrt(2) * sqrt(2) = 4 * 2 * 2 = 16 Therefore the area of the rectangle is 16.