This figure is made up of a rectangle and parallelogram.

what is the area of this figure?

enter your answer in the box. do not round any side lengths.

units²

Given the coordinates of two points, P1 and P2, the distance formula between these two points is deduced.
 d = root ((x2-x1) ^ 2 + (y2-y1) ^ 2)
 To find the area of the figure we must first find the area of the rectangle and add the area of the parallelogram.
 rectangle area
 A = (L) * (w)
 L = root ((- 6-2) ^ 2 + (- 1-1) ^ 2) = 8.25
 w = root ((- 6 - (- 5)) ^ 2 + (- 1 - (- 5)) ^ 2) = 4.12
 A = (8.25) * (4.12) = 33.99
 Parallelogram area
 A = (b) * (h)
 b = root ((3-3) ^ 2 + (3 - (- 3)) ^ 2) = 6
 h = root ((3-2) ^ 2 + (3-3) ^ 2) = 1
 A = (6) * (1) = 6
 The total area is then
 Atotal = 33.99 + 6 = 39.99 units ^ 2
 Answer
 the area of this figure is 39.99 units ^ 2

40 units2˛
    Looking at the figure, the rectangle has the vertexes (2,1), (3,-3), (-5,-5) and (-6,-1). The parallelogram has the vertexes (2,7), (3,3), (3,-3), and (2,1).

The area of a parallelogram is base times height. We have 2 vertical lines at x=2 and x=3, so the height is 1. And the length of the line from (3,3) to (3,-3) is 6, so the base is 6. Therefore the area of the parallelogram is 1*6 = 6.
   The rectangle is a tad trickier since it's not aligned with either the x or y axis. But we can use the Pythagorean theorem to get the lengths.
 L = sqrt((2 - -6)^2 + (1 - -1)^2)
 L = sqrt(8^2 + 2^2)
 L = sqrt(64 + 4)
  L = sqrt(68) = 2*sqrt(17)

   W = sqrt((2-3)^2 + (1- -3)^2)
 W = sqrt((-1)^2 + 4^2)
 W = sqrt(1 + 16)
 W = sqrt(17)

   And the area is length * width, so:
 2*sqrt(17)*sqrt(17) = 2 * 17 = 34

   And the total area is the sum of the areas, so
 34 + 6 = 40
   So the area of the figure is 40 square units.

The last one bc that's a calendar