This relies on the distance formula like number two, which you answered correctly.
I think it's helpful to draw this out. I made a digital drawing on GeoGebra (free) and connected the vertices. Now, to find the area, we need the side lengths to start. This program actually gives you the side lengths as well, but pretending we don't already have that, you can plug in each vertex into the distance formula.
[tex]a = \sqrt{(-4-3)^{2}+ (2-2)^{2}} = 7[/tex]
[tex]b = \sqrt{(3-3)^{2}+(2-(-5))^{2}} = 7[/tex]
[tex]c = \sqrt{(3-(-4))^{2}+(-5-(-2))^{2}} = \sqrt{58}[/tex]
[tex]d = \sqrt{(-4-(-4))^{2}+(2-(-2))^{2}} = 4[/tex]
So, unfortunately, we don't have a regular polygon, so this isn't simply length times width. But we can easily cut this into a rectangle and a triangle, whose areas are much easier to find (see second image).
The area, A, of a triangle is [tex] \frac{1}{2} B * h[/tex] and the area of a rectangle is simply [tex]l * w[/tex]. So the base of our triangle is 7 units, and the height is 3 units, so the area is 10.5 units squared. The length of our rectangle is 7 units, and the width is 4 units, so the area is 28 units squared.
[tex]A = A_{triangle} + A_{rectangle} = 10 \frac{1}{2}+28 = 38\frac{1}{2} [/tex] units squared.
I would like to clarify that you do not actually need to find all those distances for this problem. Once it's drawn out, it's fairly easy to tell.
