Calculating Area of Irregular Polygon in C#

Using lambda expressions this becomes trivial!

var points = GetSomePoints();

points.Add(points[0]);
var area = Math.Abs(points.Take(points.Count - 1)
   .Select((p, i) => (points[i + 1].X - p.X) * (points[i + 1].Y + p.Y))
   .Sum() / 2);

The algorithm is explained here:

[This method adds] the areas of the trapezoids defined by the polygon's edges dropped to the X-axis. When the program considers a bottom edge of a polygon, the calculation gives a negative area so the space between the polygon and the axis is subtracted, leaving the polygon's area.

The total calculated area is negative if the polygon is oriented clockwise [so the] function simply returns the absolute value.

This method gives strange results for non-simple polygons (where edges cross).

public float Area(List<PointF> vertices)
{
    vertices.Add(vertices[0]);
    return Math.Abs(vertices.Take(vertices.Count - 1).Select((p, i) => (p.X * vertices[i + 1].Y) - (p.Y * vertices[i + 1].X)).Sum() / 2);
}

Something like that for a plain polygon (compiled by notepad):

static double GetDeterminant(double x1, double y1, double x2, double y2)
{
    return x1 * y2 - x2 * y1;
}

static double GetArea(IList<Vertex> vertices)
{
    if(vertices.Count < 3)
    {
        return 0;
    }
    double area = GetDeterminant(vertices[vertices.Count - 1].X, vertices[vertices.Count - 1].Y, vertices[0].X, vertices[0].Y);
    for (int i = 1; i < vertices.Count; i++)
    {
        area += GetDeterminant(vertices[i - 1].X, vertices[i - 1].Y, vertices[i].X, vertices[i].Y);
    }
    return area / 2;
}

Although your approach doesn't pay attention to Z-axis. Therefore I'd advice to apply some transformation to get rid of it: you won't be able to get area if the polygon is not plane, whereas if it is plane you are able to get rid of the third dimension.