Calculating Area of Irregular Polygon in C#

Question

I've managed to write a 'for dummies' how to calculate the area of irregular polygon in C#, but I need it to be dynamic for any amount of verticies. Can someone please help? Class: ``` public class Vertex { private int _vertexIdx; private double _coordX; private double _coordY; private double _coordZ; public Vertex() { } public Vertex(int vertexIdx, double coordX, double coordY, double coordZ) { _vertexIdx = vertexIdx; _coordX = coordX; _coordY = coordY; _coordZ = coordZ; } public int VertexIdx { get { return _vertexIdx; } set { _vertexIdx = value; } } public double X { get { return _coordX; } set { _coordX = value; } } public double Y { get { return _coordY; } set { _coordY = value; } } public double Z { get { return _coordZ; } set { _coordZ = value; } } } ``` Form_Load: ``` List verticies = new List(); verticies.Add(new Vertex(1, 930.9729, 802.8789, 0)); verticies.Add(new Vertex(2, 941.5341, 805.662, 0)); verticies.Add(new Vertex(3, 946.5828, 799.271, 0)); verticies.…

Accepted Answer

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).

Calculating Area of Irregular Polygon in C#

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