程序代码：

#include <iostream>
#include <math.h>
using namespace std;
typedef struct tagPoint //节点的X.Y坐标
{
    int iX;
    int iY;
}T_POINT;

class cTriangle
{
private:
    T_POINT * m_Point;
public:
    cTriangle();
    ~cTriangle();
    void Create(T_POINT * pPoint, int iNums);   //pPoint指向一个点数组， iNums就是点的个数
    void GetArea(int iNums);
};
cTriangle::cTriangle()
{

}
cTriangle::~cTriangle()
{
    delete []m_Point;
}
void cTriangle::Create(T_POINT * pPoint, int iNums)
{
    m_Point = new T_POINT[iNums];
    memcpy(m_Point, pPoint, sizeof(T_POINT) * iNums);
}
void cTriangle::GetArea(int iNums)
{
    int iFor = iNums - 2;//得出共有多少三角形
    double izhou , iFirst , iSecond , iAgain , m_Area=0;
    for (int i = 0; i < iFor; ++i)
    {
        iFirst = sqrt(pow((m_Point[i+1].iX - m_Point[0].iX) , 2) + pow((m_Point[i+1].iY - m_Point[0].iY) , 2));
        iSecond = sqrt(pow((m_Point[i+2].iX - m_Point[i+1].iX) , 2) + pow((m_Point[i+2].iY - m_Point[i+1].iY) , 2));
        iAgain = sqrt(pow((m_Point[i+2].iX - m_Point[0].iX) , 2) + pow((m_Point[i+2].iY - m_Point[0].iY) , 2));
        izhou = (iFirst + iSecond + iAgain) / 2;
         m_Area += sqrt(izhou * (izhou - iFirst) * (izhou - iSecond) * (izhou - iAgain));
    }
    cout << m_Area << endl;
}
int main()
{
    T_POINT tPOINT[4];
    int iTmp;
    for (int i = 0; i < sizeof(tPOINT) / sizeof(tPOINT[0]); ++i)
    {
        cout << "请输入第" << i + 1 << "个节点的横坐标：" << endl;
        cin >> iTmp;
        tPOINT[i].iX = iTmp;
        cout << "请输入第" << i + 1 << "个节点的纵坐标：" << endl;
        cin >> iTmp;
        tPOINT[i].iY = iTmp;
    }
    cTriangle cKey;
    cKey.Create(tPOINT , sizeof(tPOINT) / sizeof(tPOINT[0]));
    cKey.GetArea(sizeof(tPOINT) / sizeof(tPOINT[0]));
    return 0;
}