Question

2. If n points are connected to form a closed polygon as shown below, the area A of the polygon can be computed as n-2 Notice that although the illustrated polygon has only six distinct corners, n for this polygon is 7 because the algorithm expects that the last point, (es yo), wil be a repeat of the initial point, (ox Yo) (x0,y0) = (xi.y1) x2,y2) (x3.V3) (x5-%) (x4 Y4)

Answer 1

SOLUTION

#include <iostream>

using namespace std;

class ClosedPolygon

{

public:

ClosedPolygon();

~ClosedPolygon();

double area();

friend istream& operator>>(istream& in, ClosedPolygon& poly);

friend ostream& operator<<(ostream& out, const ClosedPolygon& poly);

private:

void resize(int nsize);

double* x;

double* y;

int size;

};

ClosedPolygon::ClosedPolygon()

{

size = 0;

x = NULL;

y = NULL;

}

ClosedPolygon::~ClosedPolygon()

{

delete[] x;

delete[] y;

}

void ClosedPolygon::resize(int nsize)

{

delete[] x;

delete[] y;

x = new double[nsize];

y = new double[nsize];

size = nsize;

}

double ClosedPolygon::area()

{

double result = 0;

for (int i = 0; i < size - 1; i++)

{

result += (x[i+1] + x[i])*(y[i+1] - y[i]);

}

if (result < 0)

result *= -1;

return result / 2;

}

istream& operator>>(istream& in, ClosedPolygon& poly)

{

cout << "Enter number of points: ";

int size;

in >> size;

poly.resize(size);

cout << "Enter x y pair: ";

for (int i = 0; i < size; i++)

in >> poly.x[i] >> poly.y[i];

return in;

}

ostream& operator<<(ostream& out, const ClosedPolygon& poly)

{

out << "x\ty" << endl;

for (int i = 0; i < poly.size; i++)

cout << poly.x[i] << "\t" << poly.y[i] << endl;

return out;

}

int main()

{

ClosedPolygon poly;

cin >> poly;

cout << "The area is: " << poly.area() << endl;

return 0;

}

// sample run

C:Windows system3zicmd.exe Enter number of points? Enter x y pair: 4 0 4 7.5 7.5 4 0 The area is: 25.5 Press any key to continue . - .