Given ordered coordinates of a polygon with n vertices. Find area of the polygon. Here ordered mean that the coordinates are given either in clockwise manner or anticlockwise from first vertex to last.

Examples:

Input : X[] = {0, 4, 4, 0}, Y[] = {0, 0, 4, 4}; Output : 16 Input : X[] = {0, 4, 2}, Y[] = {0, 0, 4} Output : 8

We can compute area of a polygon using Shoelace formula.

Area = | 1/2 [ (x_{1}y_{2}+ x_{2}y_{3}+ ... + x_{n-1}y_{n}+ x_{n}y_{1}) - (x_{2}y_{1}+ x_{3}y_{2}+ ... + x_{n}y_{n-1}+ x_{1}y_{n}) ] |

Below is implementation of above formula.

## CPP

// C++ program to evaluate area of a polygon using // shoelace formula #include <bits/stdc++.h> using namespace std; // (X[i], Y[i]) are coordinates of i'th point. double polygonArea(double X[], double Y[], int n) { // Initialze area double area = 0.0; // Calculate value of shoelace formula int j = n - 1; for (int i = 0; i < n; i++) { area += (X[j] + X[i]) * (Y[j] - Y[i]); j = i; // j is previous vertex to i } // Return absolute value return abs(area / 2.0); } // Driver program to test above function int main() { double X[] = {0, 2, 4}; double Y[] = {1, 3, 7}; int n = sizeof(X)/sizeof(X[0]); cout << polygonArea(X, Y, n); }

## Java

// Java program to evaluate area // of a polygon using shoelace formula import java.io.*; class GFG { // (X[i], Y[i]) are coordinates of i'th point. public static double polygonArea(double X[], double Y[], int n) { // Initialze area double area = 0.0; // Calculate value of shoelace formula int j = n - 1; for (int i = 0; i < n; i++) { area += (X[j] + X[i]) * (Y[j] - Y[i]); // j is previous vertex to i j = i; } // Return absolute value return Math.abs(area / 2.0); } // Driver program public static void main (String[] args) { double X[] = {0, 2, 4}; double Y[] = {1, 3, 7}; int n = 3; System.out.println(polygonArea(X, Y, n)); } } // This code is contributed by Sunnnysingh

## Python3

# python3 program to evaluate # area of a polygon using # shoelace formula # (X[i], Y[i]) are coordinates of i'th point. def polygonArea(X, Y, n): # Initialze area area = 0.0 # Calculate value of shoelace formula j = n - 1 for i in range(0,n): area += (X[j] + X[i]) * (Y[j] - Y[i]) j = i # j is previous vertex to i # Return absolute value return int(abs(area / 2.0)) # Driver program to test above function X = [0, 2, 4] Y = [1, 3, 7] n = len(X) print(polygonArea(X, Y, n)) # This code is contributed by # Smitha Dinesh Semwal

## C#

// C# program to evaluate area // of a polygon using shoelace formula using System; class GFG { // (X[i], Y[i]) are coordinates of i'th point. public static double polygonArea(double[] X, double[] Y, int n) { // Initialze area double area = 0.0; // Calculate value of shoelace formula int j = n - 1; for (int i = 0; i < n; i++) { area += (X[j] + X[i]) * (Y[j] - Y[i]); // j is previous vertex to i j = i; } // Return absolute value return Math.Abs(area / 2.0); } // Driver program public static void Main() { double[] X = { 0, 2, 4 }; double[] Y = { 1, 3, 7 }; int n = 3; Console.WriteLine(polygonArea(X, Y, n)); } } // This code is contributed by vt_m.

Output :

2

**Why is it called Shoelace Formula?**

The formula is called so because of the way we evaluate it.

Example:

Let the input vertices be (0, 1), (2, 3), and (4, 7). Evaluation procedure matches with process of tying shoelaces. We write vertices as below 0 1 2 3 4 7 0 1 [written twice] we evaluate positive terms as below 0 \ 1 2 \ 3 4 \ 7 0 1 i.e., 0*3 + 2*7 + 4*1 = 18 we evaluate negative terms as below 0 1 2 / 3 4 / 7 0 / 1 i.e., 0*7 + 4*3 + 2*1 = 14 Area = 1/2 (18 - 14) = 2 See this for a clearer image.

**How does this work?**

We can always divide a polygon into triangles. The area formula is derived by taking each edge AB, and calculating the (signed) area of triangle ABO with a vertex at the origin O, by taking the cross-product (which gives the area of a parallelogram) and dividing by 2. As one wraps around the polygon, these triangles with positive and negative area will overlap, and the areas between the origin and the polygon will be cancelled out and sum to 0, while only the area inside the reference triangle remains. [Source : Wiki]

**Related articles : **

Minimum Cost Polygon Triangulation

Find Simple Closed Path for a given set of points

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above