# Area of a polygon with given n ordered vertices

Given ordered coordinates of a polygon with n vertices. Find the area of the polygon. Here ordered means that the coordinates are given either in a clockwise manner or anticlockwise from the 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 the area of a polygon using the 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}) ] |

The above formula is derived by following the cross product of the vertices to get the Area of triangles formed in the polygon.

Below is an implementation of the 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)` `{` ` ` `// Initialize 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)` ` ` `{` ` ` `// Initialize 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):` ` ` `# Initialize 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)` ` ` `{` ` ` ` ` `// Initialize 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.` |

## PHP

`<?php` `// PHP program to evaluate area of` `// a polygon using shoelace formula` `// (X[i], Y[i]) are` `// coordinates of i'th point.` `function` `polygonArea(` `$X` `, ` `$Y` `, ` `$n` `)` `{` ` ` `// Initialize area` ` ` `$area` `= 0.0;` ` ` `// Calculate value of` ` ` `// shoelace formula` ` ` `$j` `= ` `$n` `- 1;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `{` ` ` `$area` `+= (` `$X` `[` `$j` `] + ` `$X` `[` `$i` `]) *` ` ` `(` `$Y` `[` `$j` `] - ` `$Y` `[` `$i` `]);` ` ` ` ` `// j is previous vertex to i ` ` ` `$j` `= ` `$i` `;` ` ` `}` ` ` `// Return absolute value` ` ` `return` `abs` `(` `$area` `/ 2.0);` `}` `// Driver Code` `$X` `= ` `array` `(0, 2, 4);` `$Y` `= ` `array` `(1, 3, 7);` `$n` `= sizeof(` `$X` `);` `echo` `polygonArea(` `$X` `, ` `$Y` `, ` `$n` `);` `// This code is contributed by ajit` `?>` |

## Javascript

`<script>` `// JavaScript program to evaluate area` `// of a polygon using shoelace formula` ` ` ` ` `// (X[i], Y[i]) are coordinates of i'th point.` ` ` `function` `polygonArea(X, Y, n)` ` ` `{` ` ` `// Initialize area` ` ` `let area = 0.0;` ` ` ` ` `// Calculate value of shoelace formula` ` ` `let j = n - 1;` ` ` `for` `(let 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 Code` ` ` `let X = [0, 2, 4];` ` ` `let Y = [1, 3, 7];` ` ` ` ` `let n = 3;` ` ` `document.write(polygonArea(X, Y, n));` ` ` `// This code is contributed by target_2. ` `</script>` |

**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 areas will overlap, and the areas between the origin and the polygon will be canceled out and sum to 0, while only the area inside the reference triangle remains. [Source: Wiki]

For a better understanding look at the following diagrams:

**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

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