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# Program to find area of a triangle

• Difficulty Level : Basic
• Last Updated : 05 Jul, 2021

Finding area using given sides:

Examples :

Input : a = 5, b = 7, c = 8
Output : Area of a triangle is 17.320508

Input : a = 3, b = 4, c = 5
Output : Area of a triangle is 6.000000

Area of a triangle can simply be evaluated using following formula.

Area = sqrt(s*(s-a)*(s-b)*(s-c))
where a, b and c are lengths of sides of
triangle and s = (a+b+c)/2 ## C++

 // C++ Program to find the area// of triangle#include using namespace std; float findArea(float a, float b, float c){    // Length of sides must be positive    // and sum of any two sides    // must be smaller than third side.    if (a < 0 || b < 0 || c < 0 ||       (a + b <= c) || a + c <= b ||                       b + c <= a)    {        cout << "Not a valid triangle";        exit(0);    }    float s = (a + b + c) / 2;    return sqrt(s * (s - a) *                    (s - b) * (s - c));} // Driver Codeint main(){    float a = 3.0;    float b = 4.0;    float c = 5.0;     cout << "Area is " << findArea(a, b, c);    return 0;} // This code is contributed// by rathbhupendra

## C

 #include #include  float findArea(float a, float b, float c){    // Length of sides must be positive and sum of any two sides    // must be smaller than third side.    if (a < 0 || b < 0 || c <0 || (a+b <= c) ||        a+c <=b || b+c <=a)    {        printf("Not a valid triangle");        exit(0);    }    float s = (a+b+c)/2;    return sqrt(s*(s-a)*(s-b)*(s-c));} int main(){    float a = 3.0;    float b = 4.0;    float c = 5.0;     printf("Area is %f", findArea(a, b, c));    return 0;}

## Java

 // Java program to print// Floyd's triangle     class Test{    static float findArea(float a, float b, float c)    {        // Length of sides must be positive and sum of any two sides        // must be smaller than third side.        if (a < 0 || b < 0 || c <0 || (a+b <= c) ||            a+c <=b || b+c <=a)        {            System.out.println("Not a valid triangle");            System.exit(0);        }        float s = (a+b+c)/2;        return (float)Math.sqrt(s*(s-a)*(s-b)*(s-c));    }             // Driver method    public static void main(String[] args)    {        float a = 3.0f;        float b = 4.0f;        float c = 5.0f;             System.out.println("Area is " + findArea(a, b, c));    }}

## Python

 # Python Program to find the area# of triangle # Length of sides must be positive# and sum of any two sidesdef findArea(a,b,c):     # must be smaller than third side.    if (a < 0 or b < 0 or c < 0 or (a+b <= c) or (a+c <=b) or (b+c <=a) ):        print('Not a valid triangle')        return             # calculate the semi-perimeter    s = (a + b + c) / 2         # calculate the area    area = (s * (s - a) * (s - b) * (s - c)) ** 0.5    print('Area of a triangle is %f' %area)  # Initialize first side of trianglea = 3.0# Initialize second side of triangleb = 4.0# Initialize Third side of trianglec = 5.0findArea(a,b,c) # This code is contributed by Shariq Raza

## C#

 // C# program to print// Floyd's triangleusing System; class Test {         // Function to find area    static float findArea(float a, float b,                        float c)    {                 // Length of sides must be positive        // and sum of any two sides        // must be smaller than third side.        if (a < 0 || b < 0 || c <0 ||        (a + b <= c) || a + c <=b ||            b + c <=a)        {            Console.Write("Not a valid triangle");            System.Environment.Exit(0);        }        float s = (a + b + c) / 2;        return (float)Math.Sqrt(s * (s - a) *                            (s - b) * (s - c));    }             // Driver code    public static void Main()    {        float a = 3.0f;        float b = 4.0f;        float c = 5.0f;             Console.Write("Area is " + findArea(a, b, c));    }} // This code is contributed Nitin Mittal.

## PHP

 

## Javascript

 

Output :

Area is 6

Finding area using coordinates:

If we are given coordinates of three corners, we can apply below Shoelace formula for area.

Area = | 1/2 [ (x1y2 + x2y3 + ... + xn-1yn + xny1) -
(x2y1 + x3y2 + ... + xnyn-1 + x1yn) ] | 

## C++

 // C++ program to evaluate area of a polygon using// shoelace formula#include 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 functionint main(){    double X[] = {0, 2, 4};    double Y[] = {1, 3, 7};      int n = sizeof(X)/sizeof(X);      cout << polygonArea(X, Y, n);}

## Java

 // Java program to evaluate area of// a polygon usingshoelace formulaimport java.io.*;import java.math.*; class GFG {     // (X[i], Y[i]) are coordinates of i'th point.    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 = X.length;        System.out.println(polygonArea(X, Y, n));    }}  // This code is contributed// by Nikita Tiwari.

## Python3

 # Python 3 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 = 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 functionX = [0, 2, 4]Y = [1, 3, 7] n = len(X)print(polygonArea(X, Y, n))  # This code is contributed# by Nikita Tiwari.

## C#

 // C# program to evaluate area of// a polygon usingshoelace formulausing System; class GFG {     // (X[i], Y[i]) are coordinates    // of i'th point.    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 = X.Length;        Console.WriteLine(                 polygonArea(X, Y, n));    }} // This code is contributed by anuj_67.

## PHP

 

## Javascript

 

Output:

2