Given the sides of a triangle, the task is to find the area of this triangle.
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
Approach: The area of a triangle can simply be evaluated using following formula.
where a, b and c are lengths of sides of triangle, and
s = (a+b+c)/2
Below is the implementation of the above approach:
C++
// C++ Program to find the area // of triangle #include <bits/stdc++.h> 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 Code int 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 <stdio.h> #include <stdlib.h> 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)); } } |
Python3
# Python Program to find the area # of triangle # Length of sides must be positive # and sum of any two sides def 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 triangle a = 3.0# Initialize second side of triangle b = 4.0# Initialize Third side of triangle c = 5.0findArea(a,b,c) # This code is contributed by Shariq Raza |
C#
// C# program to print // Floyd's triangle using 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
<?php function findArea($a, $b, $c) { // Length of sides must be positive // and sum of any two sides must // be smaller than third side. if ($a < 0 or $b < 0 or $c < 0 or ($a + $b <= $c) or $a + $c <= $b or $b + $c <= $a) { echo "Not a valid triangle"; exit(0); } $s = ($a + $b + $c) / 2; return sqrt($s * ($s - $a) * ($s - $b) * ($s - $c)); } // Driver Code $a = 3.0; $b = 4.0; $c = 5.0; echo "Area is ", findArea($a, $b, $c); // This code is contributed anuJ_67. ?> |
Javascript
<script> // javascript Program to find the area // of triangle function findArea( a, b, 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) { document.write( "Not a valid triangle"); return; } let s = (a + b + c) / 2; return Math.sqrt(s * (s - a) * (s - b) * (s - c)); } // Driver Code let a = 3.0; let b = 4.0; let c = 5.0; document.write( "Area is " + findArea(a, b, c)); // This code is contributed by todaysgaurav </script> |
Output
Area is 6
Time Complexity: O(log2n)
Auxiliary Space: O(1), since no extra space has been taken.
Given the coordinates of the vertices of a triangle, the task is to find the area of this triangle.
Approach: If given coordinates of three corners, we can apply the Shoelace formula for the area below.
C++
// 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 usingshoelace formula import 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 function X = [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 formula using 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
<?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); $n = count($X); echo polygonArea($X, $Y, $n); // This code is contributed by anuj_67. ?> |
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 = i; // j is previous vertex to i } // Return absolute value return Math.abs(area / 2.0); } // Driver program to test above function let X = [0, 2, 4]; let Y = [1, 3, 7]; let n = X.length; document.write(polygonArea(X, Y, n)); // This code is contributed by Mayank Tyagi </script> |
Output
2
Time Complexity: O(n)
Auxiliary Space: O(1)
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