Programa para hallar el area de un triangulo

Posted on by Rudeus Greyrat

Dados los lados de un triángulo, la tarea es encontrar el área de este triángulo.

Ejemplos: 

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

Enfoque: el área de un triángulo se puede evaluar simplemente usando la siguiente fórmula. 

Area = \sqrt(s*(s-a)*(s-b)*(s-c))

donde a, b y c son las longitudes de los lados del triángulo, y 
s = (a+b+c)/2

Program to find area of a triangle

A continuación se muestra la implementación del enfoque anterior:

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 = 5.0
findArea(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>
Producción

Area is 6

Complejidad temporal: O(log 2 n)
Espacio auxiliar: O(1)

Dadas las coordenadas de los vértices de un triángulo, la tarea es encontrar el área de este triángulo.

Enfoque: si se dan las coordenadas de tres esquinas, podemos aplicar la fórmula Shoelace para el área de abajo.  

Area = \frac{\sum_{i=1}^{n-1} (x_i y_{(i+1)}+x_ny_{1}) - \sum_{i=1}^{n-1}(x_{(i+1)}y_i-x_1y_n)}{2}

= \frac{(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})}{2}

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>
Producción

2

Complejidad temporal: O(n)
Espacio auxiliar: O(1)

Publicación traducida automáticamente

Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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