Number of triangles formed by joining vertices of n-sided polygon with one side common

Given N-sided polygon we need to find the number of triangles formed by joining the vertices of the given polygon with exactly one side being common.

Examples:

Input : 6
Output : 12
The image below is of a triangle forming inside a Hexagon by joining vertices as shown above. The two triangles formed has one side (AB) common with that of a polygon.It depicts that with one edge of a hexagon we can make two triangles with one side common. We can’t take C or F as our third vertex as it will make 2 sides common with the hexagon.

Triangle with one side common of a hexagon


No of triangles formed ie equal to 12 as there are 6 edges in a hexagon.

Input : 5
Output : 5

Approach :

  • To make a triangle with one side common with a polygon the two vertices adjacent to the chosen common vertices cannot be considered as the third vertex of a triangle.
  • First select any one edge from the polygon. Consider this edge to be the common edge. Number of ways to select an edge in a polygon would be equal to n.
  • Now ,to form a triangle ,select any of the (n-4) vertices left .Two vertices of the common edge and two vertices adjacent to the common edge cannot be considered.
  • Number of triangle formed by joining the vertices of an n-sided polygon with one side common would be equal to n * ( n – 4) .

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above problem
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the number of triangles
void findTriangles(int n)
{
    int num;
    num = n * (n - 4);
  
    // print the number of triangles
    cout << num;
}
  
// Driver code
int main()
{
    // initialize the number of sides of a polygon
    int n;
    n = 6;
  
    findTriangles(n);
  
    return 0;
}

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Java














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// Java program to implement 


// the above problem 


class GFG 


{ 


      


// Function to find the number of triangles 


static void findTriangles(int n) 


{ 


    int num; 


    num = n * (n - 4); 


  


    // print the number of triangles 


    System.out.println(num); 


} 


  


// Driver code 


public static void main(String [] args) 


{ 


    // initialize the number of sides of a polygon 


    int n; 


    n = 6; 


  


    findTriangles(n); 


  


} 


} 


  


// This code is contributed by ihritik 



















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Python3














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# Python3 program to implement 


# the above problem 


  


# Function to find the number of triangles 


def findTriangles(n): 


    num = n * (n - 4) 


  


    # prthe number of triangles 


    print(num) 


  


# Driver code  


# initialize the number of sides of a polygon 


n = 6


  


findTriangles(n) 


  


# This codee is contributed by mohit kumar 29 



















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C#














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// C# program to implement 


// the above problem 


using System; 


  


class GFG 


{ 


      


// Function to find the number of triangles 


static void findTriangles(int n) 


{ 


    int num; 


    num = n * (n - 4); 


  


    // print the number of triangles 


    Console.WriteLine(num); 


} 


  


// Driver code 


public static void Main() 


{ 


    // initialize the number of sides of a polygon 


    int n; 


    n = 6; 


  


    findTriangles(n); 


  


} 


} 


  


// This code is contributed by ihritik 



















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Output:


12





 Time Complexity:  O(1)



          
          
          
          

            

    

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Improved By :  mohit kumar 29, ihritik