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

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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++

// 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;
}

Java

// Java program to implement
// the above problem
import java.io.*;
public 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

Python3

# Python3 program to implement
# the above problem
 
# Function to find the number of triangles
def findTriangles(n):
    num = n * (n - 4)
 
    # print the number of triangles
    print(num)
 
# Driver code 
# initialize the number of sides of a polygon
n = 6
 
findTriangles(n)
 
# This code is contributed by mohit kumar 29

C#

// 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

Javascript

<script>
 
// JavaScript program to implement
// the above problem
 
     
// Function to find the number of triangles
function findTriangles(n)
{
    var num;
    num = n * (n - 4);
 
    // print the number of triangles
    document.write(num);
}
 
// Driver code
 
// initialize the number of sides of a polygon
var n;
n = 6;
 
findTriangles(n);
 
 
// This code contributed by shikhasingrajput 
 
</script>

Output:

Time Complexity: O(1)
Auxiliary Space: O(1)

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