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Ratio of area of two nested polygons formed by connecting midpoints of sides of a regular N-sided polygon

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  • Last Updated : 17 May, 2021

Given an N-sided polygon, the task is to find the ratio of the area of the Nth to (N + 1)th N-sided regular nested polygons generated by joining the midpoints of the sides of the original polygon.

Examples :

Input: N = 3
Output: 4.000000
Explanation:

Nested Triangle

Ratio of the length of the sides formed by joining the mid-points of the triangle with the length of the side of the original triangle is 0.5. Hence, R = (Area of Nth triangle) / (Area of (N + 1)th triangle) = 4

Input: N = 4
Output: 2.000000

Approach: The problem can be solved based on the following observations:

  • Consider an N-sided regular polygon as shown in the figure below.

Representation Of Nested regular polygon of N sides.

  • A = 2 * ℼ / N 
    B = ℼ / N 
    h = r * cos(B) 
    b = h * cos(B) 
    c = h((1 – cos(A)) / 2)1/2
  • Area of the Black Isosceles Triangle:

  • Area of the Red Isosceles Triangle:

  • r = s / (2 * [1 – cos(2B)])1/2 and b = r * [cos(B)]2
  • After combining the above equations:

  • Final result obtained is as follows:

Below is the implementation of the above approach:

C++




// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the ratio of
// area of N-th and (N + 1)-th nested
// polygons formed by connecting midpoints
void AreaFactor(int n)
{
    // Stores the value of PI
    double pi = 3.14159265;
 
    // Calculating area the factor
    double areaf = 1 / (cos(pi / n)
                        * cos(pi / n));
 
    // Printing the ratio
    // precise upto 6 decimal places
    cout << fixed << setprecision(6)
         << areaf << endl;
}
 
// Driver Code
int main()
{
    int n = 4;
    AreaFactor(n);
 
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to calculate the ratio of
// area of N-th and (N + 1)-th nested
// polygons formed by connecting midpoints
static void AreaFactor(int n)
{
     
    // Stores the value of PI
    double pi = 3.14159265;
  
    // Calculating area the factor
    double areaf = 1 / (Math.cos(pi / n) *
                        Math.cos(pi / n));
  
    // Printing the ratio
    // precise upto 6 decimal places
    System.out.format("%.6f", areaf);
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 4;
     
    AreaFactor(n);
}
}
 
// This code is contributed by code_hunt


Python3




# Python3 code for the above approach
import math
 
# Function to calculate the ratio of
# area of N-th and (N + 1)-th nested
# polygons formed by connecting midpoints
def AreaFactor(n):
 
    # Stores the value of PI
    pi = 3.14159265
 
    # Calculating area the factor
    areaf = 1 / (math.cos(pi / n) *
                 math.cos(pi / n))
 
    # Printing the ratio
    # precise upto 6 decimal places
    print('%.6f' % areaf)
     
# Driver Code
if __name__ == "__main__":
   
    n = 4
    AreaFactor(n)
 
# This code is contributed by ukasp


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
// Function to calculate the ratio of
// area of N-th and (N + 1)-th nested
// polygons formed by connecting midpoints
static void AreaFactor(int n)
{
      
    // Stores the value of PI
    double pi = 3.14159265;
   
    // Calculating area the factor
    double areaf = 1 / (Math.Cos(pi / n) *
                        Math.Cos(pi / n));
   
    // Printing the ratio
    // precise upto 6 decimal places
    Console.WriteLine(Math.Round(areaf));
 
}
    // Driver Code
    public static void Main(string[] args)
    {
        int n = 4;
      
        AreaFactor(n);
    }
}
 
// This code is contributed by susmitakundugoaldanga.


Javascript




<script>
 
// Javascript program implementation
// of the approach
 
// Function to calculate the ratio of
// area of N-th and (N + 1)-th nested
// polygons formed by connecting midpoints
function AreaFactor(n)
{
    // Stores the value of PI
    let pi = 3.14159265;
  
    // Calculating area the factor
    let areaf = (1 / (Math.cos(pi / n)
                        * Math.cos(pi / n)));
  
    // Printing the ratio
    // precise upto 6 decimal places
    document.write(areaf.toFixed(6));
}
 
// Driver Code
     
        let n = 4;
    AreaFactor(n);
 
// This code is contributed by splevel62.
</script>


Output: 

2.000000

 

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

 


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