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 * ? / N
B = ? / 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 midpointsvoid 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 Codeint main(){ int n = 4; AreaFactor(n); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{// Function to calculate the ratio of// area of N-th and (N + 1)-th nested// polygons formed by connecting midpointsstatic 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 Codepublic static void main(String[] args){ int n = 4; AreaFactor(n);}}// This code is contributed by code_hunt |
Python3
# Python3 code for the above approachimport math# Function to calculate the ratio of# area of N-th and (N + 1)-th nested# polygons formed by connecting midpointsdef 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 Codeif __name__ == "__main__": n = 4 AreaFactor(n)# This code is contributed by ukasp |
C#
// C# program for the above approachusing 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 midpointsstatic 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 midpointsfunction 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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