# Program to find the side of the Octagon inscribed within the square

Given a square of side length ‘a’, the task is to find the side length of the biggest octagon that can be inscribed within it.

Examples:

Input: a = 4
Output: 1.65685

Input: a = 5
Output: 2.07107

Approach:

=> From the figure, it can be seen that, side length of the Octagon = b
=> Also since the polygons are regular, therefore 2*x + b = a
=> From the right angled triangle, x^2 + x^2 = b^2
=> Hence, x = b/?2,
=> So, ?2b + b = a
=> Therefore, b = a/(?2 +1)

Below is the implementation of the above approach:

## C++

 // C++ Program to find the side of the octagon // which can be inscribed within the square   #include using namespace std;   // Function to find the side // of the octagon float octaside(float a) {       // side cannot be negative     if (a < 0)         return -1;       // side of the octagon     float s = a / (sqrt(2) + 1);     return s; }   // Driver code int main() {       // Get he square side     float a = 4;       // Find the side length of the square     cout << octaside(a) << endl;       return 0; }

## Java

 // Java Program to find the side of the octagon // which can be inscribed within the square   import java.io.*;   class GFG {       // Function to find the side // of the octagon static double octaside(double a) {       // side cannot be negative     if (a < 0)         return -1;       // side of the octagon     double s = a / (Math.sqrt(2) + 1);     return s; }   // Driver code           public static void main (String[] args) {               // Get he square side     double a = 4;       // Find the side length of the square     System.out.println( octaside(a));                           } } //This Code  is contributed by ajit

## Python3

 # Python 3 Program to find the side # of the octagon which can be # inscribed within the square from math import sqrt   # Function to find the side # of the octagon def octaside(a):           # side cannot be negative     if a < 0:         return -1       # side of the octagon     s = a / (sqrt(2) + 1)     return s   # Driver code if __name__ == '__main__':           # Get he square side     a = 4       # Find the side length of the square     print("{0:.6}".format(octaside(a)))       # This code is contributed # by Surendra_Gangwar

## C#

 // C# Program to find the side // of the octagon which can be // inscribed within the square using System;   class GFG {       // Function to find the side // of the octagon static double octaside(double a) {       // side cannot be negative     if (a < 0)         return -1;       // side of the octagon     double s = a / (Math.Sqrt(2) + 1);     return s; }   // Driver code static public void Main () {     // Get he square side     double a = 4;           // Find the side length     // of the square     Console.WriteLine( octaside(a)); } }   // This code is contributed // by akt_mit



## Javascript



Output:

1.65685

Time Complexity: O(1)  since no loop is used the algorithm takes up constant time to perform the operations

Space Complexity: O(1) since no extra array is used so the space taken by the algorithm is constant

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next