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++ Program to find the side of the octagon
// which can be inscribed within the square

#include <bits/stdc++.h>
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 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

# 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# 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

<?php
// PHP  Program to find the side of the octagon
// which can be inscribed within the square

// Function to find the side
// of the octagon
function octaside($a)
{

    // side cannot be negative
    if ($a < 0)
        return -1;

    // side of the octagon
     $s = $a / (sqrt(2) + 1);
    return $s;
}

// Driver code

    // Get he square side
    $a = 4;

    // Find the side length of the square
    echo  octaside($a);

// This code is contributed by ajit
?>

<script>
// javascript Program to find the side of the octagon 
// which can be inscribed within the square 

// Function to find the side 
// of the octagon 
function octaside(a) 
{ 

    // side cannot be negative 
    if (a < 0) 
        return -1; 

    // side of the octagon 
    var s = a / (Math.sqrt(2) + 1); 
    return s; 
} 

// Driver code 

// Get he square side 
var a = 4; 

// Find the side length of the square 
document.write( octaside(a).toFixed(5)); 

// This code is contributed by shikhasingrajput 
</script>

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