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 <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

// 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

PHP

<?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
?>

Javascript

<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>

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