Programa para hallar el lado del octágono inscrito en el cuadrado

Dado un cuadrado de longitud de lado ‘a’, la tarea es encontrar la longitud de lado del octágono más grande que se puede inscribir en él.

Ejemplos: 

Input: a = 4
Output: 1.65685

Input: a = 5
Output: 2.07107

Enfoque :

=> 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) 

A continuación se muestra la implementación del enfoque anterior:  

// 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

Complejidad de tiempo: O(1)

Complejidad espacial: O(1)