Cuadrado más grande que se puede inscribir dentro de un hexágono que está inscrito dentro de un triángulo equilátero

Aquí se da un triángulo equilátero de lado de longitud a , que inscribe un hexágono que a su vez inscribe un cuadrado. La tarea es encontrar la longitud del lado del cuadrado.
Ejemplos: 
 

Input:  a = 6
Output: 2.538

Input: a = 8
Output: 3.384

Enfoque : 
Sabemos que la longitud del lado de un hexágono inscrito dentro de un triángulo equilátero es h = a/3 . Consulte el hexágono más grande que se puede inscribir dentro de un triángulo equilátero .
Además, la longitud del lado del cuadrado que se puede inscribir dentro de un hexágono es x = 1.268h Consulte el cuadrado más grande que se puede inscribir dentro de un hexágono .
Entonces, la longitud del lado del cuadrado inscrito dentro de un hexágono que a su vez está inscrito dentro de un triángulo equilátero, x = 0.423a .
A continuación se muestra la implementación del enfoque anterior:
 

// C++ program to find the side of the largest square
// that can be inscribed within the hexagon which in return
// is incsribed within an equilateral triangle
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the side
// of the square
float squareSide(float a)
{
 
    // Side cannot be negative
    if (a < 0)
        return -1;
 
    // side of the square
    float x = 0.423 * a;
    return x;
}
 
// Driver code
int main()
{
    float a = 8;
    cout << squareSide(a) << endl;
 
    return 0;
}

// Java program to find the side of the
// largest square that can be inscribed
// within the hexagon which in return is
// incsribed within an equilateral triangle
class cfg
{
     
// Function to find the side
// of the square
static float squareSide(float a)
{
 
    // Side cannot be negative
    if (a < 0)
        return -1;
 
    // side of the square
    float x = (0.423f * a);
    return x;
}
 
// Driver code
public static void main(String[] args)
{
    float a = 8;
    System.out.println(squareSide(a));
 
}
}
 
// This code is contributed by
// Mukul Singh.

# Python 3 program to find the side of the
# largest square that can be inscribed
# within the hexagon which in return
# is incsribed within an equilateral triangle
 
# Function to find the side of the square
def squareSide(a):
     
    # Side cannot be negative
    if (a < 0):
        return -1
 
    # side of the square
    x = 0.423 * a
    return x
 
# Driver code
if __name__ == '__main__':
    a = 8
    print(squareSide(a))
 
# This code is contributed by
# Sanjit_Prasad

// C# program to find the side of the
// largest square that can be inscribed
// within the hexagon which in return is
// incsribed within an equilateral triangle
using System;
 
class GFG
{
     
// Function to find the side
// of the square
static float squareSide(float a)
{
 
    // Side cannot be negative
    if (a < 0)
        return -1;
 
    // side of the square
    float x = (0.423f * a);
    return x;
}
 
// Driver code
public static void Main()
{
    float a = 8;
    Console.WriteLine(squareSide(a));
}
}
 
// This code is contributed by
// shs

<?php
// PHP program to find the side of the
// largest square that can be inscribed
// within the hexagon which in return is
// incsribed within an equilateral triangle
 
// Function to find the side of the square
function squareSide($a)
{
 
    // Side cannot be negative
    if ($a < 0)
        return -1;
 
    // side of the square
    $x = 0.423 * $a;
    return $x;
}
 
// Driver code
$a = 8;
echo squareSide($a);
 
// This code is contributed by ajit.
?>

<script>
// javascript program to find the side of the
// largest square that can be inscribed
// within the hexagon which in return is
// incsribed within an equilateral triangle
 
// Function to find the side
// of the square
function squareSide(a)
{
 
    // Side cannot be negative
    if (a < 0)
        return -1;
 
    // side of the square
    var x = (0.423 * a);
    return x;
}
 
// Driver code
var a = 8;
document.write(squareSide(a));
 
// This code is contributed by Princi Singh
</script>

3.384

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)