Radius of the biggest possible circle inscribed in rhombus which in turn is inscribed in a rectangle

Give a rectangle with length l & breadth b, which inscribes a rhombus, which in turn inscribes a circle. The task is to find the radius of this circle.
Examples:

Input: l = 5, b = 3
Output: 1.28624

Input: l = 6, b = 4
Output: 1.6641

Approach: From the figure, it is clear that diagonals, x & y, are equal to the length and breadth of the rectangle. 
Also radius of the circle, r, inside a rhombus is = xy/2?(x^2+y^2).
So, radius of the circle in terms of l & b is = lb/2?(l^2+b^2).
Below is the implementation of the above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the radius
// of the inscribed circle
float circleradius(float l, float b)
{

    // the sides cannot be negative
    if (l < 0 || b < 0)
        return -1;

    // radius of the circle
    float r = (l * b) / (2 * sqrt((pow(l, 2) + pow(b, 2))));
    return r;
}

// Driver code
int main()
{
    float l = 5, b = 3;
    cout << circleradius(l, b) << endl;

    return 0;
}

Java

// Java implementation of above approach

import java.io.*;

class GFG {

// Function to find the radius
// of the inscribed circle
static float circleradius(float l, float b)
{

    // the sides cannot be negative
    if (l < 0 || b < 0)
        return -1;

    // radius of the circle
    float r = (float)((l * b) / (2 * Math.sqrt((Math.pow(l, 2) + Math.pow(b, 2)))));
    return r;
}

    // Driver code
    public static void main (String[] args) {
        float l = 5, b = 3;
    System.out.print (circleradius(l, b)) ;
    }
}
// This code is contributed by inder_verma..

Python3

# Python 3 implementation of
# above approach
from math import sqrt

# Function to find the radius
# of the inscribed circle
def circleradius(l, b):

    # the sides cannot be negative
    if (l < 0 or b < 0):
        return -1

    # radius of the circle
    r = (l * b) / (2 * sqrt((pow(l, 2) +
                             pow(b, 2))));
    return r

# Driver code
if __name__ == '__main__':
    l = 5
    b = 3
    print("{0:.5}" . format(circleradius(l, b)))

# This code is contribute
# by Surendra_Gagwar

C#

// C# implementation of above approach
using System;

class GFG
{

// Function to find the radius
// of the inscribed circle
static float circleradius(float l,
                          float b)
{

    // the sides cannot be negative
    if (l < 0 || b < 0)
        return -1;

    // radius of the circle
    float r = (float)((l * b) /
              (2 * Math.Sqrt((Math.Pow(l, 2) +
                   Math.Pow(b, 2)))));
    return r;
}

// Driver code
public static void Main ()
{
    float l = 5, b = 3;
    Console.WriteLine(circleradius(l, b));
}
}

// This code is contributed
// by inder_verma

PHP

<?php
// PHP implementation of above approach

// Function to find the radius
// of the inscribed circle
function circleradius($l, $b)
{

    // the sides cannot be negative
    if ($l < 0 || $b < 0)
        return -1;

    // radius of the circle
    $r = ($l * $b) / (2 * sqrt((pow($l, 2) +
                                pow($b, 2))));
    return $r;
}

// Driver code
$l = 5;
$b = 3;
echo circleradius($l, $b), "\n";

// This code is contributed by ajit
?>

Javascript

<script>
// javascript implementation of above approach

// Function to find the radius
// of the inscribed circle
function circleradius(l , b)
{

    // the sides cannot be negative
    if (l < 0 || b < 0)
        return -1;

    // radius of the circle
    var r = ((l * b) / (2 * Math.sqrt((Math.pow(l, 2) + Math.pow(b, 2)))));
    return r;
}

var l = 5, b = 3;
document.write(circleradius(l, b).toFixed(5)) ;

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

Output:

1.28624

Time complexity: O(logn) as it is using inbuilt sqrt  function

Auxiliary Space: O(1) since using constant variables