Largest sphere that can be inscribed within a cube which is in turn inscribed within a right circular cone

Given here is a right circular cone of radius r and perpendicular height h, which is inscribed in a cube which in turn is inscribed in a sphere, the task is to find the radius of the sphere.
Examples:

Input: h = 5, r = 6 
Output: 1.57306

Input: h = 8, r = 11
Output: 2.64156

Approach: 

• Let the side of the cube = a
• Let the radius of the sphere = R
• We know, a=h*r?2/(h+?2*r)(Please refer here)
• Also, R=a/2(Please refer here)
• So, R = (h*r?2/(h+?2*r))/2

Below is the implementation of the above approach:

C++

// C++ Program to find the biggest sphere
// which is inscribed within a cube which in turn
// inscribed within a right circular cone

#include <bits/stdc++.h>
using namespace std;

// Function to find the radius of the sphere
float sphereSide(float h, float r)
{
    // height and radius cannot be negative
    if (h < 0 && r < 0)
        return -1;

    // radius of the sphere
    float R = ((h * r * sqrt(2)) / (h + sqrt(2) * r)) / 2;

    return R;
}

// Driver code
int main()
{
    float h = 5, r = 6;

    cout << sphereSide(h, r) << endl;

    return 0;
}

Java

// Java Program to find the biggest sphere
// which is inscribed within a cube which in turn
// inscribed within a right circular cone
import java.lang.Math;

class GFG
{

// Function to find the radius of the sphere
static float sphereSide(float h, float r)
{
    // height and radius cannot be negative
    if (h < 0 && r < 0)
        return -1;

    // radius of the sphere
    float R = (float)((h * r * Math.sqrt(2)) /
                    (h + Math.sqrt(2) * r)) / 2;

    return R;
}

// Driver code
public static void main(String[] args)
{
    float h = 5, r = 6;

    System.out.println(sphereSide(h, r));

}
}

// This code is contributed by Code_Mech.

Python3

# Program to find the biggest sphere
# which is inscribed within a cube which in turn
# inscribed within a right circular cone
import math

# Function to find the radius of the sphere
def sphereSide(h, r):

    # height and radius cannot be negative
    if h < 0 and r < 0:
        return -1

    # radius of the sphere
    R = (((h * r * math.sqrt(2))) /
              (h + math.sqrt(2) * r) / 2)

    return R

# Driver code
h = 5; r = 6
print(sphereSide(h, r))

# This code is contributed by Shrikant13

C#

// C# Program to find the biggest sphere
// which is inscribed within a cube which in turn
// inscribed within a right circular cone
using System;

class GFG
{

// Function to find the radius of the sphere
static float sphereSide(float h, float r)
{
    // height and radius cannot be negative
    if (h < 0 && r < 0)
        return -1;

    // radius of the sphere
    float R = (float)((h * r * Math.Sqrt(2)) /
                      (h + Math.Sqrt(2) * r)) / 2;

    return R;
}

// Driver code
public static void Main()
{
    float h = 5, r = 6;

    Console.WriteLine(sphereSide(h, r));
}
}

// This code is contributed by Code_Mech

PHP

<?php
// PHP Program to find the biggest sphere
// which is inscribed within a cube which in turn
// inscribed within a right circular cone

// Function to find the radius of the sphere
function sphereSide($h, $r)
{
    // height and radius cannot be negative
    if ($h < 0 && $r < 0)
        return -1;

    // radius of the sphere
    $R = (($h * $r * sqrt(2)) /
          ($h + sqrt(2) * $r)) / 2;

    return $R;
}

// Driver code
$h = 5; $r = 6;

echo(sphereSide($h, $r));

// This code is contributed by Code_Mech.
?>

Javascript

<script>

// javascript Program to find the biggest sphere
// which is inscribed within a cube which in turn
// inscribed within a right circular cone

// Function to find the radius of the sphere
function sphereSide(h , r)
{
    // height and radius cannot be negative
    if (h < 0 && r < 0)
        return -1;

    // radius of the sphere
    var R = ((h * r * Math.sqrt(2)) /
                    (h + Math.sqrt(2) * r)) / 2;

    return R;
}

// Driver code
var h = 5, r = 6;

document.write(sphereSide(h, r).toFixed(5));


// This code is contributed by Amit Katiyar

</script>

Output

1.57306

Time Complexity: O(1)
Auxiliary Space: O(1)