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

Given a cube of side length a, which inscribes a sphere which in turn inscribes a right circular cone. The task is to find the largest possible volume of this cone.

Examples:

Input:  a = 5
Output: 58.1481

Input: a = 8
Output: 238.175



Approach:
Let, the height of right circular cone = h.
Radius of the cone = r
Radius of the sphere = R
We, know radius of the sphere inside the cube, r = a/2. Please refer ( Largest sphere that can be inscribed inside a cube).
Also, height of cone inside the sphere, h = 4r/3.
radius of cone inside the sphere, r = 2√2r/3. Please refer (Largest right circular cone that can be inscribed within a sphere).
So, height of the cone inside the sphere which in turn is inscribed within a cube, h = 2a/3.
Radius of the cone inside the sphere which in turn is inscribed within a cube, r = √2a/3.

Below is the implementation of the above approach:

C++

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// C++ Program to find the biggest right circular cone
// that can be inscribed within a right circular cone
// which in turn is inscribed within a cube
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the biggest right circular cone
float cone(float a)
{
  
    // side cannot be negative
    if (a < 0)
        return -1;
  
    // radius of right circular cone
    float r = (a * sqrt(2)) / 3;
  
    // height of right circular cone
    float h = (2 * a) / 3;
  
    // volume of right circular cone
    float V = 3.14 * pow(r, 2) * h;
  
    return V;
}
  
// Driver code
int main()
{
    float a = 5;
    cout << cone(a) << endl;
  
    return 0;
}

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Java

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// Java Program to find the biggest right circular cone
// that can be inscribed within a right circular cone
// which in turn is inscribed within a cube
import java.io.*;
  
class GFG 
{
      
// Function to find the biggest right circular cone
static float cone(float a)
{
  
    // side cannot be negative
    if (a < 0)
        return -1;
  
    // radius of right circular cone
    float r = (float) (a * Math.sqrt(2)) / 3;
  
    // height of right circular cone
    float h = (2 * a) / 3;
  
    // volume of right circular cone
    float V = (float)(3.14 *Math. pow(r, 2) * h);
  
    return V;
}
  
// Driver code
public static void main (String[] args) 
{
    float a = 5;
    System.out.println( cone(a));
}
}
  
// This code is contributed by anuj_67..

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Python3

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# Python3 Program to find the biggest right 
# circular cone that can be inscribed within 
# a right circular cone which in turn is 
# inscribed within a cube
import math
  
# Function to find the biggest 
# right circular cone
def cone(a):
  
    # side cannot be negative
    if (a < 0):
        return -1;
  
    # radius of right circular cone
    r = (a * math.sqrt(2)) / 3;
  
    # height of right circular cone
    h = (2 * a) / 3;
  
    # volume of right circular cone
    V = 3.14 * math.pow(r, 2) * h;
  
    return V;
  
# Driver code
a = 5;
print(cone(a));
  
# This code is contributed by 
# Shivi_Aggarwal 

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C#

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// C# Program to find the biggest 
// right circular cone that can be 
// inscribed within a right circular cone
// which in turn is inscribed within a cube
using System;
  
class GFG 
{
      
// Function to find the biggest 
// right circular cone
static double cone(double a)
{
  
    // side cannot be negative
    if (a < 0)
        return -1;
  
    // radius of right circular cone
    double r = (double) (a * Math.Sqrt(2)) / 3;
  
    // height of right circular cone
    double h = (2 * a) / 3;
  
    // volume of right circular cone
    double V = (double)(3.14 * Math.Pow(r, 2) * h);
  
    return Math.Round(V,4);
}
  
// Driver code
static void Main () 
{
    double a = 5;
    Console.WriteLine(cone(a));
}
}
  
// This code is contributed by chandan_jnu

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PHP

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<?php
// PHP Program to find the biggest right 
// circular cone that can be inscribed 
// within a right circular cone which in
// turn is inscribed within a cube 
  
// Function to find the biggest
// right circular cone 
function cone($a
  
    // side cannot be negative 
    if ($a < 0) 
        return -1; 
  
    // radius of right circular cone 
    $r = ($a * sqrt(2)) / 3; 
  
    // height of right circular cone 
    $h = (2 * $a) / 3; 
  
    // volume of right circular cone 
    $V = 3.14 * pow($r, 2) * $h
  
    return $V
  
// Driver code 
$a = 5; 
echo round(cone($a), 4); 
  
// This code is contributed by Ryuga
?>

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Output:

58.1481


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