Longest rod that can be inserted within a right circular cylinder

Given a right circular cylinder of height , & radius . The task is to find the length of the longest rod that can be inserted within it.

Examples: 

Input : h = 4, r = 1.5
Output : 5

Input : h= 12, r = 2.5
Output : 13

Approach: 
From the figure, it is clear that we can get the length of the rod by using pythagoras theorem, by treating the height of cylinder as perpendicular, diameter as base and length of rod as hypotenuse.
So, l² = h² + 4*r².
Therefore, 

l = ?(h2 + 4*r2)

Follow the below steps to implement the above idea:

• Check if the height h and radius r is negative. If either is negative, return -1 to indicate an error.
• Calculate the length of the rod using the formula sqrt(h^2 + 4r^2), and assign the result to the float variable l.
• Return the value of l.

Below is the implementation of the above approach: 

C++

// C++ Program to find the longest rod
// that can be fit within a right circular cylinder
#include <bits/stdc++.h>
using namespace std;

// Function to find the side of the cube
float rod(float h, float r)
{

    // height and radius cannot be negative
    if (h < 0 && r < 0)
        return -1;

    // length of rod
    float l = sqrt(pow(h, 2) + 4 * pow(r, 2));
    return l;
}

// Driver code
int main()
{
    float h = 4, r = 1.5;

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

    return 0;
}

Java

// Java Program to find the longest rod
// that can be fit within a right circular cylinder

import java.io.*;

class GFG {


// Function to find the side of the cube
static float rod(float h, float r)
{

    // height and radius cannot be negative
    if (h < 0 && r < 0)
        return -1;

    // length of rod
    float l = (float)(Math.sqrt(Math.pow(h, 2) + 4 * Math.pow(r, 2)));
    return l;
}

// Driver code


    public static void main (String[] args) {
            float h = 4;
            float r = 1.5f;
            System.out.print(rod(h, r));
    }
}
// This code is contributed by anuj_67..

Python 3

# Python 3 Program to find the longest
# rod that can be fit within a right
# circular cylinder
import math

# Function to find the side of the cube
def rod(h, r):

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

    # length of rod
    l = (math.sqrt(math.pow(h, 2) +
               4 * math.pow(r, 2)))
    return float(l)

# Driver code
h , r = 4, 1.5
print(rod(h, r))

# This code is contributed
# by PrinciRaj1992

C#

// C# Program to find the longest
// rod that can be fit within a
// right circular cylinder
using System;

class GFG
{

// Function to find the side
// of the cube
static float rod(float h, float r)
{

    // height and radius cannot
    // be negative
    if (h < 0 && r < 0)
        return -1;

    // length of rod
    float l = (float)(Math.Sqrt(Math.Pow(h, 2) +
                            4 * Math.Pow(r, 2)));
    return l;
}

// Driver code
public static void Main ()
{
    float h = 4;
    float r = 1.5f;
    Console.WriteLine(rod(h, r));
}
}

// This code is contributed by shs

PHP

<?php
// PHP Program to find the longest
// rod that can be fit within a
// right circular cylinder

// Function to find the side
// of the cube
function rod($h, $r)
{

    // height and radius cannot
    // be negative
    if ($h < 0 && $r < 0)
        return -1;

    // length of rod
    $l = sqrt(pow($h, 2) + 4 * pow($r, 2));
    return $l;
}

// Driver code
$h = 4; $r = 1.5;

echo rod($h, $r) . "\n";

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>

// javascript Program to find the longest rod
// that can be fit within a right circular cylinder

// Function to find the side of the cube
function rod(h , r)
{

    // height and radius cannot be negative
    if (h < 0 && r < 0)
        return -1;

    // length of rod
    var l = (Math.sqrt(Math.pow(h, 2) + 4 * Math.pow(r, 2)));
    return l;
}

// Driver code
var h = 4;
var r = 1.5;
document.write(rod(h, r));

// This code contributed by shikhasingrajput

</script>

Output:

5

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