# 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 ``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

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## Javascript

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Output:
`5`

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

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