Find the hypotenuse of a right angled triangle with given two sides

Given the other two sides of a right angled triangle, the task is to find it’s hypotenuse.
Examples:

Input: side1 = 3, side2 = 4 
Output: 5.00 
3² + 4² = 5²
Input: side1 = 12, side2 = 15 
Output: 19.21 

Approach: Pythagoras theorem states that the square of hypotenuse of a right angled triangle is equal to the sum of squares of the other two sides.
Below is the implementation of the above approach:

C++

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

// Function to return the hypotenuse of the
// right angled triangle
double findHypotenuse(double side1, double side2)
{
    double h = sqrt((side1 * side1) + (side2 * side2));
    return h;
}

// Driver code
int main()
{
    int side1 = 3, side2 = 4;
    cout << fixed << showpoint;
    cout << setprecision(2);
    cout << findHypotenuse(side1, side2);
}

// This code is contributed by
// Surendra_Gangwar

Java

// Java implementation of the approach
class GFG {

    // Function to return the hypotenuse of the
    // right angled triangle
    static double findHypotenuse(double side1, double side2)
    {
        double h = Math.sqrt((side1 * side1) + (side2 * side2));
        return h;
    }

    // Driver code
    public static void main(String s[])
    {
        int side1 = 3, side2 = 4;
        System.out.printf("%.2f", findHypotenuse(side1, side2));
    }
}

Python3

# Python implementation of the approach

# Function to return the hypotenuse of the
# right angled triangle
def findHypotenuse(side1, side2):

    h = (((side1 * side1) + (side2 * side2))**(1/2));
    return h;

# Driver code
side1 = 3;
side2 = 4;

print(findHypotenuse(side1, side2));

# This code contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the hypotenuse
    // of the right angled triangle
    static double findHypotenuse(double side1,
                                 double side2)
    {
        double h = Math.Sqrt((side1 * side1) +
                             (side2 * side2));
        return h;
    }

    // Driver code
    public static void Main()
    {
        int side1 = 3, side2 = 4;
        Console.Write("{0:F2}", findHypotenuse(side1,
                                               side2));
    }
}

// This code is contributed
// by Princi Singh

Javascript

<script>
// java script  implementation of the approach

// Function to return the hypotenuse of the
//right angled triangle
function findHypotenuse(side1, side2){

    let h = (((side1 * side1) + (side2 * side2))**(1/2));
    return h;
}

// Driver code
let side1 = 3;
let side2 = 4;

document.write(findHypotenuse(side1, side2).toFixed(2));

// This code is contributed by Gottumukkala Bobby
</script>

Output:

5.00

Time Complexity: O(log(2*(s²)) where s is the side of the rectangle. because time complexity of inbuilt sqrt function is O(log(n))

Auxiliary Space: O(1)