Hallar la hipotenusa de un triangulo rectangulo de dos lados dados

Dados los otros dos lados de un triángulo rectángulo, la tarea es encontrar su hipotenusa.
Ejemplos: 
 

Entrada: lado1 = 3, lado2 = 4 
Salida: 5,00 
3 ² + 4 ² = 5 ²
Entrada: lado1 = 12, lado2 = 15 
Salida: 19,21 
 

Planteamiento: El teorema de Pitágoras establece que el cuadrado de la hipotenusa de un triángulo rectángulo es igual a la suma de los cuadrados de los otros dos lados.
A continuación se muestra la implementación del enfoque anterior:
 

// 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 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));
    }
}

# 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# 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

<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>

5.00

Complejidad de tiempo : O(log(2*(s ² )) donde s es el lado del rectángulo.

Espacio Auxiliar: O(1)