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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++
#include<bits/stdc++.h>
#include <iostream>
#include <iomanip>
using namespace std;
double findHypotenuse(double side1, double side2)
{
double h = sqrt((side1 * side1) + (side2 * side2));
return h;
}
int main()
{
int side1 = 3, side2 = 4;
cout << fixed << showpoint;
cout << setprecision(2);
cout << findHypotenuse(side1, side2);
}
Java
class GFG {
static double findHypotenuse(double side1, double side2)
{
double h = Math.sqrt((side1 * side1) + (side2 * side2));
return h;
}
public static void main(String s[])
{
int side1 = 3, side2 = 4;
System.out.printf("%.2f", findHypotenuse(side1, side2));
}
}
Python3
def findHypotenuse(side1, side2):
h = (((side1 * side1) + (side2 * side2))**(1/2));
return h;
side1 = 3;
side2 = 4;
print(findHypotenuse(side1, side2));
C#
using System;
class GFG
{
static double findHypotenuse(double side1,
double side2)
{
double h = Math.Sqrt((side1 * side1) +
(side2 * side2));
return h;
}
public static void Main()
{
int side1 = 3, side2 = 4;
Console.Write("{0:F2}", findHypotenuse(side1,
side2));
}
}
Javascript
<script>
function findHypotenuse(side1, side2){
let h = (((side1 * side1) + (side2 * side2))**(1/2));
return h;
}
let side1 = 3;
let side2 = 4;
document.write(findHypotenuse(side1, side2).toFixed(2));
</script>
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)