Python Program to Find Hypotenuse of Right Triangle

You can find the hypotenuse of a right triangle in Python using the formula hypotenuse = (side1**2 + side2**2)**0.5, where side1 and side2 are the lengths of the other two sides.

Examples

Inputside1 = 3, side2 = 4

Output5.0

Inputside1 = 5, side2 = 12

Output13.0

Inputside1 = 0, side2 = 0

Output0.0

How to Think About It

To find the hypotenuse, think of the right triangle as having two shorter sides. The hypotenuse is the longest side opposite the right angle. Use the Pythagorean theorem which says the square of the hypotenuse equals the sum of the squares of the other two sides. So, square both sides, add them, then take the square root of that sum to get the hypotenuse.

Algorithm

1

Get the lengths of the two shorter sides of the triangle.

2

Square each side length.

3

Add the two squared values together.

4

Calculate the square root of the sum to find the hypotenuse.

5

Return or print the hypotenuse value.

Code

python

side1 = float(input('Enter length of side 1: '))
side2 = float(input('Enter length of side 2: '))
hypotenuse = (side1**2 + side2**2)**0.5
print('Hypotenuse:', hypotenuse)

Output

Enter length of side 1: 3
Enter length of side 2: 4
Hypotenuse: 5.0

Dry Run

Let's trace the example where side1 = 3 and side2 = 4 through the code.

1

Input sides

side1 = 3, side2 = 4

2

Square sides

3**2 = 9, 4**2 = 16

3

Sum squares

9 + 16 = 25

4

Square root

sqrt(25) = 5.0

5

Output hypotenuse

Hypotenuse: 5.0

Step	Operation	Value
1	Input sides	side1=3, side2=4
2	Square sides	9, 16
3	Sum squares	25
4	Square root	5.0
5	Output	Hypotenuse=5.0

Why This Works

Step 1: Square the sides

We use side1**2 and side2**2 to get the squares of the two sides.

Step 2: Add the squares

Adding the squares with side1**2 + side2**2 gives the square of the hypotenuse.

Step 3: Calculate the square root

Taking the square root with **0.5 gives the length of the hypotenuse.

Alternative Approaches

Using math.sqrt() function

python

import math
side1 = float(input('Enter side 1: '))
side2 = float(input('Enter side 2: '))
hypotenuse = math.sqrt(side1**2 + side2**2)
print('Hypotenuse:', hypotenuse)

This uses the built-in math module for clearer square root calculation.

Using a function to reuse code

python

def find_hypotenuse(a, b):
    return (a**2 + b**2)**0.5

print('Hypotenuse:', find_hypotenuse(3, 4))

Encapsulates logic in a function for reuse and cleaner code.

Complexity: O(1) time, O(1) space

Time Complexity

The calculation involves a fixed number of arithmetic operations, so it runs in constant time.

Space Complexity

Only a few variables are used, so the space needed is constant.

Which Approach is Fastest?

Both direct exponentiation and using math.sqrt() are equally fast for this simple calculation.

Approach	Time	Space	Best For
Direct exponentiation (**0.5)	O(1)	O(1)	Simple quick calculation
math.sqrt() function	O(1)	O(1)	Clearer code and readability
Function encapsulation	O(1)	O(1)	Reusable code in larger programs