Hypotenuse Calculator

10 min · The Basics

🎯 Project: Hypotenuse Calculator

Let's apply what we've learned! We'll build a program that calculates the hypotenuse of a right triangle using the Pythagorean theorem. This project combines variables, type conversion, the math module, and string formatting.

The Pythagorean theorem states: c² = a² + b², which means: c = √(a² + b²)

Where:

a and b are the two shorter sides (legs) of the right triangle
c is the hypotenuse (the longest side, opposite the right angle)

Project Requirements

Your program should:

Get two sides of a right triangle (a and b) - for this exercise, use provided values
Calculate the hypotenuse using the Pythagorean theorem
Round the result to 2 decimal places for clean output
Display the result in a formatted, user-friendly message

Step-by-Step Guide

Here's how to approach this problem:

steps.py
# Step 1: Import the math module
import math

# Step 2: Define the two sides
a = 3
b = 4

# Step 3: Calculate a² and b²
a_squared = a ** 2  # 9
b_squared = b ** 2  # 16

# Step 4: Add them together
sum_of_squares = a_squared + b_squared  # 25

# Step 5: Take the square root
hypotenuse = math.sqrt(sum_of_squares)  # 5.0

# Step 6: Round to 2 decimal places
hypotenuse_rounded = round(hypotenuse, 2)

# Step 7: Display the result
print(f"For a right triangle with sides {a} and {b},")
print(f"the hypotenuse is: {hypotenuse_rounded}")

Optimized Solution

You can combine steps to make the code more concise:

optimized.py
import math

# Define the sides
a = 3
b = 4

# Calculate hypotenuse in one line
hypotenuse = math.sqrt(a ** 2 + b ** 2)

# Round and display
hypotenuse_rounded = round(hypotenuse, 2)
print(f"The hypotenuse is: {hypotenuse_rounded}")

Testing Your Solution

Test your program with known values to verify it works correctly:

test_cases.py
import math

# Test case 1: Famous 3-4-5 triangle
a, b = 3, 4
c = round(math.sqrt(a**2 + b**2), 2)
print(f"a={a}, b={b}, c={c}")  # Should be 5.0

# Test case 2: 5-12-13 triangle
a, b = 5, 12
c = round(math.sqrt(a**2 + b**2), 2)
print(f"a={a}, b={b}, c={c}")  # Should be 13.0

# Test case 3: Non-integer result
a, b = 1, 1
c = round(math.sqrt(a**2 + b**2), 2)
print(f"a={a}, b={b}, c={c}")  # Should be 1.41 (√2)

Hints and Tips

💡 Key Functions to Use

• import math - Import the math module

• math.sqrt(x) - Calculate square root

• x ** 2 - Square a number (x²)

• round(value, 2) - Round to 2 decimal places

• f"text {variable}" - Format strings for output

💡 Common Mistakes to Avoid

• Forgetting to import math

• Not squaring the sides before adding

• Taking square root before adding a² and b²

• Forgetting to round the result

Challenge: Extended Version

Once you complete the basic version, try enhancing it:

Calculate and display the area of the triangle (area = 0.5 × a × b)
Calculate and display the perimeter (perimeter = a + b + c)
Test with different values and verify results
Format all numbers to 2 decimal places for consistency

💡 Famous Example

For a triangle with sides a=3 and b=4, the hypotenuse should be 5.0. This is the famous 3-4-5 right triangle - one of the most well-known Pythagorean triples! Try it: 3² + 4² = 9 + 16 = 25, and √25 = 5.

🎉

Lesson Complete!

Great work! Continue to the next lesson.