Pythagorean Theorem Calculator

How to use

1. Choose which two sides you know (the two legs, or the hypotenuse and one leg).
2. Enter the lengths.
3. Read the third side — an exact integer, or a simplified surd with its decimal value.
4. Check the squares relation and the common-triples table for context.

How it works

Two uses of a² + b² = c²

The Pythagorean theorem relates the sides of a right triangle: a² + b² = c², where c is the hypotenuse (the side opposite the right angle). Given the two legs, square-root the sum to get the hypotenuse; given the hypotenuse and one leg, subtract and square-root — the hypotenuse must exceed the known leg, or no solution exists.

Non-integer results: radicals and decimals

Results are not always integers. Legs 3 and 4 give 5; 6 and 8 give 10. But 1 and 1 give √2 ≈ 1.414214, and 2 and 3 give √13 ≈ 3.605551. The tool provides both the simplified radical (extracting perfect-square factors, e.g. √20 = 2√5) and a decimal approximation — exams and drawings usually want the former, quick distance estimates the latter.

Common Pythagorean triples and the right-angle test

The test also works in reverse: sort the three sides; if a² + b² = c² the angle opposite the longest side is right (like 3-4-5); if the sum is smaller the triangle is acute, if larger obtuse. Integer solutions are Pythagorean triples — 3-4-5, 5-12-13, 8-15-17, 7-24-25 — and any scale-up stays valid (6-8-10, 9-12-15).

Calculation basis: the Pythagorean theorem a² + b² = c². Integer results are given exactly; otherwise the sum or difference of squares is simplified by extracting perfect-square factors, producing the form √m or k√m plus a decimal approximation. The right-angle check compares the square of the longest side with the sum of the squares of the other two.

Code example

function hypot(a, b) { return Math.hypot(a, b); }
function leg(c, a) { return Math.sqrt(c * c - a * a); }

function simplifySqrt(n) {          // √n → a√b
  let a = 1;
  for (let i = 2; i * i <= n; i++)
    while (n % (i * i) === 0) { n /= i * i; a *= i; }
  return a === 1 ? "√" + n : a + "√" + n;
}

hypot(3, 4);            // 5
simplifySqrt(12);       // "2√3"

import math

math.hypot(3, 4)     # 5.0
math.sqrt(4 ** 2 - 3 ** 2)   # 2.6457... (finding a leg)

# For exact radical simplification use sympy:
from sympy import sqrt, simplify
sqrt(12)             # 2*sqrt(3) (reduced automatically)

FAQ