The Pythagorean theorem — a² + b² = c² — is the most useful single equation in geometry. It relates the two legs (a, b) and the hypotenuse (c) of any right triangle. This guide walks through 10 worked examples in increasing order of difficulty, from “find the hypotenuse” up to applied 3D distance and Pythagorean triples.
In any right triangle, the square of the hypotenuse equals the sum of the squares of the two legs:
a² + b² = c²
where a and b are the legs (the two sides forming the right angle) and c is the hypotenuse (the side opposite the right angle, always the longest).
Legs are 3 and 4. Find the hypotenuse.
c² = 3² + 4² = 9 + 16 = 25, so c = 5.
This is the famous 3-4-5 triangle — the smallest Pythagorean triple.
Legs are 2 and 3.
c² = 4 + 9 = 13, so c = √13 ≈ 3.61.
Most real-world right triangles produce irrational hypotenuses. Leave answers as √n unless asked to decimalize.
Hypotenuse is 13, one leg is 5. Find the other.
b² = c² − a² = 169 − 25 = 144, so b = 12.
This is the 5-12-13 triple. Memorize it — it appears constantly in textbooks.
One leg is 6.5, the other is 7.5. Find the hypotenuse.
c² = 42.25 + 56.25 = 98.5, so c ≈ 9.92.
A 10-foot ladder leans against a wall. The base is 6 feet from the wall. How high up the wall does it reach?
The ladder is the hypotenuse, the wall and ground are the legs.
h² = 10² − 6² = 100 − 36 = 64, so h = 8 feet.
A TV is advertised as “55 inch” (diagonal). Its width is 48 inches. What’s its height?
h² = 55² − 48² = 3025 − 2304 = 721, so h ≈ 26.85 inches.
Find the distance from (1, 2) to (4, 6).
Horizontal leg = 4 − 1 = 3.
Vertical leg = 6 − 2 = 4.
Distance² = 3² + 4² = 25, so distance = 5.
This is the distance formula in disguise — it IS the Pythagorean theorem applied to coordinate geometry.
Triangle ABC has a right angle at C. AC = 5, BC = 12. Point D is on AB such that CD ⊥ AB. Find CD.
First find AB using Pythagoras: AB² = 25 + 144 = 169, so AB = 13.
Now use the area-equality trick: ½ · AC · BC = ½ · AB · CD
5 · 12 = 13 · CD
CD = 60/13 ≈ 4.615.
A rectangular box has length 4, width 3, height 12. Find its space diagonal (corner to opposite corner).
First find the floor diagonal d: d² = 4² + 3² = 25, so d = 5.
Now apply Pythagoras again, with d as one leg and height as the other:
D² = d² + h² = 25 + 144 = 169, so D = 13.
The pattern: D = √(l² + w² + h²). The 4-3-12 box gives a clean 13 because 3-4-5 and 5-12-13 are both triples that chain.
For any integers m > n > 0, the formulas a = m² − n², b = 2mn, c = m² + n² produce a Pythagorean triple.
With m = 3, n = 2: a = 5, b = 12, c = 13 → the (5, 12, 13) triple.
With m = 4, n = 1: a = 15, b = 8, c = 17 → the (8, 15, 17) triple.
The first 8 primitive Pythagorean triples (a, b, c) where gcd(a,b,c)=1: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (20,21,29), (9,40,41), (12,35,37), (11,60,61). Recognizing these saves you from running Pythagoras every time.
The Right Triangle Calculator handles all of these patterns automatically. For 3D distance problems (like Example 9), the 3D Pythagorean Theorem Calculator chains the theorem twice in one step.
Does the Pythagorean theorem work for non-right triangles? No. For non-right triangles use the Law of Cosines, which generalizes Pythagoras: c² = a² + b² − 2ab·cos(C). When C = 90°, cos(C) = 0 and the formula reduces back to a² + b² = c².
What’s the converse of the Pythagorean theorem? If a² + b² = c² for the three sides of a triangle, then the triangle IS a right triangle (with c being the hypotenuse). Useful for verifying right angles when you only have side lengths.
How do I memorize the 3-4-5 and 5-12-13 triples? Just check: 3² + 4² = 9 + 16 = 25 = 5². And 5² + 12² = 25 + 144 = 169 = 13². Once you’ve worked them once, they stick. Other useful triples: (8,15,17), (7,24,25), (9,40,41).