Pythagorean Theorem: Complete Lesson, Proofs and Solved Exercises

In one sentence

In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

It’s probably the most-used geometry theorem in the world: it lets you compute a length as soon as you know two lengths of a right triangle, without needing to measure.

Formal statement

Let ABC be a right triangle with the right angle at A. Write:

• a = BC (the hypotenuse — the side opposite the right angle, always the longest)
• b = AC
• c = AB

Then: $a^2 = b^2 + c^2$

Equivalently: $\text{BC}^2 = \text{AB}^2 + \text{AC}^2$

Why is it true? Two proofs

1. Visual proof (by areas)

Build a square with side $(a+b)$. Inside, you can fill it in two different ways:

Way 1: 4 identical right triangles + a square in the middle with side c (the hypotenuse). Total area = $4 \times \frac{ab}{2} + c^2 = 2ab + c^2$

Way 2: 4 identical right triangles + two squares, one of side a, one of side b. Total area = $4 \times \frac{ab}{2} + a^2 + b^2 = 2ab + a^2 + b^2$

Both areas are equal (same big square): $2ab + c^2 = 2ab + a^2 + b^2$

Simplify: $c^2 = a^2 + b^2$

Q.E.D.

2. Proof by the altitude

Draw the altitude from the right angle. It splits the triangle into two smaller right triangles, both similar to the big one.

Call h this altitude and p, q the segments it cuts on the hypotenuse (so $p + q = a$).

By similarity: $\frac{b}{a} = \frac{q}{b} \quad \text{and} \quad \frac{c}{a} = \frac{p}{c}$

So $b^2 = aq$ and $c^2 = ap$. Adding: $b^2 + c^2 = a(p + q) = a \cdot a = a^2$

Q.E.D.

The converse (essential in middle school)

The converse lets you prove a triangle is right:

If in a triangle the square of the longest side equals the sum of the squares of the other two sides, then the triangle is right (and the right angle is opposite the longest side).

Example: A triangle has sides 6, 8, 10. Is it right?

$10^2 = 100$ and $6^2 + 8^2 = 36 + 64 = 100$.

Equality holds ⇒ the triangle is right, with the right angle opposite the side of length 10.

Step-by-step method (find a missing length)

Step 1: Confirm the triangle is right (from the statement or the diagram).

Step 2: Identify the hypotenuse — the side opposite the right angle.

Step 3: Write Pythagoras with the actual lengths.

Step 4: Isolate the unknown.

Step 5: Compute, then take the square root.

Example 1 — Find the hypotenuse

Right triangle at A with AB = 3 cm and AC = 4 cm. Find BC.

• Hypotenuse: BC (opposite the right angle A).
• $\text{BC}^2 = 3^2 + 4^2 = 9 + 16 = 25$
• $\text{BC} = \sqrt{25} = 5$ cm.

Example 2 — Find a leg

Right triangle at A with BC = 13 cm and AB = 5 cm. Find AC.

• Here the hypotenuse is BC (given); we want a leg.
• $\text{BC}^2 = \text{AB}^2 + \text{AC}^2$
• $13^2 = 5^2 + \text{AC}^2$
• $169 = 25 + \text{AC}^2$
• $\text{AC}^2 = 144$
• $\text{AC} = 12$ cm.

Pythagorean triples to memorize

Trios $(a, b, c)$ of integers with $a^2 + b^2 = c^2$:

Triple	Check
(3, 4, 5)	9 + 16 = 25
(5, 12, 13)	25 + 144 = 169
(8, 15, 17)	64 + 225 = 289
(7, 24, 25)	49 + 576 = 625
(20, 21, 29)	400 + 441 = 841

Spotting a triple saves time on tests: if you see 5-12-?, you know ? = 13.

Common mistakes

1. Mixing up hypotenuse and legs: the hypotenuse is always the longest, always opposite the right angle.
2. Adding without squaring: $\text{BC} \neq \text{AB} + \text{AC}$. Pythagoras uses squares, not raw lengths.
3. Forgetting the square root: if $\text{BC}^2 = 25$, then $\text{BC} = 5$, not 25.
4. Applying Pythagoras to non-right triangles: for those, use the Law of Cosines.

Solved exercises

Exercise 1 (easy)

A triangle is right at R with RA = 8 and RB = 6. Find AB.

Show solution

AB is the hypotenuse. $\text{AB}^2 = 64 + 36 = 100$ $\text{AB} = 10$

Exercise 2 (medium)

A 5 m ladder is leaning against a wall, with its foot 3 m from the wall. How high does it reach on the wall?

Show solution

Model: right triangle with hypotenuse 5 m (the ladder), one leg 3 m (ground), find the other leg (height). $5^2 = 3^2 + h^2$ $h^2 = 16$ $h = 4$ m

Exercise 3 (use the converse)

A triangle has sides 9, 12, 15. Is it right?

Show solution

Longest side: 15. $15^2 = 225$ $9^2 + 12^2 = 81 + 144 = 225$ Equality ⇒ the triangle is right, with the right angle opposite 15.

(It’s the (3, 4, 5) triple times 3.)

Exercise 4 (exact value)

Right triangle at A with AB = 5 cm and AC = 5 cm. Find BC exactly.

Show solution

$\text{BC}^2 = 50$ $\text{BC} = \sqrt{50} = 5\sqrt{2}$ cm.

This is the diagonal of a square of side 5 — a very common case.

Takeaways

• Pythagoras only works in a right triangle.
• The hypotenuse is the longest side, opposite the right angle.
• The equation uses squares, not raw lengths.
• The converse proves a triangle is right.
• Memorize 3-4-5, 5-12-13, 8-15-17.

Going further

• The Law of Cosines generalizes Pythagoras to any triangle: $a^2 = b^2 + c^2 - 2bc\cos(A)$.
• The identity $\cos^2(x) + \sin^2(x) = 1$ is Pythagoras on the unit circle.
• In 3D, distance is a double-Pythagoras: $d = \sqrt{x^2 + y^2 + z^2}$.