One right angle, the longest side opposite it, and the most famous equation in geometry: a² + b² = c².
Point 5 of 6 in this lesson: 15.6.5 Pythagorean triples
Of all triangles, the right triangle — the one with a single square corner — is the one the world is built on. Walls meet floors at right angles; screens, bricks, doorframes, and graph paper are full of them. Its three sides obey the most famous equation in all of mathematics, known for 2500 years and proven hundreds of different ways: square the two short sides, add the results, and you land exactly on the square of the long one. With that one fact you can find a distance you could never reach a tape measure across — the slant of a ramp, the height a ladder climbs, the straight-line gap between two points on a map. This lesson meets the right triangle's parts, proves the theorem by area, runs it forward to find lengths and backward to test triangles, and ends with the whole-number triples and a congruence test made just for right triangles.
A right triangle has exactly one angle equal to 90° — the right angle, drawn with a small square in its corner. The other two angles must be acute (less than 90°), and here is a tidy fact: since all three angles add to 180° and one of them is already 90°, the remaining two share exactly 90° between them. They are complementary — they add to 90° (recall complementary & supplementary angles).
The two sides that form the right angle are the legs. The side opposite the right angle — the one facing it across the triangle — is the hypotenuse, and it is always the longest side. (That makes sense: the biggest angle always faces the biggest side, and 90° is the biggest angle here.)
In a right triangle: one 90° angle; the side opposite it is the hypotenuse (the longest side); the other two sides are the legs; and the two acute angles are complementary (∠A + ∠B = 90°).
Here is the heart of the lesson. Call the two legs a and b, and the hypotenuse c. Build a square on each side, pointing outward. The Pythagorean theorem says the area of the big square (on the hypotenuse) is exactly the sum of the areas of the two smaller squares (on the legs):
a2 + b2 = c2
Read it as areas first, not symbols. A square on a side of length a has area a2. The theorem promises that if you could pour the two leg-squares into the hypotenuse-square, they would fill it to the brim — no gaps, no overflow. With legs 3 and 4, that is 9 + 16 = 25, and 25 is exactly 52 — so the hypotenuse is 5. Drag the slider below and watch the two small areas always add up to the big one.
Take four copies of the right triangle and pack them into a big square of side (a + b) two different ways. One way leaves a tilted square of side c in the middle; the other leaves two squares, of sides a and b. Same big square, same four triangles removed — so what's left over must be equal: c2 = a2 + b2. That is one of the hundreds of known proofs.
The theorem is a machine for finding a side you can't measure. Rearrange it for whichever side is unknown.
To find the hypotenuse from the two legs, undo the square with a square root:
c = √(a2 + b2)
To find a missing leg from the hypotenuse and the other leg, subtract before you take the root (the hypotenuse-square is the big one, so the leg-square is what's left):
b = √(c2 − a2)
Legs 3 and 4: c = √(32 + 42) = √(9 + 16) = √25 = 5.
Legs 6 and 8: c = √(62 + 82) = √(36 + 64) = √100 = 10.
Hypotenuse 13, one leg 5: b = √(132 − 52) = √(169 − 25) = √144 = 12.
Not every answer is whole: legs 2 and 3 → c = √(4 + 9) = √13 ≈ 3.61. Leaving it as the surd √13 is exact; the decimal is a handy estimate.
To find a leg, you subtract the squares (c2 − a2), never add. Adding gives a number bigger than the hypotenuse — impossible, since a leg is always shorter than the hypotenuse. Always identify the hypotenuse (opposite the right angle) first.
Read the theorem backward and it becomes a test. The converse of the Pythagorean theorem says: if the three sides of a triangle satisfy a2 + b2 = c2 (with c the longest side), then the triangle has a right angle — sitting opposite that longest side. So you can decide whether a corner is square using only a tape measure.
Check the longest side's square against the sum of the other two:
3, 4, 5: 32 + 42 = 9 + 16 = 25 = 52 ✓ → a right triangle.
4, 5, 6: 42 + 52 = 16 + 25 = 41, but 62 = 36, and 41 ≠ 36 → not right (here 41 > 36, so it is acute).
A Pythagorean triple is a set of three whole numbers that fit a2 + b2 = c2 exactly. Memorize a handful and many problems become instant — you'll spot the answer without reaching for a calculator.
| a | b | c | 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 |
Every multiple of a triple is also a triple, because scaling all three sides keeps the equation balanced. From 3-4-5 you instantly get 6-8-10, 9-12-15, 15-20-25, and so on. So if a triangle has sides 9, 12, 15, you can recognize "three times 3-4-5" and know it's right without squaring a thing.
The four families worth knowing by heart: 3-4-5, 5-12-13, 8-15-17, 7-24-25 — plus all their multiples.
Back in congruent triangles we warned that SSA — two sides and a non-included angle — is not a general congruence test, because the third side can swing two ways. But there is one safe special case, and it's tailor-made for right triangles.
The HL test (Hypotenuse–Leg) says: two right triangles are congruent if their hypotenuses are equal and one pair of legs is equal. Why is it safe when ordinary SSA is not? Because once you fix the hypotenuse c and one leg a, the other leg is forced: b = √(c2 − a2) has only one answer. So HL is really SSS in disguise — all three sides are pinned down — and the "non-included angle" it relies on is the 90° angle, which never swings.
HL works only for right triangles. Without the right angle, "two sides and a non-included angle" is the SSA trap and proves nothing. The 90° corner is exactly what rescues it.
A right triangle has one 90° angle; the side opposite it is the hypotenuse (longest), the other two are the legs, and the two acute angles are complementary.
The Pythagorean theorem: a2 + b2 = c2 — the squares on the two legs fill the square on the hypotenuse. Use it forward to find a side: c = √(a2+b2), or a leg b = √(c2−a2).
The converse tests a triangle: a2+b2 = c2 ⇒ right; > ⇒ acute; < ⇒ obtuse. Whole-number fits are Pythagorean triples (3-4-5, 5-12-13, 8-15-17, 7-24-25, and multiples). For right triangles, HL (hypotenuse + one leg) proves congruence — safe SSA, because the third side is forced.
A right triangle has legs 3 and 4. Find the hypotenuse.
c = √(32 + 42) = √(9 + 16) = √25 = 5.
A right triangle has legs 6 and 8. Find the hypotenuse.
c = √(62 + 82) = √(36 + 64) = √100 = 10. (It's 2 × the 3-4-5 triple.)
A right triangle has hypotenuse 13 and one leg 5. Find the other leg.
A leg, so subtract: b = √(132 − 52) = √(169 − 25) = √144 = 12. (The 5-12-13 triple.)
Is a triangle with sides 5, 6, 7 a right triangle? Explain.
No. By the converse of the Pythagorean theorem, compare a2+b2 with the longest side's square c2. The longest side is 7: 52 + 62 = 25 + 36 = 61, but 72 = 49, and 61 ≠ 49. Since 61 > 49, it is in fact acute — not right.
Is a triangle with sides 9, 12, 15 a right triangle? Explain.
Yes. By the converse of the Pythagorean theorem, compare a2+b2 with the longest side's square c2: 92 + 122 = 81 + 144 = 225 = 152 ✓. (It's 3 × the 3-4-5 triple, so you could spot it instantly.)
A ladder 13 ft long leans against a wall with its foot 5 ft from the base. How high up the wall does it reach?
The ladder is the hypotenuse; the ground distance is one leg; the wall height is the other leg. height = √(132 − 52) = √(169 − 25) = √144 = 12 ft.
Six questions to lock it in. Tap the answer you think is right.
The big idea of this lesson is that a single right angle locks the three sides into a fixed relationship — a2 + b2 = c2 — and that this relationship runs both ways: forward it computes an unmeasurable distance, backward it certifies a right angle. Anchor it in area (the squares-on-the-sides picture) before it becomes a formula, so students see why the squares appear at all.
The misconception to watch for is applying a2 + b2 = c2 to a non-right triangle, or mislabeling which side is the hypotenuse. Insist on a habit: find the right angle, name the side opposite it as c, and only then write the equation. A close cousin is over-trusting SSA as a congruence test; remind students that the special HL case is safe only because the 90° angle forces the third side — it is SSS in disguise, not a license to use SSA in general.
This lesson supports CCSS 8.G.B.6 (explain a proof of the theorem and its converse), 8.G.B.7 (apply it to find unknown side lengths), 8.G.B.8 (apply it to find the distance between two points), HS G-SRT.C.8 (use it to solve right-triangle problems), and HS G-CO.B.8 (HL congruence explained via rigid motions / SSS).