Pythagorean Theorem

Watch a² + b² = c² proved visually with squares on the sides of a right triangle

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Squares on the three sides

a² = 9b² = 16c² = 25

a² + b² = 9 + 16 = 25

c = √25 = 5

All three sides are whole numbers: a Pythagorean triple.

Proof by rearrangement

abc²a²b²

The big square has side a + b in both pictures and holds the same four triangles. So the area the triangles leave uncovered is the same too: at the start it is c², after rearranging it is a² + b².

What the theorem says

In a right triangle the side opposite the right angle is the hypotenuse, c. The other two sides are the legs, a and b. The Pythagorean theorem says the square built on the hypotenuse has exactly the same area as the two squares built on the legs put together: a² + b² = c².

The converse is also true: if the sides of a triangle satisfy a² + b² = c², the angle opposite c is a right angle. If a² + b² > c² that angle is acute, and if a² + b² < c² it is obtuse.

Why the rearrangement proof works

Take a square of side a + b and place four copies of the triangle inside it, one in each corner. The space left in the middle is a square whose sides are all hypotenuses, so its area is c². Now slide the same four triangles into two a × b rectangles. The space left is now an a × a square and a b × b square.

Nothing changed size: the big square is the same and the four triangles are the same, so the leftover areas must be equal. Hence c² = a² + b². You can also see it with algebra:

Algebra behind the picture

(a + b)² = 4 × (½ab) + c²

a² + 2ab + b² = 2ab + c²

a² + b² = c²

Key formulas

Pythagorean theorem

a² + b² = c²

a, b
the legs (the sides that meet at the right angle)
c
the hypotenuse (the longest side)

Finding a missing side

c = √(a² + b²)

a = √(c² − b²)

Distance formula

d = √((x₂ − x₁)² + (y₂ − y₁)²)

Δx = x₂ − x₁
horizontal leg
Δy = y₂ − y₁
vertical leg

Connection: the distance formula

Any two points on a grid are the ends of the hypotenuse of a right triangle whose legs run along the axes.

02468102468|Δx| = 6|Δy| = 5dP₁P₂

P₁ = (1, 1)

P₂ = (7, 6)

Δx = 7 − 1 = 6

Δy = 6 − 1 = 5

d = √(6² + 5²) = √61 ≈ 7.81

Drag P₁ and P₂. The horizontal and vertical gaps are the legs of a right triangle, so the distance between the points is its hypotenuse: d² = Δx² + Δy².

Common Pythagorean triples

a, b, ca² + b² = c²
3, 4, 59 + 16 = 25
5, 12, 1325 + 144 = 169
8, 15, 1764 + 225 = 289
7, 24, 2549 + 576 = 625
6, 8, 1036 + 64 = 100 (3, 4, 5 doubled)

Making more triples

Any multiple of a triple is a triple. And for whole numbers m > n > 0, the numbers m² − n², 2mn and m² + n² always form one: m = 2, n = 1 gives 3, 4, 5.

How to use this visualizer

  1. 1
    Move the a and b sliders, or pick a preset triple. The squares on all three sides resize and show their areas.
  2. 2
    Check the readout: a² + b² always equals c², even when c is not a whole number.
  3. 3
    Press Rearrange to slide the four triangles. Watch the c² square turn into the a² and b² squares.
  4. 4
    Drag the progress slider to pause the proof halfway and see that every triangle only slides, never stretches.
  5. 5
    In the distance formula grid, drag P₁ and P₂ to see Δx and Δy become the legs of a right triangle.

Real-world applications

Construction

Builders mark 3, 4 and 5 units along two walls and a diagonal to check that a corner is exactly square.

Navigation

Walk 3 km east and 4 km north and you are 5 km from where you started, as the crow flies.

Screen sizes

A TV’s size is its diagonal: a 16 × 9 panel has a diagonal of √(16² + 9²) ≈ 18.4 units.

Distance in any dimension

In 3D the distance is √(Δx² + Δy² + Δz²), by applying the theorem twice.

Practice problems

Problem 1: Ladder against a wall

A 13 m ladder leans against a wall with its foot 5 m from the base of the wall. How high up the wall does it reach?
Show solution

The ladder is the hypotenuse: c = 13, a = 5.

b² = c² − a² = 169 − 25 = 144

b = √144

b = 12 m

Problem 2: Distance between two points

Find the distance between P₁ = (−2, 1) and P₂ = (4, 9).
Show solution

Δx = 4 − (−2) = 6

Δy = 9 − 1 = 8

d = √(6² + 8²) = √(36 + 64) = √100

d = 10

The legs 6 and 8 form the 6, 8, 10 triple.

Discussion

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