🏰 Geometry Kingdom · Geometry

The Pythagorean Theorem

Relate the three sides of a right triangle with $a 2 + b 2 = c 2$ , and use it to find distances.

In short

$a 2 + b 2 = c 2$ holds in every right triangle, with c the hypotenuse opposite the right angle.
Add the squares to find the hypotenuse; subtract to find a leg.
The converse tests for a right angle: if the squares match, the triangle is right-angled.
The theorem is the source of the distance formula and of every diagonal calculation.

What the theorem says

In a right triangle — and only in a right triangle — the two short sides (legs) and the long side opposite the right angle (hypotenuse) are linked by

$a 2 + b 2 = c 2$

where c is always the hypotenuse.

The statement is about areas, not lengths. Build a square on each side of the triangle; the two smaller squares together have exactly the same area as the big one. $A 3-4-5$ triangle gives $9 + 16 = 25$ .

Because the hypotenuse faces the largest angle, it is always the longest side. That gives a free sanity check: your c must come out bigger than either leg, and smaller than the two legs added together.

Finding a hypotenuse and finding a leg

Hypotenuse missing. Square the two legs, add, take the square root.

legs 5 and $12: c 2 = 25 + 144 = 169$ , so $c = 13$

Leg missing. The hypotenuse is known, so it goes alone on one side and you subtract.

hypotenuse 17, one leg $8: 8 2 + b 2 = 17 2 64 + b 2 = 289 b 2 = 225$ , so $b = 15$

The decision is always the same: identify the hypotenuse first (it is opposite the right angle, and it is the longest). If the hypotenuse is one of the numbers you were given, you subtract; if it is the one you want, you add.

Some triples come out whole and are worth recognising: 3-4-5, 5-12-13, 8-15-17, 7-24-25, and any multiple of them such as 6-8-10 or 9-12-15. Most triangles do not, so the answer is usually a decimal that needs rounding.

The converse, and why it matters

The theorem also runs backwards. If three sides satisfy $a 2 + b 2 = c 2$ with c the longest, then the triangle must be right-angled. This is the converse.

Sides $9, 40, 41: 81 + 1600 = 1681$ and $412 = 1681$ . They match, so the triangle has a right angle.

Sides $6, 7, 10: 36 + 49 = 85$ but $102 = 100$ . They do not match, so there is no right angle.

Builders use exactly this to square up a foundation: measure $3 m$ along one wall, $4 m$ along the other, and adjust until the diagonal is $5 m$ .

Where it turns up

The theorem is really a distance formula in disguise, and it appears whenever something vertical meets something horizontal.

A ladder against a wall: the ladder is the hypotenuse, the wall and the ground are the legs.
The diagonal of a rectangle: the diagonal splits it into two right triangles.
Walking 8 km east then 6 km north: you end up sqrt(64 $+ 36) = 10$ km from the start, even though you walked 14 km.
The distance between two points on a grid: the horizontal and vertical gaps are the legs.

It even works in three dimensions. For the longest rod that fits in a box, first find the diagonal across the base, then use that diagonal and the height as the legs of a second right triangle.

Worked examples

Example 1

A right triangle has legs of 9 cm and 12 cm. Find the hypotenuse.

The hypotenuse is the unknown, so the two squares are added.
Square each leg: $92 = 81$ and $122 = 144$ .
Add them: $c 2 = 81 + 144 = 225$ .
Take the square root: $c = 15$ cm. Check: 15 is longer than 12 but less than $9 + 12 = 21$ .

Example 2

A ladder $10 m$ long leans against a wall with its foot $6 m$ from the wall. How far up does it reach?

The ladder is the slanted side, so the ladder is the hypotenuse: $c = 10$ .
One leg is 6 (along the ground); the height h is the other leg, so subtract.
$62 + h 2 = 10 2$ , that is $36 + h 2 = 100$ .
$h 2 = 64$ , so $h = 8 m$ .

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

A right triangle has legs of 5 cm and 12 cm. How long is the hypotenuse?

Answer: 13 cm

$a 2 + b 2 = c 2$
$52 + 12 2 = c 2$
$25 + 144 = 169$
$c =$ sqrt(169) $= 13$ cm

Problem 2

Difficulty 3 of 5

A right triangle has hypotenuse 14 cm and one leg of 12 cm. How long is the other leg? Round to 1 decimal place.

Answer: 7.2 cm

$a 2 + b 2 = c 2$ with $c = 14$ (the hypotenuse).
$122 + b 2 = 14 2$
$b 2 = 196$ - $144 = 52$
$b =$ sqrt(52) $= 7.2$ cm

Problem 3

Difficulty 4 of 5

Zara walks 5 km east, then turns and walks 18 km north. How far is Zara from the starting point, in a straight line? Round to 1 decimal place.

Answer: 18.7 km

East and north meet at 90°, so the legs are the two parts of the walk.
$c 2 = 5 2 + 18 2 = 349$
$c =$ sqrt(349) $= 18.7$ km

Common mistakes

Adding the legs without squaring them — $3 + 4$ is 7, not 5.
Stopping at $c 2$ and forgetting the square root.
Adding when the hypotenuse was already given, instead of subtracting.
Applying the theorem to a triangle that has no right angle.

What you should be able to do

Find the hypotenuse given two legs.
Find a leg given the hypotenuse and the other leg.
Decide whether a triangle is right from its side lengths.
Apply the theorem to ladders, diagonals and distance problems.

Where this fits in the curriculum

Common Core

8.G.B.6
Grade 8 — Explain a proof of the Pythagorean theorem and its converse.
8.G.B.7
Grade 8 — Apply the Pythagorean theorem to find unknown side lengths in right triangles in two and three dimensions.
8.G.B.8
Grade 8 — Apply the Pythagorean theorem to find the distance between two points in the coordinate plane.

Ontario

G8.E2.4
Grade 8 — Describe the Pythagorean relationship using geometric models, and apply the theorem to find an unknown side length of a right triangle.

SAT

Additional Topics in Math
Right triangles, the Pythagorean theorem and distance.