Pythagoras' theorem
Introduced| English | Meaning |
|---|---|
| Pythagoras/paɪˈθæɡərəs/ | Pythagoras |
| right-angled triangle/raɪt ˈæŋɡld ˈtraɪæŋɡl/ | A triangle containing one 90° angle. |
| hypotenuse/haɪˈpɒtənjuːs/ | The side opposite the right angle in a right-angled triangle. |
| Pythagorean triples/ˈpaɪθəɡɔːrɪən ˈtrɪplz/ | Pythagorean triples |
The rope that made a right angle
- Ancient Egyptian builders used a rope with 12 equally-spaced knots to create a perfect right angle: stretch it into a triangle with sides 3, 4, and 5 — the corner opposite the longest side is always $90^{\circ}$.
- Pythagoras 勾股定理 proved why this works, and his theorem is still the most-used result in all of mathematics.
The theorem
- In a right-angled triangle 直角三角形, the square on the hypotenuse 斜边 equals the sum of the squares on the other two sides:
- $c$ is the hypotenuse — the longest side, opposite the right angle.
The classic 3-4-5 triangle: $3^2 + 4^2 = 5^2$. The theorem works for every right-angled triangle.
$c$ must be the hypotenuse. The formula $a^2 + b^2 = c^2$ only works when $c$ is the side opposite the right angle. If you label the wrong side as $c$, you'll get the wrong answer.
Pythagoras' theorem
a² + b² = c²
The squares on the two legs always add up to the square on the hypotenuse.
A right-angled triangle has short sides 3 cm and 4 cm. Find the hypotenuse (cm).
Difficulty 2/5 · 140 coins + a small bonus on first completion today
√(3² + 4²) = √25 = 5 cm.
The hypotenuse of a right-angled triangle is:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
The hypotenuse is the longest side and lies opposite the 90° angle.
In Pythagoras theorem, a² + b² = c², the side c must be the ______.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
c must always be the hypotenuse (the side opposite the right angle).
Finding a missing side
- Finding the hypotenuse: $c = \sqrt{a^2 + b^2}$ (add and square-root).
- Finding a short side: $a = \sqrt{c^2 - b^2}$ (subtract and square-root).
Hypotenuse $13$, short side $5$: $b = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$.
Pythagoras as areas: the squares on the two shorter sides ($9+16$) add up to the square on the hypotenuse ($25$)
The hypotenuse is 13 cm and one short side is 5 cm. Find the other short side (cm).
Difficulty 2/5 · 140 coins + a small bonus on first completion today
√(13² − 5²) = √(169 − 25) = √144 = 12 cm.
A right-angled triangle has hypotenuse 10 cm and one short side 6 cm. Find the other short side (cm).
Difficulty 2/5 · 140 coins + a small bonus on first completion today
√(10² − 6²) = √(100 − 36) = √64 = 8 cm.
Pythagorean triples 勾股数 worth knowing
- $3, 4, 5$ — the smallest triple.
- $5, 12, 13$ — the next most common.
- Multiples work too: $6, 8, 10$ and $9, 12, 15$ are also right-angled.
- Spotting a triple saves calculation time in exams.
A triangle with sides 6, 8, 10 is right-angled.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
6² + 8² = 36 + 64 = 100 = 10². It is a multiple of the 3-4-5 triple.
You've got it
- $a^2 + b^2 = c^2$, with $c$ the hypotenuse (opposite the right angle)
- to find a short side, subtract: $b = \sqrt{c^2 - a^2}$
- $3$-$4$-$5$ and $5$-$12$-$13$ are worth remembering