In a right triangle, the two sides that form the right angle are the legs, and the side across from the right angle is the hypotenuse, always the longest side. Their lengths are tied together by a rule that uses squares and square roots.
Why it works
Take four copies of a right triangle with legs $a$ and $b$ and hypotenuse $c$, and place them inside a big square with sides $a + b$. Arranged one way, they leave a tilted square of area $c^{2}$ uncovered in the middle. Slide them into two rectangles instead, and the uncovered part becomes two squares, of areas $a^{2}$ and $b^{2}$. The big square and the four triangles never changed, so the uncovered areas must be equal: $a^{2} + b^{2} = c^{2}$.
Key idea
Pythagorean theorem: in a right triangle with legs $a$ and $b$ and hypotenuse $c$, $$a^{2} + b^{2} = c^{2}$$
Worked examples
Worked example: Find the hypotenuse
A right triangle has legs 6 and 8. Find the hypotenuse.
- $c^{2} = 6^{2} + 8^{2} = 36 + 64 = 100$.
- $c = \sqrt{100} = 10$.
Answer: $10$
Worked example: Find a leg
A right triangle has hypotenuse 13 and one leg 5. Find the other leg.
- $5^{2} + b^{2} = 13^{2}$, so $25 + b^{2} = 169$.
- $b^{2} = 144$, so $b = 12$.
Answer: $12$
Worked example: An irrational length
A right triangle has legs 2 and 3. Find the hypotenuse.
- $c^{2} = 4 + 9 = 13$.
- 13 is not a perfect square, so $c = \sqrt{13}$, about 3.6.
Answer: $\sqrt{13} \approx 3.6$
Common mistake
Two slips. Using a leg as $c$: $c$ is always the side across from the right angle, the longest side, so when you know the hypotenuse, you subtract to find a leg. And adding before squaring: for legs 6 and 8, $(6 + 8)^{2} = 196$ is not $6^{2} + 8^{2} = 100$. Square each leg first.