Two right triangles show up so often that their side ratios are worth knowing by heart. Both come straight from the Pythagorean theorem of Proving the Pythagorean Theorem with Similarity, and their answers use the simplified square roots from Simplifying Square Roots.
Por qué funciona
45-45-90: cut a square with side $s$ along a diagonal. Each half is an isosceles right triangle with legs $s$, so its hypotenuse is $\sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt{2}$.
30-60-90: cut an equilateral triangle with side $2s$ by its altitude. Each half has angles $30^\circ$, $60^\circ$ and $90^\circ$, hypotenuse $2s$, and shorter leg $s$ (half of the base). The longer leg is $\sqrt{(2s)^2 - s^2} = \sqrt{3s^2} = s\sqrt{3}$.
30-60-90: cut an equilateral triangle with side $2s$ by its altitude. Each half has angles $30^\circ$, $60^\circ$ and $90^\circ$, hypotenuse $2s$, and shorter leg $s$ (half of the base). The longer leg is $\sqrt{(2s)^2 - s^2} = \sqrt{3s^2} = s\sqrt{3}$.
Idea clave
45-45-90: sides $s : s : s\sqrt{2}$. 30-60-90: sides $s : s\sqrt{3} : 2s$, with $s$ opposite $30^\circ$, $s\sqrt{3}$ opposite $60^\circ$ and $2s$ (the hypotenuse) opposite $90^\circ$.
Ejemplo resuelto: A 45-45-90 triangle
(a) The legs are $7$. Find the hypotenuse. (b) The hypotenuse is $10$. Find a leg.
- (a) Hypotenuse $=$ leg $\cdot \sqrt{2} = 7\sqrt{2}$.
- (b) Leg $=$ hypotenuse $\div \sqrt{2} = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$.
Respuesta: (a) $7\sqrt{2}$ (b) $5\sqrt{2}$
Ejemplo resuelto: From the hypotenuse of a 30-60-90 triangle
The hypotenuse is $14$. Find both legs.
- Shorter leg (opposite $30^\circ$): half the hypotenuse, $7$.
- Longer leg (opposite $60^\circ$): $7\sqrt{3}$.
Respuesta: $7$ and $7\sqrt{3}$
Ejemplo resuelto: From the longer leg
The longer leg of a 30-60-90 triangle is $9$. Find the shorter leg and the hypotenuse.
- Longer leg $= s\sqrt{3}$, so $s = \frac{9}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}$.
- Hypotenuse $= 2s = 6\sqrt{3}$.
Respuesta: Shorter leg $3\sqrt{3}$, hypotenuse $6\sqrt{3}$
Error común
Putting $\sqrt{3}$ on the hypotenuse, or on the leg next to the $60^\circ$ angle. Match sizes: the longest side ($2s$) faces the biggest angle ($90^\circ$), and the shortest side ($s$) faces the smallest angle ($30^\circ$). Since $\sqrt{3} \approx 1.73$ is between 1 and 2, $s\sqrt{3}$ is the middle side, across from $60^\circ$.