Special right triangles (45-45-90 and 30-60-90): 3 worked examples

A 45-45-90 triangle is an isosceles right triangle where the sides follow a specific ratio. If the legs of the triangle have length $s$, then the hypotenuse has length $s\sqrt{2}$. Given that the legs are of length 6, we set $s = 6$. Substituting this into the formula, the length of the hypotenuse is $6\sqrt{2}$.

In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, the sides follow a specific ratio based on the shorter leg $s$. The side opposite the $30^{\circ}$ angle is the shorter leg, which has length $s$. The side opposite the $60^{\circ}$ angle is the longer leg, which has length $s\sqrt{3}$. The side opposite the $90^{\circ}$ angle is the hypotenuse, which has length $2s$.

Given that the shorter leg $s = 5$:

1. The longer leg is $s\sqrt{3} = 5\sqrt{3}$.
2. The hypotenuse is $2s = 2 \times 5 = 10$.

In a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, the relationship between the legs ($s$) and the hypotenuse ($h$) is given by the formula $h = s\sqrt{2}$. We are given that the hypotenuse $h = 10$. To find the length of each leg $s$, we set up the equation $10 = s\sqrt{2}$. Solving for $s$, we divide both sides by $\sqrt{2}$, which gives $s = \frac{10}{\sqrt{2}}$. To rationalize the denominator, we multiply the numerator and the denominator by $\sqrt{2}$, resulting in $s = \frac{10\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$.