## Geometry: Common Core (15th Edition)

$x = \sqrt 2$ $y = 2$
In a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, we know that the legs are equal to one another. Since one leg measures $\sqrt 2$, $x$ also equals $\sqrt 2$. The hypotenuse is $\sqrt 2$ times each leg. Let's write an equation to solve for $y$, the length of the hypotenuse: $y = \sqrt 2(\sqrt 2)$ Multiply the radicals: $y = \sqrt 4$ Take the positive square root to solve for $y$: $y = 2$