## Elementary Geometry for College Students (6th Edition)

$EF=12$
$\triangle DEF$ is a 30-60-90 triangle. We are given the length of $FD$, which is $6\sqrt3$. The longest leg of the triangle, or the leg opposite the $60^{\circ}$ angle equals $(short leg)\sqrt3$. Since this side equals $6\sqrt3$, the shortest leg is equal to $6$. So $DE=6$. The hypotenuse is twice the length of the shortest leg. So the hypotenuse is $12$, or $EF=12$.