Answer
$DE=6\sqrt 3 \approx 10.39$
$DF=6\sqrt 3\times \sqrt 3 = 18$
Work Step by Step
By the given Figure
$ m\angle E = 2 .m\angle F$
So $ m\angle E = 60^{\circ} $ and
$ m\angle F =30^{\circ}$
$EF=12\sqrt 3$ given
Now by Theorem 5.5.2 / (30-60-90 Theorem)
$DE=6\sqrt 3 \approx 10.39$
$DF=6\sqrt 3\times \sqrt 3 = 18$
