text: ## Elementary Geometry for College Students (7th Edition)
  
  $DF = 6\sqrt 3 \approx 10.39$ $DE= 6$
  By the given Figure $\angle F= 30^{\circ}$ So remaining $\angle F= 60^{\circ}$ FE= 12 given Now by Theorem 5.5.2 / (30-60-90 Theorem) $DF = 6\sqrt 3 \approx 10.39$ $DE= 6$