#### Answer

$DF = 6\sqrt 3 \approx 10.39$
$DE= 6$

#### Work Step by Step

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)
$DF = 6\sqrt 3 \approx 10.39$
$DE= 6$

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$DF = 6\sqrt 3 \approx 10.39$
$DE= 6$

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