In triangle 45, 45, 90 DEF triangle, angle D is congruent to angle F. The perimeter of triangle DEF is 35 mm, and DF = 14mm, how do you determine DE?

A 45, 45, 90 triangle is isosceles by definition, so its two legs have equal length. Because it is given that $D \cong F$, side DF must be the hypotenuse. Side DF must also be $\sqrt{2}$ times the legs; see the picture below.
Side DE could be expressed easily as $\frac{14}{\sqrt{2}}$, or properly, $7 \sqrt{2}$, which is approximately the decimal value 9.89949493661.
However, the triangle's perimeter in this case wouldn't be 35. That's where the problem is wrong. When working with radicals your answer will hardly ever be rational. If you use this method, the triangle's perimeter is $14 + 14 \sqrt{2}$. If you use the other method, the two legs would have to be 10.5, making this an acute triangle.