Answer:
[tex]8\sqrt{2}[/tex]
Step-by-step explanation:
We can sum the angles in a triangle to 180°:
m<D + m<E + m<F = 180°
m<D + 90° + 45° = 180°
m<D = 45°
Therefore, the triangle is an isosceles right triangle, since the base angles, <F and <D are the same measure. Because the triangle is isosceles, the legs are congruent, meaning that
DE = EF
We can use the Pythagorean Theorem([tex]a^{2} + b^{2} = c^{2}[/tex]) and plug in 16 as c, and DE and EF as a and b:
[tex]DE^{2} + EF^{2} = 16^{2}[/tex]
Since DE is the same as EF, we can substitute it in:
[tex]DE^{2} + DE^{2} = 256\\2DE^{2} = 256\\DE^{2} = 128\\DE = \sqrt{128}[/tex]
The square root of 128 is the same as [tex]8\sqrt{2}[/tex] since 128 can be factored into 64 x 2. So [tex]8\sqrt{2}[/tex] is the answer.
If you are looking for an easier way to do this, once you are able to recognize DEF is an isosceles triangle, you can say [tex]DE \times \sqrt{2} = 16[/tex] as this applies to all isosceles right triangles. This simplifies to:
[tex]DE = \frac{16}{\sqrt{2}} \\DE = \frac{16}{\sqrt{2} } \times \frac{\sqrt{2}}{\sqrt{2} } \\DE = \frac{16\sqrt{2} }{2} \\DE = 8\sqrt{2}[/tex]
