Question

DEF, DF = 16 and m angle F = 45

Answer 1

Answer:  $8\sqrt{2}$

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Work Shown:

x = length of each leg

Use the pythagorean theorem to get...

$a^2+b^2 = c^2\\\\x^2+x^2 = 16^2\\\\2x^2 = 256\\\\x^2 = 256/2\\\\x^2 = 128\\\\x = \sqrt{128}\\\\x = \sqrt{64*2}\\\\x = \sqrt{64}*\sqrt{2}\\\\x = 8\sqrt{2}$

Or as an alternative path, you can divide the hypotenuse by $\sqrt{2}$ and then rationalize the denominator.

Both of these methods apply to 45-45-90 triangles only.

Answer 2

Answer:

$8\sqrt{2}$

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($a^{2} + b^{2} = c^{2}$) and plug in 16 as c, and DE and EF as a and b:

$DE^{2} + EF^{2} = 16^{2}$

Since DE is the same as EF, we can substitute it in:

$DE^{2} + DE^{2} = 256\\2DE^{2} = 256\\DE^{2} = 128\\DE = \sqrt{128}$

The square root of 128 is the same as $8\sqrt{2}$ since 128 can be factored into 64 x 2. So $8\sqrt{2}$ 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 $DE \times \sqrt{2} = 16$ as this applies to all isosceles right triangles. This simplifies to:

$DE = \frac{16}{\sqrt{2}} \\DE = \frac{16}{\sqrt{2} } \times \frac{\sqrt{2}}{\sqrt{2} } \\DE = \frac{16\sqrt{2} }{2} \\DE = 8\sqrt{2}$