Chapter 8.4, Problem 32WE

To determine

To find: As many lengths as possible

Answer to Problem 32WE

  $\begin{array}{l}FG=22\\ FD=22\sqrt{2}\\ ED=11\sqrt{6}\\ FE=11\sqrt{2}\end{array}$

Explanation of Solution

Given:

The figure:

Concept Used:

Theorem 8-6: in a $45°-45°-90°$ triangle, the hypotenuse is $\sqrt{2}$ times as long as a leg.

Theorem 8-7: In a $30°-60°-90°$ triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is $\sqrt{3}$ times as long as the shorter leg.

Consider the given figure:

It can be observed that the triangle $△DFG$ is a $45°-45°-90°$ .

Her $\overline{FD}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{FG}$ and $\overline{GD}$ are the legs.

Now, a $45°-45°-90°$ is an isosceles triangle and thus both its legs are congruent. So,

  $FG=GD=22$

Now, by theorem 8-6, the hypotenuse is $\sqrt{2}$ times as long as a leg, so

  $\begin{array}{c}FD=\sqrt{2}\left(GD\right)\\ =\sqrt{2}\left(22\right)\\ =22\sqrt{2}\end{array}$

Also the triangle $△DFE$ is a $30°-60°-90°$ . Here $\overline{FD}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{FE}$ is the shorter leg and $\overline{ED}$ are the longer leg.

Now, by theorem 8-7, the hypotenuse is twice as long as the shorter leg, so

  $\begin{array}{c}FD=2FE\\ 22\sqrt{2}=2FE\\ \frac{22\sqrt{2}}{2}=FE\\ 11\sqrt{2}=FE\end{array}$

Also, by theorem 8-7, the longer leg is $\sqrt{3}$ times as long as the shorter leg.

Thus,

  $\begin{array}{c}ED=\sqrt{3}FE\\ =\sqrt{3}\left(11\sqrt{2}\right)\\ =11\sqrt{6}\end{array}$