Chapter 8.4, Problem 25WE

To determine

To find: x and y.

Answer to Problem 25WE

  $x=8\sqrt{2}$

  $y=4\sqrt{6}$

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.

Label the given figure as shown below:

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

Her $\overline{AB}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{AC}$ and $\overline{BC}$ are the legs.

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

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

  $\begin{array}{c}AB=\sqrt{2}\left(AC\right)\\ 8=\sqrt{2}\left(AC\right)\\ \frac{8}{\sqrt{2}}=AC\\ 4\sqrt{2}=AC\end{array}$

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

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

  $\begin{array}{c}AD=2AC\\ x=2\left(4\sqrt{2}\right)\\ x=8\sqrt{2}\end{array}$

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

Thus,

  $\begin{array}{c}CD=\sqrt{3}AC\\ y=\sqrt{3}\left(4\sqrt{2}\right)\\ y=4\sqrt{6}\end{array}$