Chapter 8.4, Problem 26WE

To determine

To find: x and y.

Answer to Problem 26WE

  $x=10$

  $y=4\sqrt{2}$

Explanation of Solution

Given:

The figure:

Concept Used:

Theorem 3-11: The sum of measures of all angles of a triangle is 180.

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

Label the given figure as shown below:

And from B draw a line perpendicular to $\overline{AD}$ as shown above.

It can be observed that the quadrilateral $BCDE$ is a rectangle.

And it is known that opposite sides of a rectangle are congruent.

So,

  $BE=CD=4$

And

  $ED=BC=6$

Apply theorem 3-11 on the triangle $△ABE$ ,

  $\begin{array}{c}m\angle ABE+m\angle BEA+m\angle EAB=180\\ m\angle ABE+90+45=180\\ m\angle ABE+135=180\end{array}$

  $\begin{array}{c}m\angle ABE=180-135\\ m\angle ABE=45\end{array}$

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

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

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

So,

  $AE=BE=4$

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

  $\begin{array}{c}AB=\sqrt{2}\left(BE\right)\\ y=\sqrt{2}\left(4\right)\\ y=4\sqrt{2}\end{array}$

Lastly,

  $\begin{array}{c}AD=AE+ED\\ x=4+6\\ x=10\end{array}$