Chapter 8.2, Problem 26WE

To determine

To find: x

Answer to Problem 26WE

  $x=8\sqrt{2}$

Explanation of Solution

Given:

The figure:

Concept Used:

Theorem 5-1: Parallelogram’s opposite sides are equal.

Postulate 10: Corresponding angles are equal.

Theorem 8-2: In right triangle, square of hypogenous is equal to the square of base and perpendicular.

Label the given figure as shown below:

The quadrilateral $ABCD$ is a trapezium.

So,

  $\overline{AB}\parallel \overline{CD}$

Draw a line $\overline{BE}$ parallel to $\overline{AC}$

Now for the quadrilateral $ABCE$ , both pair of opposite sides are parallel, so $ABCE$ is a parallelogram.

By theorem 5-1, opposite sides of a parallelogram are congruent. So,

  $BE=AC=x$

  $CE=AB=6$

So,

  $\begin{array}{c}ED=CD-CE\\ =10-6\\ =4\end{array}$

Here $\overline{BE}\parallel \overline{AC}$ , and $\overline{CD}$ is a transversal, so

  $\angle BED\cong \angle ACE$

That is,

  $m\angle BED=m\angle ACE=90$

Now, consider the right triangle $△BED$ .

Now, $\overline{BD}$ is the hypotenuse and has length 12 and the lengths of the legs are $ED=4$ and $BE=x$ .

So, by theorem 8-2,

  $\begin{array}{c}B{D}^2=E{D}^2+B{E}^2\\ {12}^2=4^2+x^2\\ 144=16+x^2\end{array}$

  $\begin{array}{c}144-16=x^2\\ 128=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{128}\\ x=8\sqrt{2}\end{array}$