Chapter 5.1, Problem 9WE

To determine

To find: the values of a, b, x and y.

Answer to Problem 9WE

  $a=8$

  $b=8$

  $x=56°$

  $y=68°$

Explanation of Solution

Given:

The diagram:

Consider the figure shown below:

Here it is given that $PQRS$ is a parallelogram.

So, by theorem 5-1, its opposite sides are congruent.

Here $\overline{PS}$ , $\overline{QR}$ and $\overline{PQ}$ , $\overline{SR}$ are the opposite sides of the parallelogram $PQRS$ . So,

$\overline{PS}\cong \overline{QR}$ And $\overline{PQ}\cong \overline{SR}$

Thus,

  $\begin{array}{c}PQ=QR\\ b=8\end{array}$

Also, by theorem 5-3, $\overline{PR}$ and $\overline{QS}$ bisect each other.

That is, O is the midpoint of $\overline{PR}$ and $\overline{QS}$ .

Thus,

  $\begin{array}{c}PO=OR\\ a=8\end{array}$

Now, both pair of opposite sides of a parallelogram are parallel. So, $\overline{PS}\parallel \overline{QR}$ and $\overline{QS}$ is a transversal, so by theorem 3-2,

  $\angle PSQ\cong \angle SQR$   (Alternate interior angles)

That is,

  $\begin{array}{c}m\angle SQR=m\angle PSQ\\ m\angle SQR=56\end{array}$

Now consider the $△QOR$ , here the sides $\overline{OR}$ and $\overline{QR}$ are congruent.

So by theorem 4-1, the angles $\angle OQR$ and $\angle ROQ$ are congruent. That is,

  $\begin{array}{c}m\angle ROQ=m\angle OQR\\ x=56\end{array}$

Now apply theorem 3-11 on the triangle $△QOR$ ,

  $\begin{array}{c}m\angle QRO+m\angle ROQ+m\angle OQR=180\\ y+56+56=180\\ y+112=180\end{array}$

  $\begin{array}{c}y=180-112\\ y=68\end{array}$