Chapter 5.1, Problem 10WE

To determine

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

Answer to Problem 10WE

  $a=10$

  $b=4$

  $x=90°$

  $y=45°$

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, $\overline{PQ}\cong \overline{SR}$ and $\overline{PS}\cong \overline{QR}$

Thus,

  $\begin{array}{c}PQ=SR\\ PQ=SO+SR\\ a=6+4\\ a=10\end{array}$

And,

$PS=QR$ ….. (1)

It is known that, both pair of opposite sides of a parallelogram are parallel. So, $\overline{PS}\parallel \overline{QR}$ and $\overline{SR}$ is a transversal.

Now, it can be observed from the given figure that the transversal $\overline{SR}$ is perpendicular to the line $\overline{PS}$ , then by theorem 3-4, $\overline{SR}$ is perpendicular to the line $\overline{QR}$ . That is,

  $\begin{array}{c}m\angle SRQ=90\\ x=90\end{array}$

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

  $\begin{array}{c}m\angle QRO+m\angle ROQ+m\angle OQR=180\\ 90+45+y=180\\ 135+y=180\end{array}$

  $\begin{array}{c}y=180-135\\ y=45\end{array}$

Now consider the $△QOR$ , here the angles $\angle OQR$ and $\angle ROQ$ are congruent.

So by theorem 4-1, the sides $\overline{OR}$ and $\overline{QR}$ are congruent. That is,

  $\begin{array}{c}QR=OR\\ QR=4\end{array}$

Now from equation (1).

  $\begin{array}{c}PS=QR\\ b=4\end{array}$