Chapter 6.4, Problem 1BCYP

To determine

To calculate: the measure of angle $\angle SQR$ .

Answer to Problem 1BCYP

  $\angle SQR={26}^{\circ }$

Explanation of Solution

Given: Measure of $\angle PRS$ is ${64}^{\circ }$

in the below figure:

  

Formula Used:

In a parallelogram, opposite sides are equal

Two figures are congruent when the figures have same shape and size

Calculation:

Figure is given as follows:

  

Consider triangle $\Delta PRS$

Here, $\angle PSR={90}^{\circ }$ , as $▭PQSR$ is a rectangle

Thus, $\angle SPR$ is calculated as follows:

  $\angle SPR+\angle PSR+\angle PRS={180}^{\circ }$

On solving the above equation,

  $\begin{array}{l}\angle SPR+{90}^{\circ }+{64}^{\circ }={180}^{\circ }\\ \angle SPR={180}^{\circ }-\left({90}^{\circ }+{64}^{\circ }\right)\\ \angle SPR={180}^{\circ }-{154}^{\circ }\\ \angle SPR={26}^{\circ }\end{array}$

Now consider triangles $\Delta PRS$ and $\Delta QSR$

  $\angle PSR=\angle QRS={90}^{\circ }$

  $PS=QR$ (opposite side of rectangle are equal)

  $SR=SR$ (common)

Thus,

  $\Delta PRS\cong \Delta QSR$ (By RHS congruency)

Since corresponding sides and angles of congruent triangles are equal

Therefore,

  $\begin{array}{l}\angle SQR=\angle SPR\\ \angle SQR={26}^{\circ }\end{array}$

Conclusion:

Hence, $\angle SQR={26}^{\circ }$