Chapter 5.7, Problem 26PSC

To determine

Tofind: the descriptive name for the quadrilaterals $SQUA$ and $Q\text{'}QAA\text{'}$

Answer to Problem 26PSC

  $SQUA$ is a square and $Q\text{'}QAA\text{'}$ is an isosceles trapezoid

Explanation of Solution

Given information:

  $Q\text{'}$ is the reflection of $Q$ over the $y-axis$ and $A\text{'}$ is the reflection of $A$ over the $x-axis$ .

Formula used:

Distance between $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\text{ units}$

Slope of the line joining $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$

Calculation:

We have $Q\left(3,10\right),U\left(10,10\right),\text{ A}\left(10,3\right),\text{ S}\left(3,3\right)$ .

  $\begin{array}{l}QU=\sqrt{{\left(10-3\right)}^2+{\left(10-10\right)}^2}=\sqrt{7^2+0}=7\text{ units}\\ UA=\sqrt{{\left(10-10\right)}^2+{\left(3-10\right)}^2}=\sqrt{0+7^2}=7\text{ units}\\ AS=\sqrt{{\left(3-10\right)}^2+{\left(3-3\right)}^2}=\sqrt{7^2+0}=7\text{ units}\\ SQ=\sqrt{{\left(3-3\right)}^2+{\left(10-3\right)}^2}=\sqrt{0+7^2}=7\text{ units}\end{array}$

Since lengths of all the sides are equal, $SQUA$ is a square.

Now,

  

  $Q\text{'}$ is the reflection of $Q$ over the $y-axis$ implies that $Q\text{'}=\left(-3,10\right)$

  $A\text{'}$ is the reflection of $A$ over the $x-axis$ implies that $A\text{'}=\left(10,-3\right)$

  $Q\text{'}Q=\sqrt{{\left(3+3\right)}^2+\left(10-10\right)}=\sqrt{6^2+0}=6\text{ units}$

  $A\text{'}A=\sqrt{{\left(10-10\right)}^2+\left(3+3\right)}=\sqrt{0+6^2}=6\text{ units}$

  $Q\text{'}Q=A\text{'}A=6\text{ units}\dots \left(i\right)$

Slope of $\overline{QA}$ = $\frac{3-10}{10-3}=\frac{-7}{7}=-1$

Slope of $Q\text{'}A\text{'}$ = $\frac{-3-10}{10+3}=\frac{-13}{13}=-1$

Therefore, $\text{Slope of } \overline{QA}=\text{Slope of } \overline{Q\text{'}A\text{'}}$ implies that $\overline{QA}\parallel \overline{Q\text{'}A\text{'}}\dots \left(ii\right)$

From $\left(i\right)\text{ and }\left(ii\right)$ ,

One pair of opposite sides is parallel and another pairs are equal. This implies $Q\text{'}QAA\text{'}$ is an isosceles trapezoid.