Question

in the diagram at the right, qrst is a rectangle with rs = 2ts.
a. copy the diagram. then sketch ( r_{overleftrightarrow{qt}} (qrst) ).
b. what figure results from the reflection? use properties of reflections
to justify your solution.

a. which diagram below shows ( r_{overleftrightarrow{qt}} (qrst) )?
a.
( \begin{array}{c} \text{r} \\ \text{s} end{array} ) (quad) ( \begin{array}{c} \text{q} \\ \text{} end{array} ) (quad) ( \begin{array}{c} \text{r} \\ \text{s} end{array} )

b.
( \begin{array}{c} \text{r} \\ \text{s} end{array} ) (quad) ( \begin{array}{c} \text{q} \\ \text{t} end{array} ) (quad) ( \begin{array}{c} \text{r} \\ \text{s} end{array} )

c.
( \begin{array}{c} \text{t} \\ \text{s} end{array} ) (quad) ( \begin{array}{c} \text{q} \\ \text{r} end{array} ) (quad) ( \begin{array}{c} \text{r} \\ \text{s} end{array} )

b. what figure results from the reflection?
the figure resulting from the reflection ( r_{overleftrightarrow{qt}} (qrst) ) is a square.
justify the solution to the previous step. select the correct choice below and, if necessary, fill in the answer boxes
within your choice.
a. since reflections preserve distance, ( r_{overleftrightarrow{qt}} (square) = \text{ts} ) and ( r_{overleftrightarrow{qt}} (square) = \text{qs} ). thus, there are two pair of line
segments with the same length.
b. since ( r_{overleftrightarrow{qt}} (t) = square ), ( r_{overleftrightarrow{qt}} (square) = \text{s} ), and reflections preserve distance, ( r_{overleftrightarrow{qt}} (overleftrightarrow{ts}) = overleftrightarrow{ts} ). so, ( \text{ts} = \text{ts} ) and
( \text{ss} = 2\text{ts} ). using ( \text{rs} = 2\text{ts} ), ( square = \text{ss} ). therefore, the length equals the width in the new figure.
c. since reflections preserve distance, ( r_{overleftrightarrow{qt}} (square) = overleftrightarrow{ts} ). there is no useful relationship between ( square ) and ( \text{sq} )
that can be used. thus, only 2 sides have the same length.

Explanation:

Step 1: Analyze Reflection Properties

Reflection \( R_{\overline{QT}} \) maps points over line \( \overline{QT} \). For rectangle \( QRST \) with \( RS = 2TS \), we use reflection's distance - preserving property (\( R_{\overline{QT}}(\overline{TS})=\overline{TS'} \), so \( TS = TS' \)) and the given \( RS = 2TS \). Also, \( SS'=TS + TS'=2TS \), so \( RS = SS' \).

Step 2: Evaluate Option B

• First, \( R_{\overline{QT}}(T)=Q \) (since reflection over \( \overline{QT} \) maps \( T \) to \( Q \) as \( \overline{QT} \) is a vertical line in the diagram).
• Second, \( R_{\overline{QT}}(S)=S' \) (reflection of \( S \) over \( \overline{QT} \) is \( S' \)).
• By reflection preserving distance, \( R_{\overline{QT}}(\overline{TS})=\overline{TS'} \), so \( TS = TS' \). Then \( SS'=TS + TS' = 2TS \). Given \( RS = 2TS \), we have \( RS=SS' \). Since \( QRST \) is a rectangle, \( QR = TS \) and \( RS \) is a side. After reflection, the new figure has sides where length (e.g., \( RS \)) equals width (e.g., \( SS' \)) and all angles are right angles (since reflection preserves angle measure, and original figure is a rectangle with right angles), so it's a square. Option A's reasoning is not specific to showing a square, and Option C is incorrect as there is a useful relationship.

Answer:

b. The figure resulting from the reflection \( R_{\overline{QT}}(QRST) \) is a square.
For the justification, the correct option is B. Filling in the boxes: \( R_{\overline{QT}}(T)=\boldsymbol{Q} \), \( R_{\overline{QT}}(\boldsymbol{S}) = S' \), and \( \boldsymbol{RS}=SS' \).