Question

consider the reflection of parallelogram pqrs across the line of reflection, $overleftrightarrow{wy}$. if $rr = 14$, then $rz=square$. if $sx = 5$, then $square = 5$. options: $ss$, $xs$, $xz$

Explanation:

Step1: Recall reflection property

In a reflection, the line of reflection is the perpendicular - bisector of the segment joining a point and its image. So, if a point \(R\) is reflected across the line \(WY\) to \(R'\), then the line \(WY\) bisects the segment \(RR'\).

Step2: Calculate \(RZ\)

Since \(WY\) bisects \(RR'\), we have \(RZ=\frac{RR'}{2}\). Given \(RR' = 14\), then \(RZ=\frac{14}{2}=7\).

Step3: Recall reflection property for the second part

In a reflection, the distance from a point to the line of reflection is equal to the distance from its image to the line of reflection. Point \(S\) is reflected across the line \(WY\) to \(S'\), and the distance from \(S\) to the line of reflection (measured as \(SX\)) is equal to the distance from \(S'\) to the line of reflection (measured as \(XS'\)). So if \(SX = 5\), then \(XS'=5\).

Answer:

If \(RR' = 14\), then \(RZ = 7\).
If \(SX = 5\), then \(XS'=5\).