Question

which rigid transformation would map △mzk to △qzk? a rotation about point k a reflection across the line containing mz a reflection across the line containing zk a rotation about point z

Explanation:

Step1: Analyze rotation about point K

If we rotate about point K, the position of \( \triangle MZK \) won't map to \( \triangle QZK \) as the orientation and position relative to K won't match directly for a simple rotation to achieve the mapping.

Step2: Analyze reflection across line \( \overline{MZ} \)

Reflecting across \( \overline{MZ} \) will not map \( \triangle MZK \) to \( \triangle QZK \) because the line \( \overline{MZ} \) is not the axis of symmetry for the two triangles.

Step3: Analyze reflection across line \( \overline{ZK} \)

Reflecting across \( \overline{ZK} \) will not map \( \triangle MZK \) to \( \triangle QZK \) as \( \overline{ZK} \) is a common side but not the axis of symmetry for the required mapping.

Step4: Analyze rotation about point Z

Since \( Z \) is a common vertex, and if we rotate \( \triangle MZK \) about point \( Z \), we can map \( M \) to \( Q \) (as \( ZM = ZQ \) and \( \angle MZK=\angle QZK \)) and keep \( Z \) and \( K \) in place.

Answer:

a rotation about point \( Z \)