Question

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

Explanation:

A reflection across the line containing $\overline{ZK}$ will map $\triangle MZK$ to $\triangle QZK$ because the line $\overline{ZK}$ is the perpendicular - bisector of $\overline{MQ}$. Points $M$ and $Q$ are equidistant from the line $\overline{ZK}$, and $\angle MZK=\angle QZK = 90^{\circ}$, and $ZK$ is common to both triangles. Rotations about points $K$ or $Z$ will not map the triangles onto each other in the required way, and a reflection across $\overline{MZ}$ is incorrect as well.

Answer:

a reflection across the line containing $\overline{ZK}$