Question

which rigid transformation would map △mzk to △ qzk? a rotation about point k a reflection across the line containing \\(\overline{mz}\\) a reflection across the line containing \\(\overline{zk}\\) a rotation about point z

Explanation:

Step1: Analyze the diagram

The diagram shows \( \triangle MZK \) and \( \triangle QZK \) with \( Z \) on \( MQ \), \( KZ \perp MQ \), and \( MZ = ZQ \) (implied by the right angle and the structure). So \( KZ \) is the perpendicular bisector of \( MQ \).

Step2: Evaluate each transformation

• Rotation about \( K \): Rotating about \( K \) wouldn't map \( M \) to \( Q \) directly as the symmetry is over \( KZ \), not a rotation center at \( K \).
• Reflection over \( MZ \): Reflecting over \( MZ \) would not map \( M \) to \( Q \) since \( MZ \) is part of the base, not the axis of symmetry here.
• Reflection over \( ZK \): Since \( KZ \) is the perpendicular bisector (and the axis of symmetry), reflecting \( \triangle MZK \) over \( KZ \) would map \( M \) to \( Q \) and \( Z \) to \( Z \), \( K \) to \( K \), thus mapping \( \triangle MZK \) to \( \triangle QZK \).
• Rotation about \( Z \): Rotating about \( Z \) would not align the triangles as the symmetry is reflective over \( KZ \), not a rotation about \( Z \).

Answer:

C. a reflection across the line containing \( \overline{ZK} \)