Question

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

Explanation:

Step1: Analyze reflection across line \( ZK \)

Since \( \angle M K Z=\angle Q K Z \), \( KZ = KZ \) (common side), and \( \angle M Z K=\angle Q Z K = 90^{\circ} \). By the Angle - Side - Angle (ASA) congruence criterion, \( \triangle MZK\cong\triangle QZK \). A reflection across the line containing \( \overline{ZK} \) will map \( \triangle MZK \) to \( \triangle QZK \).

Step2: Eliminate other options

• A rotation about point \( Z \): Rotation about \( Z \) would change the orientation in a way that does not map \( \triangle MZK \) to \( \triangle QZK \) as the key congruence is based on the angle - side - angle relation across the line \( ZK \).
• A reflection across the line containing \( \overline{MZ} \): The line \( MZ \) is perpendicular to \( ZK \), and reflecting across \( MZ \) will not map \( \triangle MZK \) to \( \triangle QZK \) because the angles and side - length relations with respect to \( MZ \) do not satisfy the congruence.
• A rotation about point \( K \): Rotation about \( K \) would not preserve the right - angle and the equal - angle relations in the way required for mapping \( \triangle MZK \) to \( \triangle QZK \) as the key is the line of symmetry \( ZK \).

Answer:

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