Question

which rigid transformation would map (\triangle mzk) to (\triangle 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 rotation about K

A rotation about K would change the position of points M and Q in a circular path around K. But in the given triangles \( \triangle MZK \) and \( \triangle QZK \), the relationship is more of a mirror - like across a line.

Step2: Analyze reflection across \( MZ \)

Reflecting across \( MZ \) would not map \( \triangle MZK \) to \( \triangle QZK \) as the orientation with respect to the line \( MZ \) is not correct for a reflection to achieve the mapping.

Step3: Analyze reflection across \( ZK \)

A reflection across the line \( ZK \) (the altitude of the isosceles triangle \( \triangle MKQ \) with \( MZ = QZ \) and \( \angle MZK=\angle QZK = 90^{\circ}\)) will map point \( M \) to \( Q \) (since \( ZK \) is the perpendicular bisector of \( MQ \)) and keep \( Z \) and \( K \) in their positions. So, \( \triangle MZK\) is mapped to \( \triangle QZK\) by reflection across \( ZK \).

Step4: Analyze rotation about Z

A rotation about \( Z \) would rotate the triangle around \( Z \). But the sides \( MZ \) and \( QZ \) are equal and \( \angle MZK=\angle QZK \), a rotation is not the correct transformation as it would change the orientation in a non - mirror like way.

Answer:

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