Question

1. $overline{rt}$ perpendicularly bisects $overline{su}$. find the indicated measures.

$ru = $
$st = $
$st = $

Explanation:

Step1: Analyze RU

Since \( \overline{RT} \) perpendicularly bisects \( \overline{SU} \), \( V \) is the midpoint of \( SU \), so \( SV = VU = 5 \). Also, \( \triangle RVU \) and \( \triangle RV S \) are right triangles. Given \( RS = 8 \), \( VU = 5 \), but wait, actually, \( RU \) can be found? Wait, no, looking at the diagram, \( TU = 23 \), and since \( RT \) is the perpendicular bisector, \( RU = TU \)? Wait, no, \( RT \) bisects \( SU \) at \( V \), and is perpendicular, so \( \triangle RVS \cong \triangle RVU \) (SAS: \( RV \) common, \( SV = VU \), right angles), so \( RU = RS = 8 \)? Wait, no, maybe I misread. Wait, the diagram has \( RS = 8 \), \( TU = 23 \), and \( VU = 5 \). Wait, actually, since \( RT \) is the perpendicular bisector of \( SU \), then any point on \( RT \) is equidistant from \( S \) and \( U \). So \( RU = RS = 8 \)? Wait, no, \( TU = 23 \), so \( ST = TU = 23 \)? Wait, let's correct:

1. For \( RU \): Since \( RT \) perpendicularly bisects \( SU \), \( \triangle RVS \) and \( \triangle RVU \) are congruent (right triangles, \( SV = VU \), \( RV \) common). So \( RU = RS = 8 \)? Wait, no, \( RS \) is 8, so \( RU = 8 \)? Wait, maybe.

1. For \( ST \): Since \( RT \) is the perpendicular bisector of \( SU \), \( T \) is on \( RT \), so \( ST = TU \). Given \( TU = 23 \), so \( ST = 23 \).

Wait, let's re-express:

• Perpendicular bisector theorem: Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. So \( T \) is on \( RT \) (the perpendicular bisector of \( SU \)), so \( ST = TU \). Since \( TU = 23 \), \( ST = 23 \).

• For \( RU \): Similarly, \( R \) is on \( RT \), so \( RU = RS \). \( RS = 8 \), so \( RU = 8 \)? Wait, no, \( RS \) is 8, so \( RU = 8 \).

Wait, maybe the diagram has \( RS = 8 \), \( TU = 23 \), \( VU = 5 \). So:

• \( RU \): Since \( RT \perp SU \) and \( SV = VU \), \( \triangle RSU \) is isoceles with \( RS = RU \)? Wait, no, \( RS = 8 \), so \( RU = 8 \).

• \( ST \): Since \( T \) is on the perpendicular bisector, \( ST = TU = 23 \).

Step2: Confirm

1. \( RU \): By perpendicular bisector, \( RU = RS = 8 \) (since \( R \) is on \( RT \), the bisector, so equidistant from \( S \) and \( U \)).

1. \( ST \): \( T \) is on \( RT \), so \( ST = TU = 23 \).

Answer:

\( RU = 8 \), \( ST = 23 \) (assuming the repeated \( ST \) is a typo, and the first \( ST \) is maybe a different measure, but based on the diagram, \( ST = TU = 23 \), \( RU = RS = 8 \))