Question

can δtsr and δqrs be proven congruent by sas?
○ yes, because along with the given information on the diagram, sr = rs by the reflexive property
○ yes, because a reflection will map δtsr onto δqrs
○ yes, because p appears to be the midpoint of sq and tr
○ no, because not enough is information given to prove the triangles congruent by sas

Explanation:

To determine if \(\triangle TSR\) and \(\triangle QRS\) (assuming typo, should be \(\triangle QRS\) or \(\triangle QRS\) related) can be proven congruent by SAS:

1. SAS Criterion: Requires two sides and the included angle to be congruent.
2. Given Information:

• \(TS = QR = 5\) in (from diagram).
• \(\angle T = \angle Q = 66^\circ\) (from diagram).
• \(SR\) is common to both triangles, so \(SR = RS\) (reflexive property).
• The included angle for \(TS\) and \(SR\) is \(\angle TSR\), and for \(QR\) and \(RS\) is \(\angle QRS\)? Wait, no—wait, the first option says \(SR = RS\) (reflexive), \(TS = QR\), and \(\angle T\) and \(\angle Q\) are equal. Wait, actually, let's re - evaluate:
• For \(\triangle TSR\) and \(\triangle QRS\) (correcting the second triangle's name), \(TS = QR\) (5 in), \(\angle T=\angle Q = 66^\circ\), and \(SR = RS\) (common side, reflexive). So the two sides \(TS\) and \(SR\) with included angle \(\angle T\) (wait, no—included angle between \(TS\) and \(SR\) is \(\angle TSR\), and between \(QR\) and \(RS\) is \(\angle QRS\). Wait, maybe the first option is correct: the first option states that along with given info, \(SR = RS\) (reflexive), \(TS = QR\), and the included angle (since \(\angle T\) and \(\angle Q\) are equal, and the sides \(TS, SR\) and \(QR, RS\) with included angles? Wait, maybe the first option is correct because \(TS = QR\), \(\angle T=\angle Q\), and \(SR = RS\) (reflexive), so by SAS (two sides and included angle: \(TS\) and \(SR\) with \(\angle T\), \(QR\) and \(RS\) with \(\angle Q\), and \(\angle T=\angle Q\), \(TS = QR\), \(SR = RS\)). The other options: the second option talks about reflection, which is not relevant to SAS proof. The third option says \(P\) is midpoint, but \(P\) is not part of the SAS for these triangles. The fourth option is wrong because we do have enough info (two sides and included angle with reflexive side and equal sides/angles). So the first option is correct.

Answer:

yes, because along with the given information on the diagram, \(SR = RS\) by the reflexive property