Question

question 23 of 25 according to the diagram below, which similarity statements are true? check all that apply. a. △pqr~△psq b. △pqr~△prs c. △pqr~△qsr d. △pqs~△qrs

Explanation:

Step1: Recall similarity - criteria

In a right - triangle with an altitude drawn to the hypotenuse, we use the geometric mean theorem and AA (angle - angle) similarity criterion. In right - triangle \(PQS\) with altitude \(QR\) drawn to the hypotenuse \(PS\), we have three similar right - triangles: \(\triangle PQR\), \(\triangle QSR\), and \(\triangle PQS\).

Step2: Analyze option A

In \(\triangle PQR\) and \(\triangle PSQ\), \(\angle P=\angle P\) (common angle) and \(\angle PQR=\angle PSQ = 90^{\circ}\). By AA similarity, \(\triangle PQR\sim\triangle PSQ\).

Step3: Analyze option B

In \(\triangle PQR\) and \(\triangle PRS\), \(\angle P=\angle P\) (common angle), but \(\angle PQR
eq\angle PRS\) (since \(\angle PQR = 90^{\circ}\) and \(\angle PRS
eq90^{\circ}\)), so \(\triangle PQR\) and \(\triangle PRS\) are not similar.

Step4: Analyze option C

In \(\triangle PQR\) and \(\triangle QSR\), \(\angle PQR=\angle QSR = 90^{\circ}\) and \(\angle RQS+\angle SQR = 90^{\circ}\), \(\angle RQS+\angle QPR=90^{\circ}\), so \(\angle QPR=\angle RQS\). By AA similarity, \(\triangle PQR\sim\triangle QSR\).

Step5: Analyze option D

In \(\triangle PQS\) and \(\triangle QRS\), \(\angle PQS=\angle QRS = 90^{\circ}\) and \(\angle S\) is common. By AA similarity, \(\triangle PQS\sim\triangle QRS\).

Answer:

A. \(\triangle PQR\sim\triangle PSQ\)
C. \(\triangle PQR\sim\triangle QSR\)
D. \(\triangle PQS\sim\triangle QRS\)