Question

in the figure shown, ∠prs and ∠spq are right angles, and m∠spr = 40°. choose all true statements. a. △ spr~ △ pqr by aa b. △ spq~ △ qrp by aa c. △ spq~ △ pqr by aa d. △ spq~ △ srp by aa e. △ pqr~ △ spq by aa

Explanation:

Step1: Analyze Triangle Similarity (AA Criterion)

The AA (Angle - Angle) similarity criterion states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. We know that \(\angle PRS=\angle SPQ = 90^{\circ}\), and we will use this along with other angle relationships.

Step2: Analyze Option A

For \(\triangle SPR\) and \(\triangle PQR\):
- \(\angle PRS = 90^{\circ}\) and \(\angle PQR\) is not necessarily equal to an angle in \(\triangle SPR\) in a way that satisfies AA. Let's check angles. \(\angle SPR = 40^{\circ}\), \(\angle PRS=90^{\circ}\), so \(\angle PSR = 50^{\circ}\). For \(\triangle PQR\), \(\angle PRQ = 90^{\circ}\), but we don't have a second angle congruent to \(\triangle SPR\)'s angles. So A is false.

Step3: Analyze Option B

For \(\triangle SPQ\) and \(\triangle QRP\):
- \(\angle SPQ=90^{\circ}\), \(\angle QRP = 90^{\circ}\). Also, \(\angle PQS\) and \(\angle QPR\): Let's see, \(\angle SPR = 40^{\circ}\), \(\angle SPQ = 90^{\circ}\), so \(\angle QPR=90^{\circ}- 40^{\circ}=50^{\circ}\). In \(\triangle SPQ\), \(\angle PSQ = 50^{\circ}\) (since \(\angle SPQ = 90^{\circ}\) and \(\angle SPR = 40^{\circ}\), \(\angle PSR=50^{\circ}\) and \(\angle PSQ=\angle PSR\) as they are the same angle? Wait, no, \(S - R - Q\) is a straight line. Wait, \(\triangle SPQ\) has \(\angle SPQ = 90^{\circ}\), \(\triangle QRP\) has \(\angle QRP=90^{\circ}\). Also, \(\angle PQS\) (in \(\triangle SPQ\)) and \(\angle QPR\) (in \(\triangle QRP\)): Wait, maybe a better approach. Let's check \(\triangle SPQ\) and \(\triangle QRP\): \(\angle SPQ=\angle QRP = 90^{\circ}\), and \(\angle PQS=\angle QPR\) (because \(\angle PQS + \angle QPS=90^{\circ}\) and \(\angle QPR+\angle QPS = 90^{\circ}\), so \(\angle PQS=\angle QPR\)). So by AA, \(\triangle SPQ\sim\triangle QRP\). So B is true.

Step4: Analyze Option C

For \(\triangle SPQ\) and \(\triangle PQR\):
- \(\angle SPQ = 90^{\circ}\), \(\angle PRQ=90^{\circ}\). Also, \(\angle PQS\) is common to both \(\triangle SPQ\) and \(\triangle PQR\)? Wait, \(\triangle SPQ\) has angles \(\angle SPQ = 90^{\circ}\), \(\angle PSQ\), \(\angle PQS\). \(\triangle PQR\) has angles \(\angle PRQ = 90^{\circ}\), \(\angle QPR\), \(\angle PQR\) (which is \(\angle PQS\)). So \(\angle SPQ=\angle PRQ = 90^{\circ}\) and \(\angle PQS=\angle PQR\) (common angle). So by AA, \(\triangle SPQ\sim\triangle PQR\). So C is true.

Step5: Analyze Option D

For \(\triangle SPQ\) and \(\triangle SRP\):
- \(\angle SPQ = 90^{\circ}\), \(\angle SRP = 90^{\circ}\). Also, \(\angle PSQ\) (in \(\triangle SPQ\)) and \(\angle SPR\) (in \(\triangle SRP\)): Wait, \(\angle SPR = 40^{\circ}\), \(\angle PSQ=50^{\circ}\) (since in \(\triangle SPR\), \(\angle SPR = 40^{\circ}\), \(\angle SRP = 90^{\circ}\), so \(\angle PSR=50^{\circ}\) and \(\angle PSQ=\angle PSR\)). Wait, no, \(\angle SPQ = 90^{\circ}\), \(\angle SRP = 90^{\circ}\), and \(\angle S\) is common to both \(\triangle SPQ\) and \(\triangle SRP\) ( \(\angle S=\angle S\)). So \(\angle S\) is common, \(\angle SPQ=\angle SRP = 90^{\circ}\), so by AA, \(\triangle SPQ\sim\triangle SRP\). So D is true.

Step6: Analyze Option E

For \(\triangle PQR\) and \(\triangle SPQ\):
- Similar to option C, \(\angle PRQ=\angle SPQ = 90^{\circ}\) and \(\angle PQR=\angle PQS\) (common angle). So by AA, \(\triangle PQR\sim\triangle SPQ\). So E is true. Wait, but let's re - check:

Wait, let's re - evaluate each option carefully:

- Option A: \(\triangle SPR\) and \(\triangle PQR\): \(\angle PRS = 90^{\circ}\), \(\angle PRQ = 90^{\circ}\), but \(\angle SPR = 40^{\circ}\), \(\angle PQR\): In \(\triangle PQR\), \(\angle QPR=50^{\circ}\), \(\angle PQR\): Let's see, \(\angle PQR + \angle QPR=90^{\circ}\), so \(\angle PQR = 40^{\circ}\)? Wait, no, \(\angle SPQ = 90^{\circ}\), \(\angle SPR = 40^{\circ}\), so \(\angle QPR=50^{\circ}\). Then in \(\triangle PQR\), \(\angle PQR=90^{\circ}-\angle QPR = 40^{\circ}\). In \(\triangle SPR\), \(\angle SPR = 40^{\circ}\), \(\angle PSR = 50^{\circ}\). So \(\triangle SPR\) has angles \(40^{\circ},90^{\circ},50^{\circ}\), \(\triangle PQR\) has angles \(40^{\circ},90^{\circ},50^{\circ}\)? Wait, maybe I made a mistake earlier. Wait, \(\angle PRS = 90^{\circ}\), \(\angle PRQ = 90^{\circ}\), \(\angle SPR = 40^{\circ}\), \(\angle PQR\): Let's use the fact that \(\triangle SPR\) and \(\triangle PQR\): \(\angle PRS=\angle PRQ = 90^{\circ}\), and \(\angle SPR=\angle PQR = 40^{\circ}\) (because \(\angle PQR + \angle QPR=90^{\circ}\) and \(\angle SPR+\angle QPR = 90^{\circ}\), so \(\angle SPR=\angle PQR\)). So actually, \(\triangle SPR\sim\triangle PQR\) by AA. Oh, I made a mistake earlier.

Wait, let's start over:

Given \(\angle PRS = 90^{\circ}\), \(\angle SPQ = 90^{\circ}\), \(m\angle SPR = 40^{\circ}\).

For \(\triangle SPR\) and \(\triangle PQR\):

- \(\angle PRS=\angle PRQ = 90^{\circ}\) (right angles)
- \(\angle SPR=\angle PQR\): Because \(\angle SPR+\angle QPR = 90^{\circ}\) (since \(\angle SPQ = 90^{\circ}\)) and \(\angle PQR+\angle QPR=90^{\circ}\) (since \(\angle PRQ = 90^{\circ}\)). So by AA, \(\triangle SPR\sim\triangle PQR\). So A is true.

For \(\triangle SPQ\) and \(\triangle QRP\):

- \(\angle SPQ=\angle QRP = 90^{\circ}\)
- \(\angle PQS=\angle QPR\): As \(\angle PQS+\angle QPS = 90^{\circ}\) and \(\angle QPR+\angle QPS = 90^{\circ}\), so \(\angle PQS=\angle QPR\). By AA, \(\triangle SPQ\sim\triangle QRP\). So B is true.

For \(\triangle SPQ\) and \(\triangle PQR\):

- \(\angle SPQ=\angle PRQ = 90^{\circ}\)
- \(\angle PQS=\angle PQR\) (common angle). By AA, \(\triangle SPQ\sim\triangle PQR\). So C is true.

For \(\triangle SPQ\) and \(\triangle SRP\):

- \(\angle SPQ=\angle SRP = 90^{\circ}\)
- \(\angle S=\angle S\) (common angle). By AA, \(\triangle SPQ\sim\triangle SRP\). So D is true.

For \(\triangle PQR\) and \(\triangle SPQ\):

- \(\angle PRQ=\angle SPQ = 90^{\circ}\)
- \(\angle PQR=\angle PQS\) (common angle). By AA, \(\triangle PQR\sim\triangle SPQ\). So E is true.

Wait, but this seems contradictory to my initial analysis. Let's use the AA criterion properly.

The AA similarity states that if two angles of one triangle are congruent to two angles of another triangle, the triangles are similar.

- Option A: \(\triangle SPR\) and \(\triangle PQR\): \(\angle PRS = \angle PRQ = 90^{\circ}\), \(\angle SPR=\angle PQR\) (proven above). So AA, so A is true.
- Option B: \(\triangle SPQ\) and \(\triangle QRP\): \(\angle SPQ=\angle QRP = 90^{\circ}\), \(\angle PQS=\angle QPR\) (proven above). So AA, so B is true.
- Option C: \(\triangle SPQ\) and \(\triangle PQR\): \(\angle SPQ=\angle PRQ = 90^{\circ}\), \(\angle PQS=\angle PQR\) (common angle). So AA, so C is true.
- Option D: \(\triangle SPQ\) and \(\triangle SRP\): \(\angle SPQ=\angle SRP = 90^{\circ}\), \(\angle S=\angle S\) (common angle). So AA, so D is true.
- Option E: \(\triangle PQR\) and \(\triangle SPQ\): \(\angle PRQ=\angle SPQ = 90^{\circ}\), \(\angle PQR=\angle PQS\) (common angle). So AA, so E is true.

But this can't be right. Wait, maybe the figure is a right triangle with altitude \(PR\) to the hypotenuse \(SQ\). In a right triangle, when an altitude is drawn to the hypotenuse, the three triangles (\(\triangle SPQ\), \(\triangle SPR\), \(\triangle PQR\)) are all similar to each other.

So in a right triangle, the altitude to the hypotenuse creates two smaller right triangles that are similar to the original triangle and to each other.

So \(\triangle SPQ\) is the original right triangle, \(\triangle SPR\) and \(\triangle PQR\) are the two smaller right triangles.

So:

- \(\triangle SPR\sim\triangle PQR\) (A is true)
- \(\triangle SPQ\sim\triangle QRP\) (B is true, since \(\triangle QRP\) is \(\triangle PQR\))
- \(\triangle SPQ\sim\triangle PQR\) (C is true)
- \(\triangle SPQ\sim\triangle SRP\) (D is true, since \(\triangle SRP\) is \(\triangle SPR\))
- \(\triangle PQR\sim\triangle SPQ\) (E is true)

But let's check the angle measures:

In \(\triangle SPR\): \(\angle SPR = 40^{\circ}\), \(\angle SRP = 90^{\circ}\), so \(\angle PSR=50^{\circ}\)

In \(\triangle PQR\): \(\angle QPR = 50^{\circ}\) (since \(\angle SPQ = 90^{\circ}\) and \(\angle SPR = 40^{\circ}\)), \(\angle PRQ = 90^{\circ}\), so \(\angle PQR = 40^{\circ}\)

In \(\triangle SPQ\): \(\angle SPQ = 90^{\circ}\), \(\angle PSQ = 50^{\circ}\), \(\angle PQS = 40^{\circ}\)

So:

- \(\triangle SPR\) (angles \(40^{\circ},90^{\circ},50^{\circ}\)) and \(\triangle PQR\) (angles \(40^{\circ},90^{\circ},50^{\circ}\)): similar (A true)
- \(\triangle SPQ\) (angles \(90^{\circ},50^{\circ},40^{\circ}\)) and \(\triangle QRP\) (which is \(\triangle PQR\), angles \(40^{\circ},90^{\circ},50^{\circ}\)): similar (B true)
- \(\triangle SPQ\) (angles \(90^{\circ},50^{\circ},40^{\circ}\)) and \(\triangle PQR\) (angles \(40^{\circ},90^{\circ},50^{\circ}\)): similar (C true)
- \(\triangle SPQ\) (angles \(90^{\circ},50^{\circ},40^{\circ}\)) and \(\triangle SRP\) (which is \(\triangle SPR\), angles \(40^{\circ},90^{\circ},50^{\circ}\)): similar (D true)
- \(\triangle PQR\) (angles \(40^{\circ},90^{\circ},50^{\circ}\)) and \(\triangle SPQ\) (angles \(90^{\circ},50^{\circ},40^{\circ}\)): similar (E true)

But this seems like all options are true, but that might be the case because of the right - triangle altitude theorem.

Answer:

A. \(\triangle SPR\sim\triangle PQR\) by AA, B. \(\triangle SPQ\sim\triangle QRP\) by AA, C. \(\triangle SPQ\sim\triangle PQR\) by AA, D. \(\triangle SPQ\sim\triangle SRP\) by AA, E. \(\triangle PQR\sim\triangle SPQ\) by AA