Question

△qrs is a right triangle.
select the correct similarity statement.
△str ~ △tqr
△str ~ △rtq
△str ~ △rst
△str ~ △sqr

Explanation:

Step1: Recall Similar Triangles in Right Triangles

In a right triangle, when an altitude is drawn to the hypotenuse, the two smaller triangles are similar to the original triangle and to each other. Here, \(\triangle QRS\) is a right triangle with right angle at \(R\), and \(RT\) (or \(ST\)?) is an altitude? Wait, looking at the diagram, \(\angle R\) and \(\angle T\) are right angles. So \(\triangle STR\) and \(\triangle SQR\): Wait, no, let's check angles.

For \(\triangle STR\) and \(\triangle SQR\): \(\angle S\) is common. \(\angle STR = \angle SQR = 90^\circ\)? Wait, no, \(\triangle QRS\) is right at \(R\), so \(\angle R = 90^\circ\). \(\triangle STR\) is right at \(T\), \(\triangle SQR\) is right at \(R\). Wait, maybe better to use AA similarity.

Wait, let's list the triangles:

- \(\triangle STR\): right at \(T\), angles: \(\angle S\), \(\angle T = 90^\circ\), \(\angle R\) (of \(\triangle STR\))? Wait, no, vertices: \(S\), \(T\), \(R\). So \(\angle at T\) is right.

- \(\triangle SQR\): right at \(R\), vertices \(S\), \(Q\), \(R\). \(\angle at R\) is right.

Now, \(\angle S\) is common to both \(\triangle STR\) and \(\triangle SQR\). And both have a right angle (\(\angle T = 90^\circ\) and \(\angle R = 90^\circ\)). So by AA similarity, \(\triangle STR \sim \triangle SQR\) because they share \(\angle S\) and both have a right angle. Wait, but let's check the other options.

Option 2: \(\triangle STR \sim \triangle RST\). But \(\triangle RST\) is the same as \(\triangle STR\) (same vertices), so not similar in a non-trivial way.

Option 3: \(\triangle STR \sim \triangle RTQ\). Let's check angles. \(\triangle RTQ\) is right at \(T\)? Wait, \(\angle T\) in \(\triangle RTQ\): the diagram shows \(\angle at T\) is right? Wait, original triangle \(\triangle QRS\) is right at \(R\), and \(T\) is a point on \(QS\) (since \(T\) is on \(QS\), as the right angle at \(T\) is on \(QS\)). So \(QS\) is the hypotenuse of \(\triangle QRS\), and \(T\) is the foot of the altitude from \(R\) to \(QS\)? Wait, no, the right angle at \(T\) is between \(ST\) and \(RT\)? Wait, maybe I misread the diagram.

Wait, the diagram: \(\triangle QRS\) is right at \(R\) (so \(QR \perp RS\)). Then \(T\) is a point on \(QS\) such that \(RT \perp QS\) (since \(\angle at T\) is right). So then, in right triangle \(QRS\), with altitude \(RT\) to hypotenuse \(QS\), we have the three similar triangles: \(\triangle QRS \sim \triangle STR \sim \triangle RTQ\)? Wait, no, the three similar triangles are \(\triangle QRS\), \(\triangle QRT\), and \(\triangle RST\). Wait, maybe I messed up the labels.

Wait, let's correct: In right triangle \(QRS\) (right at \(R\)), when we draw an altitude from \(R\) to hypotenuse \(QS\), meeting at \(T\), then:

- \(\triangle QRS \sim \triangle QRT\) (AA: \(\angle Q\) common, right angles at \(R\) and \(T\))

- \(\triangle QRS \sim \triangle RST\) (AA: \(\angle S\) common, right angles at \(R\) and \(T\))

- \(\triangle QRT \sim \triangle RST\) (AA: right angles and \(\angle Q + \angle S = 90^\circ\), so \(\angle Q = \angle RST\) and \(\angle S = \angle QRT\))

Now, \(\triangle STR\) is \(\triangle RST\) (same as \(\triangle RST\)). Wait, the options:

1. \(\triangle STR \sim \triangle SQR\): \(\triangle SQR\) is \(\triangle QRS\) (same as \(\triangle QRS\)). So \(\triangle RST \sim \triangle QRS\) (which is true, as \(\triangle RST \sim \triangle QRS\) by AA: \(\angle S\) common, right angles). So \(\triangle STR\) (which is \(\triangle RST\)) is similar to \(\triangle SQR\) (which is \(\triangle QRS\)). So that's correct.

Wait, maybe the first option is correct. Let's check the other options:

Option 4: \(\triangle STR \sim \triangle TQR\). \(\triangle TQR\) is \(\triangle QRT\). So \(\triangle RST \sim \triangle QRT\) (which is true, as they are both similar to \(\triangle QRS\)). But the label: \(\triangle STR\) and \(\triangle TQR\): vertices \(S,T,R\) and \(T,Q,R\). So \(\angle T\) is right in both? \(\triangle STR\) right at \(T\), \(\triangle TQR\) right at \(T\). Then \(\angle R\) in \(\triangle STR\) and \(\angle Q\) in \(\triangle TQR\): \(\angle RST + \angle Q = 90^\circ\), \(\angle RTS + \angle RST = 90^\circ\), so \(\angle Q = \angle RTS\). So maybe, but the first option: \(\triangle STR \sim \triangle SQR\): \(\angle S\) common, right angles at \(T\) and \(R\). So AA similarity. So that's correct.

Wait, but maybe I made a mistake. Let's re-express:

\(\triangle STR\): angles at \(S\), \(T\) (right), \(R\) (of \(\triangle STR\)): \(\angle STR = 90^\circ\), \(\angle S\), \(\angle SRT\).

\(\triangle SQR\): angles at \(S\), \(Q\), \(R\) (right): \(\angle SQR = 90^\circ\), \(\angle S\), \(\angle SQR\) (wait, no, \(\angle at R\) is 90°, so \(\angle QRS = 90^\circ\), \(\angle S\), \(\angle Q\)).

So \(\angle S\) is common, \(\angle STR = \angle SQR = 90^\circ\) (wait, \(\angle SQR\) is not 90°, \(\angle QRS\) is 90°. Oh! Here's the mistake. \(\triangle SQR\) is right at \(R\), so \(\angle QRS = 90^\circ\), not \(\angle SQR\). So \(\angle at R\) is 90°, so \(\triangle SQR\) has right angle at \(R\), \(\triangle STR\) has right angle at \(T\). So \(\angle S\) is common, and \(\angle STR = \angle SRQ = 90^\circ\)? No, \(\angle STR\) is 90°, \(\angle SRQ\) is 90° (since \(\triangle QRS\) is right at \(R\)). So \(\angle S\) is common, and both have a right angle (90°), so by AA similarity, \(\triangle STR \sim \triangle SQR\) (because \(\angle S\) is common, and \(\angle STR = \angle SRQ = 90^\circ\)). Yes, that works.

So the correct option is the first one: \(\triangle STR \sim \triangle SQR\).

Answer:

A. \(\triangle STR \sim \triangle SQR\)