Question

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1. given that ( pq = st ),

( qr = tu ), and ( rp = us ), show that
there is a rigid motion that maps
( \triangle pqr ) to ( \triangle stu ).
use the tool on the right to map
( \triangle pqr ) to ( \triangle stu ).
enter your answer.
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Explanation:

Step1: Recall the SSS (Side - Side - Side) congruence criterion

If \(PQ = ST\), \(QR=TU\), and \(RP = US\), then \(\triangle PQR\cong\triangle STU\) by the SSS congruence criterion.

Step2: Understand rigid motion

A rigid motion (also known as an isometry) preserves the shape and size of a figure. Since \(\triangle PQR\) and \(\triangle STU\) are congruent (because of SSS), we can use a combination of translations, rotations, and reflections.
First, translate \(\triangle PQR\) so that point \(P\) maps to point \(S\).
Let the translation vector be \(\overrightarrow{PS}\). After translation, the image of \(P\) is \(S\).
Let \(P'\) be the image of \(P\) after translation. So \(P'=S\).
Since \(PQ = ST\), after translation, we can then rotate the translated triangle (the image of \(\triangle PQR\) after translation) around point \(S\) (the image of \(P\)) so that the image of \(Q\) (say \(Q'\)) maps to \(T\).
Since \(RP=US\) and \(QR = TU\), the image of \(R\) (say \(R'\)) will map to \(U\).

Answer:

First, translate \(\triangle PQR\) by the vector \(\overrightarrow{PS}\). Then rotate the translated triangle around point \(S\) so that the translated \(Q\) maps to \(T\) (and the translated \(R\) maps to \(U\)) (or other valid combinations of translation, rotation and/or reflection based on the SSS - congruence of the two triangles)