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With the aid of a ruler, the side lengths of both polygons can be found.

Additionally, with the help of a protractor, the angle measures of both polygons can be determined.

It is beneficial to summarize the information about the side lengths and angle measures of both polygons in a table.

Dimensions of $\mathcal{P}$	Dimensions of ${\mathcal{P}}^{\prime }$
$AB=2.1$ cm	$A^{\prime }{B}^{\prime }=2.1$ cm
$BC=2.4$ cm	$B^{\prime }{C}^{\prime }=2.4$ cm
$CD=2$ cm	$C^{\prime }{D}^{\prime }=2$ cm
$AD=1.5$ cm	$A^{\prime }{D}^{\prime }=1.5$ cm
$m\angle A=74^{\circ }$	$m\angle A^{\prime }=74^{\circ }$
$m\angle B=92^{\circ }$	$m\angle B^{\prime }=92^{\circ }$
$m\angle C=60^{\circ }$	$m\angle C^{\prime }=60^{\circ }$
$m\angle D=134^{\circ }$	$m\angle D^{\prime }=134^{\circ }$

As it can be seen in the table, both polygons $\mathcal{P}$ and ${\mathcal{P}}^{\prime }$ have the same side lengths and angle measures. Therefore, it can be concluded that $T$ does not affect the shape of $\mathcal{P}.$ In fact, $T$ only affects the position of the polygon.

It is important to note that the conclusion does not depend on the polygon but on the effect of $T$ on the polygon. By transforming different polygons, the same conclusion can be obtained.

Using a ruler, the side lengths of both quadrilaterals can be found.

Next, with a protractor, the angle measures can be found.

In the table, all the dimensions found are summarized putting on the same row the dimensions of corresponding parts. For example, $AB$ and $A^{\prime }{B}^{\prime }$ will be in the same row.

Dimensions of $ABCD$	Dimensions of $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$
$AB=2.8$ cm	$A^{\prime }{B}^{\prime }=2.8$ cm
$BC=2.2$ cm	$B^{\prime }{C}^{\prime }=2.2$ cm
$CD=1.8$ cm	$C^{\prime }{D}^{\prime }=1.8$ cm
$AD=1.9$ cm	$A^{\prime }{D}^{\prime }=1.9$ cm
$m\angle A=72^{\circ }$	$m\angle A^{\prime }=72^{\circ }$
$m\angle B=80^{\circ }$	$m\angle B^{\prime }=80^{\circ }$
$m\angle C=88^{\circ }$	$m\angle C^{\prime }=88^{\circ }$
$m\angle D=121^{\circ }$	$m\angle D^{\prime }=121^{\circ }$

As the table shows, both quadrilaterals $ABCD$ and $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ have the same side lengths and angle measures. This means that $T$ does not affect the shape of the polygon. Consequently, Ali is correct.

Although $T$ is a rigid motion, notice that the orientation of the preimage and the image are not the same. In the preimage, the vertices from $A$ to $D$ are positioned counterclockwise, while in the image, they are positioned clockwise.

This fact could be what made Magdalena think that $T$ is not a rigid motion. Notice that the transformations studied before preserve orientations.

Original Vertex	Add $3$ to Each $y\text{-}$Coordinate	New Vertex
$A(\text{-}2,\text{-}1)$	$(\text{-}2,\text{-}1+3)$	$A^{\prime }(\text{-}2,2)$
$B(2,1)$	$(2,1+3)$	$B^{\prime }(2,4)$
$C(0,2)$	$(0,2+3)$	$C^{\prime }(0,5)$

As can be seen, each vertex is translated $3$ units up. The same will happen with all the points of $△ABC.$ This can be seen in the diagram below. Consequently, the transformation performed on $△ABC$ represents a translation $3$ units up. It can be checked that both $△ABC$ and $△A^{\prime }{B}^{\prime }{C}^{\prime }$ have the same dimensions. Therefore, this operation represents a rigid motion.

Original Vertex	Double each $x\text{-}$Coordinate	New Vertex
$A(\text{-}2,\text{-}1)$	$(2(\text{-}2),\text{-}1)$	$A^{\prime }(\text{-}4,\text{-}1)$
$B(2,1)$	$(2(2),1)$	$B^{\prime }(4,1)$
$C(0,2)$	$(2(0),2)$	$C^{\prime }(0,2)$

Notice that $A$ and $B$ moved away from the $y\text{-}$axis while $C$ stayed in the same place. In the following diagram, it can be seen the effect of this transformation. The transformation performed on $△ABC$ is a horizontal stretch. It can be seen that the sides of $△A^{\prime }{B}^{\prime }{C}^{\prime }$ are longer than the sides of $△ABC.$ Also, the angle measures have changed. Consequently, this transformation is not a rigid motion.

Original Vertex	Multiply Each $y\text{-}$Coordinate by $\text{-}1$	New Vertex
$A(\text{-}2,\text{-}1)$	$(\text{-}2,\text{-}1(\text{-}1))$	$A^{\prime }(\text{-}2,1)$
$B(2,1)$	$(2,1(\text{-}1))$	$B^{\prime }(2,\text{-}1)$
$C(0,2)$	$(0,2(\text{-}1))$	$C^{\prime }(0,\text{-}2)$

Notice that each vertex was reflected across the $x\text{-}$axis. The same happens for all the points of $△ABC.$ This can be seen in the diagram below. In conclusion, the transformation performed on $△ABC$ is a reflection in the $x\text{-}$axis. It can be checked that both $△ABC$ and $△A^{\prime }{B}^{\prime }{C}^{\prime }$ have the same dimensions. Therefore, this transformation is a rigid motion.

Original Vertex	Taking the Absolute Value of Each $y\text{-}$Coordinate	New Vertex
$A(\text{-}2,\text{-}1)$	$(\text{-}2,|\text{-}1|)$	$A^{\prime }(\text{-}2,1)$
$B(2,1)$	$(2,|1|)$	$B^{\prime }(2,1)$
$C(0,2)$	$(0,|2|)$	$C^{\prime }(0,2)$

Notice that the vertices $B$ and $C$ remained at the same place, while $A$ was reflected across the $x\text{-}$axis. This means that every point of $△ABC$ whose $y\text{-}$coordinate is negative is reflected in the $x\text{-}$axis. Conversely, if the $y\text{-}$coordinate is positive, the point stays at the same place. As the graph above illustrates, the shape of $△ABC$ was changed. Therefore, this operation is not a rigid motion. The operation folded the part of the triangle that is below the $x\text{-}$axis along the $x\text{-}$axis. Consider also the points $P(\text{-}1,\text{-}0.5)$ and $Q(\text{-}1,0.5).$

These two points have the same image under this operation — $P^{\prime }(\text{-}0.5,0.5)$ and $Q^{\prime }(\text{-}0.5,0.5).$ This means that this is not a one-to-one operation. Consequently, it is not even a transformation!