Question

Identify the transformation that occurs from $\Delta ABC$ to $\Delta DEF$ and the correct congruence relation between the triangles.

• A. Reflection along the $y$-axis, $\Delta EFD\cong \Delta BCA$
• B. Clockwise rotation of $90°$ about the origin, $\Delta ABC\cong \Delta DEF$
• C. Reflection along the $y$-axis, $\Delta DEF\cong \Delta ACB$
• D. Rightward translation by $6$ units, $\Delta FED\cong \Delta BAC$

Answer 1

The correct option is A Reflection along the $y$-axis, $\Delta EFD\cong \Delta BCA$

First, let’s check for the correct transformation relation, and then for the correct congruence relation.

In option A: Clockwise rotation of $90°,\Delta ABC\cong \Delta DEF$
As rotation by $90°$ involves turning the image, triangle $DEF$ doesn’t look like the rotated image of $ABC$, which means rotation is not the transformation that occurred.

In option C: Rightward translation by $6$ units, $\Delta FED\cong \Delta BAC$
Translation involves sliding, so the orientation of the image doesn't change. But here, $DEF$ has a different orientation than $ABC$, which means this also is not the transformation that occurred.

In option B: Reflection along the $y$-axis, $\Delta DEF\cong \Delta ACB$


The above figure shows a reflection along the y-axis that turns $\Delta ABC$ into $\Delta DEF$.
Therefore, the measure of the corresponding sides and angles will be the same.
$AB=DE,BC=EF,CA=FD,\angle A=\angle D,\angle B=\angle E$, and $\angle C=\angle F$
The congruence relation $\Delta DEF\cong \Delta ACB$ implies that $DE=AC$, which is incorrect, as $AB=DE$.
The congruence relation $\Delta DEF\cong \Delta ACB$ is incorrect.
So, in option B, the transformation is correct, but the congruence is not correct. Thus, this option is also incorrect.

In option D: Reflection along the $y$-axis, $\Delta EFD\cong \Delta BCA$
We know that $\Delta ABC$ is reflected along the $y$-axis to produce image $\Delta DEF$.
According to the relation $\Delta EFD\cong \Delta BCA$, we check if the corresponding sides are congruent.
Here, $EF=BC,FD=CA,andED=BA$, which is correct.
$\Delta EFD\cong \Delta BCA$
Here, the transformation and the congruence relation is correct. Thus, this option is correct.