Which statement correctly describes the relationship between △DEF and △D′E′F′ ?

△DEF is not congruent to △D′E′F′ because there is no sequence of rigid motions that maps △DEF to △D′E′F′.

△DEF is congruent to △D′E′F′ because you can map △DEF to △D′E′F′ using a reflection across the y-axis, which is a rigid motion.

△DEF is congruent to △D′E′F′ because you can map △DEF to △D′E′F′ using a rotation of 180° about the origin, which is a rigid motion.

△DEF is congruent to △D′E′F′ because you can map △DEF to △D′E′F′ using a reflection across the x-axis, which is a rigid motion.

1) Reflections, rotations and translations are rigid transformations, because they do not modify the lengths of the segments nor the angles, so the images and the preimages are congruents.

2) Let's see what transformation map △DEF is to △D'E'F' by analyzing the vertices of preimage and image:

Preimage Image

D (-3, -1) D' (-3, 1)

E (2, -4) E' (2, 4)

F (4, -4) F' (4, 4)

As you see when the image is formed, the coordinate x of the image is kept, and the coordinate y is negated. This rule is (x, y) → (x, - y), which is the rigid transformation reflection across the x-axis.

AkhileshT AkhileshT · Answer 2 · 2019-09-20T14:20:35+02:00

The correct option is [tex]\boxed{\bf option D}[/tex]

Further explanation:

Given:

The given figure is redrawn and attched in the end as Figure 1.

The two triangles are [tex]\triangle\text{MNO}[/tex] and [tex]\triangle\text{M'N'O'}[/tex] as shown in attached Figure 1.

Calculation:

A reflection of line is defined as a mirror line or axis of symmetry.

In reflection, if the shape and size of an image is exactly same the original object, then it is called isometric reflection.

Reflection in a coordinate plane:

If we reflect a point across the [tex]x[/tex]-axis, then the [tex]x[/tex] coordinate remain same but the value of [tex]y[/tex] coordinate will change and transformed into its opposite that means its sign changes.

If we reflect a point across the [tex]y[/tex]-axis, then the value of [tex]y[/tex] coordinate remain same but the value of [tex]x[/tex] coordinate will change into its opposite sign.

Consider the attached Figure 1, and the coordinate point of the [tex]\triangle\text{MNO}[/tex] and [tex]\triangle\text{M'N'O'}[/tex] are as follows:

[tex]\boxed{\begin{aligned}\text{M}(-3,-1)\text{ and }\text{M'}(-3,1)\\\text{N}(2,-4)\text{ and }\text{N}(2,4)\\\text{O}(4,-4)\text{ and }\text{O}(4,4)\end{aligned}}[/tex]

From the above points we can see that there is only change in the value of [tex]y[/tex] coordinate and no change in the value of [tex]x[/tex] coordinate in the reflection.

So it can be said that the [tex]\triangle\text{MNO}[/tex] is congruent to [tex]\triangle\text{M'N'O'}[/tex] and we can map [tex]\triangle\text{MNO}[/tex] to [tex]\triangle\text{M'N'O'}[/tex] using a reflection across the [tex]x[/tex]-axis, which is a rigid motion.

Therefore, the correct option is [tex]\fbox{\begin{minispace}\\ \bf option D\\\end{minispace}}[/tex].

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Answer detail:

Grade: Middle school

Subject: Mathematics

Chapter: Triangles

Keywords: Triangles, DEF, D’E’F’, MNO, M'N'O', reflection, congruent triangle, mapping, x axis, rigid motion, angles, coordinate points, congruency, isometric reflection.