Question

1. the triangles are congruent. which sequence of rigid motions will take triangle xyz onto triangle bca?

a. translate xyz using directed line segment yc. rotate xyz using c as the center so that x coincides with b. reflect xyz across line cb.
b. translate xyz using directed line segment yc. rotate xyz using c as the center so that x coincides with b. reflect xyz across line ac.
c. translate xyz using directed line segment yc. rotate xyz using c as the center so that x coincides with a. reflect xyz across line cb.
d. translate xyz using directed line segment yc. rotate xyz using c as the center so that x coincides with a. reflect xyz across line ac.

1. triangle hef is the image of triangle fgh after a 180 degree rotation around point k. select all statements that must be true.

a. triangle hgf is congruent to triangle feh.
b. triangle gfh is congruent to triangle efh.
c. angle khe is congruent to angle khg.
d. angle ghk is congruent to angle efk.
e. segment eh is congruent to segment gh.
f. segment hg is congruent to segment fe.
g. segment fh is congruent to segment hf.
(from unit 2, lesson 2.)

Explanation:

Question 3

Step1: Analyze translation

Translate \(XYZ\) using directed line segment \(YC\). This moves \(Y\) to \(C\).

Step2: Analyze rotation

Rotate \(X'Y'Z'\) using \(C\) as the center. We want \(X'\) (original \(X\)) to coincide with \(B\).

Step3: Analyze reflection

Reflect \(X''Y''Z''\) across line \(CB\). This will map the triangle onto \(BCA\) as required.

• A 180 - degree rotation is a rigid motion. Rigid motions preserve congruence. So, \(\triangle GFH\cong\triangle EFH\) (by the property of rotation, pre - image and image are congruent).
• For angles: When we rotate \(\triangle FGH\) 180 degrees around \(K\) to get \(\triangle HEF\), \(\angle KHE\) and \(\angle KHG\) are vertical angles (formed by the rotation) and are congruent.
• For segments: \(EH\) and \(GH\) are corresponding sides of congruent triangles (since rotation is a rigid motion, \(EH = GH\)), \(HG\) and \(FE\) are corresponding sides of congruent triangles (\(HG=FE\)), and \(FH\) and \(HF\) are the same segment (so \(FH = HF\)).

Answer:

A. Translate \(XYZ\) using directed line segment \(YC\). Rotate \(X'Y'Z'\) using \(C\) as the center so that \(X'\) coincides with \(B\). Reflect \(X''Y''Z''\) across line \(CB\).

Question 4