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lesson 3 practice problems 1. triangle abc is congruent to triangle edf…

Question

lesson 3 practice problems

  1. triangle abc is congruent to triangle edf. so, kiran knows that there is a sequence

of rigid motions that takes abc to edf.
select all true statements after the transformations:
a. angle a coincides with angle f.
b. angle b coincides with angle d.
c. segment ac coincides with segment ef.
d. segment bc coincides with segment ed.
e. segment ab coincides with segment ed.

  1. a rotation by angle ace using point c as

the center takes triangle cba onto
triangle cde.
a. explain why the image of ray ca lines up with ray ce.
b. explain why the image of a coincides with e.
c. is triangle cba congruent to triangle cde? explain your reasoning.

Explanation:

1.

Step1: Analyze congruent triangles

Since \(\triangle ABC\cong\triangle EDF\), corresponding parts are equal. Corresponding angles: \(\angle A=\angle E\), \(\angle B = \angle D\), \(\angle C=\angle F\). Corresponding sides: \(AB = ED\), \(BC=DF\), \(AC=EF\)

Step2: Check each option

  • Option A: \(\angle A\) corresponds to \(\angle E\), not \(\angle F\). So, A is false.
  • Option B: \(\angle B\) corresponds to \(\angle D\). So, B is true.
  • Option C: \(AC\) corresponds to \(EF\). So, C is true.
  • Option D: \(BC\) corresponds to \(DF\), not \(ED\). So, D is false.
  • Option E: \(AB\) corresponds to \(ED\). So, E is true.
Brief Explanations

A rotation about point \(C\) by angle \(ACE\) means that the measure of the angle between \(CA\) and \(CE\) is the angle of rotation. When we rotate ray \(CA\) around point \(C\) by angle \(ACE\), it will land on ray \(CE\) because the rotation is defined by the angle between \(CA\) and \(CE\) with center \(C\).

Brief Explanations

Since the rotation is about point \(C\) (a rigid - motion), and \(CA = CE\) (because in a rotation, the distance from the center of rotation to a point and its image is the same). Also, as we saw in part (a) ray \(CA\) maps to ray \(CE\). So, the point \(A\) (which is on ray \(CA\) at a distance \(CA\) from \(C\)) maps to point \(E\) (which is on ray \(CE\) at a distance \(CE = CA\) from \(C\))

Answer:

B. Angle \(B\) coincides with angle \(D\), C. Segment \(AC\) coincides with segment \(EF\), E. Segment \(AB\) coincides with segment \(ED\)

2.
a.