## Engage NY Eureka Math Geometry Module 2 Lesson 14 Answer Key

### Eureka Math Geometry Module 2 Lesson 14 Example Answer Key

Example 1.

We said that for a figure A in the plane, it must be true that A ~ A. Describe why this must be true.

Answer:

→ Remember, to show that for two figures to be similar, there must be a similarity transformation that maps one to the other. Are there such transformations to show that A maps to A?

Take multiple suggestions of transformations that map A to A:

→ There are several different transformations that map A onto itself such as a rotation of 0° or a rotation of 360°.

→ A reflection of A across a line and a reflection right back achieves the same result.

→ A translation with a vector of length 0 also maps A to A.

→ A dilation with scale factor 1 maps A to A, and any combination of these transformations also maps A to A.

→ Therefore, A must be similar to A because there are many similarity transformations that map A to A.

→ This condition is labeled as reflexive because every figure is similar to itself.

Example 2.

We said that for figures A and B in the plane so that A ~ B, then it must be true that B ~ A. Describe why this must be true.

Answer:

Now that students have completed Example 1, allow them time to discuss Example 2 among themselves.

→ This condition must be true because for any composition of transformations that maps A to B, there is a composition of transformations that can undo the first composition. For example, if a translation by vector \(\overrightarrow{X Y}\) maps A to B, then the vector \(\overrightarrow{Y X}\) will undo the transformation and map B to A.

→ A counterclockwise rotation of 90° can be undone by a clockwise rotation of 90°.

→ A dilation by a scale factor of r = 2 can be undone with a dilation by a scale factor of r = \(\frac{1}{2}\).

→ A reflection across a line can be undone by a reflection back across the same line.

→ Therefore, it must be true that if a figure A is similar to a figure B in the plane (i.e., if there is a similarity transformation that maps A to B), then there must also be a composition of transformations that can undo that similarity transformation and map B back to A.

→ This condition is labeled as symmetric because of its likeness to the symmetric property of equality where if one number is equal to another number, then they both must have the same value (if a = b, then b = a).

We leave the third condition, that similarity is transitive, for the Problem Set.

Example 3.

Based on the definition of similar, how would you show that any two circles are similar?

Answer:

Based on their discussions, provide students with the following cases to help them along:

→ Consider the different cases you must address in showing that any two circles are similar to each other:

a. Circles with different centers but radii of equal length

If two circles have different centers but have radii of equal length, then the circles are congruent, and a translation along a vector that brings one center to the other will map one circle onto the other.

b. Circles with the same center but radii of different lengths

If two circles have the same center, but one circle has radius R and the

other has radius R’, then a dilation about the center with a scale factor of r = \(\frac{R^{\prime}}{R}\) maps one circle onto the other so that rR = R’.

c. Circles with different centers and radii of different lengths

If two circles have different centers and radii of different lengths, then

the composition of a dilation described in (b) and the translation

described in (a) maps one onto the other.

Students may notice that two circles with different centers and different radii length may alternatively be mapped onto each other by a single dilation as shown in the following image.

Example 4

Suppose ∆ ABC ↔ ∆ DEF and that, under this correspondence, corresponding angles are equal and corresponding sides are proportional. Does this guarantee that ∆ ABC and ∆ DEF are similar?

Answer:

→ We have already shown that if two figures (e.g., triangles) are similar, then corresponding angles are of equal measurement and corresponding sides are proportional in length.

→ This question is asking whether the converse of the theorem is true. We know that a correspondence exists between ∆ ABC and ∆ DEF. What does the correspondence imply?

The correspondence 1m plies that m∠A = m∠D, m∠B = m∠E, m∠C = m∠F, and \(\frac{D E}{A B}=\frac{E F}{B C}=\frac{D F}{A C}\).

→ We will show there is a similarity transformation taking ∆ ABC to ∆ DEF that starts with a dilation. How can we use what we know about the correspondence to express the scale factor r?

→ Since the side lengths are proportional under the correspondence, then r = \(\frac{D E}{A B}=\frac{E F}{B C}=\frac{D F}{A C}\)

→ The dilation that takes ∆ ABC to ∆ DEF can have any point for a center. The dilation maps A to A’, B to B’, and C to C’ so that ∆ ABC ~ ∆ A’B’C’.

→ How would you describe the length A’B’?

A’B’ = rAB

→ We take this one step further and say that A’B’ = rAB = \(\frac{D E}{A B}\)AB = DE. Similarly, B’C’ = EF and A’C’ = DF.

So, all the sides and all the angles of ∆ A’B’C’ and ∆ DEF match up, and the triangles are congruent.

→ Since they are congruent, a sequence of basic rigid motions takes ∆ A’B’C’ to ∆ DEF.

→ So, a dilation takes ∆ ABC to ∆ A’B’C’, and a congruence transformation takes ∆ A’B’C’ to ∆ DEF, and we conclude that ∆ ABC ~ ∆ DEF.

Example 5.

a. In the diagram below, ∆ ABC ~ ∆ A’B’C’. Describe a similarity transformation that maps ∆ ABC to ∆ A’B’C’.

Answer:

b. Joel says the sequence must require a dilation and three rigid motions, but Sharon is sure there is a similarity transformation composed of just a dilation and two rigid motions. Who is right?

Answer:

→ Step 1: Dilate by a scale factor of r and center O so that (1) the resulting ∆ ABC is congruent to ∆ A’B’C’ and (2) one pair of corresponding vertices coincide.

→ Step 2: Reflect the dilated triangle across \(\overline{B^{\prime} C^{\prime}}\).

→ Step 3: Rotate the reflected triangle until it coincides with ∆ A’B’C’.

→ We have found a similarity transformation that maps ∆ ABC to ∆ A’B’C’ with just one dilation and two rigid motions instead of three.

### Eureka Math Geometry Module 2 Lesson 14 Problem Set Answer Key

Question 1.

If you are given any two congruent triangles, describe a sequence of basic rigid motions that takes one to the other.

Answer:

Translate one triangle to the other by a vector \(\overrightarrow{X Y}\) so that the triangles coincide at a vertex.

Case 1: If both triangles are of the same orientation, simply rotate about the common vertex until the triangles coincide.

Case 2: If the triangles are of opposite orientations, reflect the one triangle over one of its two sides that include the common vertex, and then rotate around the common vertex until the triangles coincide.

Question 2.

If you are given two similar triangles that are not congruent triangles, describe a sequence of dilations and basic rigid motions that takes one to the other.

Answer:

Dilate one triangle with a center O such that when the lengths of its sides are equal to the corresponding lengths of the other triangle, one pair of corresponding vertices coincide. Then, follow one of the sequences described in Case 1 or Case 2 of Problem 1.

Students may have other similarity transformations that map one triangle to the other consisting of translations, reflections, rotations, and dilations. For example, they might dilate one triangle until the corresponding lengths are equal, then translate one triangle to the other by a vector \(\overrightarrow{X Y}\) so that the triangles coincide at a vertex, and then follow one of the sequences described in Case 1 or Case 2 of Problem 1.

Question 3.

Given two line segments, \(\overline{A B}\) and \(\overline{C D}\), of different lengths, answer the following questions:

a. It Is always possible to find a similarity transformation that maps \(\overline{A B}\) to \(\overline{C D}\) sending A to C and B to D. Describe one such similarity transformation.

Answer:

Rotate \(\overline{A B}\) so that it is parallel to \(\overline{C D}\) with corresponding points A and C (and likewise B and D) on the same side of each line segment. Then, translate the image of \(\overline{A B}\) by a vector \(\overrightarrow{X Y}\) so that the midpoint of \(\overline{A B}\) coincides with the midpoint of \(\overline{C D}\). Finally, dilate \(\overline{A B}\) until the two segments are equal in length.

b. If you are given \(\overline{A B}\) that \(\overline{C D}\) and are not parallel, are not congruent, do not share any points, and do not lie in the same line, what is the fewest number of transformations needed In a sequence to map \(\overline{A B}\) to \(\overline{C D}\)? Which transformations make this work?

Answer:

Rotate \(\overline{A B}\) to \(\overline{A^{\prime} B^{\prime}}\) so that \(\overline{A^{\prime} B^{\prime}}\) and \(\overline{C D}\) are parallel and oriented in the same direction; then, use the fact that there is a dilation that takes \(\overline{A^{\prime} B^{\prime}}\) to \(\overline{C D}\)

c. If you performed a similarity transformation that instead takes A to D and B to C, either describe what mistake was made In the similarity transformation, or describe what additional transformation is needed to fix the error so that A maps to C and B maps to D.

Answer:

The rotation in the similarity transformation was not sufficient to orient the directed line segments in the same direction, resulting in mismatched corresponding points. This error could be fixed by changing the rotation such that the desired corresponding endpoints lie on the same end of each segment.

If the desired endpoints do not coincide after a similarity transformation, rotate the line segment about its midpoint by 180°.

Question 4.

We claim that similarity Is transitive (i.e., if A, B, and C are figures in the plane such that A~B and B~C, then A~C). Describe why this must be true.

Answer:

If similarity transformation T_{1} maps A to B and similarity transformation T_{2} maps B to C, then the composition of basic rigid motions and dilations that takes A to B together with the composition of basic rigid motions and dilations that takes B to C shows that A ~ C.

Question 5.

Given two line segments, \(\overline{A B}\) and \(\overline{C D}\), of different lengths, we have seen that It is always possible to find a similarity transformation that maps \(\overline{A B}\) to \(\overline{C D}\), sending A to C and B to D with one rotation and one dilation. Can you do this with one reflection and one dilation?

Answer:

Yes. Reflect \(\overline{A B}\) to \(\overline{A^{\prime} B^{\prime}}\) so that \(\overline{A^{\prime} B^{\prime}}\) and \(\overline{C D}\) are parallel, and then use the fact that there is a dilation that takes \(\overline{A^{\prime} B^{\prime}}\) to \(\overline{C D}\).

Question 6.

Given two triangles, ∆ ABC ~ ∆ DEF, is it always possible to rotate ∆ ABC so that the sides of ∆ ABC are parallel to the corresponding sides in A DEF (e.g., \(\overline{A B}\) || \(\overline{D E}\))?

Answer:

No, it is not always possible. Sometimes a reflection is necessary. If you consider the diagram to the right, ∆ DEF can be rotated in various ways such that \(\overline{D E}\) either coincides with or is parallel to \(\overline{A B}\), where corresponding points A and D are located at the same end of the segment. However, in each case, F lies on one side of \(\overline{D E}\) but is opposite the side on which C lies with regard to \(\overline{A B}\). This means that a reflection is necessary to reorient the third vertex of the triangle.

### Eureka Math Geometry Module 2 Lesson 14 Exit Ticket Answer Key

Question 1.

In the diagram, ∆ ABC ~ ∆ DEF by the dilation with center O and a scale factor of r. Explain why ∆ DEF ~ ∆ ABC.

Answer:

We know that ∆ ABC ~ ∆ DEF by a dilation with center o and a scale factor of r. A dilation with the same center O but a scale factor of \(\frac{1}{r}\) maps ∆ DEF onto ∆ ABC; this means ∆ DEF ~ ∆ ABC.

Question 2.

Radii \(\overline{C A}\) and \(\overline{T S}\) are parallel. Is circle C_{C,CA} similar to circle C_{T,TS}? Explain.

Answer:

Yes, the circles are similar because a dilation with center O and a scale factor of r exists that maps C_{C,CA} onto C_{T,TS}.

Question 3.

Two triangles, ∆ ABC and ∆ DEF, are in the plane so that m∠A = m∠D, m∠B = m∠E, m∠C = m∠F, and \(\frac{D E}{A B}=\frac{E F}{B C}=\frac{D F}{A C}\). Summarize the argument that proves that the triangles must be similar.

Answer:

A dilation exists such that the lengths of one triangle can be made equal to the lengths of the other triangle. Once the triangles have lengths and angles of equal measurement, or are congruent, then a sequence of rigid motions maps one triangle to the other. Therefore, a similarity transformation exists that maps one triangle onto the other, and the triangles must be similar.