Eureka Math Homework Helper 2015–2016 Geometry Module 2€¦ · M2 GEOMETRY Lesson 2: Making Scale Drawings Using the Ratio Method I should remember that a ruler is critical when

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Eureka Math™ Homework Helper

2015–2016

GeometryModule 2

Lessons 1–24

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2015-16

M2 GEOMETRY

Lesson 1: Scale Drawings

I should remember that a scale drawing of a figure is made without respect to the orientation of the original figure.


1. What is a well-scaled drawing?

A well-scaled drawing of a figure is one where corresponding angles are equal in measure, and corresponding lengths are all in the same proportion.

2. Triangle 𝐷𝐷𝐷𝐷𝐷𝐷 is a scale drawing of triangle 𝐴𝐴𝐴𝐴𝐴𝐴 with scale factor 𝑟𝑟. Describe each of the following statements as sometimes, always, or never true, and justify your response. a. 𝐷𝐷𝐷𝐷 > 𝐴𝐴𝐴𝐴

Sometimes but only when 𝒓𝒓 > 𝟏𝟏

b. 𝑚𝑚∠𝐷𝐷 < 𝑚𝑚∠𝐴𝐴

Never because corresponding angles between scale drawings are equal in measure

c. 𝐸𝐸𝐸𝐸𝐷𝐷𝐸𝐸

= 𝐵𝐵𝐵𝐵𝐴𝐴𝐵𝐵

Always because distances in a scale drawing are equal to their corresponding distances in the original drawing times the scale factor: 𝑬𝑬𝑬𝑬

𝑫𝑫𝑬𝑬= 𝒓𝒓(𝑬𝑬𝑬𝑬)

𝒓𝒓(𝑫𝑫𝑬𝑬)= 𝑩𝑩𝑩𝑩

𝑨𝑨𝑩𝑩

© 2015 Great Minds eureka-math.orgGEO-M2-HWH-1.3.0-09.2015

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M2 GEOMETRY


I should remember that a scale drawing of a figure is made without respect to the orientation of the original figure.

3. Triangle 𝐴𝐴𝐴𝐴𝐴𝐴 is provided below, and one angle of scale drawing △ 𝐴𝐴′𝐴𝐴′𝐴𝐴′ is also provided. Use construction tools to complete the scale drawing so that the scale factor is 𝑟𝑟 = 3. Explain your construction. How do you know that △ 𝐴𝐴′𝐴𝐴′𝐴𝐴′ is in fact a scale drawing of △ 𝐴𝐴𝐴𝐴𝐴𝐴 with scale factor 𝑟𝑟 = 3?

Extend both rays from 𝑩𝑩′. Use the compass to mark off a length equal to 𝟑𝟑(𝑨𝑨𝑩𝑩) on one ray and a length equal to 𝟑𝟑(𝑩𝑩𝑩𝑩) on the other. Label the ends of the two lengths 𝑨𝑨′ and 𝑩𝑩′, respectively. Join 𝑨𝑨′ to 𝑩𝑩′.

By measurement, each side is three times the length of the corresponding side of the original figure, and all three angles are equal in measurement to the three corresponding angles in the original figure.


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Lesson 2: Making Scale Drawings Using the Ratio Method

I should remember that a ruler is critical when applying the ratio method in order to measure and accurately locate image vertices along each ray from the center.

I must measure the lengths 𝑂𝑂𝑂𝑂, 𝑂𝑂𝑂𝑂, and 𝑂𝑂𝑂𝑂 and double each of those lengths along the respective rays to locate 𝑂𝑂′, 𝑂𝑂′, and 𝑂𝑂′.


1. What does the ratio method rely on?

To apply the ratio method, dilate key vertices of the provided figure by the scale factor. The locations of the image points depend on the scale factor and the distance of each point from the center. If the scale factor 𝒓𝒓 is greater than 𝟏𝟏, the points will move 𝒓𝒓 times the current distance away from the center; a scale factor 𝒓𝒓 that is less than 𝟏𝟏 will mean the vertices move 𝒓𝒓 times the current distance toward the center.

2. Create a scale drawing of the figure below using the ratio method about center 𝑂𝑂 and scale factor 𝑟𝑟 = 2.


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I must measure the lengths 𝑂𝑂𝑂𝑂 and 𝑂𝑂𝑂𝑂 and take one third of each of those lengths along the respective rays to find 𝑂𝑂′ and 𝑂𝑂′.

3. Create a scale drawing of the figure below using the ratio method about center 𝑂𝑂 and scale factor 𝑟𝑟 = 13.


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Lesson 3: Making Scale Drawings Using the Parallel Method


1. Use the step-by-step guide below, a setsquare, and ruler to construct parallelogram 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸, provided side 𝐸𝐸𝐸𝐸�� and vertex 𝐸𝐸.

Step 1

Step 2

Align the setsquare and ruler; one leg of the setsquare should line up with side 𝑬𝑬𝑬𝑬��, and the perpendicular leg should be flush against the ruler.


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Step 3

Slide the setsquare up until the leg that was lined up with 𝑬𝑬𝑬𝑬�� passes through 𝑯𝑯. Draw a line that passes through 𝑯𝑯.

Step 4

Use a compass to mark off the length of 𝑬𝑬𝑬𝑬�� beginning at 𝑯𝑯, and label this point as 𝑮𝑮.

Step 5

Join 𝑬𝑬 to 𝑯𝑯 and 𝑬𝑬 to 𝑮𝑮.

In order to apply the parallel method, I must be able to use a setsquare and straightedge.


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When using the parallel method, I should remember that one point will be marked for me (e.g., 𝑍𝑍′), or I will be provided with a scale factor.

Step 1. Draw a ray beginning at 𝑂𝑂 through each vertex of the figure. Step 2. Align the setsquare and ruler; one leg of the setsquare should line up

with side 𝑌𝑌𝑍𝑍��, and the perpendicular leg should be flush against the ruler.

Step 3. Slide the setsquare along the ruler until the edge of the setsquare passes through 𝑍𝑍′. Then, along the perpendicular leg of the setsquare, draw the segment through 𝑍𝑍′ that is parallel to 𝑌𝑌𝑍𝑍�� until it intersects with 𝑂𝑂𝑌𝑌��⃗ , and label this point 𝑌𝑌′.

Step 4. Continue to create parallel segments to determine each successive vertex point.

Step 5. Use your ruler to join the final two unconnected vertices (in the case of a triangle, Step 4 is done just once before moving to Step 5, but with a figure with more vertices, Step 4 may be done several times).

2. Use the parallel method to create a scale drawing of △ 𝑋𝑋𝑌𝑌𝑍𝑍 about center 𝑂𝑂, provided 𝑍𝑍′.


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Lesson 4: Comparing the Ratio Method with the Parallel Method


1. △ 𝑃𝑃𝑃𝑃𝑃𝑃 and △ 𝑄𝑄𝑃𝑃𝑃𝑃 share the same base 𝑃𝑃𝑃𝑃�� so that points 𝑃𝑃 and 𝑄𝑄 lie on a line parallel to 𝑃𝑃𝑃𝑃�⃖��⃗ . Demonstrate why the two triangles have the same area; make additions to the diagram as necessary.

Draw altitude 𝑷𝑷𝑷𝑷′ for △𝑷𝑷𝑷𝑷𝑷𝑷. Draw altitude 𝑸𝑸𝑸𝑸′ for △ 𝑸𝑸𝑷𝑷𝑷𝑷.

Quadrilateral 𝑷𝑷𝑷𝑷′𝑸𝑸′𝑸𝑸 is a parallelogram. This implies tha𝒕𝒕 𝑷𝑷𝑷𝑷′ = 𝑸𝑸𝑸𝑸′. Since the triangles share the same base and have altitudes of equal length, then the areas of the triangles are equal.

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△𝑷𝑷𝑷𝑷𝑷𝑷) =𝟏𝟏𝟐𝟐𝑷𝑷𝑷𝑷 ⋅ 𝑷𝑷𝑷𝑷′ =

𝟏𝟏𝟐𝟐𝑷𝑷𝑷𝑷 ⋅ 𝑸𝑸𝑸𝑸′ = 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△𝑸𝑸𝑷𝑷𝑷𝑷)

In order to show that the triangles have the same area, I need to consider the elements needed to find the area such as the height of each triangle.


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2. In the following diagram, 𝐺𝐺𝐺𝐺𝐺𝐺′��. △ 𝐹𝐹𝐺𝐺𝐺𝐺 and △ 𝐹𝐹𝐺𝐺𝐺𝐺′ have different-length bases, 𝐺𝐺𝐺𝐺�� and 𝐺𝐺𝐺𝐺′��, and share vertex 𝐹𝐹. Demonstrate that the value of the ratio of their areas is equal to the value of the ratio of the lengths of their bases, that is,

Area(△ 𝐹𝐹𝐺𝐺𝐺𝐺)Area(△ 𝐹𝐹𝐺𝐺𝐺𝐺′)

=𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺′.

Make additions to the diagram as necessary.

Draw perpendicular 𝑭𝑭𝑭𝑭′�⃖��⃗ to 𝑮𝑮𝑮𝑮�⃖��⃗ ; 𝑭𝑭𝑭𝑭′�� is altitude to both △ 𝑭𝑭𝑮𝑮𝑮𝑮 and △ 𝑭𝑭𝑮𝑮𝑮𝑮′.

Use the area formula for triangles:

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△𝑭𝑭𝑮𝑮𝑮𝑮)𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑭𝑭𝑮𝑮𝑮𝑮′)

=𝟏𝟏𝟐𝟐𝑮𝑮𝑮𝑮 ⋅ 𝑭𝑭𝑭𝑭′

𝟏𝟏𝟐𝟐𝑮𝑮𝑮𝑮

′ ⋅ 𝑭𝑭𝑭𝑭′=𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮′.

3. In △ 𝑂𝑂𝑂𝑂′𝐵𝐵′, 𝑂𝑂 lies on 𝑂𝑂𝑂𝑂′��, and 𝐵𝐵 lies on 𝑂𝑂𝐵𝐵′��.

a. If 𝑂𝑂𝐵𝐵�� splits 𝑂𝑂𝑂𝑂′�� and 𝑂𝑂𝐵𝐵′�� proportionally, what is implied?

𝑨𝑨𝑨𝑨�� ∥ 𝑨𝑨′𝑨𝑨′��

b. If 𝑂𝑂𝐵𝐵�� ∥ 𝑂𝑂′𝐵𝐵′��, what is implied? Be specific.

𝑨𝑨𝑨𝑨�� is a proportional side splitter, or 𝑶𝑶𝑨𝑨′

𝑶𝑶𝑨𝑨= 𝑶𝑶𝑨𝑨′

𝑶𝑶𝑨𝑨.

By declaring “𝐺𝐺𝐺𝐺𝐺𝐺′��”, we know that 𝐺𝐺, 𝐺𝐺, and 𝐺𝐺′ are collinear.


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Lesson 5: Scale Factors


1. Point 𝑂𝑂 lies on 𝑃𝑃𝑃𝑃�⃖��⃗ . Describe the transformation that occurs based on the diagram below, and state the relationship between pre-image 𝑃𝑃𝑃𝑃�� and image 𝑃𝑃′𝑃𝑃′��.

A dilation is applied to 𝑷𝑷𝑷𝑷�� about 𝑶𝑶 by a scale factor 𝒓𝒓 > 𝟏𝟏. Since 𝑷𝑷, 𝑶𝑶, and 𝑷𝑷 are collinear, the image points remain on 𝑷𝑷𝑷𝑷�⃖��⃗ . By the dilation theorem, 𝑷𝑷′𝑷𝑷′ = 𝒓𝒓 ∙ 𝑷𝑷𝑷𝑷.

2. A dilation is applied to 𝑆𝑆𝑆𝑆�� with a center 𝑂𝑂 that does not lie on 𝑆𝑆𝑆𝑆�⃖��⃗ . Use a scale factor 𝑟𝑟 < 1 to draw a figure that describes this scenario. Use the dilation theorem to describe two facts about 𝑆𝑆′𝑆𝑆′��.

By the dilation theorem, 𝑺𝑺𝑺𝑺�⃖��⃗ ∥ 𝑺𝑺′𝑺𝑺′�⃖��⃗ , and 𝑺𝑺′𝑺𝑺′ = 𝒓𝒓 ∙ 𝑺𝑺𝑺𝑺.


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3. Two different points 𝐷𝐷 and 𝐸𝐸 are dilated from 𝑂𝑂. a. If 𝐷𝐷𝐸𝐸 = 6, and a scale factor 𝑟𝑟 is 1.5, and 𝑂𝑂 is collinear

with 𝐷𝐷 and 𝐸𝐸, use the dilation theorem to describe two facts that are known about 𝐷𝐷′𝐸𝐸′��.

𝑫𝑫′𝑬𝑬′ = 𝟏𝟏.𝟓𝟓 ∙ 𝟔𝟔 = 𝟗𝟗, and 𝑫𝑫𝑬𝑬�⃖��⃗ = 𝑫𝑫′𝑬𝑬′�⃖��⃗ .

b. If a different center, 𝐶𝐶, which is not collinear with 𝐷𝐷 and 𝐸𝐸 is used, what changes about the facts known regarding 𝐷𝐷′𝐸𝐸′��?

𝑫𝑫′𝑬𝑬′ is still 𝟗𝟗, but now 𝑫𝑫𝑬𝑬�⃖��⃗ ∥ 𝑫𝑫′𝑬𝑬′�⃖��⃗ .

I must remember that, by the dilation theorem, a segment that does not contain the center of dilation will map to a parallel segment while a segment that contains the center will map to a segment belonging to the same line.


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Lesson 6: Dilations as Transformations of the Plane


1. We have studied two categories of transformations. What are they, and what distinguishes each one?

We have studied transformations that are distance-preserving (rigid) and a kind of transformation that is not distance-preserving (not rigid)—dilations. Distance-preserving transformations—rotations, reflections, and translations—are such that the distance between any two points of a figure is the same as the distance between the images of those two points. Under a dilation from 𝑶𝑶, a point 𝑷𝑷 is assigned to a point 𝑭𝑭(𝑷𝑷) so that 𝑶𝑶𝑷𝑷′ = 𝒓𝒓 ∙ 𝑶𝑶𝑷𝑷; this means that distances under a dilation are scaled and are, therefore, not preserved.

2. What property do dilations have in common with basic rigid motions?

Both rigid motions and dilations preserve angle measure.

3. Use either a ruler or a compass to dilate ∠𝐴𝐴𝐴𝐴𝐴𝐴 from 𝑂𝑂 by a scale factor of 𝑟𝑟 = 2. Use a protractor to confirm that the dilated angle still has the same angle measure.

Any transformations map lines to lines, rays to rays, segments to segments, and angles to angles.


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4. Write the inverse dilation that will map the image point 𝑃𝑃′ back to 𝑃𝑃.

a. 𝐷𝐷𝑂𝑂,4(𝑃𝑃) 𝑫𝑫𝑶𝑶,𝟏𝟏𝟒𝟒�𝑫𝑫𝑶𝑶,𝟒𝟒(𝑷𝑷)� = 𝑷𝑷

b. 𝐷𝐷𝑂𝑂,34(𝑃𝑃) 𝑫𝑫𝑶𝑶,𝟒𝟒𝟑𝟑

�𝑫𝑫𝑶𝑶,34(𝑷𝑷)� = 𝑷𝑷

c. 𝐷𝐷𝑂𝑂,16(𝑃𝑃) 𝑫𝑫𝑶𝑶,𝟔𝟔 �𝑫𝑫𝑶𝑶,16

(𝑷𝑷)� = 𝑷𝑷

Dilations have inverse functions that return each dilated point back to itself.


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Lesson 7: How Do Dilations Map Segments?


1. Consider a dilation of a segment, 𝐴𝐴𝐴𝐴��, where the center of dilation, 𝑂𝑂, lies on 𝐴𝐴𝐴𝐴�⃖��⃗ . Explain why 𝐴𝐴𝐴𝐴�⃖��⃗ = 𝐴𝐴′𝐴𝐴′�⃖��⃗ . Do all the points between 𝐴𝐴 and 𝐴𝐴 map to all the points between 𝐴𝐴′ and 𝐴𝐴′?

By definition of dilation, 𝑨𝑨 maps to 𝑨𝑨′ along 𝑶𝑶𝑨𝑨��⃗ , and 𝑩𝑩 maps to 𝑩𝑩′ along 𝑶𝑶𝑩𝑩��⃗ ; hence, 𝑨𝑨𝑩𝑩�⃖��⃗ = 𝑨𝑨′𝑩𝑩′�⃖��⃗ . The dilation also maps all of the points between 𝑨𝑨 and 𝑩𝑩 to all of the points between 𝑨𝑨′ and 𝑩𝑩′.

2. Fill in the missing steps of the proof:

Let 𝑂𝑂 be a point not on 𝑃𝑃𝑃𝑃�⃖��⃗ and 𝐷𝐷𝑂𝑂,𝑟𝑟 be a dilation with center 𝑂𝑂 and scale factor 𝑟𝑟 > 1 that sends point 𝑃𝑃

to 𝑃𝑃′ and 𝑃𝑃 to 𝑃𝑃′. If 𝑅𝑅 is another point that lies on 𝑃𝑃𝑃𝑃�⃖��⃗ , then 𝐷𝐷𝑂𝑂,𝑟𝑟(𝑅𝑅) is a point that lies on 𝑃𝑃′𝑃𝑃′�⃖��⃗ .

1. A dilation 𝐷𝐷𝑂𝑂,𝑟𝑟 with center 𝑂𝑂 and scale factor 𝑟𝑟 sends point 𝑃𝑃 to 𝑃𝑃′ and 𝑃𝑃 to 𝑃𝑃′.

Given

2. 𝑂𝑂𝑃𝑃′

𝑂𝑂𝑃𝑃= 𝑂𝑂𝑄𝑄′

𝑂𝑂𝑄𝑄= 𝑟𝑟 By definition of dilation

3. 𝑷𝑷𝑷𝑷 �⃖��⃗ || 𝑷𝑷′𝑷𝑷′�⃖��⃗ By the triangle side splitter theorem

My plan is to use the given definition of dilation, triangle side splitter theorem, and parallel postulate to finish the proof.


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4. Let 𝑅𝑅 be a point on segment 𝑃𝑃𝑃𝑃 between 𝑃𝑃 and 𝑃𝑃. Let 𝑅𝑅′ be the dilation of 𝑅𝑅 by 𝐷𝐷𝑂𝑂,𝑟𝑟.

Given

5. 𝑶𝑶𝑷𝑷′

𝑶𝑶𝑷𝑷= 𝑶𝑶𝑹𝑹′

𝑶𝑶𝑹𝑹= 𝒓𝒓 By definition of dilation

6. 𝑃𝑃′𝑅𝑅′�⃖��⃗ || 𝑃𝑃𝑃𝑃�⃖��⃗ By the triangle side splitter theorem as applied to △ 𝑂𝑂𝑃𝑃𝑅𝑅 and the fact that 𝑃𝑃𝑃𝑃�⃖��⃗ and 𝑃𝑃𝑅𝑅�⃖��⃗ are the same line

7. 𝑃𝑃′𝑅𝑅′�⃖��⃗ = 𝑃𝑃′𝑃𝑃′�⃖��⃗

By the parallel postulate which states that through a given external point 𝑷𝑷′, there is at most one line parallel to a given line 𝑷𝑷𝑷𝑷

8. The point 𝑹𝑹′ lies on 𝑷𝑷′𝑷𝑷′�⃖��⃗ . By Step 7


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3. Parallel segments, 𝑇𝑇𝑇𝑇�� and 𝑉𝑉𝑉𝑉��, in the plane are of different lengths. Will a dilation map 𝑇𝑇𝑇𝑇�� to 𝑉𝑉𝑉𝑉��?

Rays 𝑽𝑽𝑽𝑽��⃗ and 𝑾𝑾𝑾𝑾��⃗ intersect at a point, 𝑶𝑶. Since 𝑽𝑽𝑾𝑾�� ∥ 𝑽𝑽𝑾𝑾�� in △ 𝑶𝑶𝑽𝑽𝑾𝑾, 𝑽𝑽𝑾𝑾�� is a proportional side splitter. If the sides of the triangle are split proportionally, there must be a dilation with the center at 𝑶𝑶 and a scale factor 𝒓𝒓 = 𝑽𝑽𝑶𝑶

𝑽𝑽𝑶𝑶= 𝑾𝑾𝑶𝑶

𝑾𝑾𝑶𝑶 that maps 𝑽𝑽𝑾𝑾�� to 𝑽𝑽𝑾𝑾��.


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M2 GEOMETRY

Lesson 8: How Do Dilations Map Lines, Rays, and Circles?


1. In the diagram below, 𝐴𝐴′𝐵𝐵′��⃗ is the image of 𝐴𝐴𝐵𝐵��⃗ under a dilation from point 𝑂𝑂 with an unknown scale factor; 𝐴𝐴 maps to 𝐴𝐴′, and 𝐵𝐵 maps to 𝐵𝐵′. Use direct measurement to determine the scale factor 𝑟𝑟, and then find the center of dilation 𝑂𝑂.

By the definition of dilation, 𝑨𝑨′𝑩𝑩′ = 𝒓𝒓 (𝑨𝑨𝑩𝑩), 𝑶𝑶𝑨𝑨′ = 𝒓𝒓(𝑶𝑶𝑨𝑨), and 𝑶𝑶𝑩𝑩′ = 𝒓𝒓(𝑶𝑶𝑩𝑩).

By direct measurement 𝑨𝑨′𝑩𝑩′

𝑨𝑨𝑩𝑩= 𝟓𝟓

𝟐𝟐.𝟓𝟓= 𝟐𝟐 = 𝒓𝒓.

The images of 𝑨𝑨 and 𝑩𝑩 are pushed to the left on 𝑨𝑨𝑩𝑩�⃖��⃗ under the dilation, and 𝑨𝑨′𝑩𝑩′ > 𝑨𝑨𝑩𝑩, so the center of dilation must lie on 𝑨𝑨𝑩𝑩�⃖��⃗ to the right of points 𝑨𝑨 and 𝑩𝑩.

By the definition of dilation,

𝑶𝑶𝑨𝑨′ = 𝟐𝟐(𝑶𝑶𝑨𝑨) �𝑶𝑶𝑨𝑨+𝑨𝑨𝑨𝑨′� = 𝟐𝟐�𝑶𝑶𝑨𝑨� 𝑶𝑶𝑨𝑨+𝑨𝑨𝑨𝑨′

𝑶𝑶𝑨𝑨 = 𝟐𝟐 𝑶𝑶𝑨𝑨𝑶𝑶𝑨𝑨+𝑨𝑨𝑨𝑨′

𝑶𝑶𝑨𝑨 = 𝟐𝟐

𝟏𝟏 +𝑨𝑨𝑨𝑨′𝑶𝑶𝑨𝑨

= 𝟐𝟐

𝑨𝑨𝑨𝑨′𝑶𝑶𝑨𝑨 = 𝟏𝟏

𝑨𝑨𝑨𝑨′ = 𝟏𝟏(𝑶𝑶𝑨𝑨)

𝑨𝑨𝑨𝑨′ = 𝑶𝑶𝑨𝑨

I remember from the lesson that the length of 𝑂𝑂𝐴𝐴′�� is the union of the lengths of 𝑂𝑂𝐴𝐴�� and 𝐴𝐴𝐴𝐴′��.

I need to use the definition of dilation and my ruler. I will use centimeters, but I could use inches.


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2. Dilate circle 𝐴𝐴 with radius 𝐴𝐴𝐵𝐵 from center 𝑂𝑂 with a scale factor 𝑟𝑟 = 14.

After I have drawn the rays 𝑂𝑂𝐴𝐴 and 𝑂𝑂𝐵𝐵, I can use my setsquare and the parallel method to dilate the points 𝐴𝐴 and 𝐵𝐵. The segment 𝐴𝐴𝐵𝐵 is the radius of circle 𝐴𝐴, the segment 𝐴𝐴′𝐵𝐵′ will be the dilated radius of the circle 𝐴𝐴′.

I can use my compass to draw a circle with the dilated radius and center 𝐴𝐴′.


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Lesson 9: How Do Dilations Map Angles?


1. Shown below is △ 𝐴𝐴𝐴𝐴𝐴𝐴 and its image △ 𝐴𝐴′𝐴𝐴′𝐴𝐴′ after it has been dilated from center 𝑂𝑂 by scale factor 𝑟𝑟 =

23. Prove that the dilation maps △ 𝐴𝐴𝐴𝐴𝐴𝐴 to △ 𝐴𝐴′𝐴𝐴′𝐴𝐴′ so that 𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐴𝐴′, 𝑚𝑚∠𝐴𝐴 = 𝑚𝑚∠𝐴𝐴′, and 𝑚𝑚∠𝐴𝐴 =𝑚𝑚∠𝐴𝐴′.

Locate the center of dilation 𝑶𝑶 by drawing rays through each of the pairs of corresponding points. The intersection of the rays is the center of dilation 𝑶𝑶. Since dilations map segments to segments, and the dilated segments must either coincide with their pre-image or be parallel, then we know that 𝑨𝑨𝑨𝑨�⃖��⃗ ∥𝑨𝑨′𝑨𝑨′�⃖��⃗ , 𝑨𝑨𝑨𝑨�⃖��⃗ ∥ 𝑨𝑨′𝑨𝑨′�⃖��⃗ , and 𝑨𝑨𝑨𝑨�⃖��⃗ ∥ 𝑨𝑨′𝑨𝑨′�⃖��⃗ .

Then:

- ∠𝑨𝑨′𝑨𝑨′𝑨𝑨′ is congruent to ∠𝑨𝑨𝑨𝑨𝑨𝑨. If parallel lines (𝑨𝑨𝑨𝑨�⃖��⃗ ∥ 𝑨𝑨′𝑨𝑨′�⃖��⃗ ) are cut by a transversal (𝑨𝑨𝑨𝑨�⃖��⃗ ), then corresponding angles are congruent.

- ∠𝑨𝑨′𝑨𝑨′𝑨𝑨′ is congruent to ∠𝑨𝑨𝑨𝑨𝑨𝑨. If parallel lines (𝑨𝑨𝑨𝑨�⃖��⃗ ∥ 𝑨𝑨′𝑨𝑨′�⃖��⃗ ) are cut by a transversal (𝑨𝑨𝑨𝑨�⃖��⃗ ), then corresponding angles are congruent.

- ∠𝑨𝑨′𝑨𝑨′𝑨𝑨 is congruent to ∠𝑨𝑨𝑨𝑨𝑨𝑨. If parallel lines (𝑨𝑨𝑨𝑨�⃖��⃗ ∥ 𝑨𝑨′𝑨𝑨′�⃖��⃗ ) are cut by a transversal (𝑨𝑨𝑨𝑨′�⃖��⃗ ), then alternating interior angles are congruent.

- ∠𝑨𝑨′𝑨𝑨′𝑨𝑨 is congruent to ∠𝑨𝑨′𝑨𝑨′𝑨𝑨′. If parallel lines (𝑨𝑨𝑨𝑨�⃖��⃗ ∥ 𝑨𝑨′𝑨𝑨′�⃖��⃗ ) are cut by a transversal (𝑨𝑨′𝑨𝑨′�⃖��⃗ ), then alternate interior angles are congruent.

By the transitive property, ∠𝑨𝑨𝑨𝑨𝑨𝑨 ≅ ∠𝑨𝑨′𝑨𝑨′𝑨𝑨 ≅ ∠𝑨𝑨′𝑨𝑨′𝑨𝑨′ and ∠𝑨𝑨′𝑨𝑨′𝑨𝑨′ ≅ ∠𝑨𝑨𝑨𝑨𝑨𝑨. Since congruent angles are equal in measure, 𝒎𝒎∠𝑨𝑨′𝑨𝑨′𝑨𝑨′ = 𝒎𝒎∠𝑨𝑨′, and 𝒎𝒎∠𝑨𝑨𝑨𝑨𝑨𝑨 = 𝒎𝒎∠𝑨𝑨; then, 𝒎𝒎∠𝑨𝑨 = 𝒎𝒎∠𝑨𝑨′. Similar reasoning shows that 𝒎𝒎∠𝑨𝑨 = 𝒎𝒎∠𝑨𝑨′, and 𝒎𝒎∠𝑨𝑨 = 𝒎𝒎∠𝑨𝑨′.

I will use what I know about corresponding angles and alternate interior angles being equal in measure when parallel lines are cut by a transversal.


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2. Dilate pentagon 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 from center 𝑂𝑂 using a scale factor of 𝑟𝑟 = 1

2.

A dilation maps a pentagon to a pentagon so that the ratio of corresponding sides is the same and corresponding interior angles are the same measure.

3. Dilate quadrilateral 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 from center 𝑂𝑂 using a scale factor 𝑟𝑟 = 2 12.

A dilation maps a quadrilateral to a quadrilateral so that the ratio of corresponding sides is the same and corresponding interior angles are the same measure.

I know the ratio of the corresponding sides will be the same and the corresponding interior angles will be the same measure in the dilated pentagon.

D′

Since the center of dilation is inscribed in the quadrilateral, and the scale factor is 𝑟𝑟 > 1, I know the dilated quadrilateral will circumscribe the pre-image.


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Lesson 10: Dividing the King’s Foot into 12 Equal Pieces


1. In the following diagram, 𝐴𝐴1,𝐴𝐴2, … ,𝐴𝐴5 have been constructed to be equally spaced, and segments 𝐴𝐴1𝐵𝐵1,𝐴𝐴2𝐵𝐵2, … , 𝐴𝐴5𝐵𝐵 are parallel by construction. Is segment 𝐴𝐴𝐵𝐵 divided into five equal lengths? Justify your answer.

Segments 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏 and 𝑨𝑨𝟐𝟐𝑩𝑩𝟐𝟐 are parallel by construction. By the triangle side splitter theorem, both segments are proportional side splitters of triangle 𝑨𝑨𝑩𝑩𝑨𝑨𝟓𝟓. We also know that 𝑨𝑨𝟏𝟏,𝑨𝑨𝟐𝟐, . . . ,𝑨𝑨𝟓𝟓 are equally spaced by construction. This means that 𝑨𝑨𝑨𝑨𝟏𝟏

𝑨𝑨𝑨𝑨𝟓𝟓= 𝑨𝑨𝑩𝑩𝟏𝟏

𝑨𝑨𝑩𝑩= 𝟏𝟏

𝟓𝟓 and that 𝑨𝑨𝑨𝑨𝟐𝟐

𝑨𝑨𝑨𝑨𝟓𝟓= 𝑨𝑨𝑩𝑩𝟐𝟐

𝑨𝑨𝑩𝑩= 𝟐𝟐

𝟓𝟓. By similar reasoning,

we can show that 𝑨𝑨𝑩𝑩𝟑𝟑𝑨𝑨𝑩𝑩

= 𝟑𝟑𝟓𝟓, and 𝑨𝑨𝑩𝑩𝟒𝟒

𝑨𝑨𝑩𝑩= 𝟒𝟒

𝟓𝟓. Then, 𝑨𝑨𝑩𝑩𝟏𝟏 = 𝟏𝟏

𝟓𝟓𝑨𝑨𝑩𝑩, 𝑨𝑨𝑩𝑩𝟐𝟐 = 𝟐𝟐𝟓𝟓𝑨𝑨𝑩𝑩, 𝑨𝑨𝑩𝑩𝟑𝟑 = 𝟑𝟑

𝟓𝟓𝑨𝑨𝑩𝑩, and 𝑨𝑨𝑩𝑩𝟒𝟒 =𝟒𝟒𝟓𝟓𝑨𝑨𝑩𝑩. Then, segment 𝑨𝑨𝑩𝑩 is divided into five equal lengths.

I did a similar problem like this in class.


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2. Draw a rectangle that is exactly 35 shaded.

Possible solution:

One solution might be to first use the side splitter method to divide a segment into five equal lengths and then build a rectangle around the endpoints of the divided segment.


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3. Use the dilation method to divide 𝑆𝑆𝑆𝑆�� into three equal lengths.

When using the dilation method, I remember to make the ray parallel to the segment clearly longer or shorter than the segment.


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Lesson 11: Dilations from Different Centers


1. Dilate the point 𝑃𝑃 from centers 𝑂𝑂1 and 𝑂𝑂2 by a scale factor of 𝑟𝑟 = 2. What do you notice about the line that passes through 𝑃𝑃1 and 𝑃𝑃2 with respect to the line that passes through 𝑂𝑂1 and 𝑂𝑂2?

Lines 𝑷𝑷𝟏𝟏𝑷𝑷𝟐𝟐 and 𝑶𝑶𝟏𝟏𝑶𝑶𝟐𝟐 are parallel.

The dilation from two different centers with the same scale factor will result in the line passing through the two centers and the line passing through the two dilated points being parallel.


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2. △ 𝐴𝐴1𝐵𝐵1𝐶𝐶1 is a scale drawing of △ 𝐴𝐴𝐵𝐵𝐶𝐶 with scale factor 𝑟𝑟1, and △ 𝐴𝐴2𝐵𝐵2𝐶𝐶2 is a scale drawing of △ 𝐴𝐴1𝐵𝐵1𝐶𝐶1 with scale factor 𝑟𝑟2. a. Determine the scale factor 𝑟𝑟1 for △ 𝐴𝐴1𝐵𝐵1𝐶𝐶1 (relative to △ 𝐴𝐴𝐵𝐵𝐶𝐶).

𝒓𝒓𝟏𝟏 = 𝟑𝟑

b. Determine the scale factor 𝑟𝑟2 for △ 𝐴𝐴2𝐵𝐵2𝐶𝐶2 (relative to △ 𝐴𝐴1𝐵𝐵1𝐶𝐶1).

𝒓𝒓𝟐𝟐 =𝟏𝟏𝟒𝟒

c. Show the calculation to find the value of the scale factor, 𝑟𝑟, of the dilation that takes △ 𝐴𝐴𝐵𝐵𝐶𝐶 to

△ 𝐴𝐴2𝐵𝐵2𝐶𝐶2.

The scale factor that takes △ 𝑨𝑨𝑨𝑨𝑨𝑨 to △ 𝑨𝑨𝟐𝟐𝑨𝑨𝟐𝟐𝑨𝑨𝟐𝟐 is a product of the individual scale factors of the composition of dilations.

𝒓𝒓 = 𝒓𝒓𝟏𝟏𝒓𝒓𝟐𝟐 = 𝟑𝟑𝟒𝟒

I can measure and use the definition of dilation to determine the scale factors.


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3. Segment 𝐴𝐴1𝐵𝐵1 is dilated by a scale factor 𝑟𝑟1 from 𝑂𝑂1 and results in segment 𝐴𝐴2𝐵𝐵2; segment 𝐴𝐴2𝐵𝐵2 is dilated by a scale factor 𝑟𝑟2 from 𝑂𝑂2 and results in segment 𝐴𝐴3𝐵𝐵3. a. Determine the scale factor 𝑟𝑟1 of the dilation that results in scale drawing 𝐴𝐴2𝐵𝐵2.

𝒓𝒓𝟏𝟏 =𝟏𝟏𝟐𝟐

b. Determine the scale factor 𝑟𝑟2 of the dilation that results in scale drawing 𝐴𝐴3𝐵𝐵3.

𝒓𝒓𝟏𝟏 =𝟐𝟐𝟑𝟑

c. Determine the center of the dilation, 𝑂𝑂3, that would map 𝐴𝐴1𝐵𝐵1�� to 𝐴𝐴3𝐵𝐵3��. What is the value of the scale factor, 𝑟𝑟, of this dilation?

𝒓𝒓 = 𝒓𝒓𝟏𝟏𝒓𝒓𝟐𝟐 =𝟏𝟏𝟑𝟑

d. What is significant about the position of 𝑂𝑂3 with respect to 𝑂𝑂1 and 𝑂𝑂2?

𝑶𝑶𝟑𝟑 is collinear with 𝑶𝑶𝟏𝟏 and 𝑶𝑶𝟐𝟐.


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Lesson 12: What Are Similarity Transformations, and Why Do We Need Them?

Lesson 12: What Are Similarity Transformations, and Why Do We

Need Them?

1. What is a similarity transformation?

A similarity transformation is a composition of a finite number of dilations or basic rigid motions. The scale factor of a similarity transformation is the product of the scale factors of the dilations in the composition. If there are no dilations in the composition, the scale factor is defined to be 𝟏𝟏.

A similarity transformation does not need to have a dilation. A square is a special type of rectangle, but the relationship does not work in reverse; a similar such relationship exists between congruence transformations and similarity transformations. A congruence transformation is a similarity transformation where the scale factor is 𝟏𝟏. Similarity transformations generalize the notion of congruency.

2. Figure 𝑃𝑃 is similar to figure 𝑃𝑃′. Describe a similarity transformation that maps 𝑃𝑃 onto 𝑃𝑃′ with as much detail as possible.

Figure 𝑷𝑷 is reflected over a line 𝓵𝓵, yielding 𝑷𝑷𝟏𝟏, and then dilated from a center, 𝑶𝑶, by a scale factor of 𝒓𝒓 > 𝟏𝟏, resulting in 𝑷𝑷′.

I can see that I will have to reflect figure 𝑃𝑃 and then dilate the reflected image to the right size.


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Lesson 12: What Are Similarity Transformations, and Why Do We Need Them?

3. Figure 𝐹𝐹 is similar to figure 𝐹𝐹′. Describe a similarity transformation that maps 𝐹𝐹 onto 𝐹𝐹′ with as much detail as possible.

Figure 𝑭𝑭 is rotated about 𝑪𝑪, yielding the image 𝑭𝑭𝟏𝟏. Then 𝑭𝑭𝟏𝟏 is dilated from a center, 𝑶𝑶, by a scale factor of 𝒓𝒓 < 𝟏𝟏, resulting in 𝑭𝑭𝟐𝟐. Finally, 𝑭𝑭𝟐𝟐 is reflected over a line 𝓵𝓵, resulting as the figure 𝑭𝑭′.

4. Is there a similarity transformation that maps rhombus 𝐹𝐹 to rhombus 𝐺𝐺? Use measurements to justify your answer.

There is no similarity transformation that maps figure 𝑭𝑭 to figure 𝑮𝑮 since any dilation involved in such a transformation would create a scale drawing of figure 𝑭𝑭, which 𝑮𝑮 is not. This can be proven by comparing corresponding sides to corresponding diagonals, which are not in the same proportion.

I will have to use basic rigid motions and dilations to describe the similarity transformation.


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Lesson 13: Properties of Similarity Transformations


Use a compass, protractor, and straightedge for the following problems.

1. Describe the properties shared by transformations.

1) Distinct points are mapped to distinct points.

2) Each point 𝑷𝑷′ in the plane has a pre-image.

3) There is a scale factor 𝒓𝒓 for 𝑮𝑮 so that for any pair of points 𝑷𝑷 and 𝑸𝑸 with images 𝑷𝑷′ = 𝑮𝑮(𝑷𝑷) and 𝑸𝑸′ = 𝑮𝑮(𝑸𝑸), then 𝑷𝑷′𝑸𝑸′ = 𝒓𝒓𝑷𝑷𝑸𝑸.

4) A similarity transformation sends lines to lines, rays to rays, line segments to line segments, and parallel lines to parallel lines.

5) A similarity transformation sends angles to angles of equal measure.

6) A similarity transformation maps a circle of radius 𝑹𝑹 to a circle of radius 𝒓𝒓𝑹𝑹, where 𝒓𝒓 is the scale factor of the similarity transformation.

7) All of the properties are satisfied by a similarity transformation consisting of a single translation, reflection, rotation, or dilation. If the similarity transformation consists of more than one such transformation, then the properties still hold because they hold one step at a time.

I must remember that these properties are true for dilations, rotations, reflections, and translations. Some properties, like property 3, may seem to apply only to dilations, but property 3 is applicable for any transformation because in a congruence, for example, the scale factor is 1, which makes the property true.


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2. The image of segment 𝐴𝐴𝐴𝐴 under a similarity transformation is segment 𝐴𝐴′′𝐴𝐴′′. Apply the similarity transformation if the composition is a dilation from 𝑂𝑂 with scale factor 𝑟𝑟 = 1

3 followed by a reflection over

line ℓ.


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3. Locate point 𝑃𝑃′′′ by applying the following similarity transformation: 𝑟𝑟ℓ �𝐷𝐷𝑂𝑂,2 �𝑅𝑅𝐶𝐶,60˚(𝑃𝑃)��.

The order in which I apply the similarity transformation is rotation, dilation, and then reflection.


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Lesson 14: Similarity



1. Similarity is reflexive. What does this mean? Sketch or describe two examples that support your answer.

Possible solutions:

Example 1

𝑨𝑨𝑨𝑨�� = 𝒓𝒓𝓵𝓵(𝒓𝒓𝓵𝓵(𝑨𝑨𝑨𝑨��))

A sequence of two reflections over line 𝓵𝓵 will map 𝑨𝑨𝑨𝑨�� to itself.

Example 2

𝑨𝑨𝑨𝑨�� = 𝑻𝑻𝑨𝑨″𝑨𝑨��⃗ (𝑻𝑻𝑨𝑨 ′𝑨𝑨″��⃗ �𝑻𝑻𝑨𝑨𝑨𝑨′��⃗ (𝑨𝑨𝑨𝑨��)�)

A sequence of three translations maps 𝑨𝑨𝑨𝑨�� back onto itself.

We learned that similarity is reflexive. That means that a figure in a plane can be mapped to itself by a similarly transformation.


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2. Similarity is symmetric. What does this mean? Describe an example that supports your answer.

When we say similarity is symmetric, we mean if a figure 𝑨𝑨 in the plane is similar to a figure 𝑨𝑨, then it must be true that 𝑨𝑨 is similar to 𝑨𝑨 or that there exists a similarity transformation that will map 𝑨𝑨 to 𝑨𝑨.

This is true because for every composition of transformations, there is a composition that will undo or reverse the first composition. For example, a composition of a dilation by a scale factor of 𝟑𝟑 followed by a rotation of 𝟑𝟑𝟑𝟑° can be undone by a rotation of −𝟑𝟑𝟑𝟑° followed by a dilation by a scale factor of 𝟏𝟏

𝟑𝟑.

3. Similarity is transitive. What does this mean? Describe an example that supports your answer.

When we say similarity is transitive, we mean if a figure 𝑨𝑨 in the plane is similar to a figure 𝑨𝑨, and if 𝑨𝑨 is similar to a third figure 𝑪𝑪, then it must be true that 𝑨𝑨 is similar to 𝑪𝑪 or that there exists a similarity transformation that will map 𝑨𝑨 to 𝑪𝑪.

This is true because if there is a composition of transformations 𝑻𝑻𝟏𝟏 that maps 𝑨𝑨 to 𝑨𝑨 and a composition of transformations 𝑻𝑻𝟐𝟐 that maps 𝑨𝑨 to 𝑪𝑪, then the composition of 𝑻𝑻𝟏𝟏 and 𝑻𝑻𝟐𝟐 together will map 𝑨𝑨 to 𝑪𝑪.

For example, if 𝑻𝑻𝟏𝟏 is a composition of a reflection across a line 𝓵𝓵, followed by a rotation of 𝟗𝟗𝟑𝟑° about a center 𝑪𝑪 and 𝑻𝑻𝟐𝟐 is a composition of a translation by vector 𝑨𝑨𝑨𝑨��⃗ followed by a dilation from 𝑶𝑶 by a scale factor of 𝟐𝟐, then the similarity transformation that will map 𝑨𝑨 to 𝑪𝑪 is

𝑪𝑪 = 𝑫𝑫𝑶𝑶,𝟐𝟐 �𝑻𝑻𝑨𝑨𝑨𝑨��⃗ �𝑹𝑹𝑪𝑪,𝟗𝟗𝟑𝟑°�𝒓𝒓𝓵𝓵(𝑨𝑨)��.

Once a similarity transformation is determined to take a figure to another, there is an inverse transformation that will take the figure back to the original.


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4. A correspondence exists between △ 𝐸𝐸𝐸𝐸𝐸𝐸 and △ 𝑋𝑋𝑋𝑋𝑋𝑋 so that △ 𝐸𝐸𝐸𝐸𝐸𝐸 ↔ △ 𝑋𝑋𝑋𝑋𝑋𝑋. Under this correspondence, 𝑚𝑚∠𝐸𝐸 = 𝑚𝑚∠𝑋𝑋, 𝑚𝑚∠𝐸𝐸 = 𝑚𝑚∠𝑋𝑋, 𝑚𝑚∠𝐸𝐸 = 𝑚𝑚∠𝑋𝑋, and 𝑋𝑋𝑋𝑋

𝐸𝐸𝐸𝐸= 𝑋𝑋𝑌𝑌

𝐸𝐸𝐹𝐹= 𝑌𝑌𝑋𝑋

𝐹𝐹𝐸𝐸. Demonstrate why

△ 𝐸𝐸𝐸𝐸𝐸𝐸 ~ △ 𝑋𝑋𝑋𝑋𝑋𝑋.

Since corresponding side lengths of similar triangles are proportional, a dilation exists so that the scale factor is 𝒓𝒓 = 𝑿𝑿𝑿𝑿

𝑬𝑬𝑬𝑬= 𝑿𝑿𝒀𝒀

𝑬𝑬𝑭𝑭= 𝒀𝒀𝑿𝑿

𝑭𝑭𝑬𝑬. This means that the dilation with scale factor 𝒓𝒓 maps △ 𝑬𝑬𝑬𝑬𝑭𝑭 to

△ 𝑬𝑬′𝑬𝑬′𝑭𝑭′so that 𝑬𝑬′𝑬𝑬′ = 𝒓𝒓𝑬𝑬𝑬𝑬 or 𝑬𝑬′𝑬𝑬′ = 𝑿𝑿𝑿𝑿𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬, 𝑬𝑬′𝑭𝑭′ = 𝑿𝑿𝒀𝒀

𝑬𝑬𝑭𝑭𝑬𝑬𝑭𝑭, and 𝑭𝑭′𝑬𝑬′ = 𝒀𝒀𝑿𝑿

𝑭𝑭𝑬𝑬𝑭𝑭𝑬𝑬. This means that

△ 𝑬𝑬′𝑬𝑬′𝑭𝑭′ ≅ △𝑿𝑿𝑿𝑿𝒀𝒀. Since the triangles are congruent, there must exist a congruence that maps △ 𝑬𝑬′𝑬𝑬′𝑭𝑭′ to △𝑿𝑿𝑿𝑿𝒀𝒀. Then together, the dilation and the congruence imply that a similarity transformation exists that maps △ 𝑬𝑬𝑬𝑬𝑭𝑭 to △ 𝑿𝑿𝑿𝑿𝒀𝒀, and so △ 𝑬𝑬𝑬𝑬𝑭𝑭 ~ △ 𝑿𝑿𝑿𝑿𝒀𝒀.

Similarity is symmetric because once a similarity transformation is determined to take a figure to another, there are inverse transformations that will take the figure back to the original.

I will use my definition of similarity to show why △ 𝐸𝐸𝐸𝐸𝐸𝐸 ~ △ 𝑋𝑋𝑋𝑋𝑋𝑋.


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Lesson 15: The Angle-Angle (AA) Criterion for Two Triangles to Be Similar

Lesson 15: The Angle-Angle (AA) Criterion for Two Triangles to Be

Similar


1. State the angle-angle (AA) criterion for similarity. Why is it necessary to establish that only two pairs of angles are equal in measure?

Two triangles with two pairs of corresponding angles of equal measure are similar. If two corresponding pairs of angles of two triangles are equal in measure, then, by the triangle sum theorem, the third pair of corresponding angles must also be equal in measure. Therefore, it is only necessary to establish that two pairs of angles are equal in measure.

2. Are the following triangles similar?

The triangles are not similar. They each have one angle with a measure of 𝟗𝟗𝟗𝟗°, but the remaining two angles are not equal in measure between the two triangles. By the triangle sum theorem, the remaining angle measures of △ 𝑬𝑬𝑬𝑬𝑬𝑬 are 𝟓𝟓𝟓𝟓° and 𝟓𝟓𝟑𝟑°, and the remaining angle measures of △ 𝑿𝑿𝑿𝑿𝑿𝑿 are 𝟓𝟓𝟓𝟓° and 𝟓𝟓𝟑𝟑°.

I can use my answer from Problem 1 to help me with this problem.


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3. Are the following triangles similar? If so, explain why, and solve for the values of 𝑥𝑥 and 𝑦𝑦.

Vertical angles ∠𝑬𝑬𝑮𝑮𝑮𝑮 and ∠𝑲𝑲𝑮𝑮𝑲𝑲 are equal in measure, and since 𝑬𝑬𝑮𝑮�� ∥ 𝑲𝑲𝑲𝑲��, then 𝒎𝒎∠𝑮𝑮𝑬𝑬𝑮𝑮 = 𝒎𝒎∠𝑮𝑮𝑲𝑲𝑲𝑲, and 𝒎𝒎∠𝑮𝑮𝑲𝑲𝑲𝑲 = 𝒎𝒎∠𝑮𝑮𝑮𝑮𝑬𝑬 because if parallel lines are cut by a transversal, alternate interior angles are equal in measure. Then, by the AA criterion, the triangles are similar.

Since the triangles are similar, I can now solve for the values of 𝒙𝒙 and 𝒚𝒚.

𝒚𝒚𝟗𝟗.𝟓𝟓

=𝟗𝟗𝟓𝟓

𝒚𝒚 = 𝟗𝟗.𝟕𝟕𝟓𝟓

𝟕𝟕.𝟓𝟓𝒙𝒙

=𝟗𝟗𝟓𝟓

𝒙𝒙 = 𝟓𝟓

My plan is to use the AA criterion. I will use what I know about angles formed when parallel lines are cut by a transversal and vertical angles.


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4. Is △ 𝐸𝐸𝐸𝐸𝐸𝐸 similar to △ 𝐸𝐸𝐸𝐸𝐸𝐸? Explain. If the triangles are similar, find the values of 𝑥𝑥 and 𝑦𝑦.

△ 𝑬𝑬𝑬𝑬𝑬𝑬 is similar to △ 𝑬𝑬𝑬𝑬𝑬𝑬 by the AA criterion for similarity. Both triangles share ∠𝑬𝑬, and since 𝑬𝑬𝑬𝑬�� ∥ 𝑬𝑬𝑬𝑬��, 𝒎𝒎∠𝑬𝑬𝑬𝑬𝑬𝑬 = 𝒎𝒎∠𝑬𝑬𝑬𝑬𝑬𝑬, and 𝒎𝒎∠𝑬𝑬𝑬𝑬𝑬𝑬 = 𝒎𝒎∠𝑬𝑬𝑬𝑬𝑬𝑬 (if parallel lines are cut by a transversal, then corresponding angles are equal in measure).

𝟐𝟐𝟑𝟑𝟐𝟐

= 𝒙𝒙(𝒙𝒙 + 𝟑𝟑𝟓𝟓.𝟕𝟕𝟓𝟓)

𝟐𝟐𝒙𝒙 + 𝟐𝟐𝟕𝟕.𝟓𝟓 = 𝟑𝟑𝟐𝟐𝒙𝒙

𝟐𝟐𝟕𝟕.𝟓𝟓 = 𝟑𝟑𝟑𝟑𝒙𝒙

𝒙𝒙 = 𝟐𝟐.𝟕𝟕𝟓𝟓

𝟐𝟐𝟑𝟑𝟐𝟐

= 𝟑𝟑.𝟖𝟖𝒚𝒚

𝟐𝟐𝒚𝒚 = 𝟐𝟐𝟑𝟑.𝟗𝟗

𝒚𝒚 = 𝟑𝟑𝟑𝟑.𝟖𝟖


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Lesson 16: Between-Figure and Within-Figure Ratios


1. △ 𝐷𝐷𝐷𝐷𝐷𝐷 ~ △ 𝐴𝐴𝐴𝐴𝐴𝐴 All side length measurements are in centimeters. Use between-figure ratios to

determine the unknown side lengths.

Using the given similarity statement, ∠𝑫𝑫 corresponds with ∠𝑨𝑨, and ∠𝑪𝑪 corresponds with ∠𝑭𝑭, so it follows that 𝑨𝑨𝑨𝑨�� corresponds with 𝑫𝑫𝑫𝑫��, 𝑨𝑨𝑪𝑪�� with 𝑫𝑫𝑭𝑭��, and 𝑨𝑨𝑪𝑪�� with 𝑫𝑫𝑭𝑭��.

𝑨𝑨𝑨𝑨𝑫𝑫𝑫𝑫=𝑨𝑨𝑪𝑪

𝑫𝑫𝑭𝑭

𝟑𝟑.𝟒𝟒𝟏𝟏.𝟕𝟕= 𝟒𝟒.𝟕𝟕

𝑫𝑫𝑭𝑭

𝟑𝟑.𝟒𝟒(𝑫𝑫𝑭𝑭) = 𝟕𝟕.𝟗𝟗𝟗𝟗 𝑫𝑫𝑭𝑭 = 𝟐𝟐.𝟑𝟑𝟑𝟑

𝑨𝑨𝑨𝑨𝑫𝑫𝑫𝑫= 𝑨𝑨𝑪𝑪

𝑫𝑫𝑭𝑭

𝟑𝟑.𝟒𝟒𝟏𝟏.𝟕𝟕= 𝑨𝑨𝑪𝑪

𝟐𝟐.𝟑𝟑

𝟏𝟏.𝟕𝟕(𝑨𝑨𝑪𝑪) = 𝟕𝟕.𝟖𝟖𝟐𝟐 𝑨𝑨𝑪𝑪 = 𝟒𝟒.𝟔𝟔

The lengths of 𝑫𝑫𝑭𝑭�� and 𝑨𝑨𝑪𝑪�� are 𝟐𝟐.𝟑𝟑𝟑𝟑 centimeters and 𝟒𝟒.𝟔𝟔 centimeters, respectively.

2. Given △ 𝐴𝐴𝐴𝐴𝐴𝐴 ~ △ 𝑋𝑋𝑋𝑋𝑋𝑋, answer the following questions: a. Write and find the value of the ratio that compares

the height 𝐴𝐴𝐴𝐴�� to the hypotenuse of △ 𝐴𝐴𝐴𝐴𝐴𝐴.

𝟔𝟔𝟏𝟏𝟏𝟏

=𝟑𝟑𝟑𝟑

b. Write and find the value of the ratio that compares the base 𝐴𝐴𝐴𝐴�� to the hypotenuse of △ 𝐴𝐴𝐴𝐴𝐴𝐴.

𝟖𝟖𝟏𝟏𝟏𝟏

=𝟒𝟒𝟑𝟑

c. Write and find the value of the ratio that compares the height 𝐴𝐴𝐴𝐴�� to the base 𝐴𝐴𝐴𝐴�� of △ 𝐴𝐴𝐴𝐴𝐴𝐴.

𝟔𝟔𝟖𝟖

=𝟑𝟑𝟒𝟒

I know that with between-figure ratios, one number comes from one triangle, and the other number in the ratio comes from a different similar triangle.


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d. Use within-figure ratios to find the corresponding height of △ 𝑋𝑋𝑋𝑋𝑋𝑋.

𝑨𝑨𝑨𝑨𝑨𝑨𝑪𝑪 =𝑿𝑿𝑿𝑿

𝑿𝑿𝒀𝒀

𝟑𝟑𝟒𝟒

=𝑿𝑿𝑿𝑿𝟒𝟒.𝟖𝟖

𝟒𝟒(𝑿𝑿𝑿𝑿) = 𝟏𝟏𝟒𝟒.𝟒𝟒 𝑿𝑿𝑿𝑿 = 𝟑𝟑.𝟔𝟔

e. Use within-figure ratios to find the hypotenuse of △ 𝑋𝑋𝑋𝑋𝑋𝑋.

𝑨𝑨𝑨𝑨𝑨𝑨𝑪𝑪 =𝑿𝑿𝑿𝑿

𝑿𝑿𝒀𝒀

𝟑𝟑𝟑𝟑

=𝟑𝟑.𝟔𝟔𝑿𝑿𝒀𝒀

𝟑𝟑(𝑿𝑿𝒀𝒀) = 𝟏𝟏𝟖𝟖

𝑿𝑿𝒀𝒀 = 𝟔𝟔

I can use my answer from part (d) to determine the hypotenuse of △ 𝑋𝑋𝑋𝑋𝑋𝑋.

I know that for within-figure ratios, one ratio contains numbers that represent side lengths from one triangle, and the second ratio contains numbers that represent the side lengths from a second similar triangle.


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3. Pete is wondering how tall the big screen is that rests on a 100-foot tall platform. Pete asks his friend Marci, who is 5 ft. 9 in. tall, to stand approximately 18 yards from the base of the platform. Lying on the ground, Pete visually aligns the top of Marci’s head with the top of the big screen and marks his location on the ground approximately 2 ft. 3 in. from his friend. Use Pete’s measurements to approximate the height of the big screen.

Pete’s location on the ground is approximately 𝟑𝟑𝟔𝟔.𝟐𝟐𝟑𝟑 𝐟𝐟𝐟𝐟. from the base of the platform. His visual line forms two similar right triangles with the height of the big screen and platform and the height of his friend. Let 𝒙𝒙 represent the height of the big screen.

𝟑𝟑𝟔𝟔.𝟐𝟐𝟑𝟑𝟐𝟐.𝟐𝟐𝟑𝟑

=𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟑𝟑.𝟕𝟕𝟑𝟑

𝟑𝟑𝟐𝟐𝟑𝟑.𝟒𝟒𝟑𝟑𝟕𝟕𝟑𝟑 = 𝟐𝟐.𝟐𝟐𝟑𝟑𝒙𝒙 + 𝟐𝟐𝟐𝟐𝟑𝟑 𝟗𝟗𝟖𝟖.𝟒𝟒𝟑𝟑𝟕𝟕𝟑𝟑 = 𝟐𝟐.𝟐𝟐𝟑𝟑𝒙𝒙 𝟒𝟒𝟑𝟑.𝟕𝟕𝟑𝟑 = 𝒙𝒙

The height of the big screen is about 𝟒𝟒𝟑𝟑.𝟕𝟕𝟑𝟑 𝐟𝐟𝐟𝐟. (or 𝟒𝟒𝟑𝟑 𝐟𝐟𝐟𝐟.𝟗𝟗 𝐢𝐢𝐢𝐢.).

The height of the larger triangle is 100 added to the height of the big screen.

I need to work in the same units. I use feet.

I know the two nested triangles formed are similar by AA criterion.

I can use between-figure ratios or within-figure ratios.


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Lesson 17: The Side-Angle-Side (SAS) and Side-Side-Side (SSS) Criteria for Two Triangles to Be Similar

Lesson 17: The Side-Angle-Side (SAS) and Side-Side-Side (SSS)

Criteria for Two Triangles to Be Similar

1. State which of the three triangles, if any, are similar and why.

Triangles 𝑨𝑨 and 𝑪𝑪 are similar because they share three pairs of corresponding sides that are in the same ratio.

𝟑𝟑.𝟔𝟔𝟐𝟐

=𝟏𝟏𝟏𝟏.𝟖𝟖𝟔𝟔

=𝟏𝟏𝟐𝟐.𝟔𝟔𝟕𝟕

=𝟗𝟗𝟓𝟓

Triangles 𝑨𝑨 and 𝑩𝑩 are similar because the ratios of their corresponding sides are in the same ratio.

𝟑𝟑.𝟔𝟔𝟑𝟑

=𝟏𝟏𝟏𝟏.𝟖𝟖𝟗𝟗

=𝟏𝟏𝟐𝟐.𝟔𝟔𝟏𝟏𝟏𝟏.𝟓𝟓

=𝟔𝟔𝟓𝟓

Further, if triangle 𝑨𝑨 is similar to triangle 𝑪𝑪, and triangle 𝑨𝑨 is similar to 𝑩𝑩, then triangle 𝑩𝑩 is similar to triangle 𝑪𝑪.

I need to check the corresponding sides of all the triangles (i.e., 𝐴𝐴 and 𝐵𝐵, 𝐴𝐴 and 𝐶𝐶, and 𝐵𝐵 and 𝐶𝐶) to determine if I can use the SSS theorem.

This is true by the transitive property of similarity.


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2. State which of the four triangles, if any, are similar and why.

Triangles 𝑨𝑨 and 𝑩𝑩 are the only similar triangles because they have the same angle measures. Using the angle sum of a triangle, both triangles 𝑨𝑨 and 𝑩𝑩 have angles of 𝟑𝟑𝟏𝟏°, 𝟒𝟒𝟓𝟓°, and 𝟏𝟏𝟏𝟏𝟓𝟓°.

3. State which of the three triangles, if any, are similar and why.

Triangles 𝑨𝑨 and 𝑪𝑪 are similar because they have two pairs of corresponding sides that are in the same ratio, and their included angles are equal measures. Triangle 𝑩𝑩 cannot be shown to be similar because, even though it has two sides that are the same length as two sides of triangle 𝑨𝑨, the 𝟒𝟒𝟏𝟏° angle in triangle 𝑩𝑩 is not the included angle and, therefore, does not correspond to the 𝟒𝟒𝟏𝟏° angle in triangle 𝑨𝑨.

If I have two known sides, the angle must be the included angle. That means the ratio of the corresponding adjacent sides of the congruent angle must be the same to use the SAS theorem.

I will use the angle sum of a triangle to determine the missing angle in each triangle. I will use the AA criterion for similarity where applicable.


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4. For the pair of similar triangles below, determine the unknown lengths of the sides labeled with letters.

The ratios of corresponding sides must be equal, so 𝟒𝟒𝟏𝟏𝟐𝟐𝒏𝒏

= 𝟖𝟖𝟔𝟔, giving 𝒏𝒏 = 𝟑𝟑 𝟑𝟑

𝟖𝟖. Likewise, 𝒎𝒎

𝟑𝟑𝟑𝟑𝟒𝟒= 𝟖𝟖

𝟔𝟔, giving

𝒎𝒎 = 𝟓𝟓.

I will use the marked congruent angles to help determine the corresponding sides to set up my ratios.


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Lesson 18: Similarity and the Angle Bisector Theorem


1. In the following diagram, 𝐵𝐵𝐵𝐵��⃗ bisects ∠𝐵𝐵, and 𝐴𝐴𝐴𝐴�⃖��⃗ is parallel to 𝐵𝐵𝐵𝐵�⃖��⃗ . By the angle bisector theorem, since 𝐵𝐵𝐵𝐵��⃗ bisects ∠𝐵𝐵, the value of the ratio 𝐴𝐴𝐵𝐵:𝐵𝐵𝐵𝐵 should be equal to the value of the ratio 𝐴𝐴𝐵𝐵:𝐵𝐵𝐵𝐵. Answer the following guided questions that help illustrate why this is true.

a. We can get close to showing that 𝐴𝐴𝐵𝐵:𝐵𝐵𝐵𝐵 = 𝐴𝐴𝐵𝐵:𝐵𝐵𝐵𝐵 by showing that 𝐵𝐵𝐵𝐵:𝐵𝐵𝐵𝐵 = 𝐴𝐴𝐴𝐴:𝐴𝐴𝐵𝐵, a relationship that can be established once it is shown that △ 𝐵𝐵𝐵𝐵𝐵𝐵 ~ △ 𝐴𝐴𝐵𝐵𝐴𝐴. What information establishes that △ 𝐵𝐵𝐵𝐵𝐵𝐵 ~ △ 𝐴𝐴𝐵𝐵𝐴𝐴? ∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑬𝑬𝑩𝑩𝑬𝑬 are equal in measure since they are vertical angles, and ∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑩𝑩𝑬𝑬𝑬𝑬 are equal in measure since they are alternate interior angles along parallel lines cut by a transversal. Then by the AA theorem, △ 𝑩𝑩𝑩𝑩𝑩𝑩 ~ △ 𝑬𝑬𝑩𝑩𝑬𝑬. Since the triangles are similar, their corresponding sides have proportional lengths.

b. Why is it important to establish the relationship 𝐵𝐵𝐵𝐵:𝐵𝐵𝐵𝐵 = 𝐴𝐴𝐴𝐴:𝐴𝐴𝐵𝐵?

Out of the four lengths in the relationship, three lengths (𝑩𝑩𝑩𝑩, 𝑩𝑩𝑩𝑩, and 𝑬𝑬𝑩𝑩) are in the relationship we are trying to demonstrate (𝑬𝑬𝑩𝑩:𝑩𝑩𝑩𝑩 = 𝑬𝑬𝑩𝑩:𝑩𝑩𝑩𝑩).

c. How can we establish a relationship that yields information on the length of 𝐴𝐴𝐵𝐵��? We know that ∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑩𝑩𝑬𝑬𝑬𝑬 are equal in measure (alternate interior angles) and that the measures of ∠𝑬𝑬𝑩𝑩𝑩𝑩 and ∠𝑩𝑩𝑩𝑩𝑩𝑩 are equal since they are the bisected halves of ∠𝑩𝑩. This means that ∠𝑬𝑬𝑩𝑩𝑩𝑩 and ∠𝑩𝑩𝑬𝑬𝑬𝑬 are equal in measure. By the converse of the base angles of an isosceles triangle theorem, it follows that △ 𝑬𝑬𝑩𝑩𝑬𝑬 is isosceles, and 𝑬𝑬𝑩𝑩 = 𝑬𝑬𝑬𝑬.

It is helpful to clearly mark any angles of equal measure in such a diagram.


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d. Is it now possible to show that 𝐴𝐴𝐵𝐵:𝐵𝐵𝐵𝐵 = 𝐴𝐴𝐵𝐵:𝐵𝐵𝐵𝐵? Since 𝑬𝑬𝑩𝑩 = 𝑬𝑬𝑬𝑬, 𝑬𝑬𝑩𝑩 can be substituted into the relationship in part (b) so that 𝑩𝑩𝑩𝑩:𝑩𝑩𝑩𝑩 = (𝑬𝑬𝑩𝑩):𝑬𝑬𝑩𝑩. This can be rewritten as 𝑬𝑬𝑩𝑩:𝑩𝑩𝑩𝑩 = 𝑬𝑬𝑩𝑩:𝑩𝑩𝑩𝑩.

2. In the following triangle, 𝐵𝐵𝐴𝐴��⃗ bisects ∠𝐵𝐵 and intersects 𝐴𝐴𝐵𝐵�� at point 𝐴𝐴. Find the lengths of 𝑥𝑥 and 𝑦𝑦.

𝒙𝒙 + 𝒚𝒚 = 𝟖𝟖

𝒙𝒙 = 𝟖𝟖 − 𝒚𝒚

𝟕𝟕𝟏𝟏𝟏𝟏

=𝒙𝒙𝒚𝒚

𝟕𝟕𝒚𝒚 = 𝟏𝟏𝟏𝟏𝒙𝒙 𝟕𝟕𝒚𝒚 = 𝟏𝟏𝟏𝟏(𝟖𝟖 − 𝒚𝒚) 𝟕𝟕𝒚𝒚 = 𝟖𝟖𝟏𝟏 − 𝟏𝟏𝟏𝟏𝒚𝒚 𝟏𝟏𝟕𝟕𝒚𝒚 = 𝟖𝟖𝟏𝟏

𝒚𝒚 =𝟖𝟖𝟏𝟏𝟏𝟏𝟕𝟕

= 𝟒𝟒𝟏𝟏𝟏𝟏𝟏𝟏𝟕𝟕

𝒙𝒙 = 𝟖𝟖 − �𝟒𝟒𝟏𝟏𝟏𝟏𝟏𝟏𝟕𝟕� = 𝟑𝟑

𝟓𝟓𝟏𝟏𝟕𝟕

The length of 𝒙𝒙 is 𝟑𝟑 𝟓𝟓𝟏𝟏𝟕𝟕

, and the length of 𝒚𝒚 is 𝟒𝟒 𝟏𝟏𝟏𝟏𝟏𝟏𝟕𝟕

.

Rewriting this relationship will require several algebraic properties.


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Lesson 19: Families of Parallel Lines and the Circumference of the Earth

Lesson 19: Families of Parallel Lines and the Circumference of the

Earth

1. In the following triangle, 𝐷𝐷𝐷𝐷�� ∥ 𝐴𝐴𝐴𝐴��.

a. Does 𝐷𝐷𝐷𝐷�� split the sides 𝐴𝐴𝐴𝐴�� and 𝐴𝐴𝐴𝐴�� proportionally? By the triangle side splitter theorem, since 𝑫𝑫𝑫𝑫�� is parallel to 𝑨𝑨𝑨𝑨��, it also splits sides 𝑨𝑨𝑨𝑨�� and 𝑨𝑨𝑨𝑨�� proportionally.

b. Show that 𝑚𝑚:𝑚𝑚′ = 𝑛𝑛:𝑛𝑛′ is equivalent to 𝑚𝑚:𝑛𝑛 = 𝑚𝑚′:𝑛𝑛′.

𝒎𝒎𝒎𝒎′

=𝒏𝒏𝒏𝒏′

𝒎𝒎𝒏𝒏′ = 𝒎𝒎′𝒏𝒏 𝒎𝒎𝒏𝒏′𝒏𝒏𝒏𝒏′

=𝒎𝒎′𝒏𝒏𝒏𝒏𝒏𝒏′

𝒎𝒎𝒏𝒏

=𝒎𝒎′𝒏𝒏′


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2. In the following figure, lines that appear to be parallel are in fact parallel. Show two different solutions to solve for the length 𝑥𝑥.

𝟏𝟏𝟏𝟏𝟓𝟓

=𝟏𝟏𝟏𝟏𝒙𝒙

𝟏𝟏𝟏𝟏𝒙𝒙 = 𝟔𝟔𝟔𝟔

𝒙𝒙 =𝟑𝟑𝟔𝟔𝟕𝟕

𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

=𝟓𝟓𝒙𝒙

𝟏𝟏𝟏𝟏𝒙𝒙 = 𝟔𝟔𝟔𝟔

𝒙𝒙 =𝟑𝟑𝟔𝟔𝟕𝟕

The length 𝒙𝒙 is equal to 𝟑𝟑𝟔𝟔𝟕𝟕

.

I must remember the following theorem: If parallel lines are intersected by two transversals, then the ratios of the segments determined along each transversal between the parallel lines are equal.


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3. In the following figure, lines that appear to be parallel are in fact parallel. 𝑉𝑉𝑉𝑉 = 4.5, 𝑉𝑉𝑊𝑊 = 8, and 𝑆𝑆𝑆𝑆 = 10. Find 𝑆𝑆𝑆𝑆 and 𝑆𝑆𝑆𝑆.

Let 𝑻𝑻𝑻𝑻 be 𝒙𝒙, and 𝑺𝑺𝑻𝑻 be (𝟏𝟏𝟔𝟔 − 𝒙𝒙).

𝟏𝟏.𝟓𝟓𝟖𝟖

=𝟏𝟏𝟔𝟔 − 𝒙𝒙𝒙𝒙

𝟖𝟖𝟔𝟔 − 𝟖𝟖𝒙𝒙 = 𝟏𝟏.𝟓𝟓𝒙𝒙 𝟏𝟏𝟏𝟏.𝟓𝟓𝒙𝒙 = 𝟖𝟖𝟔𝟔

𝒙𝒙 = 𝟔𝟔.𝟏𝟏

𝑻𝑻𝑻𝑻 is equal to 𝟔𝟔.𝟏𝟏, and 𝑺𝑺𝑻𝑻 is equal to �𝟏𝟏𝟔𝟔 − (𝟔𝟔.𝟏𝟏)�, or 𝟑𝟑.𝟔𝟔.


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4. In Eratosthenes’ quest to determine the circumference of the earth, we hypothesized that he found the angle between the sun’s rays and the perpendicular to the ground by using a pole (assumed to also be perpendicular to the ground). Using the pole’s shadow and a tool like a protractor, he measured the angle to be 7.2°. Why is the height of the pole in this measurement irrelevant? Use a diagram to support your answer.

The height of the pole used in the measurement is irrelevant because the combination of the way the sun’s rays meet the ground and the perpendicularity of a pole to the ground together guarantee similar triangles formed by the rays, poles, and shadows of poles. As shown in the diagram, △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨𝑫𝑫𝑨𝑨 each share an angle at ∠𝑨𝑨, and each triangle has a 𝟗𝟗𝟔𝟔° angle. Then, △ 𝑨𝑨𝑨𝑨𝑨𝑨 and △ 𝑨𝑨𝑫𝑫𝑨𝑨 are similar by the AA theorem. The respective shadows and poles are proportional in length. Therefore, regardless of the length of the pole, the use of its shadow will help determine the 𝟕𝟕.𝟏𝟏° angle formed by the sun’s ray and the pole.


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Lesson 20: How Far Away Is the Moon?


1. In order for a spherical object to appear as though it is just blocking out the sun, the object must be held from the eye at a length of 108 times the diameter of the spherical object. Approximately how many feet from the eye would the following objects need to be to just block out the sun: a. A basketball?

A basketball has a diameter of approximately 𝟗𝟗.𝟓𝟓𝟓𝟓 inches. To just block out the sun, it must be a distance of 𝟏𝟏𝟏𝟏𝟏𝟏(𝟗𝟗.𝟓𝟓𝟓𝟓) inches from the eye, or approximately 𝟏𝟏𝟖𝟖 feet from the eye.

b. A ping pong ball?

A ping pong ball has a diameter of approximately 𝟏𝟏.𝟓𝟓𝟓𝟓 inches. To just block out the sun, it must be a distance of 𝟏𝟏𝟏𝟏𝟏𝟏(𝟏𝟏.𝟓𝟓𝟓𝟓) inches from the eye, or approximately 𝟏𝟏𝟏𝟏 feet from the eye.

c. Describe what would happen if you held the basketball a distance less than the answer you found in part (a). Describe what would happen if held it a distance greater than what you found in part (a).

If the basketball were held less than 𝟏𝟏𝟖𝟖 feet from the eye, it would more than block out the sun, while holding it at a distance greater than 𝟏𝟏𝟖𝟖 feet would not block out the sun. Instead, the basketball would appear as a dark disk against the light of the sun. Do not attempt this!

d. Since the moon just blocks out the sun during a solar eclipse, what can be concluded about the distance of the moon from Earth?

The distance of the moon from Earth must be 𝟏𝟏𝟏𝟏𝟏𝟏 times the moon’s diameter.

During a solar eclipse, the moon passes between the earth and the sun. During a lunar eclipse, the earth passes between the sun and the moon.


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2. Based on a few assumptions made and a clever model, the ancient Greeks were able to approximate the distance between the moon and the earth. Use what you learned in class to answer the following questions on the model the Greeks developed.

a. What does the term “1 unit” refer to?

The diameter of the moon

b. How was it established that 𝐷𝐷𝐷𝐷 = 2.5 units?

Observation of the time needed for the moon to pass through the earth’s shadow during a total lunar eclipse yielded that the diameter of the cross section of the earth’s conical shadow at the distance of the moon is about 𝟐𝟐 𝟏𝟏

𝟐𝟐 moon diameters.

c. It is critical to establish 𝐴𝐴𝐴𝐴𝐴𝐴𝐷𝐷 as a parallelogram in order to approximate the distance between the

moon and the earth. Explain how the quadrilateral can be determined as a parallelogram.

It can be shown that the opposite sides of the quadrilateral are both parallel.

𝑨𝑨𝑨𝑨�� ∥ 𝑩𝑩𝑩𝑩�� since the diameter of the earth is parallel to the segment formed by the moon’s diameter and the diameter of the shadow at the distance of the moon.

It is assumed that the conical shadows of celestial bodies are similar; therefore, the two-dimensional profile view, which are isosceles triangles, are also similar. Then, 𝒎𝒎∠𝑨𝑨𝑩𝑩𝑨𝑨 = 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 since the shadows are similar triangles, which implies that 𝑨𝑨𝑩𝑩�� ∥ 𝑩𝑩𝑨𝑨�� (alternate interior angles).

d. How was the earth’s diameter (roughly 8,000 miles) used to approximate the distance to the moon?

Once it was established that 𝑨𝑨𝑩𝑩𝑩𝑩𝑨𝑨 was a parallelogram, and 𝑨𝑨𝑨𝑨 = 𝟑𝟑.𝟓𝟓 units, the following could be written:

𝟑𝟑.𝟓𝟓 units ≈ 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 miles 𝟏𝟏 unit ≈ 𝟐𝟐𝟑𝟑𝟏𝟏𝟏𝟏 miles

This means that the diameter of the moon was approximated at 𝟐𝟐,𝟑𝟑𝟏𝟏𝟏𝟏 miles, and the distance to the earth was 𝟏𝟏𝟏𝟏𝟏𝟏 times that, or 𝟏𝟏𝟏𝟏𝟏𝟏(𝟐𝟐𝟑𝟑𝟏𝟏𝟏𝟏) miles, or approximately 𝟐𝟐𝟏𝟏𝟏𝟏,𝟏𝟏𝟏𝟏𝟏𝟏 miles.


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Lesson 21: Special Relationships Within Right Triangles—Dividing into Two Similar Sub-Triangles

Lesson 21: Special Relationships Within Right Triangles—Dividing

into Two Similar Sub-Triangles

1. Use similar triangles to find the length of the altitudes labeled with variables in each triangle below.

△ 𝑨𝑨𝑨𝑨𝑨𝑨 ~ △ 𝑨𝑨𝑨𝑨𝑩𝑩 by AA criterion, so corresponding sides are proportional.

𝒙𝒙𝟑𝟑

=𝟏𝟏𝟏𝟏𝒙𝒙

𝒙𝒙𝟏𝟏 = 𝟑𝟑𝟑𝟑 𝒙𝒙 = √𝟑𝟑𝟑𝟑

𝒙𝒙 = 𝟑𝟑

2. Given right triangle 𝐸𝐸𝐸𝐸𝐸𝐸 with altitude 𝐹𝐹𝐸𝐸�� drawn to the hypotenuse, find the lengths of 𝐹𝐹𝐸𝐸��, 𝐹𝐹𝐸𝐸��, and 𝐹𝐹𝐸𝐸��. Note: 𝐸𝐸𝐸𝐸 = 18.

The altitude drawn from 𝑭𝑭 to 𝑯𝑯 cuts triangle 𝑬𝑬𝑬𝑬𝑯𝑯 into two similar sub-triangles providing the following correspondence:

△ 𝑬𝑬𝑯𝑯𝑬𝑬 ~ △ 𝑬𝑬𝑭𝑭𝑯𝑯 ~△𝑯𝑯𝑭𝑭𝑬𝑬

Using the shorter leg:hypotenuse ratio for the similar triangles,

𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

=𝑭𝑭𝑯𝑯𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏(𝑭𝑭𝑯𝑯) 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

= 𝑭𝑭𝑯𝑯

𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

= 𝑭𝑭𝑯𝑯

𝟕𝟕𝟕𝟕𝟗𝟗

= 𝑭𝑭𝑯𝑯

𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

=𝑭𝑭𝑬𝑬𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏(𝑭𝑭𝑬𝑬) 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

= 𝑭𝑭𝑬𝑬

𝟓𝟓𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

= 𝑭𝑭𝑬𝑬

𝟓𝟓𝟓𝟓𝟗𝟗

= 𝑭𝑭𝑬𝑬

By addition: 𝑭𝑭𝑬𝑬 + 𝑭𝑭𝑬𝑬 = 𝑬𝑬𝑬𝑬

𝟓𝟓𝟓𝟓𝟗𝟗

+ 𝑭𝑭𝑬𝑬 = 𝟏𝟏𝟏𝟏

𝑭𝑭𝑬𝑬 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟗𝟗

I can use the cutouts we made in class to help determine the correct ratios.

When writing a similarity statement, I like to keep the right angle as the middle letter.


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3. Given triangle 𝐼𝐼𝐼𝐼𝐼𝐼 with altitude 𝐼𝐼𝐽𝐽� , 𝐼𝐼𝐽𝐽 = 24, and 𝐼𝐼𝐽𝐽 = 10, find 𝐼𝐼𝐼𝐼, 𝐼𝐼𝐼𝐼, 𝐽𝐽𝐼𝐼, and 𝐼𝐼𝐼𝐼.

Altitude 𝑱𝑱𝑱𝑱�� cuts △ 𝑰𝑰𝑱𝑱𝑰𝑰 into two similar sub-triangles such that △ 𝑰𝑰𝑱𝑱𝑰𝑰 ~ △ 𝑰𝑰𝑱𝑱𝑱𝑱~ △ 𝑱𝑱𝑱𝑱𝑰𝑰.

By the Pythagorean theorem:

𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝑰𝑰𝑱𝑱𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟓𝟓𝟕𝟕𝟑𝟑 = 𝑰𝑰𝑱𝑱𝟏𝟏

𝟑𝟑𝟕𝟕𝟑𝟑 = 𝑰𝑰𝑱𝑱𝟏𝟏

√𝟑𝟑𝟕𝟕𝟑𝟑 = 𝑰𝑰𝑱𝑱 𝟏𝟏𝟑𝟑 = 𝑰𝑰𝑱𝑱

Using the shorter leg:longer leg ratio,

𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

=𝟏𝟏𝟑𝟑𝑱𝑱𝑰𝑰

𝟏𝟏𝟏𝟏(𝑱𝑱𝑰𝑰) = 𝟑𝟑𝟏𝟏𝟏𝟏

𝑱𝑱𝑰𝑰 =𝟑𝟑𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

𝑱𝑱𝑰𝑰 = 𝟑𝟑𝟏𝟏𝟏𝟏𝟓𝟓

Using the shorter leg:hypotenuse ratio,

𝟏𝟏𝟏𝟏𝟏𝟏𝟑𝟑

=𝟏𝟏𝟑𝟑𝑰𝑰𝑰𝑰

𝟏𝟏𝟏𝟏(𝑰𝑰𝑰𝑰) = 𝟑𝟑𝟕𝟕𝟑𝟑

𝑰𝑰𝑰𝑰 =𝟑𝟑𝟕𝟕𝟑𝟑𝟏𝟏𝟏𝟏

𝑰𝑰𝑰𝑰 = 𝟑𝟑𝟕𝟕𝟑𝟑𝟓𝟓

Using addition,

𝑰𝑰𝑱𝑱 + 𝑱𝑱𝑰𝑰 = 𝑰𝑰𝑰𝑰

𝟏𝟏𝟏𝟏 + 𝑱𝑱𝑰𝑰 = 𝟑𝟑𝟕𝟕𝟑𝟑𝟓𝟓

𝑱𝑱𝑰𝑰 = 𝟓𝟓𝟕𝟕𝟑𝟑𝟓𝟓

Since they are all similar right triangles, I can use the Pythagorean theorem to double check my answers.


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4. Given right triangle 𝑀𝑀𝑀𝑀𝑀𝑀 with altitude 𝑀𝑀𝑁𝑁��, find 𝑀𝑀𝑁𝑁, 𝑀𝑀𝑁𝑁, 𝑀𝑀𝑁𝑁, and 𝑀𝑀𝑀𝑀.

Using the Pythagorean theorem,

�𝟑𝟑√𝟕𝟕�𝟏𝟏

+ �√𝟏𝟏𝟏𝟏�𝟏𝟏

= 𝑴𝑴𝑷𝑷𝟏𝟏 𝟑𝟑𝟑𝟑 + 𝟏𝟏𝟏𝟏 = 𝑴𝑴𝑷𝑷𝟏𝟏

𝟕𝟕𝟑𝟑 = 𝑴𝑴𝑷𝑷𝟏𝟏 √𝟕𝟕𝟑𝟑 = 𝑴𝑴𝑷𝑷

An altitude from the right angle in a right triangle to the hypotenuse cuts the triangle into two similar right triangles, △𝑴𝑴𝑴𝑴𝑷𝑷 ~ △𝑴𝑴𝑴𝑴𝑴𝑴~ △𝑴𝑴𝑴𝑴𝑷𝑷.

Using the shorter leg:hypotenuse ratio,

√𝟏𝟏𝟏𝟏√𝟕𝟕𝟑𝟑

=𝑴𝑴𝑴𝑴𝟑𝟑√𝟕𝟕

𝟑𝟑√𝟕𝟕𝟏𝟏 = 𝑴𝑴𝑴𝑴�√𝟕𝟕𝟑𝟑�

𝟑𝟑√𝟕𝟕𝟏𝟏√𝟕𝟕𝟑𝟑

= 𝑴𝑴𝑴𝑴

√𝟏𝟏𝟏𝟏√𝟕𝟕𝟑𝟑

=𝑴𝑴𝑷𝑷√𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏 = 𝑴𝑴𝑷𝑷�√𝟕𝟕𝟑𝟑� 𝟏𝟏𝟏𝟏√𝟕𝟕𝟑𝟑

= 𝑴𝑴𝑷𝑷

Using addition,

𝑴𝑴𝑴𝑴 + 𝑴𝑴𝑷𝑷 = 𝑴𝑴𝑷𝑷

𝑴𝑴𝑴𝑴 +𝟏𝟏𝟏𝟏√𝟕𝟕𝟑𝟑

= √𝟕𝟕𝟑𝟑

𝑴𝑴𝑴𝑴 = √𝟕𝟕𝟑𝟑 −𝟏𝟏𝟏𝟏√𝟕𝟕𝟑𝟑

𝑴𝑴𝑴𝑴 =𝟑𝟑𝟑𝟑√𝟕𝟕𝟑𝟑

√10

3√7 Since 73 is not a perfect square and has no perfect square factors, the answer is √73.


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Lesson 22: Multiplying and Dividing Expressions with Radicals


Express each number in its simplest radical form.

1. √3 ⋅ √15 =

√𝟑𝟑 ⋅ √𝟏𝟏𝟏𝟏 = √𝟑𝟑 ⋅ √𝟑𝟑 ⋅ √𝟏𝟏

= 𝟑𝟑√𝟏𝟏

2. √450 =

√𝟒𝟒𝟏𝟏𝟒𝟒 = √𝟐𝟐𝟏𝟏 ∙ √𝟗𝟗 ∙ √𝟐𝟐

= 𝟏𝟏 ∙ 𝟑𝟑√𝟐𝟐

= 𝟏𝟏𝟏𝟏√𝟐𝟐

3. √24𝑥𝑥5 =

�𝟐𝟐𝟒𝟒𝒙𝒙𝟏𝟏 = √𝟒𝟒√𝟔𝟔�𝒙𝒙𝟒𝟒√𝒙𝒙

= 𝟐𝟐𝒙𝒙𝟐𝟐√𝟔𝟔𝒙𝒙

4. √105 =

The number 𝟏𝟏𝟒𝟒𝟏𝟏 can be factored, but none of the factors are perfect squares, which are necessary to simplify. Therefore, √𝟏𝟏𝟒𝟒𝟏𝟏 cannot be simplified.

5. Show and explain that 𝟐𝟐√𝟑𝟑

and2√𝟑𝟑𝟑𝟑

are equivalent.

𝟐𝟐√𝟑𝟑

=𝟐𝟐√𝟑𝟑

×√𝟑𝟑√𝟑𝟑

=𝟐𝟐√𝟑𝟑𝟑𝟑

By rationalizing the denominator, I can show the two expressions are equivalent.

Using Rule 1, I can rewrite √15 as √3 ∙ √5 so that I can simplify.

I want to rewrite the number under the radical sign, 450, as products of perfect squares, if possible.

I could rewrite √𝑥𝑥4 as �(𝑥𝑥2)2.

I need to rationalize the denominator to write the second expression in simplest radical form.


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6. � 𝟑𝟑𝟑𝟑𝟐𝟐

=

� 𝟑𝟑𝟑𝟑𝟐𝟐

=√𝟑𝟑√𝟑𝟑𝟐𝟐

×√𝟐𝟐√𝟐𝟐

=√𝟔𝟔√𝟔𝟔𝟒𝟒

=√𝟔𝟔𝟖𝟖

7. Determine the exact area of the shaded region that consists of atriangle and a semicircle, shown to the right.

The area of the triangle is

𝑨𝑨 = 𝟏𝟏𝟐𝟐�√𝟏𝟏��√𝟏𝟏� = 𝟏𝟏

𝟐𝟐(𝟏𝟏) = 𝟏𝟏

𝟐𝟐.

Let 𝒅𝒅 be the length of the hypotenuse.

By special triangles or the Pythagorean theorem, 𝒅𝒅 = √𝟏𝟏𝟒𝟒.

The radius of the semicircle is 𝟏𝟏𝟐𝟐 √𝟏𝟏𝟒𝟒 = √𝟏𝟏𝟒𝟒

𝟐𝟐.

The area of the semicircle is

𝑨𝑨 =𝟏𝟏𝟐𝟐

𝝅𝝅�𝟏𝟏𝟐𝟐√

𝟏𝟏𝟒𝟒�𝟐𝟐

=𝟏𝟏𝟐𝟐𝝅𝝅�

𝟏𝟏𝟒𝟒� (𝟏𝟏𝟒𝟒)

=𝟏𝟏𝝅𝝅𝟒𝟒

.

𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓(𝐀𝐀𝐀𝐀𝐀𝐀𝐓𝐓) = 𝟏𝟏𝟐𝟐

+𝟏𝟏𝝅𝝅𝟒𝟒

=𝟏𝟏𝟒𝟒𝟒𝟒

+𝟏𝟏𝝅𝝅𝟒𝟒

=𝟏𝟏𝟒𝟒 + 𝟏𝟏𝝅𝝅

𝟒𝟒

The sum of the areas of the triangle and the semicircle is 𝟏𝟏𝟒𝟒+𝟏𝟏𝝅𝝅𝟒𝟒

square units.

I could multiply √3√32

by √32√32

or √8√8

or √2√2

to get the result of a perfect square under the radical sign in the denominator.

√5

√5


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Lesson 23: Adding and Subtracting Expressions with Radicals


Express each answer in simplified radical form.

1. 22√3 − 18√3 =

𝟐𝟐𝟐𝟐√𝟑𝟑 − 𝟏𝟏𝟏𝟏√𝟑𝟑 = (𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟏𝟏)√𝟑𝟑

= 𝟒𝟒√𝟑𝟑

2. √18 + 5√2 =

�𝟏𝟏𝟏𝟏+𝟓𝟓�𝟐𝟐= 𝟑𝟑�𝟐𝟐+𝟓𝟓�𝟐𝟐 = (𝟑𝟑 + 𝟓𝟓)√𝟐𝟐

= 𝟏𝟏√𝟐𝟐

3. 4√5 + 11√125 =

𝟒𝟒√𝟓𝟓 + 𝟏𝟏𝟏𝟏√𝟏𝟏𝟐𝟐𝟓𝟓 = 𝟒𝟒√𝟓𝟓 + 𝟏𝟏𝟏𝟏√𝟐𝟐𝟓𝟓√𝟓𝟓

= 𝟒𝟒√𝟓𝟓 + 𝟏𝟏𝟏𝟏(𝟓𝟓)√𝟓𝟓

= 𝟒𝟒√𝟓𝟓 + 𝟓𝟓𝟓𝟓√𝟓𝟓

= (𝟒𝟒 + 𝟓𝟓𝟓𝟓)√𝟓𝟓

= 𝟓𝟓𝟓𝟓√𝟓𝟓

I will use the distributive property to add or subtract expressions with radicals.

I need to rewrite √18 so I can apply the distributive property.


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3√7

2√28

√175

4. Determine the area and perimeter of the triangle shown. Simplify as much as possible.

Perimeter:

�𝟏𝟏𝟏𝟏𝟓𝟓+𝟐𝟐�𝟐𝟐𝟏𝟏+𝟑𝟑�𝟏𝟏= �𝟐𝟐𝟓𝟓�𝟏𝟏+𝟐𝟐�𝟒𝟒�𝟏𝟏+𝟑𝟑�𝟏𝟏 = 𝟓𝟓√𝟏𝟏 + 𝟐𝟐 ∙ 𝟐𝟐√𝟏𝟏 + 𝟑𝟑√𝟏𝟏

= 𝟓𝟓√𝟏𝟏 + 𝟒𝟒√𝟏𝟏 + 𝟑𝟑√𝟏𝟏

= (𝟓𝟓 + 𝟒𝟒 + 𝟑𝟑)√𝟏𝟏

= 𝟏𝟏𝟐𝟐√𝟏𝟏

Area:

𝟐𝟐�𝟐𝟐𝟏𝟏�𝟑𝟑�𝟏𝟏�𝟐𝟐 =

𝟒𝟒�𝟏𝟏�𝟑𝟑�𝟏𝟏�𝟐𝟐 = 𝟏𝟏𝟒𝟒

𝟐𝟐 = 𝟒𝟒𝟐𝟐

The perimeter is 𝟏𝟏𝟐𝟐√𝟏𝟏 units, and the area is 𝟒𝟒𝟐𝟐 square units.

5. Parallelogram 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 has an area of 12√5 square units. 𝐴𝐴𝐴𝐴 = 3√5, and 𝐺𝐺 and 𝐻𝐻 are midpoints of 𝐴𝐴𝐷𝐷�� and 𝐴𝐴𝐷𝐷��, respectively. Find the area of the shaded region. Write your answer in simplest radical form.

Using the area of a parallelogram,

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨) = 𝒃𝒃𝒃𝒃

𝟏𝟏𝟐𝟐√𝟓𝟓 = 𝟑𝟑√𝟓𝟓 ⋅ 𝒃𝒃 𝟒𝟒 = 𝒃𝒃

The height of the parallelogram is 𝟒𝟒 units.

The area of the shaded region is the sum of the areas of △ 𝑬𝑬𝑬𝑬𝑬𝑬 and △ 𝑭𝑭𝑬𝑬𝑬𝑬.

The given points 𝑬𝑬 and 𝑬𝑬 are midpoints of 𝑨𝑨𝑬𝑬�� and 𝑨𝑨𝑬𝑬��; therefore, by the triangle side splitter theorem, 𝑬𝑬𝑬𝑬�� must be parallel to 𝑨𝑨𝑨𝑨��, and thus, also parallel to 𝑨𝑨𝑨𝑨��. Furthermore, 𝑬𝑬𝑬𝑬 = 𝟏𝟏

𝟐𝟐𝑨𝑨𝑨𝑨 = 𝟏𝟏

𝟐𝟐𝑨𝑨𝑨𝑨 = 𝟑𝟑

𝟐𝟐√𝟓𝟓.

I learned about the side splitter theorem in Lesson 4 and have been using it ever since.

3√5


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△ 𝑬𝑬𝑬𝑬𝑬𝑬 ~△ 𝑬𝑬𝑨𝑨𝑨𝑨 by AA criterion with a scale factor of 𝟏𝟏𝟐𝟐. The areas of scale drawings are related by the

square of the scale factor; therefore, 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) = �𝟏𝟏𝟐𝟐�𝟐𝟐⋅ 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑨𝑨𝑨𝑨).

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑨𝑨𝑨𝑨) =𝟏𝟏𝟐𝟐⋅ 𝟑𝟑√𝟓𝟓 ⋅ 𝟒𝟒

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑨𝑨𝑨𝑨) = 𝟔𝟔√𝟓𝟓

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) = �𝟏𝟏𝟐𝟐�𝟐𝟐⋅ 𝟔𝟔√𝟓𝟓

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) =𝟏𝟏𝟒𝟒⋅ 𝟔𝟔√𝟓𝟓

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) =𝟑𝟑𝟐𝟐√𝟓𝟓

By a similar argument,

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑭𝑭𝑬𝑬𝑬𝑬) =𝟑𝟑𝟐𝟐√𝟓𝟓

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑬𝑬𝑬𝑬𝑭𝑭𝑬𝑬) = 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) + 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑭𝑭𝑬𝑬𝑬𝑬)

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑬𝑬𝑬𝑬𝑭𝑭𝑬𝑬) =𝟑𝟑𝟐𝟐√𝟓𝟓 +

𝟑𝟑𝟐𝟐√𝟓𝟓

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑬𝑬𝑬𝑬𝑭𝑭𝑬𝑬) = 𝟑𝟑√𝟓𝟓

The area of the shaded region is 𝟑𝟑√𝟓𝟓 square units.

My plan is to determine 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(𝑬𝑬𝑬𝑬𝑭𝑭𝑬𝑬) =𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑬𝑬𝑬𝑬𝑬𝑬) + 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀(△ 𝑭𝑭𝑬𝑬𝑬𝑬).


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Lesson 24: Prove the Pythagorean Theorem Using Similarity


1. In right triangle 𝐸𝐸𝐸𝐸𝐸𝐸, an altitude has been drawn from vertex 𝐸𝐸 and meets 𝐸𝐸𝐸𝐸�� at 𝐻𝐻. Answer the following questions based on this figure. 𝐻𝐻 divides 𝐸𝐸𝐸𝐸�� into the lengths 𝑥𝑥 and 𝑦𝑦.

a. △ 𝐸𝐸𝐸𝐸𝐸𝐸 ~ △ 𝐻𝐻𝐸𝐸𝐸𝐸 ~ △𝐻𝐻𝐸𝐸𝐸𝐸. Explain why.

△ 𝑬𝑬𝑬𝑬𝑬𝑬 ~ △𝑯𝑯𝑬𝑬𝑬𝑬 by the AA criterion, as both triangles have a 𝟗𝟗𝟗𝟗° angle, and both triangles share ∠𝑬𝑬. △ 𝑬𝑬𝑬𝑬𝑬𝑬 ~ △𝑯𝑯𝑬𝑬𝑬𝑬, also by the AA criterion, as both triangles have a 𝟗𝟗𝟗𝟗° angle, and both triangles share ∠𝑬𝑬. Similarity is transitive; therefore, △𝑯𝑯𝑬𝑬𝑬𝑬 ~ △𝑯𝑯𝑬𝑬𝑬𝑬.

b. Sketch each of the three triangles separately so that their corresponding sides and angles are easy to see.

In order to prove the Pythagorean theorem, we want to show that the sum of each leg length, squared, is equal to the hypotenuse, squared, or based on this figure, 𝑓𝑓2 + 𝑔𝑔2 = 𝑒𝑒2. To prove the Pythagorean theorem using similarity, we will employ the fact that the ratios of corresponding proportional lengths are equal in value between similar triangles.

To help guide your sketches, consider beginning with the right angle for each triangle to help orient you in the sketching process.


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c. △ 𝐸𝐸𝐸𝐸𝐸𝐸 and △ 𝐻𝐻𝐸𝐸𝐸𝐸 each have ∠𝐸𝐸 as an acute angle. Write the equivalent shorter leg:hypotenuse ratios for these triangles.

𝒙𝒙𝒈𝒈

=𝒈𝒈𝒆𝒆

𝒈𝒈𝟐𝟐 = 𝒙𝒙𝒆𝒆

d. △ 𝐸𝐸𝐸𝐸𝐸𝐸 and △ 𝐻𝐻𝐸𝐸𝐸𝐸 each have ∠𝐸𝐸 as an acute angle. Write the equivalent longer leg:hypotenuse ratios for these triangles.

𝒚𝒚𝒇𝒇

=𝒇𝒇𝒆𝒆

𝒇𝒇𝟐𝟐 = 𝒚𝒚𝒆𝒆

e. Use parts (c) and (d) to show that 𝑓𝑓2 + 𝑔𝑔2 = 𝑒𝑒2.

𝒇𝒇𝟐𝟐 + 𝒈𝒈𝟐𝟐 = 𝒙𝒙𝒆𝒆 + 𝒚𝒚𝒆𝒆

= 𝒆𝒆(𝒙𝒙 + 𝒚𝒚)

= 𝒆𝒆(𝒆𝒆)

= 𝒆𝒆𝟐𝟐

2. What is the leg:hypotenuse ratio of a 45–45–90 triangle? How do you know?

Take for example a 𝟒𝟒𝟒𝟒–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗 triangle with leg lengths of 𝟏𝟏 each. Let the hypotenuse be a length 𝒄𝒄. Then by the Pythagorean theorem:

𝟏𝟏𝟐𝟐 + 𝟏𝟏𝟐𝟐 = 𝒄𝒄𝟐𝟐

𝟐𝟐 = 𝒄𝒄𝟐𝟐

𝒄𝒄 = √𝟐𝟐

The leg:hypotenuse ratio is 𝟏𝟏:√𝟐𝟐. By the AA criterion, triangles with angle measures 𝟒𝟒𝟒𝟒–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗 are similar and, therefore, corresponding lengths are proportional. This means that all 𝟒𝟒𝟒𝟒–𝟒𝟒𝟒𝟒–𝟗𝟗𝟗𝟗 triangles will have a leg:hypotenuse ratio of 𝟏𝟏:√𝟐𝟐.

3. Consider any 30–60–90 triangle.

a. Write the three ratios that compare the lengths of each pair of sides in a 30–60–90 triangle.

The shorter leg:hypotenuse ratio is 𝟏𝟏:𝟐𝟐. The longer leg:hypotenuse ratio is √𝟑𝟑:𝟐𝟐. The shorter leg:hypotenuse ratio is 𝟏𝟏:√𝟑𝟑.


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b. By the AA criterion, all triangles with angle measures of 30–60–90 are similar. Use the ratios from part (a) to find the leg lengths of a 30–60–90 triangle with a hypotenuse of 100. Let 𝒂𝒂 represent the length of the shorter leg. Since the shorter leg:hypotenuse ratio is 𝟏𝟏:𝟐𝟐, then

𝒂𝒂𝟏𝟏𝟗𝟗𝟗𝟗

=𝟏𝟏𝟐𝟐

𝒂𝒂 = 𝟒𝟒𝟗𝟗.

Let 𝒃𝒃 represent the length of the longer leg. Since the longer leg:hypotenuse ratio is √𝟑𝟑:𝟐𝟐, then

𝒃𝒃𝟏𝟏𝟗𝟗𝟗𝟗

=√𝟑𝟑𝟐𝟐

𝟐𝟐𝒃𝒃 = 𝟏𝟏𝟗𝟗𝟗𝟗√𝟑𝟑

𝒃𝒃 = 𝟒𝟒𝟗𝟗√𝟑𝟑.


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Eureka Math Homework Helper 2015–2016 Geometry Module 2€¦ · M2 GEOMETRY Lesson 2: Making Scale Drawings Using the Ratio Method I should remember that a ruler is critical when