A Story of Ratios: Grade 8 Module 3 Lesson Excerpts
Lesson 1, Exercises 2-Ββ6
Use the diagram below to answer Exercises 2β6. Let there be a dilation from center π. Then π·ππππ‘πππ(π) = π! and π·ππππ‘ππ(π) = π!. In the diagram below, ππ = 3 cm and ππ = 4 cm as shown.
1. If the scale factor is π = 3, what is the length of segment ππ! ?
2. Use the definition of dilation to show that your answer to Exercise 2 is correct.
3. If the scale factor is π = 3, what is the length of segment ππ!?
4. Use the definition of dilation to show that your answer to Exercise 4 is correct.
5. If you know that ππ = 3, ππ! = 9, how could you use that information to determine the scale factor?
Grade 8 Module 3 Lesson Excerpts
Lesson 2, Problem Set 1
1. Use a ruler to dilate the following figure from center π, with scale factor π = !!.
Grade 8 Module 3 Lesson Excerpts
Lesson 3, Example 2
Β§ In the picture below, we have a triangle π΄π΅πΆ, that has been dilated from center π, by a scale factor of π = !
!. It is noted by π΄βπ΅βπΆβ.
Ask students what we can do to map this new triangle, β³ π΄β²π΅β²πΆβ², back to the original. Tell them to be as specific as possible. Students should write their conjectures or share with a partner.
Β§ Letβs use the definition of dilation and some side lengths to help us figure out how to map β³ π΄β²π΅β²πΆβ² back onto β³ π΄π΅πΆ. How are the lengths ππ΄! and ππ΄ related?
ΓΊ We know by the definition of dilation that ππ΄! = π ππ΄ .
Β§ We know that π = !!. Letβs say that the length of ππ΄ is 6 units (we can pick any number, but 6
will make it easy for us to compute). What is the length of ππ΄β?
ΓΊ Since ππ΄β² = !!ππ΄ , and we are saying that the length of ππ΄ is 6, then ππ΄β² = !
!Γ6 =
2, and ππ΄β² = 2 units. Β§ Now since we want to dilate triangle π΄βπ΅βπΆβ to the size of triangle π΄π΅πΆ, we need to know what
scale factor π is required so that ππ΄ = π ππ΄β² . What scale factor should we use and why?
ΓΊ We need a scale factor π = 3 because we want ππ΄ = π ππ΄β² . Using the lengths from before, we have 6 = πΓ2. Therefore, π = 3.
Β§ Now that we know the scale factor, what precise dilation would map triangle π΄βπ΅βπΆβ onto triangle π΄π΅πΆ?
ΓΊ A dilation from center π with scale factor π = 3. Lesson 4, Discussion
Theorem: Given a dilation with center π and scale factor π, then for any two points π, π in the plane so the π, π, π are not collinear, the lines ππ and πβπβ are parallel, where πβ = πππππ‘πππ(π) and πβ = πππππ‘πππ (π), and furthermore, πβ²πβ² = π ππ .
Ask students to paraphrase the theorem in their own words or offer them the following version of the theorem: FTS states that given a dilation from center π, and points π and π (points π,π,π are not on the same line), the segments formed when you connect π to π, and πβ² to πβ², are parallel. More surprising is the fact that the segment ππ, even though it was not dilated as points π and π were, dilates to segment πβ²πβ² and the length of πβ²πβ² is the length of ππ multiplied by the scale factor.
Grade 8 Module 3 Lesson Excerpts
Lesson 5, Exercise 3
1. In the diagram below, you are given center π and ray ππ΄. Point π΄ is dilated by a scale factor π = !
!". Use what you know about FTS to find the location of point π΄β.
Lesson 6, Example 1 In Lesson 5 we found the location of a dilated point by using the knowledge of dilation and scale factor, as well as the lines of the coordinate plane to ensure equal angles to find the coordinates of the dilated point. For example, we were given the point π΄ = (5, 2) and told the scale factor of dilation was π = 2. We created the following picture and determined the location of π΄β² to be 10, 4 .
Β§ We can use this information, and the observations we made at the beginning of class, to
develop a shortcut for finding the coordinates of dilated points when the center of dilation is
Grade 8 Module 3 Lesson Excerpts
the origin.
Β§ Notice that the horizontal distance from the π¦-Ββaxis to point π΄ was multiplied by a scale factor of 2. That is, the π₯-Ββcoordinate of point π΄ was multiplied by a scale factor of 2. Similarly, the vertical distance from the π₯-Ββaxis to point π΄ was multiplied by a scale factor of 2.
Β§ Here are the coordinates of point π΄ = 5, 2 and the dilated point π΄! = 10, 4 . Since the scale factor was 2, we can more easily see what happened to the coordinates of π΄ after the dilation if we write the coordinates of π΄β² as (2Γ5, 2Γ2). That is, the scale factor of 2 multiplied each of the coordinates of point π΄ to get π΄!.
Β§ The reasoning goes back to our understanding of dilation. The length π ππ΅ = |ππ΅!|, and the length π π΄π΅ = |π΄!π΅!|, therefore
π =ππ΅!
ππ΅=
π΄!B'π΄π΅
where the length of the segment |ππ΅!| is the π₯-Ββcoordinate of the dilated point, i.e., 10, and the length of the segment |π΄!π΅!| is the π¦-Ββcoordinate of the dilated point, i.e., 4. In other words, based on what we know about the lengths of dilated segments, when the center of dilation is the origin, we can determine the coordinates of a dilated point by multiplying each of the coordinates in the original point by the scale factor.
Lesson 7, Discussion Theorem: Dilations preserve the degrees of angles.
We know that dilations map angles to angles. Let there be a dilation from center π and scale factor π. Given β πππ , we want to show that if πβ = π·ππππ‘πππ(π), πβ = π·ππππ‘πππ(π), and π β = π·ππππ‘πππ(π ), then β πππ = β πβ²πβ²π β². In other words, when we dilate an angle, the measure of the angle remains unchanged. Take a moment to draw a picture of the situation.
Β§ Could Line πβπβ be parallel to Line ππ?
Β§ Could Line πβπβ intersect Line ππ ?
Grade 8 Module 3 Lesson Excerpts
Β§ Could Line πβπβ to be parallel to Line ππ ? Β§ Now that we are sure that Line πβπ β intersects Line ππ , mark that point of intersection on your
drawing (extend rays if necessary). Letβs call that point of intersection point π΅. Β§ At this point, we have all the information that we need to show that β πππ = β πβ²πβ²π β² .
(Give students several minutes in small groups to discuss possible proofs for why β πππ =β πβ²πβ²π β² .)
Β§ Using FTS, and our knowledge of angles formed by parallel lines cut by a transversal, we have proven that dilations are degree-Ββpreserving transformations.
Lesson 8, Exercise 1
1. Triangle π¨π©πͺ was dilated from center πΆ by scale factor π = ππ. The dilated triangle is noted
by π¨βπ©βπͺβ. Another triangle π¨βπ©βπͺβ is congruent to triangle π¨βπ©βπͺβ ( i.e., βπ¨"π©"πͺ" β βπ¨β²π©β²πͺβ²). Describe the dilation followed by the basic rigid motion that would map triangle π¨βπ©βπͺβ onto triangle π¨π©πͺ.
Grade 8 Module 3 Lesson Excerpts
Lesson 9, Exploratory Challenge 2 1. The goal is to show that if β³ π΄π΅πΆ is similar to β³ π΄β²π΅β²πΆβ², and β³ π΄β²π΅β²πΆβ² is similar to β³
π΄β²β²π΅β²β²πΆβ²β², then β³ π΄π΅πΆ is similar to β³ π΄!!π΅!!πΆ!!. Symbolically, if β³ π΄π΅πΆ βΌβ³ π΄!π΅!πΆ! and β³ π΄!π΅!πΆ! βΌβ³ π΄!!π΅!!πΆ!!, then β³ π΄π΅πΆ βΌβ³ π΄!!π΅!!πΆ!!.
a. Describe the similarity that proves β³ π΄π΅πΆ βΌβ³ π΄!π΅!πΆ!.
b. Describe the similarity that proves β³ π΄!π΅!πΆ! βΌβ³ π΄!!π΅!!πΆ!!.
c. Verify that in fact β³ π΄π΅πΆ βΌβ³ π΄β²β²π΅!!πΆ!! by checking corresponding angles and corresponding side lengths. Then describe the sequence that would prove the similarity β³ π΄π΅πΆ βΌβ³ π΄!!π΅!!πΆ!!.
d. Is it true that if β³ π΄π΅πΆ βΌβ³ π΄!π΅!πΆ! and β³ π΄!π΅!πΆ! βΌβ³ π΄"π΅"πΆ", that β³ π΄π΅πΆ βΌβ³ π΄"π΅"πΆ"? Why do you think this is so?
Grade 8 Module 3 Lesson Excerpts
Lesson 10, Exercises 4 and 5 1. Are the triangles shown below similar? Present an informal argument as to why they are
or why they are not.
2. Are the triangles shown below similar? Present an informal argument as to why they are or why they are not.
Grade 8 Module 3 Lesson Excerpts
Lesson 11, Exercise 1 1. In the diagram below, you have β³ π΄π΅πΆ and β³ π΄π΅!πΆ!. Use this information to answer parts
(a) β (d).
a. Based on the information given, is β³ π΄π΅πΆ~ β³ π΄π΅!πΆ!? Explain.
b. Assume line π΅πΆ is parallel to line π΅β²πΆβ². With this information, can you say that β³ π΄π΅πΆ~ β³ π΄π΅!πΆ!? Explain.
c. Given that β³ π΄π΅πΆ~ β³ π΄π΅!πΆ!, determine the length of π΄πΆβ².
d. Given that β³ π΄π΅πΆ~ β³ π΄π΅!πΆ!, determine the length of π΄π΅.
Grade 8 Module 3 Lesson Excerpts
Lesson 12, Exercise 1 1. You want to determine the approximate height of one of the tallest buildings in the city.
You are told that if you place a mirror some distance from yourself so that you can see the top of the building in the mirror, then you can indirectly measure the height using similar triangles. Let πΆ be the location of the mirror so that the figure shown can see the top of the building.
a. Explain why β³ π΄π΅π~ β³ πππ.
b. Label the diagram with the following information: The distance from eye-Ββlevel straight down to the ground is 5.3 feet. The distance from the figure to the mirror is 7.2 feet. The distance from the figure to the base of the building is 1,750 feet. The height of the building will be represented by π₯.
c. What is the distance from the mirror to the building?
d. Do you have enough information to determine the approximate height of the building? If yes, determine the approximate height of the building. If not, what additional information is needed?
Grade 8 Module 3 Lesson Excerpts
Lesson 13, Discussion of Proof What we are going to do in this lesson is take a right triangle, β³ π΄π΅πΆ, and use what we know about similarity of triangles to prove π! + π! = π!.
For the proof, we will draw a line from vertex πΆ to a point π· so that the line is perpendicular to side π΄π΅.
We draw this particular line, line πΆπ·, because it divides the original triangle into three similar triangles. Before we move on, can you name the three triangles?
Letβs look at the triangles in a different orientation in order to see why they are similar. We can use our basic rigid motions to separate the three triangles. Doing so ensures that the lengths of segments and degrees of angles are preserved.
In order to have similar triangles, they must have two common angles, by the AA criterion. Which angles prove that β³ π΄π·πΆ and β³ π΄πΆπ΅ similar?
What must that mean about β πΆ from β³ π΄π·πΆ and β π΅ from β³ π΄πΆπ΅?
Grade 8 Module 3 Lesson Excerpts
Which angles prove that β³ π΄πΆπ΅ and β³ πΆπ·π΅ similar?
What must that mean about β π΄ from β³ π΄πΆπ΅ and β πΆ from β³ πΆπ·π΅?
If β³ π΄π·πΆ β½ β³ π΄πΆπ΅ and β³ π΄πΆπ΅ β½ β³ πΆπ·π΅, is it true that β³ π΄π·πΆ β½ β³ πΆπ·π΅? How do you know?
When we have similar triangles, we know that their side lengths are proportional. Therefore, if we consider β³ π΄π·πΆ and β³ π΄πΆπ΅, we can write
π΄πΆπ΄π΅
=π΄π·π΄πΆ
.
By the cross-Ββmultiplication algorithm,
π΄πΆ ! = π΄π΅ β π΄π· .
By considering β³ π΄πΆπ΅ and β³ πΆπ·π΅, we can write π΅π΄π΅πΆ
=π΅πΆπ΅π·
.
Which again by the cross-Ββmultiplication algorithm,
π΅πΆ ! = π΅π΄ β π΅π· . If we add the two equations together, we get
π΄πΆ ! + π΅πΆ ! = π΄π΅ β π΄π· + π΅π΄ β π΅π· . By the distributive property, we can rewrite the right side of the equation because there is a common factor of π΄π΅ . Now we have
π΄πΆ ! + π΅πΆ ! = π΄π΅ π΄π· + π΅π· . Keeping our goal in mind, we want to prove that π! + π! = π! ; letβs see how close we are. Using our diagram where three triangles are within one, (shown below), what side lengths are represented by π΄πΆ ! + π΅πΆ !?
Now letβs examine the right side of our equation: π΄π΅ π΄π· + π΅π· . We want this to be equal to π!; does it?
We have just proven the Pythagorean theorem using what we learned about similarity. At this point we have seen the proof of the theorem in terms of congruence and now similarity.
Grade 8 Module 3 Lesson Excerpts
Lesson 14, Discussion of Proof of Converse
The following is a proof of the converse. Assume we are given a triangle π΄π΅πΆ with sides π, π, and π. We want to show that β π΄πΆπ΅ is a right angle. To do so, we will assume that β π΄πΆπ΅ is not a right angle. Then |β π΄πΆπ΅| > 90Λ or |β π΄πΆπ΅| < 90Λ. For brevity, we will only show the case for when |β π΄πΆπ΅| > 90Λ (the proof of the other case is similar). In the diagram below, we extend π΅πΆ to a ray π΅πΆ and let the perpendicular from π΄ meet the ray at point π·.
Let π = |πΆπ·| and π = |π΄π·|.
Then by the Pythagorean theorem applied to β³ π΄πΆπ· and β³ π΄π΅π· results in
π! = π! + π! and π! = π +π ! + π!. Since we know what π! and π! are from the above equations, we can substitute those values into π! = π! + π! to get
π +π ! + π! = π! +π! + π!. Since π +π ! = π +π π +π = π! + ππ + ππ +π! = π! + 2ππ +π!, then we have
π! + 2ππ +π! + π! = π! +π! + π!. We can subtract the terms π!,π!, and π! from both sides of the equal sign. Then we have
2ππ = 0. But this cannot be true because 2ππ is a length; therefore, it cannot be equal to zero. Which means our assumption that |β π΄πΆπ΅| > 90Λ cannot be true. We can write a similar proof to show that |β π΄πΆπ΅| < 90Λ cannot be true either. Therefore, |β π΄πΆπ΅| = 90Λ.