A Story of Ratios: Grade 8 Module 3 Lesson Excerpts
Lesson 1, Exercises 2-Āā6
Use the diagram below to answer Exercises 2ā6. Let there be a dilation from center š. Then š·šššš”ššš(š) = š! and š·šššš”šš(š) = š!. In the diagram below, šš = 3 cm and šš = 4 cm as shown.
1. If the scale factor is š = 3, what is the length of segment šš! ?
2. Use the definition of dilation to show that your answer to Exercise 2 is correct.
3. If the scale factor is š = 3, what is the length of segment šš!?
4. Use the definition of dilation to show that your answer to Exercise 4 is correct.
5. If you know that šš = 3, šš! = 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 š, with scale factor š = !!.
Grade 8 Module 3 Lesson Excerpts
Lesson 3, Example 2
Ā§ In the picture below, we have a triangle š“šµš¶, that has been dilated from center š, by a scale factor of š = !
!. It is noted by š“āšµāš¶ā.
Ask students what we can do to map this new triangle, ā³ š“ā²šµā²š¶ā², back to the original. Tell them to be as specific as possible. Students should write their conjectures or share with a partner.
Ā§ Letās use the definition of dilation and some side lengths to help us figure out how to map ā³ š“ā²šµā²š¶ā² back onto ā³ š“šµš¶. How are the lengths šš“! and šš“ related?
Ćŗ We know by the definition of dilation that šš“! = š šš“ .
Ā§ We know that š = !!. Letās say that the length of šš“ is 6 units (we can pick any number, but 6
will make it easy for us to compute). What is the length of šš“ā?
Ćŗ Since šš“ā² = !!šš“ , and we are saying that the length of šš“ is 6, then šš“ā² = !
!Ć6 =
2, and šš“ā² = 2 units. Ā§ Now since we want to dilate triangle š“āšµāš¶ā to the size of triangle š“šµš¶, we need to know what
scale factor š is required so that šš“ = š šš“ā² . What scale factor should we use and why?
Ćŗ We need a scale factor š = 3 because we want šš“ = š šš“ā² . Using the lengths from before, we have 6 = šĆ2. Therefore, š = 3.
Ā§ Now that we know the scale factor, what precise dilation would map triangle š“āšµāš¶ā onto triangle š“šµš¶?
Ćŗ A dilation from center š with scale factor š = 3. Lesson 4, Discussion
Theorem: Given a dilation with center š and scale factor š, then for any two points š, š in the plane so the š, š, š are not collinear, the lines šš and šāšā are parallel, where šā = ššššš”ššš(š) and šā = ššššš”ššš (š), and furthermore, šā²šā² = š šš .
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 š, and points š and š (points š,š,š are not on the same line), the segments formed when you connect š to š, and šā² to šā², are parallel. More surprising is the fact that the segment šš, even though it was not dilated as points š and š were, dilates to segment šā²šā² and the length of šā²šā² is the length of šš multiplied by the scale factor.
Grade 8 Module 3 Lesson Excerpts
Lesson 5, Exercise 3
1. In the diagram below, you are given center š and ray šš“. Point š“ is dilated by a scale factor š = !
!". Use what you know about FTS to find the location of point š“ā.
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 š“ = (5, 2) and told the scale factor of dilation was š = 2. We created the following picture and determined the location of š“ā² to be 10, 4 .
Ā§ 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.
Ā§ Notice that the horizontal distance from the š¦-Āāaxis to point š“ was multiplied by a scale factor of 2. That is, the š„-Āācoordinate of point š“ was multiplied by a scale factor of 2. Similarly, the vertical distance from the š„-Āāaxis to point š“ was multiplied by a scale factor of 2.
Ā§ Here are the coordinates of point š“ = 5, 2 and the dilated point š“! = 10, 4 . Since the scale factor was 2, we can more easily see what happened to the coordinates of š“ after the dilation if we write the coordinates of š“ā² as (2Ć5, 2Ć2). That is, the scale factor of 2 multiplied each of the coordinates of point š“ to get š“!.
Ā§ The reasoning goes back to our understanding of dilation. The length š ššµ = |ššµ!|, and the length š š“šµ = |š“!šµ!|, therefore
š =ššµ!
ššµ=
š“!B'š“šµ
where the length of the segment |ššµ!| is the š„-Āācoordinate of the dilated point, i.e., 10, and the length of the segment |š“!šµ!| is the š¦-Āācoordinate 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 š and scale factor š. Given ā ššš , we want to show that if šā = š·šššš”ššš(š), šā = š·šššš”ššš(š), and š ā = š·šššš”ššš(š ), then ā ššš = ā šā²šā²š ā². 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.
Ā§ Could Line šāšā be parallel to Line šš?
Ā§ Could Line šāšā intersect Line šš ?
Grade 8 Module 3 Lesson Excerpts
Ā§ Could Line šāšā to be parallel to Line šš ? Ā§ Now that we are sure that Line šāš ā intersects Line šš , mark that point of intersection on your
drawing (extend rays if necessary). Letās call that point of intersection point šµ. Ā§ At this point, we have all the information that we need to show that ā ššš = ā šā²šā²š ā² .
(Give students several minutes in small groups to discuss possible proofs for why ā ššš =ā šā²šā²š ā² .)
Ā§ Using FTS, and our knowledge of angles formed by parallel lines cut by a transversal, we have proven that dilations are degree-Āāpreserving transformations.
Lesson 8, Exercise 1
1. Triangle šØš©šŖ was dilated from center š¶ by scale factor š = šš. The dilated triangle is noted
by šØāš©āšŖā. Another triangle šØāš©āšŖā is congruent to triangle šØāš©āšŖā ( i.e., āšØ"š©"šŖ" ā āšØā²š©ā²šŖā²). Describe the dilation followed by the basic rigid motion that would map triangle šØāš©āšŖā onto triangle šØš©šŖ.
Grade 8 Module 3 Lesson Excerpts
Lesson 9, Exploratory Challenge 2 1. The goal is to show that if ā³ š“šµš¶ is similar to ā³ š“ā²šµā²š¶ā², and ā³ š“ā²šµā²š¶ā² is similar to ā³
š“ā²ā²šµā²ā²š¶ā²ā², then ā³ š“šµš¶ is similar to ā³ š“!!šµ!!š¶!!. Symbolically, if ā³ š“šµš¶ ā¼ā³ š“!šµ!š¶! and ā³ š“!šµ!š¶! ā¼ā³ š“!!šµ!!š¶!!, then ā³ š“šµš¶ ā¼ā³ š“!!šµ!!š¶!!.
a. Describe the similarity that proves ā³ š“šµš¶ ā¼ā³ š“!šµ!š¶!.
b. Describe the similarity that proves ā³ š“!šµ!š¶! ā¼ā³ š“!!šµ!!š¶!!.
c. Verify that in fact ā³ š“šµš¶ ā¼ā³ š“ā²ā²šµ!!š¶!! by checking corresponding angles and corresponding side lengths. Then describe the sequence that would prove the similarity ā³ š“šµš¶ ā¼ā³ š“!!šµ!!š¶!!.
d. Is it true that if ā³ š“šµš¶ ā¼ā³ š“!šµ!š¶! and ā³ š“!šµ!š¶! ā¼ā³ š“"šµ"š¶", that ā³ š“šµš¶ ā¼ā³ š“"šµ"š¶"? 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 ā³ š“šµš¶ and ā³ š“šµ!š¶!. Use this information to answer parts
(a) ā (d).
a. Based on the information given, is ā³ š“šµš¶~ ā³ š“šµ!š¶!? Explain.
b. Assume line šµš¶ is parallel to line šµā²š¶ā². With this information, can you say that ā³ š“šµš¶~ ā³ š“šµ!š¶!? Explain.
c. Given that ā³ š“šµš¶~ ā³ š“šµ!š¶!, determine the length of š“š¶ā².
d. Given that ā³ š“šµš¶~ ā³ š“šµ!š¶!, determine the length of š“šµ.
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 š¶ be the location of the mirror so that the figure shown can see the top of the building.
a. Explain why ā³ š“šµš~ ā³ ššš.
b. Label the diagram with the following information: The distance from eye-Āālevel 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 š„.
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, ā³ š“šµš¶, and use what we know about similarity of triangles to prove š! + š! = š!.
For the proof, we will draw a line from vertex š¶ to a point š· so that the line is perpendicular to side š“šµ.
We draw this particular line, line š¶š·, because it divides the original triangle into three similar triangles. Before we move on, can you name the three triangles?
Letās 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 ā³ š“š·š¶ and ā³ š“š¶šµ similar?
What must that mean about ā š¶ from ā³ š“š·š¶ and ā šµ from ā³ š“š¶šµ?
Grade 8 Module 3 Lesson Excerpts
Which angles prove that ā³ š“š¶šµ and ā³ š¶š·šµ similar?
What must that mean about ā š“ from ā³ š“š¶šµ and ā š¶ from ā³ š¶š·šµ?
If ā³ š“š·š¶ ā½ ā³ š“š¶šµ and ā³ š“š¶šµ ā½ ā³ š¶š·šµ, is it true that ā³ š“š·š¶ ā½ ā³ š¶š·šµ? How do you know?
When we have similar triangles, we know that their side lengths are proportional. Therefore, if we consider ā³ š“š·š¶ and ā³ š“š¶šµ, we can write
š“š¶š“šµ
=š“š·š“š¶
.
By the cross-Āāmultiplication algorithm,
š“š¶ ! = š“šµ ā š“š· .
By considering ā³ š“š¶šµ and ā³ š¶š·šµ, we can write šµš“šµš¶
=šµš¶šµš·
.
Which again by the cross-Āāmultiplication algorithm,
šµš¶ ! = šµš“ ā šµš· . If we add the two equations together, we get
š“š¶ ! + šµš¶ ! = š“šµ ā š“š· + šµš“ ā šµš· . By the distributive property, we can rewrite the right side of the equation because there is a common factor of š“šµ . Now we have
š“š¶ ! + šµš¶ ! = š“šµ š“š· + šµš· . Keeping our goal in mind, we want to prove that š! + š! = š! ; letās see how close we are. Using our diagram where three triangles are within one, (shown below), what side lengths are represented by š“š¶ ! + šµš¶ !?
Now letās examine the right side of our equation: š“šµ š“š· + šµš· . We want this to be equal to š!; 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 š“šµš¶ with sides š, š, and š. We want to show that ā š“š¶šµ is a right angle. To do so, we will assume that ā š“š¶šµ is not a right angle. Then |ā š“š¶šµ| > 90Ė or |ā š“š¶šµ| < 90Ė. For brevity, we will only show the case for when |ā š“š¶šµ| > 90Ė (the proof of the other case is similar). In the diagram below, we extend šµš¶ to a ray šµš¶ and let the perpendicular from š“ meet the ray at point š·.
Let š = |š¶š·| and š = |š“š·|.
Then by the Pythagorean theorem applied to ā³ š“š¶š· and ā³ š“šµš· results in
š! = š! + š! and š! = š +š ! + š!. Since we know what š! and š! are from the above equations, we can substitute those values into š! = š! + š! to get
š +š ! + š! = š! +š! + š!. Since š +š ! = š +š š +š = š! + šš + šš +š! = š! + 2šš +š!, then we have
š! + 2šš +š! + š! = š! +š! + š!. We can subtract the terms š!,š!, and š! from both sides of the equal sign. Then we have
2šš = 0. But this cannot be true because 2šš is a length; therefore, it cannot be equal to zero. Which means our assumption that |ā š“š¶šµ| > 90Ė cannot be true. We can write a similar proof to show that |ā š“š¶šµ| < 90Ė cannot be true either. Therefore, |ā š“š¶šµ| = 90Ė.