Eureka Math Homework Helper 2015โ2016 Geometry Module 1...2015-16 Homework M1 GEOMETRY Lesson 3 :...
Transcript of Eureka Math Homework Helper 2015โ2016 Geometry Module 1...2015-16 Homework M1 GEOMETRY Lesson 3 :...
Eureka Math, A Story of Functionsยฎ
Published by the non-profit Great Minds.
Copyright ยฉ 2015 Great Minds. No part of this work may be reproduced, distributed, modified, sold, or commercialized, in whole or in part, without consent of the copyright holder. Please see our User Agreement for more information. โGreat Mindsโ and โEureka Mathโ are registered trademarks of Great Minds.
Eureka Mathโข Homework Helper
2015โ2016
GeometryModule 1
Lessons 1โ34
2015-16
Lesson 1: Construct an Equilateral Triangle
M1 GEOMETRY
Lesson 1: Construct an Equilateral Triangle
1. The segment below has a length of ๐๐. Use a compassto mark all the points that are at a distance ๐๐ frompoint ๐ถ๐ถ.
I remember that the figure formed by the set of all the points that are a fixed distance from a given point is a circle.
1
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 1: Construct an Equilateral Triangle
M1
GEOMETRY
2. Use a compass to determine which of the following points, ๐ ๐ , ๐๐, ๐๐, and ๐๐, lie on the same circle about center ๐ถ๐ถ. Explain how you know.
If I set the compass point at ๐ช๐ช and adjust the compass so that the pencil passes through each of the points ๐น๐น, ๐บ๐บ, ๐ป๐ป, and ๐ผ๐ผ, I see that points ๐บ๐บ and ๐ผ๐ผ both lie on the same circle. Points ๐น๐น and ๐ป๐ป each lie on a different circle that does not pass through any of the other points.
Another way I solve this problem is by thinking about the lengths ๐ช๐ช๐น๐น, ๐ช๐ช๐บ๐บ, ๐ช๐ช๐ผ๐ผ, and ๐ช๐ช๐ป๐ป as radii. If I adjust my compass to any one of the radii, I can use that compass adjustment to compare the other radii. If the distance between any pair of points was greater or less than the compass adjustment, I would know that the point belonged to a different circle.
2
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 1: Construct an Equilateral Triangle
M1
GEOMETRY
3. Two points have been labeled in each of the following diagrams. Write a sentence for each point that describes what is known about the distance between the given point and each of the centers of the circles.
a. Circle ๐ถ๐ถ1 has a radius of 4; Circle ๐ถ๐ถ2 has a radius of 6.
b. Circle ๐ถ๐ถ3 has a radius of 4; Circle ๐ถ๐ถ4 has a radius of 4.
Point ๐ท๐ท is a distance of ๐๐ from ๐ช๐ช๐๐ and a distance greater than ๐๐ from ๐ช๐ช๐๐. Point ๐น๐น is a distance of ๐๐ from ๐ช๐ช๐๐ and a distance ๐๐ from ๐ช๐ช๐๐.
Point ๐บ๐บ is a distance of ๐๐ from ๐ช๐ช๐๐ and a distance greater than ๐๐ from ๐ช๐ช๐๐. Point ๐ป๐ป is a distance of ๐๐ from ๐ช๐ช๐๐ and a distance of ๐๐ from ๐ช๐ช๐๐.
c. Asha claims that the points ๐ถ๐ถ1, ๐ถ๐ถ2, and ๐ ๐ are the vertices of an equilateral triangle since ๐ ๐ is the intersection of the two circles. Nadege says this is incorrect but that ๐ถ๐ถ3, ๐ถ๐ถ4, and ๐๐ are the vertices of an equilateral triangle. Who is correct? Explain.
Nadege is correct. Points ๐ช๐ช๐๐, ๐ช๐ช๐๐, and ๐น๐น are not the vertices of an equilateral triangle because the distance between each pair of vertices is not the same. The points ๐ช๐ช๐๐, ๐ช๐ช๐๐, and ๐ป๐ป are the vertices of an equilateral triangle because the distance between each pair of vertices is the same.
Since the labeled points are each on a circle, I can describe the distance from each point to the center relative to the radius of the respective circle.
3
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 1: Construct an Equilateral Triangle
M1
GEOMETRY
4. Construct an equilateral triangle ๐ด๐ด๐ด๐ด๐ถ๐ถ that has a side length ๐ด๐ด๐ด๐ด, below. Use precise language to list the steps to perform the construction.
or
1. Draw circle ๐จ๐จ: center ๐จ๐จ, radius ๐จ๐จ๐จ๐จ. 2. Draw circle ๐จ๐จ: center ๐จ๐จ, radius ๐จ๐จ๐จ๐จ. 3. Label one intersection as ๐ช๐ช. 4. Join ๐จ๐จ, ๐จ๐จ, ๐ช๐ช.
I must use both endpoints of the segment as the centers of the two circles I must construct.
4
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 2: Construct an Equilateral Triangle
M1
GEOMETRY
Lesson 2: Construct an Equilateral Triangle
1. In parts (a) and (b), use a compass to determine whether the provided points determine the vertices of an equilateral triangle.
a.
b.
Points ๐จ๐จ, ๐ฉ๐ฉ, and ๐ช๐ช do not determine the vertices of an equilateral triangle because the distance between ๐จ๐จ and ๐ฉ๐ฉ, as measured by adjusting the compass, is not the same distance as between ๐ฉ๐ฉ and ๐ช๐ช and as between ๐ช๐ช and ๐จ๐จ.
Points ๐ซ๐ซ, ๐ฌ๐ฌ, and ๐ญ๐ญ do determine the vertices of an equilateral triangle because the distance between ๐ซ๐ซ and ๐ฌ๐ฌ is the same distance as between ๐ฌ๐ฌ and ๐ญ๐ญ and as between ๐ญ๐ญ and ๐ซ๐ซ.
The distance between each pair of vertices of an equilateral triangle is the same.
5
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 2: Construct an Equilateral Triangle
M1
GEOMETRY
2. Use what you know about the construction of an equilateral triangle to recreate parallelogram ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด
below, and write a set of steps that yields this construction.
Possible steps:
1. Draw ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ. 2. Draw circle ๐จ๐จ: center ๐จ๐จ, radius ๐จ๐จ๐ฉ๐ฉ. 3. Draw circle ๐ฉ๐ฉ: center ๐ฉ๐ฉ, radius ๐ฉ๐ฉ๐จ๐จ. 4. Label one intersection as ๐ช๐ช. 5. Join ๐ช๐ช to ๐จ๐จ and ๐ฉ๐ฉ. 6. Draw circle ๐ช๐ช: center ๐ช๐ช, radius ๐ช๐ช๐จ๐จ. 7. Label the intersection of circle ๐ช๐ช with circle ๐ฉ๐ฉ as ๐ซ๐ซ. 8. Join ๐ซ๐ซ to ๐ฉ๐ฉ and ๐ช๐ช.
6
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 2: Construct an Equilateral Triangle
M1
GEOMETRY
3. Four identical equilateral triangles can be arranged so that each of three of the triangles shares a side with the remaining triangle, as in the diagram. Use a compass to recreate this figure, and write a set of steps that yields this construction.
Possible steps:
1. Draw ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ. 2. Draw circle ๐จ๐จ: center ๐จ๐จ, radius ๐จ๐จ๐ฉ๐ฉ. 3. Draw circle ๐ฉ๐ฉ: center ๐ฉ๐ฉ, radius ๐ฉ๐ฉ๐จ๐จ. 4. Label one intersection as ๐ช๐ช; label the other intersection as ๐ซ๐ซ. 5. Join ๐จ๐จ and ๐ฉ๐ฉ with both ๐ช๐ช and ๐ซ๐ซ. 6. Draw circle ๐ซ๐ซ: center ๐ซ๐ซ, radius ๐ซ๐ซ๐จ๐จ. 7. Label the intersection of circle ๐ซ๐ซ with circle ๐จ๐จ as ๐ฌ๐ฌ. 8. Join ๐ฌ๐ฌ to ๐จ๐จ and ๐ซ๐ซ. 9. Label the intersection of circle ๐ซ๐ซ with circle ๐ฉ๐ฉ as ๐ญ๐ญ. 10. Join ๐ญ๐ญ to ๐ฉ๐ฉ and ๐ซ๐ซ.
7
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16 M1 GEOMETRY
Lesson 3: Copy and Bisect an Angle
Lesson 3: Copy and Bisect an Angle
1. Krysta is copying โ ๐ด๐ด๐ด๐ด๐ด๐ด to construct โ ๐ท๐ท๐ท๐ท๐ท๐ท.a. Complete steps 5โ9, and use a compass and
straightedge to finish constructing โ ๐ท๐ท๐ท๐ท๐ท๐ท. Steps to copy an angle are as follows:
1. Label the vertex of the original angle as B.
2. Draw EG๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ as one side of the angle to be drawn.3. Draw circle B: center B, any radius.4. Label the intersections of circle with the sides of the angle as and .5. Draw circle : center , radius .
as 6. 6 Label intersection of circle with 7.7 Draw circle : center , radius .8.8 Label either intersection of circle and circle as .
9. Draw
b. Underline the steps that describe the construction ofcircles used in the copied angle.
c. Why must circle ๐ท๐ท have a radius of length ๐ด๐ด๐ด๐ด?
The intersection of circle ๐ฉ๐ฉ and circle ๐ช๐ช: center ๐ช๐ช, radius๐ช๐ช๐จ๐จ determines point ๐จ๐จ. To mirror this location for thecopied angle, circle ๐ญ๐ญ must have a radius of length ๐ช๐ช๐จ๐จ.
I must remember how the radius of each of the two circles used in this construction impacts the key points that determine the copied angle.
8
Homework Helper A Story of Functions
๐ท๐ท ๐ท๐ทBB๐ด๐ด
๐ด๐ด ๐ด๐ด
๐ท๐ท๐ท๐ท ๐ท๐ท ๐ด๐ด๐ด๐ด
๐ท๐ท ๐ท๐ท ๐ท๐ท
.EG๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ
E๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ .D
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 3: Copy and Bisect an Angle
2. ๐ด๐ด๐ท๐ท๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ is the angle bisector of โ ๐ด๐ด๐ด๐ด๐ด๐ด.
a. Write the steps that yield ๐ด๐ด๐ท๐ท๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ as the angle bisector of โ ๐ด๐ด๐ด๐ด๐ด๐ด.
Steps to construct an angle bisector are as follows:
1. Label the vertex of the angle as ๐ฉ๐ฉ. 2. Draw circle ๐ฉ๐ฉ: center ๐ฉ๐ฉ, any size radius. 3. Label intersections of circle ๐ฉ๐ฉ with the rays of the angle as ๐จ๐จ and ๐ช๐ช. 4. Draw circle ๐จ๐จ: center ๐จ๐จ, radius ๐จ๐จ๐ช๐ช. 5. Draw circle ๐ช๐ช: center ๐ช๐ช, radius ๐ช๐ช๐จ๐จ. 6. At least one of the two intersection points of circle ๐จ๐จ and circle
๐ช๐ช lies in the interior of the angle. Label that intersection point ๐ซ๐ซ.
7. Draw ๐ฉ๐ฉ๐ซ๐ซ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ .
b. Why do circles ๐ด๐ด and ๐ด๐ด each have a radius equal to length ๐ด๐ด๐ด๐ด? Point ๐ซ๐ซ is as far from ๐จ๐จ as it is from ๐ช๐ช since ๐จ๐จ๐ซ๐ซ = ๐ช๐ช๐ซ๐ซ = ๐จ๐จ๐ช๐ช. As long as ๐จ๐จ and ๐ช๐ช are equal distances from vertex ๐ฉ๐ฉ and each of the circles has a radius equal ๐จ๐จ๐ช๐ช, ๐ซ๐ซ will be an equal distance from ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ
and ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . All the points that are equidistant from the two rays lie on the angle bisector.
I have to remember that the circle I construct with center at ๐ด๐ด can have a radius of any length, but the circles with centers ๐ด๐ด and ๐ด๐ด on each of the rays must have a radius ๐ด๐ด๐ด๐ด.
9
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 4: Construct a Perpendicular Bisector
Lesson 4: Construct a Perpendicular Bisector
1. Perpendicular bisector ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ is constructed to ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ; the intersection of ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ with the segment is labeled ๐๐. Use the idea of folding to explain why ๐ด๐ด and ๐ด๐ด are symmetric with respect to ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ .
If the segment is folded along ๐ท๐ท๐ท๐ท๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ so that ๐จ๐จ coincides with ๐ฉ๐ฉ, then ๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ coincides with ๐ฉ๐ฉ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, or ๐จ๐จ๐จ๐จ = ๐ฉ๐ฉ๐จ๐จ; ๐จ๐จ is the midpoint of the segment. โ ๐จ๐จ๐จ๐จ๐ท๐ท and โ ๐ฉ๐ฉ๐จ๐จ๐ท๐ท also coincide, and since they are two identical angles on a straight line, the sum of their measures must be ๐๐๐๐๐๐ยฐ, or each has a measure of ๐๐๐๐ยฐ. Thus, ๐จ๐จ and ๐ฉ๐ฉ are symmetric with respect to ๐ท๐ท๐ท๐ท๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ .
To be symmetric with respect to ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , the portion of ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ on one side of ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ must be mirrored on the opposite side of ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ .
10
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 4: Construct a Perpendicular Bisector
2. The construction of the perpendicular bisector has been started below. Complete both the construction and the steps to the construction.
1. Draw circle : center , radius . 2. Draw circle ๐ฟ๐ฟ: center ๐ฟ๐ฟ, radius ๐ฟ๐ฟ๐ฟ๐ฟ. 3. Label the points of intersections as ๐จ๐จ and ๐ฉ๐ฉ.
4. Draw ๐จ๐จ๐ฉ๐ฉ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ .
This construction is similar to the construction of an equilateral triangle. Instead of requiring one point that is an equal distance from both centers, this construction requires two points that are an equal distance from both centers.
11
Homework Helper A Story of Functions
๐๐ ๐๐ ๐๐๐๐
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 4: Construct a Perpendicular Bisector
3. Rhombus ๐๐๐๐๐๐๐๐ can be constructed by joining the midpoints of rectangle ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด. Use the perpendicular bisector construction to help construct rhombus ๐๐๐๐๐๐๐๐.
The midpoint of ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ is vertically aligned to the midpoint of ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ. I can use the construction of the perpendicular bisector to determine the perpendicular bisector of ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
12
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 5: Points of Concurrencies
Lesson 5: Points of Concurrencies
1. Observe the construction below, and explain the significance of point ๐๐.
Lines ๐๐, ๐๐, and ๐๐, which are each perpendicular bisectors of a side of the triangle, are concurrent at point ๐ท๐ท.
a. Describe the distance between ๐ด๐ด and ๐๐ and between ๐ต๐ต and ๐๐. Explain why this true.
๐ท๐ท is equidistant from ๐จ๐จ and ๐ฉ๐ฉ. Any point that lies on the perpendicular bisector of ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ is equidistant from either endpoint ๐จ๐จ or ๐ฉ๐ฉ.
b. Describe the distance between ๐ถ๐ถ and ๐๐ and between ๐ต๐ต and ๐๐. Explain why this true.
๐ท๐ท is equidistant from ๐ฉ๐ฉ and ๐ช๐ช. Any point that lies on the perpendicular bisector of ๐ฉ๐ฉ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ is equidistant from either endpoint ๐ฉ๐ฉ or ๐ช๐ช.
c. What do the results of Problem 1 parts (a) and (b) imply about ๐๐?
Since ๐ท๐ท is equidistant from ๐จ๐จ and ๐ฉ๐ฉ and from ๐ฉ๐ฉ and ๐ช๐ช, then it is also equidistant from ๐จ๐จ and ๐ช๐ช. This is why ๐ท๐ท is the point of concurrency of the three perpendicular bisectors.
The markings in the figure imply lines ๐๐, ๐๐, and ๐๐ are perpendicular bisectors.
13
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 5: Points of Concurrencies
2. Observe the construction below, and explain the significance of point ๐๐.
Rays ๐จ๐จ๐จ๐จ, ๐ช๐ช๐จ๐จ, and ๐ฉ๐ฉ๐จ๐จ are each angle bisectors of an angle of a triangle, and are concurrent at point ๐จ๐จ, or all three angle bisectors intersect in a single point.
a. Describe the distance between ๐๐ and rays ๐ด๐ด๐ต๐ต๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐ด๐ด๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . Explain why this true.
๐จ๐จ is equidistant from ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . Any point that lies on the angle bisector of โ ๐จ๐จ is equidistant from rays ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ .
b. Describe the distance between ๐๐ and rays ๐ต๐ต๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐ต๐ต๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . Explain why this true.
๐จ๐จ is equidistant from ๐ฉ๐ฉ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐ฉ๐ฉ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . Any point that lies on the angle bisector of โ ๐ฉ๐ฉ is equidistant from rays ๐ฉ๐ฉ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐ฉ๐ฉ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ .
c. What do the results of Problem 2 parts (a) and (b) imply about ๐๐?
Since ๐จ๐จ is equidistant from ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and from ๐ฉ๐ฉ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐ฉ๐ฉ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ , then it is also equidistant from ๐ช๐ช๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐ช๐ช๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . This is why ๐จ๐จ is the point of concurrency of the three angle bisectors.
The markings in the figure imply that rays ๐ด๐ด๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ , ๐ถ๐ถ๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ , and ๐ต๐ต๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ are all angle bisectors.
14
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 6: Solve for Unknown AnglesโAngles and Lines at a Point
Lesson 6: Solve for Unknown AnglesโAngles and Lines at a Point
1. Write an equation that appropriately describes each of the diagrams below.
a. b.
๐๐ + ๐๐ + ๐๐ + ๐ ๐ = ๐๐๐๐๐๐ยฐ ๐๐ + ๐๐ + ๐๐ + ๐ ๐ + ๐๐ + ๐๐ + ๐๐ = ๐๐๐๐๐๐ยฐ
Adjacent angles on a line sum to 180ยฐ. Adjacent angles around a point sum to 360ยฐ.
15
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 6: Solve for Unknown AnglesโAngles and Lines at a Point
2. Find the measure of โ ๐ด๐ด๐ด๐ด๐ด๐ด.
๐๐๐๐ + ๐๐๐๐ยฐ + ๐๐ = ๐๐๐๐๐๐ยฐ
๐๐๐๐ + ๐๐๐๐ยฐ = ๐๐๐๐๐๐ยฐ
๐๐๐๐ = ๐๐๐๐๐๐ยฐ
๐๐ = ๐๐๐๐ยฐ
The measure of โ ๐จ๐จ๐จ๐จ๐จ๐จ is ๐๐(๐๐๐๐ยฐ), or ๐๐๐๐๐๐ยฐ.
3. Find the measure of โ ๐ท๐ท๐ด๐ด๐ท๐ท.
๐๐ + ๐๐๐๐ + ๐๐๐๐ + ๐๐๐๐ + ๐๐๐๐๐๐ยฐ = ๐๐๐๐๐๐ยฐ
๐๐๐๐๐๐+ ๐๐๐๐๐๐ยฐ = ๐๐๐๐๐๐ยฐ
๐๐๐๐๐๐ = ๐๐๐๐๐๐ยฐ
๐๐ = ๐๐๐๐ยฐ
The measure of โ ๐ซ๐ซ๐จ๐จ๐ซ๐ซ is ๐๐(๐๐๐๐ยฐ), or ๐๐๐๐ยฐ.
I must solve for ๐ฅ๐ฅ before I can find the measure of โ ๐ด๐ด๐ด๐ด๐ด๐ด.
16
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 7: Solve for Unknown AnglesโTransversals
Lesson 7: Solve for Unknown AnglesโTransversals
1. In the following figure, angle measures ๐๐, ๐๐, ๐๐, and ๐๐ are equal. List four pairs of parallel lines.
Four pairs of parallel lines: ๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ฌ๐ฌ๐ฌ๐ฌ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , ๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ช๐ช๐ช๐ช๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , ๐จ๐จ๐ช๐ช๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ช๐ช๐ซ๐ซ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , ๐ช๐ช๐ช๐ช๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ฌ๐ฌ๐ฌ๐ฌ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ
2. Find the measure of โ a.
The measure of ๐๐ is ๐๐๐๐๐๐ยฐ โ ๐๐๐๐๐๐ยฐ, or ๐๐๐๐ยฐ.
I can look for pairs of alternate interior angles and corresponding angles to help identify which lines are parallel.
I can extend lines to make the angle relationships more clear. I then need to apply what I know about alternate interior and supplementary angles to solve for ๐๐.
17
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16 M1
GEOMETRY
Lesson 7: Solve for Unknown AnglesโTransversals
3. Find the measure of ๐๐.
The measure of ๐๐ is ๐๐๐๐ยฐ + ๐๐๐๐ยฐ, or ๐๐๐๐ยฐ.
4. Find the value of ๐ฅ๐ฅ.
๐๐๐๐ + ๐๐ = ๐๐๐๐๐๐
๐๐๐๐ = ๐๐๐๐๐๐
๐๐ = ๐๐๐๐
I can draw a horizontal auxiliary line, parallel to the other horizontal lines in order to make the necessary corresponding angle pairs more apparent.
18
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 8: Solve for Unknown AnglesโAngles in a Triangle
Lesson 8: Solve for Unknown AnglesโAngles in a Triangle
1. Find the measure of ๐๐.
๐ ๐ + ๐๐๐๐๐๐ยฐ + ๐๐๐๐ยฐ = ๐๐๐๐๐๐ยฐ
๐ ๐ + ๐๐๐๐๐๐ยฐ = ๐๐๐๐๐๐ยฐ
๐ ๐ = ๐๐๐๐ยฐ
I need to apply what I know about complementary and supplementary angles to begin to solve for ๐๐.
19
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 8: Solve for Unknown AnglesโAngles in a Triangle
M1
GEOMETRY
2. Find the measure of ๐๐.
๐๐๐๐ + ๐๐๐๐ยฐ = ๐๐๐๐๐๐ยฐ
๐๐๐๐ = ๐๐๐๐๐๐ยฐ
๐๐ = ๐๐๐๐ยฐ
I need to apply what I know about parallel lines cut by a transversal and alternate interior angles in order to solve for ๐๐.
20
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 8: Solve for Unknown AnglesโAngles in a Triangle
M1
GEOMETRY
3. Find the measure of ๐ฅ๐ฅ.
(๐๐๐๐๐๐ยฐ โ ๐๐๐๐) + ๐๐๐๐๐๐ยฐ + ๐๐ = ๐๐๐๐๐๐ยฐ
๐๐๐๐๐๐ยฐ โ ๐๐๐๐ = ๐๐๐๐๐๐ยฐ
๐๐๐๐ = ๐๐๐๐๐๐ยฐ
๐๐ = ๐๐๐๐ยฐ
I need to add an auxiliary line to modify the diagram; the modified diagram has enough information to write an equation that I can use to solve for ๐ฅ๐ฅ.
21
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 9: Unknown Angle ProofsโWriting Proofs
Lesson 9: Unknown Angle ProofsโWriting Proofs
1. Use the diagram below to prove that ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
๐๐โ ๐ท๐ท + ๐๐โ ๐ธ๐ธ+ ๐๐โ ๐ท๐ท๐ท๐ท๐ธ๐ธ = ๐๐๐๐๐๐ยฐ The sum of the angle measures in a triangle is ๐๐๐๐๐๐ยฐ.
๐๐โ ๐ท๐ท๐ท๐ท๐ธ๐ธ = ๐๐๐๐ยฐ Subtraction property of equality
๐๐โ ๐ผ๐ผ+ ๐๐โ ๐ป๐ป + ๐๐โ ๐ป๐ป๐ป๐ป๐ผ๐ผ = ๐๐๐๐๐๐ยฐ The sum of the angle measures in a triangle is ๐๐๐๐๐๐ยฐ.
๐๐โ ๐ป๐ป๐ป๐ป๐ผ๐ผ = ๐๐๐๐ยฐ Subtraction property of equality
๐๐โ ๐ท๐ท๐ท๐ท๐ธ๐ธ + ๐๐โ ๐ป๐ป๐ป๐ป๐ผ๐ผ+ ๐๐โ ๐ป๐ป๐บ๐บ๐ท๐ท = ๐๐๐๐๐๐ยฐ The sum of the angle measures in a triangle is ๐๐๐๐๐๐ยฐ.
๐๐โ ๐ป๐ป๐บ๐บ๐ท๐ท = ๐๐๐๐ยฐ Subtraction property of equality
๐ป๐ป๐ป๐ป๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ธ๐ธ๐ท๐ท๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ Perpendicular lines form ๐๐๐๐ยฐ angles.
To show that ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, I need to first show that ๐๐โ ๐๐๐๐๐๐ = 90ยฐ.
22
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 9: Unknown Angle ProofsโWriting Proofs
2. Prove ๐๐โ ๐๐๐๐๐๐ = ๐๐โ ๐๐๐๐๐๐.
๐ท๐ท๐ธ๐ธ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ป๐ป๐ป๐ป๏ฟฝโ๏ฟฝ๏ฟฝโ , ๐ธ๐ธ๐ผ๐ผ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ป๐ป๐บ๐บ๏ฟฝโ๏ฟฝ๏ฟฝโ Given
๐๐โ ๐ท๐ท๐ธ๐ธ๐ป๐ป = ๐๐โ ๐ป๐ป๐ป๐ป๐ป๐ป, ๐๐โ ๐ท๐ท๐ธ๐ธ๐ป๐ป = ๐๐โ ๐บ๐บ๐ป๐ป๐ป๐ป If parallel lines are cut by a transversal, then corresponding angles are equal in measure.
๐๐โ ๐ท๐ท๐ธ๐ธ๐ท๐ท = ๐๐โ ๐ท๐ท๐ธ๐ธ๐ป๐ป โ๐๐โ ๐ท๐ท๐ธ๐ธ๐ป๐ป,
๐๐โ ๐ป๐ป๐ป๐ป๐บ๐บ = ๐๐โ ๐ป๐ป๐ป๐ป๐ป๐ปโ๐๐โ ๐บ๐บ๐ป๐ป๐ป๐ป
Partition property
๐๐โ ๐ท๐ท๐ธ๐ธ๐ท๐ท = ๐๐โ ๐ป๐ป๐ป๐ป๐ป๐ปโ๐๐โ ๐บ๐บ๐ป๐ป๐ป๐ป Substitution property of equality
๐๐โ ๐ท๐ท๐ธ๐ธ๐ท๐ท = ๐๐โ ๐ป๐ป๐ป๐ป๐บ๐บ Substitution property of equality
I need to consider how angles โ ๐๐๐๐๐๐ and โ ๐๐๐๐๐๐ are related to angles I know to be equal in measure in the diagram.
23
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 9: Unknown Angle ProofsโWriting Proofs
3. In the diagram below, ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ bisects โ ๐๐๐๐๐๐, and ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ bisects โ ๐๐๐๐๐๐. Prove that ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
๐ท๐ท๐ธ๐ธ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ท๐ท๐ป๐ป๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , ๐บ๐บ๐ป๐ป๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ bisects โ ๐ธ๐ธ๐บ๐บ๐ธ๐ธ and ๐ธ๐ธ๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ bisects โ ๐บ๐บ๐ธ๐ธ๐ท๐ท Given
๐๐โ ๐ธ๐ธ๐บ๐บ๐ธ๐ธ = ๐๐โ ๐บ๐บ๐ธ๐ธ๐ท๐ท If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
๐๐โ ๐ธ๐ธ๐บ๐บ๐ป๐ป = ๐๐โ ๐ป๐ป๐บ๐บ๐ธ๐ธ, ๐๐โ ๐บ๐บ๐ธ๐ธ๐๐ = ๐๐โ ๐๐๐ธ๐ธ๐ท๐ท Definition of bisect
๐๐โ ๐ธ๐ธ๐บ๐บ๐ธ๐ธ = ๐๐โ ๐ธ๐ธ๐บ๐บ๐ป๐ป+ ๐๐โ ๐ป๐ป๐บ๐บ๐ธ๐ธ,
๐๐โ ๐บ๐บ๐ธ๐ธ๐ท๐ท = ๐๐โ ๐บ๐บ๐ธ๐ธ๐๐ +๐๐โ ๐๐๐ธ๐ธ๐ท๐ท
Partition property
๐๐โ ๐ธ๐ธ๐บ๐บ๐ธ๐ธ = ๐๐(๐๐โ ๐ป๐ป๐บ๐บ๐ธ๐ธ), ๐๐โ ๐บ๐บ๐ธ๐ธ๐ท๐ท = ๐๐(๐๐โ ๐บ๐บ๐ธ๐ธ๐๐) Substitution property of equality
๐๐(๐๐โ ๐ป๐ป๐บ๐บ๐ธ๐ธ) = ๐๐(๐๐โ ๐บ๐บ๐ธ๐ธ๐๐) Substitution property of equality
๐๐โ ๐ป๐ป๐บ๐บ๐ธ๐ธ = ๐๐โ ๐บ๐บ๐ธ๐ธ๐๐ Division property of equality
๐บ๐บ๐ป๐ป๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ธ๐ธ๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ If two lines are cut by a transversal such that a pair of alternate interior angles are equal in measure, then the lines are parallel.
Since the alternate interior angles along a transversal that cuts parallel lines are equal in measure, the bisected halves are also equal in measure. This will help me determine whether segments ๐๐๐๐ and ๐๐๐๐ are parallel.
24
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 10: Unknown Angle ProofsโProofs with Constructions
M1
GEOMETRY
Lesson 10: Unknown Angle ProofsโProofs with Constructions
1. Use the diagram below to prove that ๐๐ = 106ยฐ โ ๐๐.
Construct ๐ผ๐ผ๐ผ๐ผ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ parallel to ๐ป๐ป๐ป๐ป๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐บ๐บ๐บ๐บ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ .
๐๐โ ๐บ๐บ๐ผ๐ผ๐ผ๐ผ = ๐๐ If parallel lines are cut by a transversal, then corresponding angles are equal in measure.
๐๐โ ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ = ๐๐๐๐๐๐ยฐ โ ๐๐ Partition property
๐๐โ ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ = ๐๐ If parallel lines are cut by a transversal, then corresponding angles are equal in measure.
๐๐ = ๐๐๐๐๐๐ยฐ โ ๐๐ Substitution property of equality
Just as if this were a numeric problem, I need to construct a horizontal line through ๐๐, so I can see the special angle pairs created by parallel lines cut by a transversal.
25
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 10: Unknown Angle ProofsโProofs with Constructions
M1
GEOMETRY
2. Use the diagram below to prove that mโ C = ๐๐ + ๐๐.
Construct ๐ฎ๐ฎ๐ฎ๐ฎ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ parallel to ๐บ๐บ๐ป๐ป๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐ญ๐ญ๐ญ๐ญ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ through ๐ช๐ช.
๐๐โ ๐ป๐ป๐ช๐ช๐ฎ๐ฎ = ๐๐, ๐๐โ ๐ญ๐ญ๐ช๐ช๐ฎ๐ฎ = ๐ ๐ If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
๐๐โ ๐ช๐ช = ๐๐ + ๐ ๐ Partition property
26
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 10: Unknown Angle ProofsโProofs with Constructions
M1
GEOMETRY
3. Use the diagram below to prove that ๐๐โ ๐ถ๐ถ๐ถ๐ถ๐ถ๐ถ = ๐๐ + 90ยฐ.
Construct ๐ซ๐ซ๐ญ๐ญ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ parallel to ๐ช๐ช๐ป๐ป๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ . Extend ๐บ๐บ๐ป๐ป๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ so that it intersects ๐ซ๐ซ๐ญ๐ญ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ ; extend ๐ช๐ช๐ป๐ป๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
๐๐โ ๐ช๐ช๐ซ๐ซ๐ญ๐ญ+ ๐๐๐๐ยฐ = ๐๐๐๐๐๐ยฐ If parallel lines are cut by a transversal, then same-side interior angles are supplementary.
๐๐โ ๐ช๐ช๐ซ๐ซ๐ญ๐ญ = ๐๐๐๐ยฐ Subtraction property of equality
๐๐โ ๐ซ๐ซ๐ญ๐ญ๐ป๐ป = ๐๐ If parallel lines are cut by a transversal, then corresponding angles are equal in measure.
๐๐โ ๐ญ๐ญ๐ซ๐ซ๐ญ๐ญ = ๐๐ If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
๐๐โ ๐ช๐ช๐ซ๐ซ๐ญ๐ญ = ๐๐ + ๐๐๐๐ยฐ Partition property
I am going to need multiple constructions to show why the measure of โ ๐ถ๐ถ๐ถ๐ถ๐ถ๐ถ = ๐๐ + 90ยฐ.
27
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 11: Unknown Angle ProofsโProofs of Known Facts
Lesson 11: Unknown Angle ProofsโProofs of Known Facts
1. Given: ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝโ intersect at ๐ ๐ . Prove: ๐๐โ 2 = ๐๐โ 4
๐ฝ๐ฝ๐ฝ๐ฝ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ intersect at ๐น๐น. Given
๐๐โ ๐๐ +๐๐โ ๐๐ = ๐๐๐๐๐๐ยฐ; ๐๐โ ๐๐+ ๐๐โ ๐๐ = ๐๐๐๐๐๐ยฐ Angles on a line sum to ๐๐๐๐๐๐ยฐ.
๐๐โ ๐๐ +๐๐โ ๐๐ = ๐๐โ ๐๐+ ๐๐โ ๐๐ Substitution property of equality
๐๐โ ๐๐ = ๐๐โ ๐๐ Subtraction property of equality
28
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 11: Unknown Angle ProofsโProofs of Known Facts
2. Given: ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ ; ๐ ๐ ๐ ๐ ๏ฟฝโ๏ฟฝ๏ฟฝโ โฅ ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ
Prove: ๐๐๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ ๐ ๐ ๐ ๏ฟฝโ๏ฟฝ๏ฟฝโ
๐ท๐ท๐ท๐ท๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ป๐ป๐ป๐ป๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ ; ๐น๐น๐น๐น๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ โฅ ๐ป๐ป๐ป๐ป๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ Given
๐๐โ ๐ท๐ท๐ฝ๐ฝ๐ธ๐ธ = ๐๐๐๐ยฐ; ๐๐โ ๐น๐น๐ธ๐ธ๐ป๐ป = ๐๐๐๐ยฐ Perpendicular lines form ๐๐๐๐ยฐ angles.
๐๐โ ๐ท๐ท๐ฝ๐ฝ๐ธ๐ธ = ๐๐โ ๐น๐น๐ธ๐ธ๐ป๐ป Transitive property of equality
โฅ If two lines are cut by a transversal such that a pair of corresponding angles are equal in measure, then the lines are parallel.
29
Homework Helper A Story of Functions
๐ท๐ท๐ท๐ท๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ ๐น๐น๐น๐น๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
Lesson 12: TransformationsโThe Next Level
GEOMETRY
Lesson 12: TransformationsโThe Next Level
1. Recall that a transformation ๐น๐น of the plane is a function that assigns to each point ๐๐ of the plane a unique
point ๐น๐น(๐๐) in the plane. Of the countless kinds of transformations, a subset exists that preserves lengths and angle measures. In other words, they are transformations that do not distort the figure. These transformations, specifically reflections, rotations, and translations, are called basic rigid motions.
Examine each pre-image and image pair. Determine which pairs demonstrate a rigid motion applied to the pre-image.
Pre-Image Image
Is this transformation an example of a rigid motion?
Explain.
a.
No, this transformation did not preserve lengths, even though it seems to have preserved angle measures.
b.
Yes, this is a rigid motionโa translation.
c.
Yes, this is a rigid motionโa reflection.
d.
No, this transformation did not preserve lengths or angle measures.
30
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
Lesson 12: TransformationsโThe Next Level
GEOMETRY
e.
Yes, this is a rigid motionโa rotation.
2. Each of the following pairs of diagrams shows the same figure as a pre-image and as a post-transformation image. Each of the second diagrams shows the details of how the transformation is performed. Describe what you see in each of the second diagrams.
The line that the pre-image is reflected over is the perpendicular bisector of each of the segments joining the corresponding vertices of the triangles.
For each of the transformations, I must describe all the details that describe the โmechanicsโ of how the transformation works. For example, I see that there are congruency marks on each half of the segments that join the corresponding vertices and that each segment is perpendicular to the line of reflection. This is essential to how the reflection works.
31
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
Lesson 12: TransformationsโThe Next Level
GEOMETRY
The segment ๐ท๐ท๐ท๐ท is rotated counterclockwise about ๐ฉ๐ฉ by ๐๐๐๐ยฐ. The path that describes how ๐ท๐ท maps to ๐ท๐ทโฒ is a circle with center ๐ฉ๐ฉ and radius ๐ฉ๐ฉ๐ท๐ท๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ; ๐ท๐ท moves counterclockwise along the circle. โ ๐ท๐ท๐ฉ๐ฉ๐ท๐ทโฒ has a measure of ๐๐๐๐ยฐ. A similar case can be made for ๐ท๐ท.
The pre-image โณ ๐จ๐จ๐ฉ๐ฉ๐จ๐จ has been translated the length and direction of vector ๐จ๐จ๐จ๐จโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ .
32
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 13: Rotations
M1
GEOMETRY
Lesson 13: Rotations
1. Recall the definition of rotation:
For 0ยฐ < ๐๐ยฐ < 180ยฐ, the rotation of ๐๐ degrees around the center ๐ถ๐ถ is the transformation ๐ ๐ ๐ถ๐ถ,๐๐ of the plane defined as follows:
1. For the center point ๐ถ๐ถ, ๐ ๐ ๐ถ๐ถ,๐๐(๐ถ๐ถ) = ๐ถ๐ถ, and
2. For any other point ๐๐, ๐ ๐ ๐ถ๐ถ,๐๐(๐๐) is the point ๐๐ that lies in the counterclockwise half-plane of ๐ถ๐ถ๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ , such that ๐ถ๐ถ๐๐ = ๐ถ๐ถ๐๐ and ๐๐โ ๐๐๐ถ๐ถ๐๐ = ๐๐ยฐ.
a. Which point does the center ๐ถ๐ถ map to once the rotation has been applied?
By the definition, the center ๐ช๐ช maps to itself: ๐น๐น๐ช๐ช,๐ฝ๐ฝ(๐ช๐ช) = ๐ช๐ช.
b. The image of a point ๐๐ that undergoes a rotation ๐ ๐ ๐ถ๐ถ,๐๐ is the image point ๐๐: ๐ ๐ ๐ถ๐ถ,๐๐(๐๐) = ๐๐. Point ๐๐ is said to lie in the counterclockwise half plane of ๐ถ๐ถ๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . Shade the counterclockwise half plane of ๐ถ๐ถ๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ .
I must remember that a half plane is a line in a plane that separates the plane into two sets.
33
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 13: Rotations
M1
GEOMETRY
c. Why does part (2) of the definition include ๐ถ๐ถ๐๐ = ๐ถ๐ถ๐๐? What relationship does ๐ถ๐ถ๐๐ = ๐ถ๐ถ๐๐ have with the circle in the diagram above?
๐ช๐ช๐ช๐ช = ๐ช๐ช๐ช๐ช describes how ๐ช๐ช maps to ๐ช๐ช. The rotation, a function, describes a path such that ๐ช๐ช โrotatesโ (let us remember there is no actual motion) along the circle ๐ช๐ช with radius ๐ช๐ช๐ช๐ช (and thereby also of radius ๐ช๐ช๐ช๐ช).
d. Based on the figure on the prior page, what is the angle of rotation, and what is the measure of the angle of rotation?
The angle of rotation is โ ๐ช๐ช๐ช๐ช๐ช๐ช, and the measure is ๐ฝ๐ฝห.
2. Use a protractor to determine the angle of rotation.
I must remember that the angle of rotation is found by forming an angle from any pair of corresponding points and the center of rotation; the measure of this angle is the angle of rotation.
34
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 13: Rotations
M1
GEOMETRY
3. Determine the center of rotation for the following pre-image and image.
I must remember that the center of rotation is located by the following steps: (1) join two pairs of corresponding points in the pre-image and image, (2) take the perpendicular bisector of each segment, and finally (3) identify the intersection of the bisectors as the center of rotation.
35
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 14: Reflections
M1
GEOMETRY
Lesson 14: Reflections
1. Recall the definition of reflection:
For a line ๐๐ in the plane, a reflection across ๐๐ is the transformation ๐๐๐๐ of the plane defined as follows:
1. For any point ๐๐ on the line ๐๐, ๐๐๐๐(๐๐) = ๐๐, and 2. For any point ๐๐ not on ๐๐, ๐๐๐๐(๐๐) is the point ๐๐
so that ๐๐ is the perpendicular bisector of the segment ๐๐๐๐.
a. Where do the points that belong to a line of reflection map to once the reflection is applied?
Any point ๐ท๐ท on the line of reflection maps to itself: ๐๐๐๐(๐ท๐ท) = ๐ท๐ท.
b. Once a reflection is applied, what is the relationship between a point, its reflected image, and the line of reflection? For example, based on the diagram above, what is the relationship between ๐ด๐ด, ๐ด๐ดโฒ, and line ๐๐?
Line ๐๐ is the perpendicular bisector to the segment that joins ๐จ๐จ and ๐จ๐จโฒ.
c. Based on the diagram above, is there a relationship between the distance from ๐ต๐ต to ๐๐ and from ๐ต๐ตโฒ to ๐๐?
Any pair of corresponding points is equidistant from the line of reflection.
๐๐
I can model a reflection by folding paper: The fold itself is the line of reflection.
36
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 14: Reflections
M1
GEOMETRY
2. Using a compass and straightedge, determine the line of reflection for pre-image โณ ๐ด๐ด๐ต๐ต๐ด๐ด and โณ
Write the steps to the construction.
1. Draw circle ๐ช๐ช: center ๐ช๐ช, radius ๐ช๐ช๐ช๐ชโฒ.
2. Draw circle ๐ช๐ชโฒ: center ๐ช๐ชโฒ, radius ๐ช๐ชโฒ๐ช๐ช.
3. Draw a line through the points of intersection between circles ๐ช๐ช and ๐ช๐ชโฒ.
37
Homework Helper A Story of Functions
๐ด๐ดโฒ๐ต๐ตโฒ๐ด๐ดโฒ.
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 14: Reflections
M1
GEOMETRY
3. Using a compass and straightedge, reflect point ๐ด๐ด over line ๐๐. Write the steps to the construction.
1. Draw circle ๐จ๐จ such that the circle intersects with line ๐๐ in two locations, ๐บ๐บ and ๐ป๐ป.
2. Draw circle ๐บ๐บ: center ๐บ๐บ, radius ๐บ๐บ๐จ๐จ.
3. Draw circle ๐ป๐ป: center ๐ป๐ป, radius ๐ป๐ป๐จ๐จ.
4. Label the intersection of circles ๐บ๐บ and ๐ป๐ป as ๐จ๐จโฒ.
38
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 15: Rotations, Reflections, and Symmetry
Lesson 15: Rotations, Reflections, and Symmetry
1. A symmetry of a figure is a basic rigid motion that maps the figure back onto itself. A figure is said to have line symmetry if there exists a line (or lines) so that the image of the figure when reflected over the line(s) is itself. A figure is said to have nontrivial rotational symmetry if a rotation of greater than 0ยฐ but less than 360ยฐ maps a figure back to itself. A trivial symmetry is a transformation that maps each point of a figure back to the same point (i.e., in terms of a function, this would be ๐๐(๐ฅ๐ฅ) = ๐ฅ๐ฅ). An example of this is a rotation of 360ยฐ. a. Draw all lines of symmetry for the equilateral hexagon below. Locate the center of rotational
symmetry.
b. How many of the symmetries are rotations (of an angle of rotation less than or equal to 360ยฐ)? What are the angles of rotation that yield symmetries?
๐๐, including the identity symmetry. The angles of rotation are: ๐๐๐๐ยฐ, ๐๐๐๐๐๐ยฐ, ๐๐๐๐๐๐ยฐ, ๐๐๐๐๐๐ยฐ, ๐๐๐๐๐๐ยฐ, and ๐๐๐๐๐๐ยฐ.
c. How many of the symmetries are reflections?
๐๐
39
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 15: Rotations, Reflections, and Symmetry
d. How many places can vertex ๐ด๐ด be moved by some symmetry of the hexagon?
๐จ๐จ can be moved to ๐๐ placesโ๐จ๐จ, ๐ฉ๐ฉ, ๐ช๐ช, ๐ซ๐ซ, ๐ฌ๐ฌ, and ๐ญ๐ญ.
e. For a given symmetry, if you know the image of ๐ด๐ด, how many possibilities exist for the image of ๐ต๐ต?
๐๐
2. Shade as few of the nine smaller sections as possible so that the resulting figure has a. Only one vertical and one horizontal line of symmetry. b. Only two lines of symmetry about the diagonals. c. Only one horizontal line of symmetry. d. Only one line of symmetry about a diagonal. e. No line of symmetry.
Possible solutions:
a. b. c.
d. e.
40
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 16: Translations
M1
GEOMETRY
Lesson 16: Translations
1. Recall the definition of translation:
For vector ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ , the translation along ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ is the transformation ๐๐๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ of the plane defined as follows:
1. For any point ๐๐ on ๐ด๐ด๐ด๐ด๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , ๐๐๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ (๐๐) is the point ๐๐ on ๐ด๐ด๐ด๐ด๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ so that ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ has the same length and the same direction as ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ , and
2. For any point ๐๐ not on ๐ด๐ด๐ด๐ด๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , ๐๐๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ (๐๐) is the point ๐๐ obtained as follows. Let ๐๐ be the line passing
through ๐๐ and parallel to ๐ด๐ด๐ด๐ด๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ . Let ๐๐ be the line passing through ๐ด๐ด and parallel to ๐ด๐ด๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ . The point ๐๐ is the intersection of ๐๐ and ๐๐.
41
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 16: Translations
M1
GEOMETRY
2. Use a compass and straightedge to translate segment ๐บ๐บ๐บ๐บ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ along vector ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ .
To find ๐บ๐บโฒ, I must mark off the length of ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ in the direction of the vector from ๐บ๐บ. I will repeat these steps to locate ๐บ๐บโฒ.
42
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 16: Translations
M1
GEOMETRY
3. Use a compass and straightedge to translate point ๐บ๐บ along vector ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . Write the steps to this construction.
1. Draw circle ๐ฎ๐ฎ: center ๐ฎ๐ฎ, radius ๐จ๐จ๐จ๐จ.
2. Draw circle ๐จ๐จ: center ๐จ๐จ, radius ๐จ๐จ๐ฎ๐ฎ.
3. Label the intersection of circle ๐ฎ๐ฎ and circle ๐จ๐จ as ๐ฎ๐ฎโฒ. (Circles ๐ฎ๐ฎ and ๐จ๐จ intersect in two locations; pick the intersection so that ๐จ๐จ and ๐ฎ๐ฎโฒ are in opposite half planes of ๐จ๐จ๐ฎ๐ฎ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ .)
To find ๐บ๐บโฒ, my construction is really resulting in locating the fourth vertex of a parallelogram.
43
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 17: Characterize Points on a Perpendicular Bisector
M1
GEOMETRY
Lesson 17: Characterize Points on a Perpendicular Bisector
1. Perpendicular bisectors are essential to the rigid motions reflections and rotations. a. How are perpendicular bisectors essential to
reflections?
The line of reflection is a perpendicular bisector to the segment that joins each pair of pre-image and image points of a reflected figure.
b. How are perpendicular bisectors essential to rotations?
Perpendicular bisectors are key to determining the center of a rotation. The center of a rotation is determined by joining two pairs of pre-image and image points and constructing the perpendicular bisector of each of the segments. Where the perpendicular bisectors intersect is the center of the rotation.
2. Rigid motions preserve distance, or in other words, the image of a figure that has had a rigid motion applied to it will maintain the same lengths as the original figure. a. Based on the following rotation, which of the following statements must be
true? i. ๐ด๐ด๐ด๐ด = ๐ด๐ดโฒ๐ด๐ดโฒ True ii. ๐ต๐ต๐ต๐ตโฒ = ๐ถ๐ถ๐ถ๐ถโฒ False iii. ๐ด๐ด๐ถ๐ถ = ๐ด๐ดโฒ๐ถ๐ถโฒ True iv. ๐ต๐ต๐ด๐ด = ๐ต๐ตโฒ๐ด๐ดโฒ True v. ๐ถ๐ถ๐ด๐ดโฒ = ๐ถ๐ถโฒ๐ด๐ด False
I can re-examine perpendicular bisectors (in regard to reflections) in Lesson 14 and rotations in Lesson 13.
44
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 17: Characterize Points on a Perpendicular Bisector
M1
GEOMETRY
b. Based on the following rotation, which of the following statements must be true? i. ๐ถ๐ถ๐ถ๐ถโฒ = ๐ต๐ต๐ต๐ตโฒ False ii. ๐ต๐ต๐ถ๐ถ = ๐ต๐ตโฒ๐ถ๐ถโฒ True
3. In the following figure, point ๐ต๐ต is reflected across line ๐๐. c. What is the relationship between ๐ต๐ต, ๐ต๐ตโฒ, and ๐๐?
Line ๐๐ is the perpendicular bisector of ๐ฉ๐ฉ๐ฉ๐ฉโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
d. What is the relationship between ๐ต๐ต, ๐ต๐ตโฒ, and any point ๐๐ on ๐๐?
๐ฉ๐ฉ and ๐ฉ๐ฉโฒ are equidistant from line ๐๐ and therefore equidistant from any point ๐ท๐ท on ๐๐.
45
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 18: Looking More Carefully at Parallel Lines
Lesson 18: Looking More Carefully at Parallel Lines
1. Given that โ ๐ต๐ต and โ ๐ถ๐ถ are supplementary and ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ต๐ต๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, prove that ๐๐โ ๐ด๐ด = ๐๐โ ๐ถ๐ถ.
โ ๐ฉ๐ฉ and โ ๐ช๐ช are supplementary. Given
๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ฉ๐ฉ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ Given
โ ๐ฉ๐ฉ and โ ๐จ๐จ are supplementary. If two parallel lines are cut by a transversal, then same-side interior angles are supplementary.
๐๐โ ๐ฉ๐ฉ + ๐๐โ ๐จ๐จ = ๐๐๐๐๐๐ยฐ Definition of supplementary angles
๐๐โ ๐ฉ๐ฉ + ๐๐โ ๐ช๐ช = ๐๐๐๐๐๐ยฐ Definition of supplementary angles
๐๐โ ๐ฉ๐ฉ + ๐๐โ ๐จ๐จ = ๐๐โ ๐ฉ๐ฉ + ๐๐โ ๐ช๐ช Substitution property of equality
๐๐โ ๐จ๐จ = ๐๐โ ๐ช๐ช Subtraction property of equality
2. Mathematicians state that if a transversal is perpendicular to two distinct lines, then the distinct lines are parallel. Prove this statement. (Include a labeled drawing with your proof.)
๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ช๐ช๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ฎ๐ฎ๐ฎ๐ฎ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ Given
๐๐โ ๐จ๐จ๐จ๐จ๐ช๐ช = ๐๐๐๐ยฐ Definition of perpendicular lines
๐๐โ ๐จ๐จ๐จ๐จ๐ฎ๐ฎ = ๐๐๐๐ยฐ Definition of perpendicular lines
๐๐โ ๐จ๐จ๐จ๐จ๐ช๐ช = ๐๐โ ๐จ๐จ๐จ๐จ๐ฎ๐ฎ Substitution property of equality
๐ช๐ช๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ฎ๐ฎ๐ฎ๐ฎ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ If a transversal cuts two lines such that corresponding angles are equal in measure, then the two lines are parallel.
If ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ต๐ต๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, then โ ๐ด๐ด and โ ๐ต๐ต are supplementary because they are same-side interior angles.
If a transversal is perpendicular to one of the two lines, then it meets that line at an angle of 90ยฐ. Since the lines are parallel, I can use corresponding angles of parallel lines to show that the transversal meets the other line, also at an angle of 90ยฐ.
46
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 18: Looking More Carefully at Parallel Lines
3. In the figure, ๐ด๐ด and ๐น๐น lie on ๐ด๐ด๐ต๐ต๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐๐โ ๐ต๐ต๐ด๐ด๐ถ๐ถ = ๐๐โ ๐ด๐ด๐น๐น๐ด๐ด, and ๐๐โ ๐ถ๐ถ = ๐๐โ ๐ด๐ด. Prove that ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ถ๐ถ๐ต๐ต๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
๐๐โ ๐ฉ๐ฉ๐จ๐จ๐ช๐ช = ๐๐โ ๐จ๐จ๐ช๐ช๐จ๐จ Given
๐๐โ ๐ช๐ช = ๐๐โ ๐จ๐จ Given
๐๐โ ๐ฉ๐ฉ๐จ๐จ๐ช๐ช+ ๐๐โ ๐ช๐ช+ ๐๐โ ๐ฉ๐ฉ = ๐๐๐๐๐๐ยฐ Sum of the angle measures in a triangle is ๐๐๐๐๐๐ยฐ.
๐๐โ ๐จ๐จ๐ช๐ช๐จ๐จ+ ๐๐โ ๐จ๐จ + ๐๐โ ๐จ๐จ = ๐๐๐๐๐๐ยฐ Sum of the angle measures in a triangle is ๐๐๐๐๐๐ยฐ.
๐๐โ ๐จ๐จ๐ช๐ช๐จ๐จ+ ๐๐โ ๐จ๐จ + ๐๐โ ๐ฉ๐ฉ = ๐๐๐๐๐๐ยฐ Substitution property of equality
๐๐โ ๐ฉ๐ฉ = ๐๐๐๐๐๐ยฐ โ๐๐โ ๐จ๐จ๐ช๐ช๐จ๐จโ๐๐โ ๐จ๐จ Subtraction property of equality
๐๐โ ๐จ๐จ = ๐๐๐๐๐๐ยฐ โ๐๐โ ๐จ๐จ๐ช๐ช๐จ๐จ โ๐๐โ ๐จ๐จ Subtraction property of equality
๐๐โ ๐จ๐จ = ๐๐โ ๐ฉ๐ฉ Substitution property of equality
๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ช๐ช๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ If two lines are cut by a transversal such that a pair of alternate interior angles are equal in measure, then the lines are parallel.
I know that in any triangle, the three angle measures sum to 180ยฐ. If two angles in one triangle are equal in measure to two angles in another triangle, then the third angles in each triangle must be equal in measure.
47
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 19: Construct and Apply a Sequence of Rigid Motions
Lesson 19: Construct and Apply a Sequence of Rigid Motions
1. Use your understanding of congruence to answer each of the following. a. Why canโt a square be congruent to a regular hexagon?
A square cannot be congruent to a regular hexagon because there is no rigid motion that takes a figure with four vertices to a figure with six vertices.
b. Can a square be congruent to a rectangle?
A square can only be congruent to a rectangle if the sides of the rectangle are all the same length as the sides of the square. This would mean that the rectangle is actually a square.
2. The series of figures shown in the diagram shows the images of โณ ๐ด๐ด๐ด๐ด๐ด๐ด under a sequence of rigid motions in the plane. Use a piece of patty paper to find and describe the sequence of rigid motions that shows โณ ๐ด๐ด๐ด๐ด๐ด๐ด โ โณ ๐ด๐ดโฒโฒโฒ๐ด๐ดโฒโฒโฒ๐ด๐ดโฒโฒโฒ. Label the corresponding image points in the diagram using prime notation.
First, a rotation of ๐๐๐๐ยฐ about point ๐ช๐ชโฒโฒโฒ in a clockwise direction takes โณ ๐จ๐จ๐จ๐จ๐ช๐ช to โณ ๐จ๐จโฒ๐จ๐จโฒ๐ช๐ชโฒ. Next, a translation along ๐ช๐ชโฒ๐ช๐ชโฒโฒโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ takes โณ ๐จ๐จโฒ๐จ๐จโฒ๐ช๐ชโฒ to โณ ๐จ๐จโฒโฒ๐จ๐จโฒโฒ๐ช๐ชโฒโฒ. Finally, a reflection over ๐จ๐จโฒโฒ๐ช๐ชโฒโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ takes โณ ๐จ๐จโฒโฒ๐จ๐จโฒโฒ๐ช๐ชโฒโฒ to โณ ๐จ๐จโฒโฒโฒ๐จ๐จโฒโฒโฒ๐ช๐ชโฒโฒโฒ.
I know that by definition, a rectangle is a quadrilateral with four right angles.
To be congruent, the figures must have a correspondence of vertices. I know that a square has four vertices, and a regular hexagon has six vertices. No matter what sequence of rigid motions I use, I cannot correspond all vertices of the hexagon with vertices of the square.
I can see that โณ ๐ด๐ดโฒโฒโฒ๐ด๐ดโฒโฒโฒ๐ด๐ดโฒโฒโฒ is turned on the plane compared to โณ ๐ด๐ด๐ด๐ด๐ด๐ด, so I should look for a rotation in my sequence. I also see a vector, so there might be a translation, too.
48
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 19: Construct and Apply a Sequence of Rigid Motions
3. In the diagram to the right, โณ ๐ด๐ด๐ด๐ด๐ด๐ด โ โณ ๐ท๐ท๐ด๐ด๐ด๐ด. a. Describe two distinct rigid motions, or sequences of
rigid motions, that map ๐ด๐ด onto ๐ท๐ท.
The most basic of rigid motions mapping ๐จ๐จ onto ๐ซ๐ซ is a reflection over ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
Another possible sequence of rigid motions includes a rotation about ๐จ๐จ of degree measure equal to ๐๐โ ๐จ๐จ๐จ๐จ๐ซ๐ซ followed by a reflection over ๐จ๐จ๐ซ๐ซ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
b. Using the congruence that you described in your response to part (a), what does ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ map to?
By a reflection of the plane over ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ maps to ๐ซ๐ซ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
c. Using the congruence that you described in your response to part (a), what does ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ map to?
By a reflection of the plane over ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ maps to itself because it lies in the line of reflection.
In the given congruence statement, the vertices of the first triangle are named in a clockwise direction, but the corresponding vertices of the second triangle are named in a counterclockwise direction. The change in orientation tells me that a reflection must be involved.
49
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 20: Applications of Congruence in Terms of Rigid Motions
Lesson 20: Applications of Congruence in Terms of Rigid Motions
1. Give an example of two different quadrilaterals and a correspondence between their vertices such that (a) all four corresponding angles are congruent, and (b) none of the corresponding sides are congruent.
The following represents one of many possible answers to this problem.
Square ๐จ๐จ๐จ๐จ๐จ๐จ๐จ๐จ and rectangle ๐ฌ๐ฌ๐ฌ๐ฌ๐ฌ๐ฌ๐ฌ๐ฌ meet the above criteria. By definition, both quadrilaterals are required to have four right angles, which means that any correspondence of vertices will map together congruent angles. A square is further required to have all sides of equal length, so as long as none of the sides of the rectangle are equal in length to the sides of the square, the criteria are satisfied.
2. Is it possible to give an example of two triangles and a correspondence between their vertices such that only two of the corresponding angles are congruent? Explain your answer.
Any triangle has three angles, and the sum of the measures of those angles is ๐๐๐๐๐๐ยฐ. If two triangles are given such that one pair of angles measure ๐๐ยฐ and a second pair of angles measure ๐๐ยฐ, then by the angle sum of a triangle, the remaining angle would have to have a measure of (๐๐๐๐๐๐ โ ๐๐ โ ๐๐)ยฐ. This means that the third pair of corresponding angles must also be congruent, so no, it is not possible.
3. Translations, reflections, and rotations are referred to as rigid motions. Explain why the term rigid is used.
Each of the rigid motions is a transformation of the plane that can be modelled by tracing a figure on the plane onto a transparency and transforming the transparency by following the given function rule. In each case, the image is identical to the pre-image because translations, rotations, and reflections preserve distance between points and preserve angles between lines. The transparency models rigidity.
I know that some quadrilaterals have matching angle characteristics such as squares and rectangles. Both of these quadrilaterals are required to have four right angles.
I know that every triangle, no matter what size or classification, has an angle sum of 180ยฐ.
The term rigid means not flexible.
50
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 21: Correspondence and Transformations
Lesson 21: Correspondence and Transformations
1. The diagram below shows a sequence of rigid motions that maps a pre-image onto a final image. a. Identify each rigid motion in the sequence, writing the composition using function notation.
The first rigid motion is a translation along ๐ช๐ช๐ช๐ชโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ to yield triangle ๐จ๐จโฒ๐ฉ๐ฉโฒ๐ช๐ชโฒ. The second rigid motion is a reflection over ๐ฉ๐ฉ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ to yield triangle ๐จ๐จโฒโฒ๐ฉ๐ฉโฒโฒ๐ช๐ชโฒโฒ. In function notation, the
sequence of rigid motions is ๐๐๐ฉ๐ฉ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ ๏ฟฝ๐ป๐ป๐ช๐ช๐ช๐ชโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ (โณ ๐จ๐จ๐ฉ๐ฉ๐ช๐ช)๏ฟฝ.
b. Trace the congruence of each set of corresponding sides and angles through all steps in the sequence, proving that the pre-image is congruent to the final image by showing that every side and every angle in the pre-image maps onto its corresponding side and angle in the image.
Sequence of corresponding sides: ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โ ๐จ๐จโฒโฒ๐ฉ๐ฉโฒโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐ฉ๐ฉ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โ ๐ฉ๐ฉโฒโฒ๐ช๐ชโฒโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, and ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โ ๐จ๐จโฒโฒ๐ช๐ชโฒโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
Sequence of corresponding angles: โ ๐จ๐จ โ โ ๐จ๐จโฒโฒ, โ ๐ฉ๐ฉ โ โ ๐ฉ๐ฉโฒโฒ, and โ ๐ช๐ช โ โ ๐ช๐ชโฒโฒ.
c. Make a statement about the congruence of the pre-image and the final image.
โณ ๐จ๐จ๐ฉ๐ฉ๐ช๐ช โ โณ ๐จ๐จโฒโฒ๐ฉ๐ฉโฒโฒ๐ช๐ชโฒโฒ
51
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 21: Correspondence and Transformations
2. Triangle ๐๐๐๐๐๐ is a reflected image of triangle ๐ด๐ด๐ด๐ด๐ด๐ด over a line โ. Is it possible for a translation or a rotation to map triangle ๐๐๐๐๐๐ back to the corresponding vertices in its pre-image, triangle ๐ด๐ด๐ด๐ด๐ด๐ด? Explain why or why not.
The orientation of three non-collinear points will change under a reflection of the plane over a line. This means that if you consider the correspondence ๐จ๐จ โ ๐ป๐ป, ๐ฉ๐ฉ โ ๐น๐น, and ๐ช๐ช โ ๐บ๐บ, if the vertices ๐จ๐จ, ๐ฉ๐ฉ, and ๐ช๐ช are oriented in a clockwise direction on the plane, then the vertices ๐ป๐ป, ๐น๐น, and ๐บ๐บ will be oriented in a counterclockwise direction. It is possible to map ๐ป๐ป to ๐จ๐จ, ๐บ๐บ to ๐ช๐ช, or ๐น๐น to ๐ฉ๐ฉ individually under a variety of translations or rotations; however, a reflection is required in order to map each of ๐ป๐ป, ๐น๐น, and ๐บ๐บ to its corresponding pre-image.
3. Describe each transformation given by the sequence of rigid motions below, in function notation, using the correct sequential order.
๐๐๐๐ ๏ฟฝ๐๐๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๐๐๐๐,60ยฐ(โณ ๐๐๐๐๐๐)๏ฟฝ๏ฟฝ
The first rigid motion is a rotation of โณ ๐ฟ๐ฟ๐ฟ๐ฟ๐ฟ๐ฟ around point ๐ฟ๐ฟ of ๐๐๐๐ยฐ. Next is a translation along ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . The final rigid motion is a reflection over a given line ๐๐.
I know that in function notation, the innermost function, in this case ๐๐๐๐,60ยฐ(โณ ๐๐๐๐๐๐), is the first to be carried out on the points in the plane.
When I look at the words printed on my t-shirt in a mirror, the order of the letters, and even the letters themselves, are completely backward. I can see the words correctly if I look at a reflection of my reflection in another mirror.
52
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16 M1
GEOMETRY
Lesson 22: Congruence Criteria for TrianglesโSAS
Lesson 22: Congruence Criteria for TrianglesโSAS
1. We define two figures as congruent if there exists a finite composition of rigid motions that maps oneonto the other. The following triangles meet the Side-Angle-Side criterion for congruence. The criteriontells us that only a few parts of two triangles, as well as a correspondence between them, is necessary todetermine that the two triangles are congruent.Describe the rigid motion in each step of the proof for the SAS criterion:
Given: โณ ๐๐๐๐๐๐ and โณ ๐๐โฒ๐๐โฒ๐๐โฒ so that ๐๐๐๐ = ๐๐โฒ๐๐โฒ (Side), ๐๐โ ๐๐ = ๐๐โ ๐๐โฒ (Angle), and ๐๐๐๐ = ๐๐โฒ๐๐โฒ (Side).
Prove: โณ ๐๐๐๐๐๐ โ โณ ๐๐โฒ๐๐โฒ๐๐โฒ
1 Given, distinct triangles โณ ๐๐๐๐๐๐ and โณ ๐๐โฒ๐๐โฒ๐๐โฒ so that ๐๐๐๐ = ๐๐โฒ๐๐โฒ, ๐๐โ ๐๐ = ๐๐โ ๐๐โฒ, and ๐๐๐๐ = ๐๐โฒ๐๐โฒ.
2 ๐ป๐ป๐ท๐ทโฒ๐ท๐ท๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ (โณ๐ท๐ทโฒ๐ธ๐ธโฒ๐น๐นโฒ) =โณ ๐ท๐ท๐ธ๐ธโฒโฒ๐น๐นโฒโฒ; โณ ๐ท๐ทโฒ๐ธ๐ธโฒ๐น๐นโฒ is translated along vector ๐ท๐ทโฒ๐ท๐ท๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . โณ ๐ท๐ท๐ธ๐ธ๐น๐น and โณ ๐ท๐ท๐ธ๐ธโฒโฒ๐น๐นโฒโฒ share common vertex ๐ท๐ท.
3 ๐น๐น๐ท๐ท,โ๐ฝ๐ฝ(โณ๐ท๐ท๐ธ๐ธโฒโฒ๐น๐นโฒโฒ) = โณ ๐ท๐ท๐ธ๐ธโฒโฒโฒ๐น๐น; โณ๐ท๐ท๐ธ๐ธโฒโฒ๐น๐นโฒโฒ is rotated about center ๐ท๐ท by ๐ฝ๐ฝห clockwise. โณ๐ท๐ท๐ธ๐ธ๐น๐น and โณ๐ท๐ท๐ธ๐ธโฒโฒโฒ๐น๐น share common side ๐ท๐ท๐น๐น๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
4 ๐๐๐ท๐ท๐น๐น๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ (โณ๐ท๐ท๐ธ๐ธโฒโฒโฒ๐น๐น) =โณ ๐ท๐ท๐ธ๐ธ๐น๐น; โณ๐ท๐ท๐ธ๐ธโฒโฒโฒ๐น๐น is reflected across โณ ๐ท๐ท๐ธ๐ธโฒโฒโฒ๐น๐น coincides with โณ๐ท๐ท๐ธ๐ธ๐น๐น.
53
Homework Helper A Story of Functions
๏ฟฝโ๏ฟฝ๐ท๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ๐น๐น
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 22: Congruence Criteria for TrianglesโSAS
M1
GEOMETRY
a. In Step 3, how can we be certain that ๐๐" will map to ๐๐?
By assumption, ๐ท๐ท๐น๐นโฒโฒ = ๐ท๐ท๐น๐น. This means that not only will ๐ท๐ท๐น๐นโฒโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ map to ๐ท๐ท๐น๐น๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ under the rotation, but ๐น๐นโฒโฒ will map to ๐น๐น.
b. In Step 4, how can we be certain that ๐๐โฒโฒโฒ will map to ๐๐?
Rigid motions preserve angle measures. This means that ๐๐โ ๐ธ๐ธ๐ท๐ท๐น๐น = ๐๐โ ๐ธ๐ธโฒโฒโฒ๐ท๐ท๐น๐น. Then the reflection maps ๐ท๐ท๐ธ๐ธโฒโฒโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ to ๐ท๐ท๐ธ๐ธ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . Since ๐ท๐ท๐ธ๐ธโฒโฒโฒ = ๐ท๐ท๐ธ๐ธ, ๐ธ๐ธโฒโฒโฒ will map to ๐ธ๐ธ.
c. In this example, we began with two distinct triangles that met the SAS criterion. Now consider triangles that are not distinct and share a common vertex. The following two scenarios both show a pair of triangles that meet the SAS criterion and share a common vertex. In a proof to show that the triangles are congruent, which pair of triangles will a translation make most sense as a next step? In which pair is the next step a rotation? Justify your response.
Pair (ii) will require a translation next because currently, the common vertex is between a pair of angles whose measures are unknown. The next step for pair (i) is a rotation, as the common vertex is one between angles of equal measure, by assumption, and therefore can be rotated so that a pair of sides of equal length become a shared side.
I must remember that if the lengths of ๐๐๐๐" and ๐๐๐๐ were not known, the rotation would result in coinciding rays ๐๐๐๐โฒโฒ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ and ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ but nothing further.
i. ii.
54
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 22: Congruence Criteria for TrianglesโSAS
M1 GEOMETRY
2. Justify whether the triangles meet the SAS congruence criteria; explicitly state which pairs of sides orangles are congruent and why. If the triangles do meet the SAS congruence criteria, describe the rigidmotion(s) that would map one triangle onto the other.a. Given: Rhombus ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด
Do โณ ๐ด๐ด๐๐๐ด๐ด and โณ ๐ด๐ด๐๐๐ด๐ด meet the SAS criterion?
Rhombus ๐จ๐จ๐จ๐จ๐จ๐จ๐จ๐จ Given
๐จ๐จ๐น๐น๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ and ๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ are perpendicular. Property of a rhombus
๐๐โ ๐จ๐จ๐น๐น๐จ๐จ = ๐๐โ ๐จ๐จ๐น๐น๐จ๐จ All right angles are equal in measure.
๐จ๐จ๐น๐น = ๐น๐น๐จ๐จ Diagonals of a rhombus bisect each other.
๐จ๐จ๐น๐น = ๐จ๐จ๐น๐น Reflexive property
โณ ๐จ๐จ๐น๐น๐จ๐จ โ โณ ๐จ๐จ๐น๐น๐จ๐จ SAS
One possible rigid motion that maps โณ ๐จ๐จ๐น๐น๐จ๐จ to โณ ๐จ๐จ๐น๐น๐จ๐จ is a reflection over the line ๐จ๐จ๐น๐น๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ .
b. Given: Isosceles triangle โณ ABC with ๐ด๐ด๐ด๐ด = ๐ด๐ด๐ด๐ด and angle bisector ๐ด๐ด๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
Do โณ ๐ด๐ด๐ด๐ด๐๐ and โณ ๐ด๐ด๐ด๐ด๐๐ meet the SAS criterion?
๐จ๐จ๐จ๐จ = ๐จ๐จ๐จ๐จ Given
๐จ๐จ๐ท๐ท๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ is an angle bisector Given
๐จ๐จ๐ท๐ท = ๐จ๐จ๐ท๐ท Reflexive property
๐๐โ ๐จ๐จ๐จ๐จ๐ท๐ท = ๐๐โ ๐จ๐จ๐จ๐จ๐ท๐ท Definition of angle bisector
โณ ๐จ๐จ๐จ๐จ๐ท๐ท โ โณ ๐จ๐จ๐จ๐จ๐ท๐ท SAS
55
Homework Helper A Story of Functions
One possible rigid motion that maps โณ ๐จ๐จ to โณ ๐จ๐จ is a reflection over the line ๐จ๐จ๐ท๐ท ๐จ๐จ๐ท๐ท ๏ฟฝโ๏ฟฝ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . ๐ท๐ท
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 23: Base Angles of Isosceles Triangles
M1
GEOMETRY
Lesson 23: Base Angles of Isosceles Triangles
1. In an effort to prove that ๐๐โ ๐ต๐ต = ๐๐โ ๐ถ๐ถ in isosceles triangle ๐ด๐ด๐ต๐ต๐ถ๐ถ by using rigid motions, the following argument is made to show that ๐ต๐ต maps to ๐ถ๐ถ:
Given: Isosceles โณ ๐ด๐ด๐ต๐ต๐ถ๐ถ, with ๐ด๐ด๐ต๐ต = ๐ด๐ด๐ถ๐ถ
Prove: ๐๐โ ๐ต๐ต = ๐๐โ ๐ถ๐ถ
Construction: Draw the angle bisector ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ of โ ๐ด๐ด, where ๐ด๐ด is the intersection of the bisector and ๐ต๐ต๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ. We need to show that rigid motions map point ๐ต๐ต to point ๐ถ๐ถ and point ๐ถ๐ถ to point ๐ต๐ต.
Since ๐จ๐จ is on the line of reflection, ๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , ๐๐๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ (๐จ๐จ) = ๐จ๐จ. Reflections preserve angle measures, so the measure of the reflected angle ๐๐๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ (โ ๐ฉ๐ฉ๐จ๐จ๐จ๐จ) equals the measure of โ ๐ช๐ช๐จ๐จ๐จ๐จ; therefore, ๐๐๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ = ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ . Reflections also preserve lengths of segments; therefore, the reflection of ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ still has the same length as ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ. By hypothesis, ๐จ๐จ๐ฉ๐ฉ = ๐จ๐จ๐ช๐ช, so the length of the reflection is also equal to ๐จ๐จ๐ช๐ช. Then ๐๐๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ (๐ฉ๐ฉ) = ๐ช๐ช.
Use similar reasoning to show that ๐๐๐ด๐ด๐ด๐ด๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ (๐ถ๐ถ) = ๐ต๐ต.
Again, we use a reflection in our reasoning. ๐จ๐จ is on the line of reflection, ๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , so ๐๐๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ (๐จ๐จ) = ๐จ๐จ.
Since reflections preserve angle measures, the measure of the reflected angle ๐๐๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ (โ ๐ช๐ช๐จ๐จ๐จ๐จ) equals the measure of โ ๐ฉ๐ฉ๐จ๐จ๐จ๐จ, implying that ๐๐๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ ๏ฟฝ = ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ .
Reflections also preserve lengths of segments. This means that the reflection of ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, or the image of ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ (๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ), has the same length as ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ. By hypothesis, ๐จ๐จ๐ฉ๐ฉ = ๐จ๐จ๐ช๐ช, so the length of the reflection is also equal to ๐จ๐จ๐ฉ๐ฉ. This implies ๐๐๐จ๐จ๐จ๐จ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ (๐ช๐ช) = ๐ฉ๐ฉ. We conclude then that ๐๐โ ๐ฉ๐ฉ = ๐๐โ ๐ช๐ช.
I must remember that proving this fact using rigid motions relies on the idea that rigid motions preserve lengths and angle measures. This is what ultimately allows me to map ๐ถ๐ถ to ๐ต๐ต.
56
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 23: Base Angles of Isosceles Triangles
M1
GEOMETRY
2. Given: ๐๐โ 1 = ๐๐โ 2; ๐๐โ 3 = ๐๐โ 4
Prove: ๐๐โ 5 = ๐๐โ 6
๐๐โ ๐๐ = ๐๐โ ๐๐
๐๐โ ๐๐ = ๐๐โ ๐๐
Given
๐จ๐จ๐จ๐จ = ๐ฉ๐ฉ๐จ๐จ
๐จ๐จ๐จ๐จ = ๐ช๐ช๐จ๐จ
If two angles of a triangle are equal in measure, then the sides opposite the angles are equal in length.
๐ฉ๐ฉ๐จ๐จ = ๐ช๐ช๐จ๐จ Transitive property of equality
๐๐โ ๐๐ = ๐๐โ ๐๐ If two sides of a triangle are equal in length, then the angles opposite them are equal in measure.
57
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 23: Base Angles of Isosceles Triangles
M1
GEOMETRY
3. Given: Rectangle ๐ด๐ด๐ต๐ต๐ถ๐ถ๐ด๐ด; ๐ฝ๐ฝ is the midpoint of ๐ด๐ด๐ต๐ต๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
Prove: โณ ๐ฝ๐ฝ๐ถ๐ถ๐ด๐ด is isosceles
๐จ๐จ๐ฉ๐ฉ๐ช๐ช๐จ๐จ is a rectangle. Given
๐๐โ ๐จ๐จ = ๐๐โ ๐ฉ๐ฉ All angles of a rectangle are right angles.
๐จ๐จ๐จ๐จ = ๐ฉ๐ฉ๐ช๐ช Opposite sides of a rectangle are equal in length.
๐ฑ๐ฑ is the midpoint of ๐จ๐จ๐ฉ๐ฉ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ. Given
๐จ๐จ๐ฑ๐ฑ = ๐ฉ๐ฉ๐ฑ๐ฑ Definition of midpoint
โณ ๐จ๐จ๐ฑ๐ฑ๐จ๐จ โ โณ ๐ฉ๐ฉ๐ฑ๐ฑ๐ช๐ช SAS
๐ฑ๐ฑ๐จ๐จ = ๐ฑ๐ฑ๐ช๐ช Corresponding sides of congruent triangles are equal in length.
โณ ๐ฑ๐ฑ๐ช๐ช๐จ๐จ is isosceles. Definition of isosceles triangle
58
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 23: Base Angles of Isosceles Triangles
M1 GEOMETRY
4. Given: ๐ ๐ ๐ ๐ = ๐ ๐ ๐ ๐ ; ๐๐โ 2 = ๐๐โ 7
Prove: โณ ๐ ๐ ๐ ๐ ๐ ๐ is isosceles
๐น๐น๐น๐น = ๐น๐น๐น๐น Given
๐๐โ ๐๐ = ๐๐โ ๐๐ Base angles of an isosceles triangle are equal in measure.
๐๐โ ๐๐ = ๐๐โ ๐๐ Given
๐๐โ ๐๐ +๐๐โ ๐๐ + ๐๐โ ๐๐ = ๐๐๐๐๐๐ยฐ
๐๐โ ๐๐ +๐๐โ ๐๐ + ๐๐โ ๐๐ = ๐๐๐๐๐๐ยฐ
The sum of angle measures in a triangle is ๐๐๐๐๐๐ห.
๐๐โ ๐๐ +๐๐โ ๐๐ + ๐๐โ ๐๐ = ๐๐โ ๐๐ + ๐๐โ ๐๐+ ๐๐โ ๐๐ Substitution property of equality
๐๐โ ๐๐ +๐๐โ ๐๐ + ๐๐โ ๐๐ = ๐๐โ ๐๐ + ๐๐โ ๐๐+ ๐๐โ ๐๐ Substitution property of equality
๐๐โ ๐๐ = ๐๐โ ๐๐ Subtraction property of equality
๐๐โ ๐๐ +๐๐โ ๐๐ = ๐๐๐๐๐๐ยฐ
๐๐โ ๐๐ +๐๐โ ๐๐ = ๐๐๐๐๐๐ยฐ
Linear pairs form supplementary angles.
๐๐โ ๐๐ +๐๐โ ๐๐ = ๐๐โ ๐๐+ ๐๐โ ๐๐ Substitution property of equality
๐๐โ ๐๐ +๐๐โ ๐๐ = ๐๐โ ๐๐+ ๐๐โ ๐๐ Substitution property of equality
๐๐โ ๐๐ = ๐๐โ ๐๐ Subtraction property of equality
๐น๐น๐น๐น = ๐น๐น๐น๐น If two angles of a triangle are equal in measure, then the sides opposite the angles are equal in length.
โณ ๐น๐น๐น๐น๐น๐น is isosceles. Definition of isosceles triangle
59
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 24: Congruence Criteria for TrianglesโASA and SSS
M1
GEOMETRY
Lesson 24: Congruence Criteria for TrianglesโASA and SSS
1. For each of the following pairs of triangles, name the congruence criterion, if any, that proves the triangles are congruent. If none exists, write โnone.โ a.
SAS
b.
SSS
c.
ASA
d.
none
e.
ASA
In addition to markings indicating angles of equal measure and sides of equal lengths, I must observe diagrams for common sides and angles, vertical angles, and angle pair relationships created by parallel lines cut by a transversal. I must also remember that AAA is not a congruence criterion.
60
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 24: Congruence Criteria for TrianglesโASA and SSS
M1
GEOMETRY
2. ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด is a rhombus. Name three pairs of triangles that are congruent so that no more than one pair is congruent to each other and the criteria you would use to support their congruency.
Possible solution: โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐จ๐จ๐จ๐จ๐จ๐จ, โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐ฉ๐ฉ๐จ๐จ๐จ๐จ, and โณ ๐จ๐จ๐จ๐จ๐ฉ๐ฉ โ โณ ๐จ๐จ๐จ๐จ๐ฉ๐ฉ. All three pairs can be supported by SAS/SSS/ASA.
3. Given: ๐๐ = ๐ ๐ and ๐๐๐๐ = ๐๐๐๐ Prove: ๐๐ = ๐๐
๐๐ = ๐๐ Given
๐ท๐ท๐ท๐ท = ๐จ๐จ๐ท๐ท Given
๐๐ = ๐๐ Vertical angles are equal in measure.
โณ ๐ท๐ท๐ท๐ท๐ท๐ท โ โณ ๐จ๐จ๐ท๐ท๐ธ๐ธ ASA
๐ท๐ท๐ท๐ท = ๐ท๐ท๐ธ๐ธ Corresponding sides of congruent triangles are equal in length.
โณ ๐ท๐ท๐ท๐ท๐ธ๐ธ is isosceles. Definition of isosceles triangle
๐๐ = ๐๐ Base angles of an isosceles triangle are equal in measure.
61
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 24: Congruence Criteria for TrianglesโASA and SSS
M1
GEOMETRY
4. Given: Isoscelesโณ ๐ด๐ด๐ด๐ด๐ด๐ด; ๐ด๐ด๐๐ = ๐ด๐ด๐ด๐ด Prove: โ ๐ด๐ด๐๐๐ด๐ด โ โ ๐ด๐ด๐ด๐ด๐ด๐ด
Isosceles โณ ๐จ๐จ๐จ๐จ๐จ๐จ Given
๐จ๐จ๐จ๐จ = ๐จ๐จ๐จ๐จ Definition of isosceles triangle
๐จ๐จ๐ท๐ท = ๐จ๐จ๐จ๐จ Given
๐๐โ ๐จ๐จ = ๐๐โ ๐จ๐จ Reflexive property
โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐จ๐จ๐ท๐ท๐จ๐จ SAS
๐ท๐ท๐จ๐จ = ๐จ๐จ๐จ๐จ Corresponding sides of congruent triangles are equal in length.
๐จ๐จ๐ท๐ท + ๐ท๐ท๐จ๐จ = ๐จ๐จ๐จ๐จ
๐จ๐จ๐จ๐จ +๐จ๐จ๐จ๐จ = ๐จ๐จ๐จ๐จ
Partition property
๐จ๐จ๐ท๐ท + ๐ท๐ท๐จ๐จ = ๐จ๐จ๐จ๐จ+ ๐จ๐จ๐จ๐จ Substitution property of equality
๐จ๐จ๐ท๐ท + ๐ท๐ท๐จ๐จ = ๐จ๐จ๐ท๐ท + ๐จ๐จ๐จ๐จ Substitution property of equality
๐ท๐ท๐จ๐จ = ๐จ๐จ๐จ๐จ Subtraction property of equality
๐จ๐จ๐จ๐จ = ๐จ๐จ๐จ๐จ Reflexive property
โณ ๐ท๐ท๐จ๐จ๐จ๐จ โ โณ๐จ๐จ๐จ๐จ๐จ๐จ SSS
โ ๐จ๐จ๐ท๐ท๐จ๐จ โ โ ๐จ๐จ๐จ๐จ๐จ๐จ Corresponding angles of congruent triangles are congruent.
I must prove two sets of triangles are congruent in order to prove โ ๐ด๐ด๐๐๐ด๐ด โ โ ๐ด๐ด๐ด๐ด๐ด๐ด.
62
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 25: Congruence Criteria for TrianglesโAAS and HL
Lesson 25: Congruence Criteria for TrianglesโAAS and HL
1. Draw two triangles that meet the AAA criterion but are not congruent.
2. Draw two triangles that meet the SSA criterion but are not congruent. Label or mark the triangles with the appropriate measurements or congruency marks.
3. Describe, in terms of rigid motions, why triangles that meet the AAA and SSA criteria are not necessarily congruent.
Triangles that meet either the AAA or SSA criteria are not necessarily congruent because there may or may not be a finite composition of rigid motions that maps one triangle onto the other. For example, in the diagrams in Problem 2, there is no composition of rigid motions that will map one triangle onto the other.
63
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 25: Congruence Criteria for TrianglesโAAS and HL
4. Given: ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โ ๐ถ๐ถ๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐ถ๐ถ๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ,๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ถ๐ถ๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ,
๐ถ๐ถ is the midpoint of ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
๐ด๐ด is the midpoint of ๐ถ๐ถ๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
Prove: โณ ๐ด๐ด๐ด๐ด๐ถ๐ถ โ โณ ๐ถ๐ถ๐ด๐ด๐ด๐ด
๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โ ๐ช๐ช๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ Given
๐ผ๐ผ is the midpoint of ๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
๐ฝ๐ฝ is the midpoint of ๐ช๐ช๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
Given
๐จ๐จ๐จ๐จ = ๐๐๐จ๐จ๐ผ๐ผ
๐ช๐ช๐จ๐จ = ๐๐๐ช๐ช๐ฝ๐ฝ
Definition of midpoint
๐๐๐จ๐จ๐ผ๐ผ = ๐๐๐ช๐ช๐ฝ๐ฝ Substitution property of equality
๐จ๐จ๐ผ๐ผ = ๐ช๐ช๐ฝ๐ฝ Division property of equality
๐ช๐ช๐ผ๐ผ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
๐จ๐จ๐ฝ๐ฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ช๐ช๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
Given
๐๐โ ๐จ๐จ๐ผ๐ผ๐จ๐จ = ๐๐โ ๐ช๐ช๐ฝ๐ฝ๐จ๐จ = ๐๐๐๐ยฐ Definition of perpendicular
๐๐โ ๐จ๐จ๐จ๐จ๐ผ๐ผ = ๐๐โ ๐ช๐ช๐จ๐จ๐ฝ๐ฝ Vertical angles are equal in measure.
โณ ๐จ๐จ๐จ๐จ๐ผ๐ผ โ โณ ๐ช๐ช๐จ๐จ๐ฝ๐ฝ AAS
64
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 25: Congruence Criteria for TrianglesโAAS and HL
5. Given: ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ด๐ด๐ต๐ต๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐ถ๐ถ๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ด๐ด๐ต๐ต๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ,
๐ด๐ด๐ด๐ด = ๐ต๐ต๐ถ๐ถ, ๐ด๐ด๐ถ๐ถ = ๐ต๐ต๐ด๐ด
Prove: โณ ๐ด๐ด๐ด๐ด๐ด๐ด โ โณ ๐ถ๐ถ๐ต๐ต๐ถ๐ถ
๐จ๐จ๐จ๐จ = ๐ซ๐ซ๐ช๐ช Given
๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐จ๐จ๐ซ๐ซ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
๐ช๐ช๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐จ๐จ๐ซ๐ซ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
Given
๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐โ ๐ช๐ช๐ช๐ช๐ซ๐ซ = ๐๐๐๐ยฐ Definition of perpendicular
๐จ๐จ๐ช๐ช = ๐ซ๐ซ๐จ๐จ Given
๐จ๐จ๐ช๐ช = ๐จ๐จ๐จ๐จ + ๐จ๐จ๐ช๐ช
๐ซ๐ซ๐จ๐จ = ๐ซ๐ซ๐ช๐ช + ๐ช๐ช๐จ๐จ
Partition property
๐จ๐จ๐จ๐จ + ๐จ๐จ๐ช๐ช = ๐ซ๐ซ๐ช๐ช+ ๐ช๐ช๐จ๐จ Substitution property of equality
๐จ๐จ๐จ๐จ = ๐ซ๐ซ๐ช๐ช Subtraction property of equality
โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐ช๐ช๐ซ๐ซ๐ช๐ช HL
65
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 26: Triangle Congruency Proofs
M1
GEOMETRY
Lesson 26: Triangle Congruency Proofs
1. Given: ๐ด๐ด๐ด๐ด = ๐ด๐ด๐ด๐ด, ๐๐โ ๐ด๐ด๐ด๐ด๐ด๐ด = ๐๐โ ๐ด๐ด๐ด๐ด๐ด๐ด = 90ยฐ
Prove: ๐ด๐ด๐ด๐ด = ๐ด๐ด๐ด๐ด
๐จ๐จ๐จ๐จ = ๐จ๐จ๐จ๐จ Given
๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐๐๐ยฐ Given
๐๐โ ๐จ๐จ = ๐๐โ ๐จ๐จ Reflexive property
โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐จ๐จ๐จ๐จ๐จ๐จ AAS
๐จ๐จ๐จ๐จ = ๐จ๐จ๐จ๐จ Corresponding sides of congruent triangles are equal in length.
66
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 26: Triangle Congruency Proofs
M1
GEOMETRY
2. Given: ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ bisects โ ๐๐๐๐๐๐. ๐ท๐ท๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐ท๐ท๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
Prove: ๐ท๐ท๐๐๐๐๐๐ is a rhombus.
๐ป๐ป๐ป๐ป๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ bisects โ ๐บ๐บ๐ป๐ป๐บ๐บ Given
๐๐โ ๐ซ๐ซ๐ป๐ป๐ป๐ป = ๐๐โ ๐ญ๐ญ๐ป๐ป๐ป๐ป Definition of bisect
๐๐โ ๐ซ๐ซ๐ป๐ป๐ป๐ป = ๐๐โ ๐ญ๐ญ๐ป๐ป๐ป๐ป
๐๐โ ๐ญ๐ญ๐ป๐ป๐ป๐ป = ๐๐โ ๐ซ๐ซ๐ป๐ป๐ป๐ป
If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
๐ป๐ป๐ป๐ป = ๐ป๐ป๐ป๐ป Reflexive property
โณ ๐ซ๐ซ๐ป๐ป๐ป๐ป โ โณ ๐ญ๐ญ๐ป๐ป๐ป๐ป ASA
๐๐โ ๐ซ๐ซ๐ป๐ป๐ป๐ป = ๐๐โ ๐ซ๐ซ๐ป๐ป๐ป๐ป
๐๐โ ๐ญ๐ญ๐ป๐ป๐ป๐ป = ๐๐โ ๐ญ๐ญ๐ป๐ป๐ป๐ป
Substitution property of equality
โณ ๐ซ๐ซ๐ป๐ป๐ป๐ป is isosceles; โณ ๐ญ๐ญ๐ป๐ป๐ป๐ป is isosceles
When the base angles of a triangle are equal in measure, the triangle is isosceles
๐ซ๐ซ๐ป๐ป = ๐ซ๐ซ๐ป๐ป
๐ญ๐ญ๐ป๐ป = ๐ญ๐ญ๐ป๐ป
Definition of isosceles
๐ซ๐ซ๐ป๐ป = ๐ญ๐ญ๐ป๐ป
๐ซ๐ซ๐ป๐ป = ๐ญ๐ญ๐ป๐ป
Corresponding sides of congruent triangles are equal in length.
๐ซ๐ซ๐ป๐ป = ๐ญ๐ญ๐ป๐ป = ๐ซ๐ซ๐ป๐ป = ๐ญ๐ญ๐ป๐ป Transitive property
๐ซ๐ซ๐ป๐ป๐ญ๐ญ๐ป๐ป is a rhombus. Definition of rhombus
I must remember that in addition to showing that each half of ๐ท๐ท๐๐๐๐๐๐ is an isosceles triangle, I must also show that the lengths of the sides of both isosceles triangles are equal to each other, making ๐ท๐ท๐๐๐๐๐๐ a rhombus.
67
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 26: Triangle Congruency Proofs
M1
GEOMETRY
3. Given: ๐๐โ 1 = ๐๐โ 2
๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐ด๐ด๐ท๐ท๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ
๐๐๐ท๐ท = ๐๐๐ด๐ด
Prove: โณ ๐๐๐๐๐๐ is isosceles.
๐๐โ ๐๐ = ๐๐โ ๐๐ Given
๐๐โ ๐ธ๐ธ๐จ๐จ๐จ๐จ = ๐๐โ ๐น๐น๐จ๐จ๐ซ๐ซ Supplements of angles of equal measure are equal in measure.
๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ธ๐ธ๐น๐น๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, ๐จ๐จ๐ซ๐ซ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ธ๐ธ๐น๐น๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ Given
๐๐โ ๐จ๐จ๐จ๐จ๐ธ๐ธ = ๐๐โ ๐จ๐จ๐ซ๐ซ๐น๐น = ๐๐๐๐ยฐ Definition of perpendicular
๐ธ๐ธ๐ซ๐ซ = ๐น๐น๐จ๐จ Given
๐ธ๐ธ๐ซ๐ซ = ๐ธ๐ธ๐จ๐จ + ๐จ๐จ๐ซ๐ซ
๐น๐น๐จ๐จ = ๐น๐น๐ซ๐ซ +๐ซ๐ซ๐จ๐จ
Partition property
๐ธ๐ธ๐จ๐จ + ๐จ๐จ๐ซ๐ซ = ๐น๐น๐ซ๐ซ + ๐ซ๐ซ๐จ๐จ Substitution property of equality
๐ธ๐ธ๐จ๐จ = ๐น๐น๐ซ๐ซ Subtraction property of equality
โณ ๐จ๐จ๐ธ๐ธ๐จ๐จ โ โณ ๐จ๐จ๐น๐น๐ซ๐ซ AAS
๐๐โ ๐จ๐จ๐ธ๐ธ๐จ๐จ = ๐๐โ ๐จ๐จ๐น๐น๐ซ๐ซ Corresponding angles of congruent triangles are equal in measure.
โณ ๐ท๐ท๐ธ๐ธ๐น๐น is isosceles. When the base angles of a triangle are equal in measure, then the triangle is isosceles.
68
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 27: Triangle Congruency Proofs
M1
GEOMETRY
Lesson 27: Triangle Congruency Proofs
1. Given: โณ ๐ท๐ท๐ท๐ท๐ท๐ท and โณ ๐บ๐บ๐ท๐ท๐บ๐บ are equilateral triangles
Prove: โณ ๐ท๐ท๐บ๐บ๐ท๐ท โ โณ ๐ท๐ท๐บ๐บ๐ท๐ท
โณ ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ and โณ ๐ฎ๐ฎ๐ซ๐ซ๐ฎ๐ฎ are equilateral triangles.
Given
๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ = ๐๐โ ๐ฎ๐ฎ๐ซ๐ซ๐ฎ๐ฎ = ๐๐๐๐ยฐ All angles of an equilateral triangle are equal in measure
๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ฎ๐ฎ = ๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ซ๐ซ+ ๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ฎ๐ฎ
๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ฎ๐ฎ = ๐๐โ ๐ฎ๐ฎ๐ซ๐ซ๐ฎ๐ฎ+ ๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ฎ๐ฎ
Partition property
๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ฎ๐ฎ = ๐๐โ ๐ฎ๐ฎ๐ซ๐ซ๐ฎ๐ฎ+ ๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ฎ๐ฎ Substitution property of equality
๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ฎ๐ฎ = ๐๐โ ๐ซ๐ซ๐ซ๐ซ๐ฎ๐ฎ Substitution property of equality
๐ซ๐ซ๐ซ๐ซ = ๐ซ๐ซ๐ซ๐ซ
๐ฎ๐ฎ๐ซ๐ซ = ๐ฎ๐ฎ๐ซ๐ซ
Property of an equilateral triangle
โณ ๐ซ๐ซ๐ฎ๐ฎ๐ซ๐ซ โ โณ๐ซ๐ซ๐ฎ๐ฎ๐ซ๐ซ SAS
69
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 27: Triangle Congruency Proofs
M1
GEOMETRY
2. Given: ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ is a square. ๐๐ is a point on ๐ ๐ ๐ ๐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, and ๐๐ is on ๐ ๐ ๐ ๐ ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ such that ๐๐๐ ๐ ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ ๐ ๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
Prove: โณ ๐ ๐ ๐ ๐ ๐๐ โ โณ๐ ๐ ๐ ๐ ๐๐
๐น๐น๐น๐น๐น๐น๐น๐น is a square. Given
๐น๐น๐น๐น = ๐น๐น๐น๐น Property of a square
๐๐โ ๐น๐น๐น๐น๐น๐น = ๐๐โ ๐น๐น๐น๐น๐น๐น = ๐๐๐๐ยฐ Property of a square
๐๐โ ๐น๐น๐น๐น๐น๐น+ ๐๐โ ๐น๐น๐น๐น๐น๐น = ๐๐๐๐๐๐ยฐ Angles on a line sum to ๐๐๐๐๐๐ยฐ.
๐๐โ ๐น๐น๐น๐น๐น๐น = ๐๐๐๐ยฐ Subtraction property of equality
๐๐โ ๐น๐น๐น๐น๐น๐น = ๐๐โ ๐น๐น๐น๐น๐น๐น If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
๐๐โ ๐น๐น๐น๐น๐น๐น = ๐๐โ ๐น๐น๐น๐น๐น๐น Complements of angles of equal measures are equal.
โณ ๐น๐น๐น๐น๐น๐น โ โณ ๐น๐น๐น๐น๐น๐น ASA
70
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
Lesson 27: Triangle Congruency Proofs
M1
GEOMETRY
3. Given: ๐ด๐ด๐ด๐ด๐ด๐ด๐ท๐ท and ๐ท๐ท๐ท๐ท๐บ๐บ๐บ๐บ are congruent rectangles.
Prove: ๐๐โ ๐ท๐ท๐ท๐ท๐ ๐ = ๐๐โ ๐ท๐ท๐ท๐ท๐ ๐
๐จ๐จ๐จ๐จ๐จ๐จ๐ซ๐ซ and ๐ซ๐ซ๐ซ๐ซ๐ฎ๐ฎ๐ฎ๐ฎ are congruent rectangles.
Given
๐ฎ๐ฎ๐ซ๐ซ = ๐จ๐จ๐ซ๐ซ
๐ซ๐ซ๐จ๐จ = ๐ซ๐ซ๐ซ๐ซ
Corresponding sides of congruent figures are equal in length.
๐๐โ ๐ฎ๐ฎ๐ซ๐ซ๐ซ๐ซ = ๐๐โ ๐จ๐จ๐ซ๐ซ๐จ๐จ Corresponding angles of congruent figures are equal in measure.
๐๐โ ๐ฎ๐ฎ๐ซ๐ซ๐จ๐จ = ๐๐โ ๐ฎ๐ฎ๐ซ๐ซ๐ซ๐ซ + ๐๐โ ๐ท๐ท๐ซ๐ซ๐จ๐จ
๐๐โ ๐จ๐จ๐ซ๐ซ๐ซ๐ซ = ๐๐โ ๐จ๐จ๐ซ๐ซ๐จ๐จ+ ๐๐โ ๐จ๐จ๐ซ๐ซ๐ท๐ท
Partition property
๐๐โ ๐ฎ๐ฎ๐ซ๐ซ๐จ๐จ = ๐๐โ ๐จ๐จ๐ซ๐ซ๐จ๐จ+ ๐๐โ ๐ท๐ท๐ซ๐ซ๐จ๐จ
Substitution property of equality
๐๐โ ๐ฎ๐ฎ๐ซ๐ซ๐จ๐จ = ๐๐โ ๐จ๐จ๐ซ๐ซ๐ซ๐ซ Substitution property of equality
โณ ๐ฎ๐ฎ๐ซ๐ซ๐จ๐จ โ โณ ๐จ๐จ๐ซ๐ซ๐ซ๐ซ SAS
๐๐โ ๐ฎ๐ฎ๐จ๐จ๐ซ๐ซ = ๐๐โ ๐จ๐จ๐ซ๐ซ๐ซ๐ซ Corresponding angles of congruent triangles are equal in measure.
๐๐โ ๐ท๐ท๐ซ๐ซ๐ท๐ท = ๐๐โ ๐ท๐ท๐ซ๐ซ๐ท๐ท Reflexive property
โณ ๐ซ๐ซ๐ท๐ท๐จ๐จ โ โณ ๐ซ๐ซ๐ท๐ท๐ซ๐ซ ASA
๐๐โ ๐ซ๐ซ๐ท๐ท๐น๐น = ๐๐โ ๐ซ๐ซ๐ท๐ท๐น๐น Corresponding angles of congruent triangles are equal in measure.
71
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 28: Properties of Parallelograms
Since the triangles are congruent, I can use the fact that their corresponding angles are equal in measure as a property of parallelograms.
Lesson 28: Properties of Parallelograms
1. Given: โณ ๐ด๐ด๐ด๐ด๐ด๐ด โ โณ ๐ถ๐ถ๐ด๐ด๐ด๐ด Prove: Quadrilateral ๐ด๐ด๐ด๐ด๐ถ๐ถ๐ด๐ด is a parallelogram
Proof:
โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐ช๐ช๐จ๐จ๐จ๐จ Given
๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐โ ๐ช๐ช๐จ๐จ๐จ๐จ; ๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐โ ๐ช๐ช๐จ๐จ๐จ๐จ
Corresponding angles of congruent triangles are equal in measure.
๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ช๐ช๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ,๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ If two lines are cut by a transversal such that alternate interior angles are equal in measure, then the lines are parallel.
Quadrilateral ๐จ๐จ๐จ๐จ๐ช๐ช๐จ๐จ is a parallelogram
Definition of parallelogram (A quadrilateral in which both pairs of opposite sides are parallel.)
72
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 28: Properties of Parallelograms
I need to use what is given to determine pairs of congruent triangles. Then, I can use the fact that their corresponding angles are equal in measure to prove that the quadrilateral is a parallelogram.
2. Given: ๐ด๐ด๐ด๐ด โ ๐ด๐ด๐ถ๐ถ;๐ด๐ด๐ด๐ด โ ๐ด๐ด๐ด๐ด Prove: Quadrilateral ๐ด๐ด๐ด๐ด๐ถ๐ถ๐ด๐ด is a parallelogram
Proof:
๐จ๐จ๐จ๐จ โ ๐ช๐ช๐จ๐จ;๐จ๐จ๐จ๐จ โ ๐จ๐จ๐จ๐จ Given
๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐โ ๐ช๐ช๐จ๐จ๐จ๐จ; ๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐โ ๐ช๐ช๐จ๐จ๐จ๐จ
Vertical angles are equal in measure.
โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐ช๐ช๐จ๐จ๐จ๐จ; โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐ช๐ช๐จ๐จ๐จ๐จ
SAS
๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐โ ๐ช๐ช๐จ๐จ๐จ๐จ; ๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐โ ๐ช๐ช๐จ๐จ๐จ๐จ
Corresponding angles of congruent triangles are equal in measure
๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐ช๐ช๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ,๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ โฅ ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ If two lines are cut by a transversal such that alternate interior angles are equal in measure, then the lines are parallel
Quadrilateral ๐จ๐จ๐จ๐จ๐ช๐ช๐จ๐จ is a parallelogram
Definition of parallelogram (A quadrilateral in which both pairs of opposite sides are parallel)
73
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 28: Properties of Parallelograms
3. Given: Diagonals ๐ด๐ด๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ and ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ bisect each other;
โ ๐ด๐ด๐ด๐ด๐ด๐ด โ โ ๐ด๐ด๐ด๐ด๐ด๐ด Prove: Quadrilateral ๐ด๐ด๐ด๐ด๐ถ๐ถ๐ด๐ด is a rhombus
Proof:
Diagonals ๐จ๐จ๐ช๐ช๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ and ๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ bisect each other
Given
๐จ๐จ๐จ๐จ = ๐จ๐จ๐ช๐ช;๐จ๐จ๐จ๐จ = ๐จ๐จ๐จ๐จ Definition of a segment bisector
โ ๐จ๐จ๐จ๐จ๐จ๐จ โ โ ๐จ๐จ๐จ๐จ๐จ๐จ Given
๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐โ ๐จ๐จ๐จ๐จ๐จ๐จ = ๐๐๐๐ห
๐๐โ ๐จ๐จ๐จ๐จ๐ช๐ช = ๐๐โ ๐จ๐จ๐จ๐จ๐ช๐ช = ๐๐๐๐ห
Angles on a line sum to ๐๐๐๐๐๐ห and since both angles are congruent, each angle measures ๐๐๐๐ห
โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐ช๐ช๐จ๐จ๐จ๐จ โ โณ๐ช๐ช๐จ๐จ๐จ๐จ
SAS
๐จ๐จ๐จ๐จ = ๐จ๐จ๐ช๐ช = ๐ช๐ช๐จ๐จ = ๐จ๐จ๐จ๐จ Corresponding sides of congruent triangles are equal in length
Quadrilateral ๐จ๐จ๐จ๐จ๐ช๐ช๐จ๐จ is a rhombus
Definition of rhombus (A quadrilateral with all sides of equal length)
In order to prove that ๐ด๐ด๐ด๐ด๐ถ๐ถ๐ด๐ด is a rhombus, I need to show that it has four sides of equal length. I can do this by showing the four triangles are all congruent.
74
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 28: Properties of Parallelograms
With one pair of opposite sides proven to be equal in length, I can look for a way to show that the other pair of opposite sides is equal in length to establish that ๐ด๐ด๐ด๐ด๐ถ๐ถ๐ด๐ด is a parallelogram.
4. Given: Parallelogram ๐ด๐ด๐ด๐ด๐ถ๐ถ๐ด๐ด, โ ๐ด๐ด๐ด๐ด๐ด๐ด โ โ ๐ถ๐ถ๐ถ๐ถ๐ด๐ด Prove: Quadrilateral ๐ด๐ด๐ด๐ด๐ด๐ด๐ถ๐ถ is a parallelogram
Proof:
Parallelogram ๐จ๐จ๐จ๐จ๐ช๐ช๐จ๐จ, โ ๐จ๐จ๐จ๐จ๐จ๐จ โ โ ๐ช๐ช๐ช๐ช๐จ๐จ Given
๐จ๐จ๐จ๐จ = ๐จ๐จ๐ช๐ช;๐จ๐จ๐จ๐จ = ๐ช๐ช๐จ๐จ Opposite sides of parallelograms are equal in length.
๐๐โ ๐จ๐จ = ๐๐โ ๐ช๐ช Opposite angles of parallelograms are equal in measure.
โณ ๐จ๐จ๐จ๐จ๐จ๐จ โ โณ ๐ช๐ช๐ช๐ช๐จ๐จ AAS
๐จ๐จ๐จ๐จ = ๐ช๐ช๐ช๐ช;๐จ๐จ๐จ๐จ = ๐ช๐ช๐จ๐จ Corresponding sides of congruent triangles are equal in length.
๐จ๐จ๐จ๐จ + ๐จ๐จ๐จ๐จ = ๐จ๐จ๐จ๐จ; Partition property
๐ช๐ช๐ช๐ช+ ๐ช๐ช๐จ๐จ = ๐ช๐ช๐จ๐จ
๐จ๐จ๐จ๐จ + ๐จ๐จ๐จ๐จ = ๐ช๐ช๐ช๐ช+ ๐ช๐ช๐จ๐จ Substitution property of equality
๐จ๐จ๐จ๐จ + ๐จ๐จ๐จ๐จ = ๐จ๐จ๐จ๐จ + ๐ช๐ช๐จ๐จ Substitution property of equality
๐จ๐จ๐จ๐จ = ๐ช๐ช๐จ๐จ Subtraction property of equality
Quadrilateral ๐จ๐จ๐จ๐จ๐จ๐จ๐ช๐ช is a parallelogram If both pairs of opposite sides of a quadrilateral are equal in length, the quadrilateral is a parallelogram
75
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 29: Special Lines in Triangles
I need to remember that a midsegment joins midpoints of two sides of a triangle.
A midsegment is parallel to the third side of the triangle. I must keep an eye out for special angles formed by parallel lines cut by a transversal.
Lesson 29: Special Lines in Triangles
In Problems 1โ4, all the segments within the triangles are midsegments.
1. ๐๐ = ๐๐๐๐ ๐๐ = ๐๐๐๐ ๐๐ = ๐๐๐๐.๐๐
2.
๐๐ = ๐๐๐๐๐๐ ๐๐ = ๐๐๐๐
The ๐๐๐๐๐๐ยฐ angle and the angle marked ๐๐ยฐ are corresponding angles; ๐๐ = ๐๐๐๐๐๐. This means the angle measures of the large triangle are ๐๐๐๐ยฐ (corresponding angles), ๐๐๐๐๐๐ยฐ, and ๐๐ยฐ; this makes ๐๐ = ๐๐๐๐ because the sum of the measures of angles of a triangle is ๐๐๐๐๐๐ยฐ.
76
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 29: Special Lines in Triangles
Mark the diagram using what you know about the relationship between the lengths of the midsegment and the side of the triangle opposite each midsegment.
Consider marking each triangle with angle measures (i.e., โ 1,โ 2,โ 3) to help identify the correspondences.
3. Find the perimeter, ๐๐, of the triangle.
๐ท๐ท = ๐๐(๐๐) + ๐๐(๐๐๐๐) + ๐๐(๐๐๐๐) = ๐๐๐๐
4. State the appropriate correspondences among the four congruent triangles within โณ ๐ด๐ด๐ด๐ด๐ด๐ด.
โณ ๐จ๐จ๐ท๐ท๐จ๐จ โ โณ๐ท๐ท๐ท๐ท๐ท๐ท โ โณ๐จ๐จ๐ท๐ท๐ธ๐ธ โ โณ๐ท๐ท๐จ๐จ๐ท๐ท
77
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 30: Special Lines in Triangles
I need to remember that a centroid divides a median into two lengths; the longer segment is twice the length of the shorter.
I can mark the diagram using what I know about the relationship between the lengths of the segments that make up the medians.
Lesson 30: Special Lines in Triangles
1. ๐น๐น is the centroid of triangle ๐ด๐ด๐ด๐ด๐ด๐ด. If the length of ๐ด๐ด๐น๐น๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ is 14, what is the length of median ๐ด๐ด๐ต๐ต๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ?
๐๐๐๐
(๐ฉ๐ฉ๐ฉ๐ฉ) = ๐ฉ๐ฉ๐ฉ๐ฉ
๐๐๐๐
(๐ฉ๐ฉ๐ฉ๐ฉ) = ๐๐๐๐
๐ฉ๐ฉ๐ฉ๐ฉ = ๐๐๐๐
2. ๐ด๐ด is the centroid of triangle ๐ ๐ ๐ ๐ ๐ ๐ . If ๐ด๐ด๐ถ๐ถ = 9 and ๐ด๐ด๐ ๐ = 13, what are the lengths of ๐ ๐ ๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ and ๐ ๐ ๐ต๐ต๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ?
๐๐๐๐
(๐น๐น๐น๐น) = ๐ช๐ช๐น๐น
๐๐๐๐
(๐น๐น๐น๐น) = ๐๐
๐น๐น๐น๐น = ๐๐๐๐ ๐๐๐๐
(๐บ๐บ๐ฉ๐ฉ) = ๐ช๐ช๐บ๐บ
๐๐๐๐
(๐บ๐บ๐ฉ๐ฉ) = ๐๐๐๐
๐บ๐บ๐ฉ๐ฉ =๐๐๐๐๐๐
= ๐๐๐๐.๐๐
78
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 30: Special Lines in Triangles
๐ต๐ต๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ and ๐พ๐พ๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ are the shorter and longer segments, respectively, along each of the medians they belong to.
3. ๐ ๐ ๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ,๐๐๐ต๐ต๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, and ๐๐๐๐๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ are medians. If ๐ ๐ ๐ถ๐ถ = 18, ๐๐๐๐ = 12 and ๐ ๐ ๐๐ = 17, what is the perimeter of โณ ๐ด๐ด๐ ๐ ๐๐?
๐ช๐ช๐ช๐ช =๐๐๐๐
(๐ช๐ช๐น๐น) = ๐๐๐๐
๐ช๐ช๐ช๐ช =๐๐๐๐
(๐ฝ๐ฝ๐ช๐ช) = ๐๐
๐ช๐ช๐ช๐ช =๐๐๐๐
(๐ช๐ช๐ป๐ป) = ๐๐.๐๐
Perimeter (โณ ๐ช๐ช๐ช๐ช๐ช๐ช) = ๐๐๐๐ + ๐๐ + ๐๐.๐๐ = ๐๐๐๐.๐๐
4. In the following figure, โณ ๐ต๐ต๐ด๐ด๐พ๐พ is equilateral. If the perimeter of โณ ๐ต๐ต๐ด๐ด๐พ๐พ is 18 and ๐ต๐ต and ๐ถ๐ถ are midpoints of ๐ฝ๐ฝ๐พ๐พ๏ฟฝ๏ฟฝ๏ฟฝ and ๐ฝ๐ฝ๐ฝ๐ฝ๏ฟฝ respectively, what are the lengths of ๐ต๐ต๐ฝ๐ฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ and ๐พ๐พ๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ?
If the perimeter of โณ ๐ฉ๐ฉ๐ช๐ช๐๐ is ๐๐๐๐, then ๐ฉ๐ฉ๐ช๐ช = ๐๐๐ช๐ช = ๐๐. ๐๐๐๐
(๐ฉ๐ฉ๐๐) = ๐ฉ๐ฉ๐ช๐ช
๐ฉ๐ฉ๐๐ = ๐๐(๐๐)
๐ฉ๐ฉ๐๐ = ๐๐๐๐
๐๐๐๐
(๐๐๐น๐น) = ๐๐๐ช๐ช
๐๐๐น๐น =๐๐๐๐
(๐๐)
๐๐๐น๐น = ๐๐
79
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 31: Construct a Square and a Nine-Point Circle
The steps I use to determine the midpoint of a segment are very similar to the steps to construct a perpendicular bisector. The main difference is that I do not need to draw in ๐ถ๐ถ๐ถ๐ถ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ , I need it as a guide to find the intersection with ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ.
Lesson 31: Construct a Square and a Nine-Point Circle
1. Construct the midpoint of segment ๐ด๐ด๐ด๐ด and write the steps to the construction.
1. Draw circle ๐จ๐จ: center ๐จ๐จ, radius ๐จ๐จ๐จ๐จ.
2. Draw circle ๐จ๐จ: center ๐จ๐จ, radius ๐จ๐จ๐จ๐จ.
3. Label the two intersections of the circles as ๐ช๐ช and ๐ซ๐ซ.
4. Label the intersection of ๐ช๐ช๐ซ๐ซ๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ with ๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ as midpoint ๐ด๐ด.
2. Create a copy of ๐ด๐ด๐ด๐ด๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ and label it as ๐ถ๐ถ๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ and write the steps to the construction. 1. Draw a segment and label one endpoint ๐ช๐ช. 2. Mark off the length of ๐จ๐จ๐จ๐จ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ along the drawn segment; label the marked point as ๐ซ๐ซ.
80
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 31: Construct a Square and a Nine-Point Circle
I must remember that the intersections ๐๐ or ๐๐ may lie outside the triangle, as shown in the example.
3. Construct the three altitudes of โณ ๐ด๐ด๐ด๐ด๐ถ๐ถ and write the steps to the construction. Label the orthocenter
as ๐๐.
1. Draw circle ๐จ๐จ: center ๐จ๐จ, with radius so that circle ๐จ๐จ intersects ๐จ๐จ๐ช๐ช๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ in two points; label these points as ๐ฟ๐ฟ and ๐๐.
2. Draw circle ๐ฟ๐ฟ: center ๐ฟ๐ฟ, radius ๐ฟ๐ฟ๐๐.
3. Draw circle ๐๐: center ๐๐, radius ๐๐๐ฟ๐ฟ.
4. Label either intersection of circles ๐ฟ๐ฟ and ๐๐ as ๐๐.
5. Label the intersection of ๐จ๐จ๐๐๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ with ๐จ๐จ๐ช๐ช๏ฟฝโ๏ฟฝ๏ฟฝ๏ฟฝโ as ๐น๐น (this is altitude ๐จ๐จ๐น๐น๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ)
6. Repeat steps 1-5 from vertices ๐จ๐จ and ๐ช๐ช.
81
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 32: Construct a Nine-Point Circle
I need to know how to construct a perpendicular bisector in order to determine the circumcenter of a triangle, which is the point of concurrency of three perpendicular bisectors of a triangle.
I must remember that the center of the circle that circumscribes a triangle is the circumcenter of that triangle, which might lie outside the triangle.
Lesson 32: Construct a Nine-Point Circle
1. Construct the perpendicular bisector of segment ๐ด๐ด๐ด๐ด.
2. Construct the circle that circumscribes โณ ๐ด๐ด๐ด๐ด๐ด๐ด. Label the center of the circle as ๐๐.
82
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 32: Construct a Nine-Point Circle
3. Construct a square ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด based on the provided segment ๐ด๐ด๐ด๐ด.
83
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 33: Review of the Assumptions
I take axioms, or assumptions, for granted; they are the basis from which all other facts can be derived.
I should remember that basic rigid motions are a subset of transformations in general.
Lesson 33: Review of the Assumptions
1. Points ๐ด๐ด, ๐ต๐ต, and ๐ถ๐ถ are collinear. ๐ด๐ด๐ต๐ต = 1.5 and ๐ต๐ต๐ถ๐ถ = 3. What is the length of ๐ด๐ด๐ถ๐ถ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ, and what assumptions do we make in answering this question?
๐จ๐จ๐ช๐ช = ๐๐.๐๐. The Distance and Ruler Axioms.
2. Find the angle measures marked ๐ฅ๐ฅ and ๐ฆ๐ฆ and justify the answer with the facts that support your reasoning.
๐๐ = ๐๐๐๐ยฐ, ๐๐ = ๐๐๐๐ยฐ
The angle marked ๐๐ and ๐๐๐๐๐๐ยฐ are a linear pair and are supplementary. The angle vertical to ๐๐ has the same measure as ๐๐, and the sum of angle measures of a triangle is ๐๐๐๐๐๐ยฐ.
3. What properties of basic rigid motions do we assume to be true?
It is assumed that under any basic rigid motion of the plane, the image of a line is a line, the image of a ray is a ray, and the image of a segment is a segment. Additionally, rigid motions preserve lengths of segments and measures of angles.
84
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 33: Review of the Assumptions
I must remember that this is referred to as โangles at a pointโ.
When parallel lines are intersected by a transversal, I must look for special angle pair relationships.
4. Find the measures of angle ๐ฅ๐ฅ.
The sum of the measures of all adjacent angles formed by three or more rays with the same vertex is ๐๐๐๐๐๐ยฐ.
๐๐(๐๐๐๐ยฐ) + ๐๐(๐๐๐๐ยฐ) + ๐๐(๐๐๐๐ยฐ) + ๐๐(๐๐) = ๐๐๐๐๐๐ยฐ
๐๐๐๐ยฐ + ๐๐๐๐๐๐ยฐ + ๐๐๐๐๐๐ยฐ + ๐๐๐๐ = ๐๐๐๐๐๐ยฐ
๐๐ = ๐๐ยฐ
5. Find the measures of angles ๐ฅ๐ฅ and ๐ฆ๐ฆ.
๐๐ = ๐๐๐๐ยฐ, ๐๐ = ๐๐๐๐ยฐ
โ ๐น๐น๐น๐น๐น๐น and โ ๐น๐น๐น๐น๐ธ๐ธ are same side interior angles and are therefore supplementary. โ ๐น๐น๐น๐น๐น๐น is vertical to โ ๐ด๐ด๐น๐น๐ด๐ด and therefore the angles are equal in measure. Finally, the angle sum of a triangle is ๐๐๐๐๐๐ยฐ.
85
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 34: Review of the Assumptions
I must remember that these criteria imply the existence of a rigid motion that maps one triangle to the other, which of course renders them congruent.
Lesson 34: Review of the Assumptions
1. Describe all the criteria that indicate whether two triangles will be congruent or not.
Given two triangles, โณ ๐จ๐จ๐จ๐จ๐จ๐จ and โณ ๐จ๐จโฒ๐จ๐จโฒ๐จ๐จโฒ:
If ๐จ๐จ๐จ๐จ = ๐จ๐จโฒ๐จ๐จโฒ (Side), ๐๐โ ๐จ๐จ = ๐๐โ ๐จ๐จโฒ (Angle), ๐จ๐จ๐จ๐จ = ๐จ๐จโฒ๐จ๐จโฒ(Side), then the triangles are congruent. (SAS)
If ๐๐โ ๐จ๐จ = ๐๐โ ๐จ๐จโฒ (Angle), ๐จ๐จ๐จ๐จ = ๐จ๐จโฒ๐จ๐จโฒ (Side), and ๐๐โ ๐จ๐จ = ๐๐โ ๐จ๐จโฒ (Angle), then the triangles are congruent. (ASA)
If ๐จ๐จ๐จ๐จ = ๐จ๐จโฒ๐จ๐จโฒ (Side), ๐จ๐จ๐จ๐จ = ๐จ๐จโฒ๐จ๐จโฒ (Side), and ๐จ๐จ๐จ๐จ = ๐จ๐จโฒ๐จ๐จโฒ (Side), then the triangles are congruent. (SSS)
If ๐จ๐จ๐จ๐จ = ๐จ๐จโฒ๐จ๐จโฒ (Side), ๐๐โ ๐จ๐จ = ๐๐โ ๐จ๐จโฒ (Angle), and โ ๐จ๐จ = โ ๐จ๐จโฒ (Angle), then the triangles are congruent. (AAS)
Given two right triangles, โณ ๐จ๐จ๐จ๐จ๐จ๐จ and โณ ๐จ๐จโฒ๐จ๐จโฒ๐จ๐จโฒ, with right angles โ ๐จ๐จ and โ ๐จ๐จโฒ, if ๐จ๐จ๐จ๐จ = ๐จ๐จโฒ๐จ๐จโฒ (Leg) and ๐จ๐จ๐จ๐จ = ๐จ๐จโฒ๐จ๐จโฒ (Hypotenuse), then the triangles are congruent. (HL)
86
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015
2015-16
M1
GEOMETRY
Lesson 34: Review of the Assumptions
I must remember that a midsegment joins midpoints of two sides of a triangle and is parallel to the third side.
The centroid of a triangle is the point of concurrency of three medians of a triangle.
2. In the following figure, ๐บ๐บ๐บ๐บ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ is a midsegment. Find ๐ฅ๐ฅ and ๐ฆ๐ฆ. Determine the perimeter of โณ ๐ด๐ด๐บ๐บ๐บ๐บ.
๐๐ = ๐๐๐๐ยฐ, ๐๐ = ๐๐๐๐ยฐ
Perimeter of โณ ๐จ๐จ๐จ๐จ๐จ๐จ is:
๐๐๐๐
(๐๐๐๐) +๐๐๐๐
(๐๐๐๐) + ๐๐๐๐ = ๐๐๐๐.๐๐
3. In the following figure, ๐บ๐บ๐บ๐บ๐บ๐บ๐บ๐บ and ๐บ๐บ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ๐ฝ are squares and ๐บ๐บ๐บ๐บ = ๐บ๐บ๐ฝ๐ฝ. Prove that ๐ ๐ ๐บ๐บ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ is an angle bisector.
4. How does a centroid divide a median?
The centroid divides a median into two parts: from the vertex to centroid, and centroid to midpoint in a ratio of ๐๐:๐๐.
Proof: ๐จ๐จ๐จ๐จ๐ฎ๐ฎ๐ฎ๐ฎ and ๐ฎ๐ฎ๐ฑ๐ฑ๐ฑ๐ฑ๐ฑ๐ฑ are squares and ๐จ๐จ๐ฎ๐ฎ = ๐ฎ๐ฎ๐ฑ๐ฑ
Given
โ ๐จ๐จ and โ ๐ฑ๐ฑ are right angles All angles of a square are right angles.
โณ ๐จ๐จ๐ฎ๐ฎ๐ฎ๐ฎ and โณ๐ฑ๐ฑ๐ฎ๐ฎ๐ฎ๐ฎ are right triangles
Definition of right triangle.
๐ฎ๐ฎ๐ฎ๐ฎ = ๐ฎ๐ฎ๐ฎ๐ฎ Reflexive Property
โณ ๐จ๐จ๐ฎ๐ฎ๐ฎ๐ฎ โ โณ๐ฑ๐ฑ๐ฎ๐ฎ๐ฎ๐ฎ HL
๐๐โ ๐จ๐จ๐ฎ๐ฎ๐ฎ๐ฎ = ๐๐โ ๐ฑ๐ฑ๐ฎ๐ฎ๐ฎ๐ฎ Corresponding angles of congruent triangles are equal in measure.
๐ฎ๐ฎ๐ฎ๐ฎ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโ is an angle bisector Definition of angle bisector.
87
Homework Helper A Story of Functions
ยฉ 2015 Great Minds eureka-math.org GEO-M1-HWH-1.3.0-08.2015