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2015-16
Lesson 1: Construct an Equilateral Triangle
M1
GEOMETRY
Lesson 1: Construct an Equilateral Triangle
1. The segment below has a length of ðð. Use a compassto mark all the points that are at a distance ðð frompoint ð¶ð¶.
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.
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Lesson 1: Construct an Equilateral Triangle
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2. Use a compass to determine which of the following points, ð ð , ðð, ðð, and ðð, lie on the same circle about center ð¶ð¶. Explain how you know.
If I set the compass point at ðªðª and adjust the compass so that the pencil passes through each of the points ð¹ð¹, ðºðº, ð»ð», and ðŒðŒ, I see that points ðºðº and ðŒðŒ both lie on the same circle. Points ð¹ð¹ and ð»ð» 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 ðªðªð¹ð¹, ðªðªðºðº, ðªðªðŒðŒ, and ðªðªð»ð» 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.
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Lesson 1: Construct an Equilateral Triangle
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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 ð¶ð¶1 has a radius of 4; Circle ð¶ð¶2 has a radius of 6.
b. Circle ð¶ð¶3 has a radius of 4; Circle ð¶ð¶4 has a radius of 4.
Point ð·ð· is a distance of ðð from ðªðªðð and a distance greater than ðð from ðªðªðð. Point ð¹ð¹ is a distance of ðð from ðªðªðð and a distance ðð from ðªðªðð.
Point ðºðº is a distance of ðð from ðªðªðð and a distance greater than ðð from ðªðªðð. Point ð»ð» is a distance of ðð from ðªðªðð and a distance of ðð from ðªðªðð.
c. Asha claims that the points ð¶ð¶1, ð¶ð¶2, and ð ð are the vertices of an equilateral triangle since ð ð is the intersection of the two circles. Nadege says this is incorrect but that ð¶ð¶3, ð¶ð¶4, and ðð are the vertices of an equilateral triangle. Who is correct? Explain.
Nadege is correct. Points ðªðªðð, ðªðªðð, and ð¹ð¹ are not the vertices of an equilateral triangle because the distance between each pair of vertices is not the same. The points ðªðªðð, ðªðªðð, and ð»ð» 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.
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Lesson 1: Construct an Equilateral Triangle
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4. Construct an equilateral triangle ðŽðŽðŽðŽð¶ð¶ that has a side length ðŽðŽðŽðŽ, below. Use precise language to list the steps to perform the construction.
or
1. Draw circle ðšðš: center ðšðš, radius ðšðšðšðš. 2. Draw circle ðšðš: center ðšðš, radius ðšðšðšðš. 3. Label one intersection as ðªðª. 4. Join ðšðš, ðšðš, ðªðª.
I must use both endpoints of the segment as the centers of the two circles I must construct.
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Lesson 2: Construct an Equilateral Triangle
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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 ðšðš, ð©ð©, and ðªðª do not determine the vertices of an equilateral triangle because the distance between ðšðš and ð©ð©, as measured by adjusting the compass, is not the same distance as between ð©ð© and ðªðª and as between ðªðª and ðšðš.
Points ð«ð«, ð¬ð¬, and ðð do determine the vertices of an equilateral triangle because the distance between ð«ð« and ð¬ð¬ is the same distance as between ð¬ð¬ and ðð and as between ðð and ð«ð«.
The distance between each pair of vertices of an equilateral triangle is the same.
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Lesson 2: Construct an Equilateral Triangle
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2. Use what you know about the construction of an equilateral triangle to recreate parallelogram ðŽðŽðŽðŽðŽðŽðŽðŽ
below, and write a set of steps that yields this construction.
Possible steps:
1. Draw ðšðšð©ð©ï¿œï¿œï¿œï¿œ. 2. Draw circle ðšðš: center ðšðš, radius ðšðšð©ð©. 3. Draw circle ð©ð©: center ð©ð©, radius ð©ð©ðšðš. 4. Label one intersection as ðªðª. 5. Join ðªðª to ðšðš and ð©ð©. 6. Draw circle ðªðª: center ðªðª, radius ðªðªðšðš. 7. Label the intersection of circle ðªðª with circle ð©ð© as ð«ð«. 8. Join ð«ð« to ð©ð© and ðªðª.
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Lesson 2: Construct an Equilateral Triangle
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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 ðšðšð©ð©ï¿œï¿œï¿œï¿œ. 2. Draw circle ðšðš: center ðšðš, radius ðšðšð©ð©. 3. Draw circle ð©ð©: center ð©ð©, radius ð©ð©ðšðš. 4. Label one intersection as ðªðª; label the other intersection as ð«ð«. 5. Join ðšðš and ð©ð© with both ðªðª and ð«ð«. 6. Draw circle ð«ð«: center ð«ð«, radius ð«ð«ðšðš. 7. Label the intersection of circle ð«ð« with circle ðšðš as ð¬ð¬. 8. Join ð¬ð¬ to ðšðš and ð«ð«. 9. Label the intersection of circle ð«ð« with circle ð©ð© as ðð. 10. Join ðð to ð©ð© and ð«ð«.
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Lesson 3: Copy and Bisect an Angle
Lesson 3: Copy and Bisect an Angle
1. Krysta is copying â ðŽðŽðŽðŽðŽðŽ to construct â ð·ð·ð·ð·ð·ð·. a. Complete steps 5â9, and use a compass and
straightedge to finish constructing â ð·ð·ð·ð·ð·ð·. 1. Steps to copy an angle are as follows: 2. Label the vertex of the original angle as B.
3. Draw EGᅵᅵᅵᅵᅵâ as one side of the angle to be drawn. 4. Draw circle B: center B, any radius. 5. Label the intersections of circle ð©ð© with the sides of theangle as ðšðš and ðªðª. 6. Draw circle ð¬ð¬: center ð¬ð¬, radius ð©ð©ðšðš.
7. Label intersection of circle ð¬ð¬ with ð¬ð¬ð¬ð¬ï¿œï¿œï¿œï¿œï¿œâ as ðð. 8. Draw circle ðð: center ðð, radius ðªðªðšðš. 9. Label either intersection of circle ð¬ð¬ and circle ðð as ð«ð«.
10. Draw ð¬ð¬ð«ð«ï¿œï¿œï¿œï¿œï¿œï¿œâ .
b. Underline the steps that describe the construction of circles used in the copied angle.
c. Why must circle ð·ð· have a radius of length ðŽðŽðŽðŽ?
The intersection of circle ð©ð© and circle ðªðª: center ðªðª, radius ðªðªðšðš determines point ðšðš. To mirror this location for the copied angle, circle ðð must have a radius of length ðªðªðšðš.
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.
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Lesson 3: Copy and Bisect an Angle
2. ðŽðŽð·ð·ï¿œï¿œï¿œï¿œï¿œï¿œâ is the angle bisector of â ðŽðŽðŽðŽðŽðŽ.
a. Write the steps that yield ðŽðŽð·ð·ï¿œï¿œï¿œï¿œï¿œï¿œâ as the angle bisector of â ðŽðŽðŽðŽðŽðŽ.
Steps to construct an angle bisector are as follows:
1. Label the vertex of the angle as ð©ð©. 2. Draw circle ð©ð©: center ð©ð©, any size radius. 3. Label intersections of circle ð©ð© with the rays of the angle as ðšðš and ðªðª. 4. Draw circle ðšðš: center ðšðš, radius ðšðšðªðª. 5. Draw circle ðªðª: center ðªðª, radius ðªðªðšðš. 6. At least one of the two intersection points of circle ðšðš and circle
ðªðª lies in the interior of the angle. Label that intersection point ð«ð«.
7. Draw ð©ð©ð«ð«ï¿œï¿œï¿œï¿œï¿œï¿œâ .
b. Why do circles ðŽðŽ and ðŽðŽ each have a radius equal to length ðŽðŽðŽðŽ? Point ð«ð« is as far from ðšðš as it is from ðªðª since ðšðšð«ð« = ðªðªð«ð« = ðšðšðªðª. As long as ðšðš and ðªðª are equal distances from vertex ð©ð© and each of the circles has a radius equal ðšðšðªðª, ð«ð« will be an equal distance from ðšðšð©ð©ï¿œï¿œï¿œï¿œï¿œï¿œâ
and ðšðšðªðªï¿œï¿œï¿œï¿œï¿œâ . 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 ðŽðŽ can have a radius of any length, but the circles with centers ðŽðŽ and ðŽðŽ on each of the rays must have a radius ðŽðŽðŽðŽ.
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Lesson 4: Construct a Perpendicular Bisector
Lesson 4: Construct a Perpendicular Bisector
1. Perpendicular bisector ððððï¿œâᅵᅵᅵâ is constructed to ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ; the intersection of ððððï¿œâᅵᅵᅵâ with the segment is labeled ðð. Use the idea of folding to explain why ðŽðŽ and ðŽðŽ are symmetric with respect to ððððï¿œâᅵᅵᅵâ .
If the segment is folded along ð·ð·ð·ð·ï¿œâᅵᅵᅵâ so that ðšðš coincides with ð©ð©, then ðšðšðšðšï¿œï¿œï¿œï¿œï¿œ coincides with ð©ð©ðšðšï¿œï¿œï¿œï¿œï¿œ, or ðšðšðšðš = ð©ð©ðšðš; ðšðš is the midpoint of the segment. â ðšðšðšðšð·ð· and â ð©ð©ðšðšð·ð· also coincide, and since they are two identical angles on a straight line, the sum of their measures must be ðððððð°, or each has a measure of ðððð°. Thus, ðšðš and ð©ð© are symmetric with respect to ð·ð·ð·ð·ï¿œâᅵᅵᅵâ .
To be symmetric with respect to ððððï¿œâᅵᅵᅵâ , the portion of ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ on one side of ððððï¿œâᅵᅵᅵâ must be mirrored on the opposite side of ððððï¿œâᅵᅵᅵâ .
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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 ð¿ð¿: center ð¿ð¿, radius ð¿ð¿ð¿ð¿. 2. Draw circle ð¿ð¿: center ð¿ð¿, radius ð¿ð¿ð¿ð¿. 3. Label the points of intersections as ðšðš and ð©ð©.
4. Draw ðšðšð©ð©ï¿œâᅵᅵᅵâ .
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.
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Lesson 4: Construct a Perpendicular Bisector
3. Rhombus ðððððððð can be constructed by joining the midpoints of rectangle ðŽðŽðŽðŽðŽðŽðŽðŽ. Use the perpendicular bisector construction to help construct rhombus ðððððððð.
The midpoint of ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ is vertically aligned to the midpoint of ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ. I can use the construction of the perpendicular bisector to determine the perpendicular bisector of ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ.
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Lesson 5: Points of Concurrencies
Lesson 5: Points of Concurrencies
1. Observe the construction below, and explain the significance of point ðð.
Lines ðð, ðð, and ðð, which are each perpendicular bisectors of a side of the triangle, are concurrent at point ð·ð·.
a. Describe the distance between ðŽðŽ and ðð and between ðµðµ and ðð. Explain why this true.
ð·ð· is equidistant from ðšðš and ð©ð©. Any point that lies on the perpendicular bisector of ðšðšð©ð©ï¿œï¿œï¿œï¿œ is equidistant from either endpoint ðšðš or ð©ð©.
b. Describe the distance between ð¶ð¶ and ðð and between ðµðµ and ðð. Explain why this true.
ð·ð· is equidistant from ð©ð© and ðªðª. Any point that lies on the perpendicular bisector of ð©ð©ðªðªï¿œï¿œï¿œï¿œ is equidistant from either endpoint ð©ð© or ðªðª.
c. What do the results of Problem 1 parts (a) and (b) imply about ðð?
Since ð·ð· is equidistant from ðšðš and ð©ð© and from ð©ð© and ðªðª, then it is also equidistant from ðšðš and ðªðª. This is why ð·ð· is the point of concurrency of the three perpendicular bisectors.
The markings in the figure imply lines ðð, ðð, and ðð are perpendicular bisectors.
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Lesson 5: Points of Concurrencies
2. Observe the construction below, and explain the significance of point ðð.
Rays ðšðšðšðš, ðªðªðšðš, and ð©ð©ðšðš are each angle bisectors of an angle of a triangle, and are concurrent at point ðšðš, or all three angle bisectors intersect in a single point.
a. Describe the distance between ðð and rays ðŽðŽðµðµï¿œï¿œï¿œï¿œï¿œâ and ðŽðŽð¶ð¶ï¿œï¿œï¿œï¿œï¿œâ . Explain why this true.
ðšðš is equidistant from ðšðšð©ð©ï¿œï¿œï¿œï¿œï¿œï¿œâ and ðšðšðªðªï¿œï¿œï¿œï¿œï¿œâ . Any point that lies on the angle bisector of â ðšðš is equidistant from rays ðšðšð©ð©ï¿œï¿œï¿œï¿œï¿œï¿œâ and ðšðšðªðªï¿œï¿œï¿œï¿œï¿œâ .
b. Describe the distance between ðð and rays ðµðµðŽðŽï¿œï¿œï¿œï¿œï¿œâ and ðµðµð¶ð¶ï¿œï¿œï¿œï¿œï¿œâ . Explain why this true.
ðšðš is equidistant from ð©ð©ðšðšï¿œï¿œï¿œï¿œï¿œï¿œâ and ð©ð©ðªðªï¿œï¿œï¿œï¿œï¿œï¿œâ . Any point that lies on the angle bisector of â ð©ð© is equidistant from rays ð©ð©ðšðšï¿œï¿œï¿œï¿œï¿œï¿œâ and ð©ð©ðªðªï¿œï¿œï¿œï¿œï¿œï¿œâ .
c. What do the results of Problem 2 parts (a) and (b) imply about ðð?
Since ðšðš is equidistant from ðšðšð©ð©ï¿œï¿œï¿œï¿œï¿œï¿œâ and ðšðšðªðªï¿œï¿œï¿œï¿œï¿œâ and from ð©ð©ðšðšï¿œï¿œï¿œï¿œï¿œï¿œâ and ð©ð©ðªðªï¿œï¿œï¿œï¿œï¿œï¿œâ , then it is also equidistant from ðªðªð©ð©ï¿œï¿œï¿œï¿œï¿œï¿œâ and ðªðªðšðšï¿œï¿œï¿œï¿œï¿œâ . This is why ðšðš is the point of concurrency of the three angle bisectors.
The markings in the figure imply that rays ðŽðŽððᅵᅵᅵᅵᅵâ , ð¶ð¶ððᅵᅵᅵᅵᅵâ , and ðµðµððᅵᅵᅵᅵᅵâ are all angle bisectors.
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GEOMETRY
Lesson 6: Solve for Unknown AnglesâAngles and Lines at a Point
Lesson 6: Solve for Unknown AnglesâAngles and Lines at a Point
1. Write an equation that appropriately describes each of the diagrams below.
a. b.
ðð + ðð + ðð + ð ð = ðððððð° ðð + ðð + ðð + ð ð + ðð + ðð + ðð = ðððððð°
Adjacent angles on a line sum to 180°. Adjacent angles around a point sum to 360°.
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Lesson 6: Solve for Unknown AnglesâAngles and Lines at a Point
2. Find the measure of â ðŽðŽðŽðŽðŽðŽ.
ðððð + ðððð° + ðð = ðððððð°
ðððð + ðððð° = ðððððð°
ðððð = ðððððð°
ðð = ðððð°
The measure of â ðšðšðšðšðšðš is ðð(ðððð°), or ðððððð°.
3. Find the measure of â ð·ð·ðŽðŽð·ð·.
ðð + ðððð + ðððð + ðððð + ðððððð° = ðððððð°
ðððððð+ ðððððð° = ðððððð°
ðððððð = ðððððð°
ðð = ðððð°
The measure of â ð«ð«ðšðšð«ð« is ðð(ðððð°), or ðððð°.
I must solve for ð¥ð¥ before I can find the measure of â ðŽðŽðŽðŽðŽðŽ.
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Lesson 7: Solve for Unknown AnglesâTransversals
Lesson 7: Solve for Unknown AnglesâTransversals
1. In the following figure, angle measures ðð, ðð, ðð, and ðð are equal. List four pairs of parallel lines.
Four pairs of parallel lines: ðšðšðšðšï¿œâᅵᅵᅵâ ⥠ð¬ð¬ð¬ð¬ï¿œâᅵᅵᅵâ , ðšðšðšðšï¿œâᅵᅵᅵâ ⥠ðªðªðªðªï¿œâᅵᅵᅵâ , ðšðšðªðªï¿œâᅵᅵᅵâ ⥠ðªðªð«ð«ï¿œâᅵᅵᅵâ , ðªðªðªðªï¿œâᅵᅵᅵâ ⥠ð¬ð¬ð¬ð¬ï¿œâᅵᅵᅵâ
2. Find the measure of â a.
The measure of ðð is ðððððð° â ðððððð°, or ðððð°.
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 ðð.
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Lesson 7: Solve for Unknown AnglesâTransversals
3. Find the measure of ðð.
The measure of ðð is ðððð° + ðððð°, or ðððð°.
4. Find the value of ð¥ð¥.
ðððð + ðð = ðððððð
ðððð = ðððððð
ðð = ðððð
I can draw a horizontal auxiliary line, parallel to the other horizontal lines in order to make the necessary corresponding angle pairs more apparent.
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Lesson 8: Solve for Unknown AnglesâAngles in a Triangle
Lesson 8: Solve for Unknown AnglesâAngles in a Triangle
1. Find the measure of ðð.
ð ð + ðððððð° + ðððð° = ðððððð°
ð ð + ðððððð° = ðððððð°
ð ð = ðððð°
I need to apply what I know about complementary and supplementary angles to begin to solve for ðð.
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Lesson 8: Solve for Unknown AnglesâAngles in a Triangle
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2. Find the measure of ðð.
ðððð + ðððð° = ðððððð°
ðððð = ðððððð°
ðð = ðððð°
I need to apply what I know about parallel lines cut by a transversal and alternate interior angles in order to solve for ðð.
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Lesson 8: Solve for Unknown AnglesâAngles in a Triangle
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3. Find the measure of ð¥ð¥.
(ðððððð° â ðððð) + ðððððð° + ðð = ðððððð°
ðððððð° â ðððð = ðððððð°
ðððð = ðððððð°
ðð = ðððð°
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 ð¥ð¥.
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Lesson 9: Unknown Angle ProofsâWriting Proofs
Lesson 9: Unknown Angle ProofsâWriting Proofs
1. Use the diagram below to prove that ððððᅵᅵᅵᅵ ⥠ððððᅵᅵᅵᅵ.
ððâ ð·ð· + ððâ ðžðž+ ððâ ð·ð·ð·ð·ðžðž = ðððððð° The sum of the angle measures in a triangle is ðððððð°.
ððâ ð·ð·ð·ð·ðžðž = ðððð° Subtraction property of equality
ððâ ðŒðŒ+ ððâ ð»ð» + ððâ ð»ð»ð»ð»ðŒðŒ = ðððððð° The sum of the angle measures in a triangle is ðððððð°.
ððâ ð»ð»ð»ð»ðŒðŒ = ðððð° Subtraction property of equality
ððâ ð·ð·ð·ð·ðžðž + ððâ ð»ð»ð»ð»ðŒðŒ+ ððâ ð»ð»ðºðºð·ð· = ðððððð° The sum of the angle measures in a triangle is ðððððð°.
ððâ ð»ð»ðºðºð·ð· = ðððð° Subtraction property of equality
ð»ð»ð»ð»ï¿œï¿œï¿œï¿œ ⥠ðžðžð·ð·ï¿œï¿œï¿œï¿œ Perpendicular lines form ðððð° angles.
To show that ððððᅵᅵᅵᅵ ⥠ððððᅵᅵᅵᅵ, I need to first show that ððâ ðððððð = 90°.
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Lesson 9: Unknown Angle ProofsâWriting Proofs
2. Prove ððâ ðððððð = ððâ ðððððð.
ð·ð·ðžðžï¿œâᅵᅵᅵâ ⥠ð»ð»ð»ð»ï¿œâᅵᅵâ , ðžðžðŒðŒï¿œâᅵᅵᅵᅵâ ⥠ð»ð»ðºðºï¿œâᅵᅵâ Given
ððâ ð·ð·ðžðžð»ð» = ððâ ð»ð»ð»ð»ð»ð», ððâ ð·ð·ðžðžð»ð» = ððâ ðºðºð»ð»ð»ð» If parallel lines are cut by a transversal, then corresponding angles are equal in measure.
ððâ ð·ð·ðžðžð·ð· = ððâ ð·ð·ðžðžð»ð» âððâ ð·ð·ðžðžð»ð»,
ððâ ð»ð»ð»ð»ðºðº = ððâ ð»ð»ð»ð»ð»ð»âððâ ðºðºð»ð»ð»ð»
Partition property
ððâ ð·ð·ðžðžð·ð· = ððâ ð»ð»ð»ð»ð»ð»âððâ ðºðºð»ð»ð»ð» Substitution property of equality
ððâ ð·ð·ðžðžð·ð· = ððâ ð»ð»ð»ð»ðºðº Substitution property of equality
I need to consider how angles â ðððððð and â ðððððð are related to angles I know to be equal in measure in the diagram.
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Lesson 9: Unknown Angle ProofsâWriting Proofs
3. In the diagram below, ððððᅵᅵᅵᅵᅵ bisects â ðððððð, and ððððᅵᅵᅵᅵ bisects â ðððððð. Prove that ððððᅵᅵᅵᅵᅵ ⥠ððððᅵᅵᅵᅵ.
ð·ð·ðžðžï¿œâᅵᅵᅵâ ⥠ð·ð·ð»ð»ï¿œâᅵᅵᅵâ , ðºðºð»ð»ï¿œï¿œï¿œï¿œï¿œ bisects â ðžðžðºðºðžðž and ðžðžððᅵᅵᅵᅵ bisects â ðºðºðžðžð·ð· Given
ððâ ðžðžðºðºðžðž = ððâ ðºðºðžðžð·ð· If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
ððâ ðžðžðºðºð»ð» = ððâ ð»ð»ðºðºðžðž, ððâ ðºðºðžðžðð = ððâ ðððžðžð·ð· Definition of bisect
ððâ ðžðžðºðºðžðž = ððâ ðžðžðºðºð»ð»+ ððâ ð»ð»ðºðºðžðž,
ððâ ðºðºðžðžð·ð· = ððâ ðºðºðžðžðð +ððâ ðððžðžð·ð·
Partition property
ððâ ðžðžðºðºðžðž = ðð(ððâ ð»ð»ðºðºðžðž), ððâ ðºðºðžðžð·ð· = ðð(ððâ ðºðºðžðžðð) Substitution property of equality
ðð(ððâ ð»ð»ðºðºðžðž) = ðð(ððâ ðºðºðžðžðð) Substitution property of equality
ððâ ð»ð»ðºðºðžðž = ððâ ðºðºðžðžðð Division property of equality
ðºðºð»ð»ï¿œï¿œï¿œï¿œï¿œ ⥠ðžðžððᅵᅵᅵᅵ 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 ðððð and ðððð are parallel.
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Lesson 10: Unknown Angle ProofsâProofs with Constructions
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Lesson 10: Unknown Angle ProofsâProofs with Constructions
1. Use the diagram below to prove that ðð = 106° â ðð.
Construct ðŒðŒðŒðŒï¿œâᅵᅵᅵᅵâ parallel to ð»ð»ð»ð»ï¿œâᅵᅵᅵâ and ðºðºðºðºï¿œâᅵᅵᅵâ .
ððâ ðºðºðŒðŒðŒðŒ = ðð If parallel lines are cut by a transversal, then corresponding angles are equal in measure.
ððâ ðŒðŒðŒðŒðŒðŒ = ðððððð° â ðð Partition property
ððâ ðŒðŒðŒðŒðŒðŒ = ðð If parallel lines are cut by a transversal, then corresponding angles are equal in measure.
ðð = ðððððð° â ðð Substitution property of equality
Just as if this were a numeric problem, I need to construct a horizontal line through ðð, so I can see the special angle pairs created by parallel lines cut by a transversal.
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2. Use the diagram below to prove that mâ C = ðð + ðð.
Construct ð®ð®ð®ð®ï¿œâᅵᅵᅵᅵâ parallel to ðºðºð»ð»ï¿œâᅵᅵᅵâ and ððððï¿œâᅵᅵᅵâ through ðªðª.
ððâ ð»ð»ðªðªð®ð® = ðð, ððâ ðððªðªð®ð® = ð ð If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
ððâ ðªðª = ðð + ð ð Partition property
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3. Use the diagram below to prove that ððâ ð¶ð¶ð¶ð¶ð¶ð¶ = ðð + 90°.
Construct ð«ð«ððï¿œâᅵᅵᅵâ parallel to ðªðªð»ð»ï¿œâᅵᅵᅵâ . Extend ðºðºð»ð»ï¿œï¿œï¿œï¿œ so that it intersects ð«ð«ððï¿œâᅵᅵᅵâ ; extend ðªðªð»ð»ï¿œï¿œï¿œï¿œ.
ððâ ðªðªð«ð«ðð+ ðððð° = ðððððð° If parallel lines are cut by a transversal, then same-side interior angles are supplementary.
ððâ ðªðªð«ð«ðð = ðððð° Subtraction property of equality
ððâ ð«ð«ððð»ð» = ðð If parallel lines are cut by a transversal, then corresponding angles are equal in measure.
ððâ ððð«ð«ðð = ðð If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
ððâ ðªðªð«ð«ðð = ðð + ðððð° Partition property
I am going to need multiple constructions to show why the measure of â ð¶ð¶ð¶ð¶ð¶ð¶ = ðð + 90°.
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Lesson 11: Unknown Angle ProofsâProofs of Known Facts
Lesson 11: Unknown Angle ProofsâProofs of Known Facts
1. Given: ððððï¿œâᅵᅵᅵᅵâ and ððððï¿œâᅵᅵâ intersect at ð ð . Prove: ððâ 2 = ððâ 4
ðœðœðœðœï¿œâᅵᅵᅵᅵᅵâ and ððððï¿œâᅵᅵᅵâ intersect at ð¹ð¹. Given
ððâ ðð +ððâ ðð = ðððððð°; ððâ ðð+ ððâ ðð = ðððððð° Angles on a line sum to ðððððð°.
ððâ ðð +ððâ ðð = ððâ ðð+ ððâ ðð Substitution property of equality
ððâ ðð = ððâ ðð Subtraction property of equality
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Lesson 11: Unknown Angle ProofsâProofs of Known Facts
2. Given: ððððï¿œâᅵᅵᅵâ ⥠ððððï¿œâᅵᅵᅵâ ; ð ð ð ð ï¿œâᅵᅵâ ⥠ððððï¿œâᅵᅵᅵâ
Prove: ððððï¿œâᅵᅵᅵâ ⥠ð ð ð ð ï¿œâᅵᅵâ
ð·ð·ð·ð·ï¿œâᅵᅵᅵâ ⥠ð»ð»ð»ð»ï¿œâᅵᅵᅵâ ; ð¹ð¹ð¹ð¹ï¿œâᅵᅵᅵâ ⥠ð»ð»ð»ð»ï¿œâᅵᅵᅵâ Given
ððâ ð·ð·ðœðœðžðž = ðððð°; ððâ ð¹ð¹ðžðžð»ð» = ðððð° Perpendicular lines form ðððð° angles.
ððâ ð·ð·ðœðœðžðž = ððâ ð¹ð¹ðžðžð»ð» Transitive property of equality
ð·ð·ð·ð· ⥠ð¹ð¹ð¹ð¹ If two lines are cut by a transversal such that a pair of corresponding angles are equal in measure, then the lines are parallel.
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Lesson 12: TransformationsâThe Next Level
GEOMETRY
Lesson 12: TransformationsâThe Next Level
1. Recall that a transformation ð¹ð¹ of the plane is a function that assigns to each point ðð of the plane a unique
point ð¹ð¹(ðð) 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 motionâa translation.
c.
Yes, this is a rigid motionâa reflection.
d.
No, this transformation did not preserve lengths or angle measures.
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e.
Yes, this is a rigid motionâa 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 triangle.
For each of the transformations, I must describe all the details that describe the âmechanicsâ 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.
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Lesson 12: TransformationsâThe Next Level
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The segment ð·ð·ð·ð· is rotated counterclockwise about ð©ð© by ðððð°. The path that describes how ð·ð· maps to ð·ð·â² is a circle with center ð©ð© and radius ð©ð©ð·ð·ï¿œï¿œï¿œï¿œ; ð·ð· moves counterclockwise along the circle. â ð·ð·ð©ð©ð·ð·â² has a measure of ðððð°. A similar case can be made for ð·ð·.
The pre-image â³ ðšðšð©ð©ðšðš has been translated the length and direction of vector ðšðšðšðšâ²ï¿œï¿œï¿œï¿œï¿œï¿œâ .
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Lesson 13: Rotations
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GEOMETRY
Lesson 13: Rotations
1. Recall the definition of rotation:
For 0° < ðð° < 180°, the rotation of ðð degrees around the center ð¶ð¶ is the transformation ð ð ð¶ð¶,ðð of the plane defined as follows:
1. For the center point ð¶ð¶, ð ð ð¶ð¶,ðð(ð¶ð¶) = ð¶ð¶, and
2. For any other point ðð, ð ð ð¶ð¶,ðð(ðð) is the point ðð that lies in the counterclockwise half-plane of ð¶ð¶ððᅵᅵᅵᅵᅵâ , such that ð¶ð¶ðð = ð¶ð¶ðð and ððâ ððð¶ð¶ðð = ðð°.
a. Which point does the center ð¶ð¶ map to once the rotation has been applied?
By the definition, the center ðªðª maps to itself: ð¹ð¹ðªðª,ðœðœ(ðªðª) = ðªðª.
b. The image of a point ðð that undergoes a rotation ð ð ð¶ð¶,ðð is the image point ðð: ð ð ð¶ð¶,ðð(ðð) = ðð. Point ðð is said to lie in the counterclockwise half plane of ð¶ð¶ððᅵᅵᅵᅵᅵâ . Shade the counterclockwise half plane of ð¶ð¶ððᅵᅵᅵᅵᅵâ .
I must remember that a half plane is a line in a plane that separates the plane into two sets.
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c. Why does part (2) of the definition include ð¶ð¶ðð = ð¶ð¶ðð? What relationship does ð¶ð¶ðð = ð¶ð¶ðð have with the circle in the diagram above?
ðªðªðªðª = ðªðªðªðª describes how ðªðª maps to ðªðª. The rotation, a function, describes a path such that ðªðª ârotatesâ (let us remember there is no actual motion) along the circle ðªðª with radius ðªðªðªðª (and thereby also of radius ðªðªðªðª).
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 â ðªðªðªðªðªðª, and the measure is ðœðœË.
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.
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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.
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Lesson 14: Reflections
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GEOMETRY
Lesson 14: Reflections
1. Recall the definition of reflection:
For a line ðð in the plane, a reflection across ðð is the transformation ðððð of the plane defined as follows:
1. For any point ðð on the line ðð, ðððð(ðð) = ðð, and 2. For any point ðð not on ðð, ðððð(ðð) is the point ðð
so that ðð is the perpendicular bisector of the segment ðððð.
a. Where do the points that belong to a line of reflection map to once the reflection is applied?
Any point ð·ð· on the line of reflection maps to itself: ðððð(ð·ð·) = ð·ð·.
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 ðŽðŽ, ðŽðŽâ², and line ðð?
Line ðð is the perpendicular bisector to the segment that joins ðšðš and ðšðšâ².
c. Based on the diagram above, is there a relationship between the distance from ðµðµ to ðð and from ðµðµâ² to ðð?
Any pair of corresponding points is equidistant from the line of reflection.
ðð
I can model a reflection by folding paper: The fold itself is the line of reflection.
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2. Using a compass and straightedge, determine the line of reflection for pre-image â³ ðŽðŽðµðµðŽðŽ and â³ ðŽðŽâ²ðµðµ
â²ðŽðŽâ².
Write the steps to the construction.
1. Draw circle ðªðª: center ðªðª, radius ðªðªðªðªâ².
2. Draw circle ðªðªâ²: center ðªðªâ², radius ðªðªâ²ðªðª.
3. Draw a line through the points of intersection between circles ðªðª and ðªðªâ².
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3. Using a compass and straightedge, reflect point ðŽðŽ over line ðð. Write the steps to the construction.
1. Draw circle ðšðš such that the circle intersects with line ðð in two locations, ðºðº and ð»ð».
2. Draw circle ðºðº: center ðºðº, radius ðºðºðšðš.
3. Draw circle ð»ð»: center ð»ð», radius ð»ð»ðšðš.
4. Label the intersection of circles ðºðº and ð»ð» as ðšðšâ².
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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 0° but less than 360° 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 ðð(ð¥ð¥) = ð¥ð¥). An example of this is a rotation of 360°. 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 360°)? What are the angles of rotation that yield symmetries?
ðð, including the identity symmetry. The angles of rotation are: ðððð°, ðððððð°, ðððððð°, ðððððð°, ðððððð°, and ðððððð°.
c. How many of the symmetries are reflections?
ðð
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Lesson 15: Rotations, Reflections, and Symmetry
d. How many places can vertex ðŽðŽ be moved by some symmetry of the hexagon?
ðšðš can be moved to ðð placesâðšðš, ð©ð©, ðªðª, ð«ð«, ð¬ð¬, and ðð.
e. For a given symmetry, if you know the image of ðŽðŽ, how many possibilities exist for the image of ðµðµ?
ðð
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.
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Lesson 16: Translations
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GEOMETRY
Lesson 16: Translations
1. Recall the definition of translation:
For vector ðŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ , the translation along ðŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ is the transformation ðððŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ of the plane defined as follows:
1. For any point ðð on ðŽðŽðŽðŽï¿œâᅵᅵᅵâ , ðððŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ (ðð) is the point ðð on ðŽðŽðŽðŽï¿œâᅵᅵᅵâ so that ððððᅵᅵᅵᅵᅵâ has the same length and the same direction as ðŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ , and
2. For any point ðð not on ðŽðŽðŽðŽï¿œâᅵᅵᅵâ , ðððŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ (ðð) is the point ðð obtained as follows. Let ðð be the line passing
through ðð and parallel to ðŽðŽðŽðŽï¿œâᅵᅵᅵâ . Let ðð be the line passing through ðŽðŽ and parallel to ðŽðŽððï¿œâᅵᅵᅵâ . The point ðð is the intersection of ðð and ðð.
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Lesson 16: Translations
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2. Use a compass and straightedge to translate segment ðºðºðºðºï¿œï¿œï¿œï¿œ along vector ðŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ .
To find ðºðºâ², I must mark off the length of ðŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ in the direction of the vector from ðºðº. I will repeat these steps to locate ðºðºâ².
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3. Use a compass and straightedge to translate point ðºðº along vector ðŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ . Write the steps to this construction.
1. Draw circle ð®ð®: center ð®ð®, radius ðšðšðšðš.
2. Draw circle ðšðš: center ðšðš, radius ðšðšð®ð®.
3. Label the intersection of circle ð®ð® and circle ðšðš as ð®ð®â². (Circles ð®ð® and ðšðš intersect in two locations; pick the intersection so that ðšðš and ð®ð®â² are in opposite half planes of ðšðšð®ð®ï¿œâᅵᅵᅵâ .)
To find ðºðºâ², my construction is really resulting in locating the fourth vertex of a parallelogram.
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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. ðŽðŽðŽðŽ = ðŽðŽâ²ðŽðŽâ² True ii. ðµðµðµðµâ² = ð¶ð¶ð¶ð¶â² False iii. ðŽðŽð¶ð¶ = ðŽðŽâ²ð¶ð¶â² True iv. ðµðµðŽðŽ = ðµðµâ²ðŽðŽâ² True v. ð¶ð¶ðŽðŽâ² = ð¶ð¶â²ðŽðŽ False
I can re-examine perpendicular bisectors (in regard to reflections) in Lesson 14 and rotations in Lesson 13.
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b. Based on the following rotation, which of the following statements must be true? i. ð¶ð¶ð¶ð¶â² = ðµðµðµðµâ² False ii. ðµðµð¶ð¶ = ðµðµâ²ð¶ð¶â² True
3. In the following figure, point ðµðµ is reflected across line ðð. c. What is the relationship between ðµðµ, ðµðµâ², and ðð?
Line ðð is the perpendicular bisector of ð©ð©ð©ð©â²ï¿œï¿œï¿œï¿œï¿œ.
d. What is the relationship between ðµðµ, ðµðµâ², and any point ðð on ðð?
ð©ð© and ð©ð©â² are equidistant from line ðð and therefore equidistant from any point ð·ð· on ðð.
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Lesson 18: Looking More Carefully at Parallel Lines
Lesson 18: Looking More Carefully at Parallel Lines
1. Given that â ðµðµ and â ð¶ð¶ are supplementary and ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ ⥠ðµðµð¶ð¶ï¿œï¿œï¿œï¿œ, prove that ððâ ðŽðŽ = ððâ ð¶ð¶.
â ð©ð© and â ðªðª are supplementary. Given
ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ð©ð©ðªðªï¿œï¿œï¿œï¿œ Given
â ð©ð© and â ðšðš are supplementary. If two parallel lines are cut by a transversal, then same-side interior angles are supplementary.
ððâ ð©ð© + ððâ ðšðš = ðððððð° Definition of supplementary angles
ððâ ð©ð© + ððâ ðªðª = ðððððð° Definition of supplementary angles
ððâ ð©ð© + ððâ ðšðš = ððâ ð©ð© + ððâ ðªðª Substitution property of equality
ððâ ðšðš = ððâ ðªðª 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.)
ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ðªðªðªðªï¿œï¿œï¿œï¿œ, ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ð®ð®ð®ð®ï¿œï¿œï¿œï¿œï¿œ Given
ððâ ðšðšðšðšðªðª = ðððð° Definition of perpendicular lines
ððâ ðšðšðšðšð®ð® = ðððð° Definition of perpendicular lines
ððâ ðšðšðšðšðªðª = ððâ ðšðšðšðšð®ð® Substitution property of equality
ðªðªðªðªï¿œï¿œï¿œï¿œ ⥠ð®ð®ð®ð®ï¿œï¿œï¿œï¿œï¿œ If a transversal cuts two lines such that corresponding angles are equal in measure, then the two lines are parallel.
If ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ ⥠ðµðµð¶ð¶ï¿œï¿œï¿œï¿œ, then â ðŽðŽ and â ðµðµ 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 90°. 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 90°.
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Lesson 18: Looking More Carefully at Parallel Lines
3. In the figure, ðŽðŽ and ð¹ð¹ lie on ðŽðŽðµðµï¿œï¿œï¿œï¿œ, ððâ ðµðµðŽðŽð¶ð¶ = ððâ ðŽðŽð¹ð¹ðŽðŽ, and ððâ ð¶ð¶ = ððâ ðŽðŽ. Prove that ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ ⥠ð¶ð¶ðµðµï¿œï¿œï¿œï¿œ.
ððâ ð©ð©ðšðšðªðª = ððâ ðšðšðªðªðšðš Given
ððâ ðªðª = ððâ ðšðš Given
ððâ ð©ð©ðšðšðªðª+ ððâ ðªðª+ ððâ ð©ð© = ðððððð° Sum of the angle measures in a triangle is ðððððð°.
ððâ ðšðšðªðªðšðš+ ððâ ðšðš + ððâ ðšðš = ðððððð° Sum of the angle measures in a triangle is ðððððð°.
ððâ ðšðšðªðªðšðš+ ððâ ðšðš + ððâ ð©ð© = ðððððð° Substitution property of equality
ððâ ð©ð© = ðððððð° âððâ ðšðšðªðªðšðšâððâ ðšðš Subtraction property of equality
ððâ ðšðš = ðððððð° âððâ ðšðšðªðªðšðš âððâ ðšðš Subtraction property of equality
ððâ ðšðš = ððâ ð©ð© Substitution property of equality
ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ðªðªð©ð©ï¿œï¿œï¿œï¿œ 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 180°. 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.
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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 canât 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 â³ ðŽðŽðŽðŽðŽðŽ 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 â³ ðŽðŽðŽðŽðŽðŽ â â³ ðŽðŽâ²â²â²ðŽðŽâ²â²â²ðŽðŽâ²â²â². Label the corresponding image points in the diagram using prime notation.
First, a rotation of ðððð° about point ðªðªâ²â²â² in a clockwise direction takes â³ ðšðšðšðšðªðª to â³ ðšðšâ²ðšðšâ²ðªðªâ². Next, a translation along ðªðªâ²ðªðªâ²â²â²ï¿œï¿œï¿œï¿œï¿œï¿œï¿œï¿œï¿œï¿œâ takes â³ ðšðšâ²ðšðšâ²ðªðªâ² to â³ ðšðšâ²â²ðšðšâ²â²ðªðªâ²â². Finally, a reflection over ðšðšâ²â²ðªðªâ²â²ï¿œï¿œï¿œï¿œï¿œï¿œï¿œ takes â³ ðšðšâ²â²ðšðšâ²â²ðªðªâ²â² to â³ ðšðšâ²â²â²ðšðšâ²â²â²ðªðªâ²â²â².
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 â³ ðŽðŽâ²â²â²ðŽðŽâ²â²â²ðŽðŽâ²â²â² is turned on the plane compared to â³ ðŽðŽðŽðŽðŽðŽ, so I should look for a rotation in my sequence. I also see a vector, so there might be a translation, too.
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Lesson 19: Construct and Apply a Sequence of Rigid Motions
3. In the diagram to the right, â³ ðŽðŽðŽðŽðŽðŽ â â³ ð·ð·ðŽðŽðŽðŽ. a. Describe two distinct rigid motions, or sequences of
rigid motions, that map ðŽðŽ onto ð·ð·.
The most basic of rigid motions mapping ðšðš onto ð«ð« is a reflection over ðšðšðªðªï¿œï¿œï¿œï¿œ.
Another possible sequence of rigid motions includes a rotation about ðšðš of degree measure equal to ððâ ðšðšðšðšð«ð« followed by a reflection over ðšðšð«ð«ï¿œï¿œï¿œï¿œï¿œ.
b. Using the congruence that you described in your response to part (a), what does ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ map to?
By a reflection of the plane over ðšðšðªðªï¿œï¿œï¿œï¿œ, ðšðšðªðªï¿œï¿œï¿œï¿œ maps to ð«ð«ðªðªï¿œï¿œï¿œï¿œ.
c. Using the congruence that you described in your response to part (a), what does ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ map to?
By a reflection of the plane over ðšðšðªðªï¿œï¿œï¿œï¿œ, ðšðšðªðªï¿œï¿œï¿œï¿œ 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.
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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 ðšðšðšðšðšðšðšðš and rectangle ð¬ð¬ð¬ð¬ð¬ð¬ð¬ð¬ 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 ðððððð°. If two triangles are given such that one pair of angles measure ðð° and a second pair of angles measure ðð°, then by the angle sum of a triangle, the remaining angle would have to have a measure of (ðððððð â ðð â ðð)°. 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 180°.
The term rigid means not flexible.
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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 ðªðªðªðªâ²ï¿œï¿œï¿œï¿œï¿œï¿œâ to yield triangle ðšðšâ²ð©ð©â²ðªðªâ². The second rigid motion is a reflection over ð©ð©ðªðªï¿œï¿œï¿œï¿œ to yield triangle ðšðšâ²â²ð©ð©â²â²ðªðªâ²â². In function notation, the
sequence of rigid motions is ððð©ð©ðªðªï¿œï¿œï¿œï¿œ ï¿œð»ð»ðªðªðªðªâ²ï¿œï¿œï¿œï¿œï¿œï¿œï¿œâ (â³ ðšðšð©ð©ðªðª)ï¿œ.
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: ðšðšð©ð©ï¿œï¿œï¿œï¿œ â ðšðšâ²â²ð©ð©â²â²ï¿œï¿œï¿œï¿œï¿œï¿œï¿œ, ð©ð©ðªðªï¿œï¿œï¿œï¿œ â ð©ð©â²â²ðªðªâ²â²ï¿œï¿œï¿œï¿œï¿œï¿œï¿œ, and ðšðšðªðªï¿œï¿œï¿œï¿œ â ðšðšâ²â²ðªðªâ²â²ï¿œï¿œï¿œï¿œï¿œï¿œï¿œ.
Sequence of corresponding angles: â ðšðš â â ðšðšâ²â², â ð©ð© â â ð©ð©â²â², and â ðªðª â â ðªðªâ²â².
c. Make a statement about the congruence of the pre-image and the final image.
â³ ðšðšð©ð©ðªðª â â³ ðšðšâ²â²ð©ð©â²â²ðªðªâ²â²
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Lesson 21: Correspondence and Transformations
2. Triangle ðððððð is a reflected image of triangle ðŽðŽðŽðŽðŽðŽ over a line â. Is it possible for a translation or a rotation to map triangle ðððððð back to the corresponding vertices in its pre-image, triangle ðŽðŽðŽðŽðŽðŽ? 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 ðšðš â ð»ð», ð©ð© â ð¹ð¹, and ðªðª â ðºðº, if the vertices ðšðš, ð©ð©, and ðªðª are oriented in a clockwise direction on the plane, then the vertices ð»ð», ð¹ð¹, and ðºðº will be oriented in a counterclockwise direction. It is possible to map ð»ð» to ðšðš, ðºðº to ðªðª, or ð¹ð¹ to ð©ð© individually under a variety of translations or rotations; however, a reflection is required in order to map each of ð»ð», ð¹ð¹, and ðºðº 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.
ðððð ï¿œðððŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ ï¿œðððð,60°(â³ ðððððð)ᅵᅵ
The first rigid motion is a rotation of â³ ð¿ð¿ð¿ð¿ð¿ð¿ around point ð¿ð¿ of ðððð°. Next is a translation along ðšðšð©ð©ï¿œï¿œï¿œï¿œï¿œï¿œâ . The final rigid motion is a reflection over a given line ðð.
I know that in function notation, the innermost function, in this case ðððð,60°(â³ ðððððð), 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.
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Lesson 22: Congruence Criteria for TrianglesâSAS
Lesson 22: Congruence Criteria for TrianglesâSAS
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: â³ ðððððð and â³ ððâ²ððâ²ððâ² so that ðððð = ððâ²ððâ² (Side), ððâ ðð = ððâ ððâ² (Angle), and ðððð = ððâ²ððâ² (Side).
Prove: â³ ðððððð â â³ ððâ²ððâ²ððâ²
1 Given, distinct triangles â³ ðððððð and â³ ððâ²ððâ²ððâ² so that ðððð = ððâ²ððâ², ððâ ðð = ððâ ððâ², and ðððð = ððâ²ððâ².
2 ð»ð»ð·ð·â²ð·ð·ï¿œï¿œï¿œï¿œï¿œï¿œï¿œï¿œâ (â³ð·ð·â²ðžðžâ²ð¹ð¹â²) =â³ ð·ð·ðžðžâ²â²ð¹ð¹â²â²; â³ ð·ð·â²ðžðžâ²ð¹ð¹â² is translated along vector ð·ð·â²ð·ð·ï¿œï¿œï¿œï¿œï¿œï¿œï¿œâ . â³ ð·ð·ðžðžð¹ð¹ and â³ ð·ð·ðžðžâ²â²ð¹ð¹â²â² share common vertex ð·ð·.
3 ð¹ð¹ð·ð·,âðœðœ(â³ð·ð·ðžðžâ²â²ð¹ð¹â²â²) = â³ ð·ð·ðžðžâ²â²â²ð¹ð¹; â³ð·ð·ðžðžâ²â²ð¹ð¹â²â² is rotated about center ð·ð· by ðœðœË clockwise. â³ð·ð·ðžðžð¹ð¹ and â³ð·ð·ðžðžâ²â²â²ð¹ð¹ share common side ð·ð·ð¹ð¹ï¿œï¿œï¿œï¿œ.
4 ððð·ð·ð¹ð¹ï¿œâᅵᅵᅵâ (â³ð·ð·ðžðžâ²â²â²ð¹ð¹) =â³ ð·ð·ðžðžð¹ð¹; â³ð·ð·ðžðžâ²â²â²ð¹ð¹ is reflected across ð·ð·ð¹ð¹ï¿œï¿œï¿œï¿œ. â³ ð·ð·ðžðžâ²â²â²ð¹ð¹ coincides with â³ð·ð·ðžðžð¹ð¹.
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a. In Step 3, how can we be certain that ðð" will map to ðð?
By assumption, ð·ð·ð¹ð¹â²â² = ð·ð·ð¹ð¹. This means that not only will ð·ð·ð¹ð¹â²â²ï¿œï¿œï¿œï¿œï¿œï¿œï¿œï¿œâ map to ð·ð·ð¹ð¹ï¿œï¿œï¿œï¿œï¿œï¿œâ under the rotation, but ð¹ð¹â²â² will map to ð¹ð¹.
b. In Step 4, how can we be certain that ððâ²â²â² will map to ðð?
Rigid motions preserve angle measures. This means that ððâ ðžðžð·ð·ð¹ð¹ = ððâ ðžðžâ²â²â²ð·ð·ð¹ð¹. Then the reflection maps ð·ð·ðžðžâ²â²â²ï¿œï¿œï¿œï¿œï¿œï¿œï¿œï¿œï¿œï¿œâ to ð·ð·ðžðžï¿œï¿œï¿œï¿œï¿œï¿œâ . Since ð·ð·ðžðžâ²â²â² = ð·ð·ðžðž, ðžðžâ²â²â² will map to ðžðž.
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 ðððð" and ðððð were not known, the rotation would result in coinciding rays ððððâ²â²ï¿œï¿œï¿œï¿œï¿œï¿œï¿œï¿œâ and ððððᅵᅵᅵᅵᅵâ but nothing further.
i. ii.
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2. Justify whether the triangles meet the SAS congruence criteria; explicitly state which pairs of sides or angles are congruent and why. If the triangles do meet the SAS congruence criteria, describe the rigid motion(s) that would map one triangle onto the other. a. Given: Rhombus ðŽðŽðŽðŽðŽðŽðŽðŽ
Do â³ ðŽðŽðððŽðŽ and â³ ðŽðŽðððŽðŽ meet the SAS criterion?
Rhombus ðšðšðšðšðšðšðšðš Given
ðšðšð¹ð¹ï¿œï¿œï¿œï¿œ and ðšðšðšðšï¿œï¿œï¿œï¿œï¿œ are perpendicular. Property of a rhombus
ððâ ðšðšð¹ð¹ðšðš = ððâ ðšðšð¹ð¹ðšðš All right angles are equal in measure.
ðšðšð¹ð¹ = ð¹ð¹ðšðš Diagonals of a rhombus bisect each other.
ðšðšð¹ð¹ = ðšðšð¹ð¹ Reflexive property
â³ ðšðšð¹ð¹ðšðš â â³ ðšðšð¹ð¹ðšðš SAS
One possible rigid motion that maps â³ ðšðšð¹ð¹ðšðš to â³ ðšðšð¹ð¹ðšðš is a reflection over the line ðšðšð¹ð¹ï¿œâᅵᅵᅵâ .
b. Given: Isosceles triangle â³ ABC with ðŽðŽðŽðŽ = ðŽðŽðŽðŽ and angle bisector ðŽðŽððᅵᅵᅵᅵ.
Do â³ ðŽðŽðŽðŽðð and â³ ðŽðŽðŽðŽðð meet the SAS criterion?
ðšðšðšðš = ðšðšðšðš Given
ðšðšð·ð·ï¿œï¿œï¿œï¿œ is an angle bisector Given
ðšðšð·ð· = ðšðšð·ð· Reflexive property
ððâ ðšðšðšðšð·ð· = ððâ ðšðšðšðšð·ð· Definition of angle bisector
â³ ðšðšðšðšð·ð· â â³ ðšðšðšðšð·ð· SAS
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Lesson 23: Base Angles of Isosceles Triangles
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GEOMETRY
Lesson 23: Base Angles of Isosceles Triangles
1. In an effort to prove that ððâ ðµðµ = ððâ ð¶ð¶ in isosceles triangle ðŽðŽðµðµð¶ð¶ by using rigid motions, the following argument is made to show that ðµðµ maps to ð¶ð¶:
Given: Isosceles â³ ðŽðŽðµðµð¶ð¶, with ðŽðŽðµðµ = ðŽðŽð¶ð¶
Prove: ððâ ðµðµ = ððâ ð¶ð¶
Construction: Draw the angle bisector ðŽðŽðŽðŽï¿œï¿œï¿œï¿œï¿œâ of â ðŽðŽ, where ðŽðŽ is the intersection of the bisector and ðµðµð¶ð¶ï¿œï¿œï¿œï¿œ. We need to show that rigid motions map point ðµðµ to point ð¶ð¶ and point ð¶ð¶ to point ðµðµ.
Since ðšðš is on the line of reflection, ðšðšðšðšï¿œâᅵᅵᅵâ , ðððšðšðšðšï¿œâᅵᅵᅵâ (ðšðš) = ðšðš. Reflections preserve angle measures, so the measure of the reflected angle ðððšðšðšðšï¿œâᅵᅵᅵâ (â ð©ð©ðšðšðšðš) equals the measure of â ðªðªðšðšðšðš; therefore, ðððšðšðšðšï¿œâᅵᅵᅵâ ï¿œðšðšð©ð©ï¿œï¿œï¿œï¿œï¿œï¿œâ ï¿œ = ðšðšðªðªï¿œï¿œï¿œï¿œï¿œâ . Reflections also preserve lengths of segments; therefore, the reflection of ðšðšð©ð©ï¿œï¿œï¿œï¿œ still has the same length as ðšðšð©ð©ï¿œï¿œï¿œï¿œ. By hypothesis, ðšðšð©ð© = ðšðšðªðª, so the length of the reflection is also equal to ðšðšðªðª. Then ðððšðšðšðšï¿œâᅵᅵᅵâ (ð©ð©) = ðªðª.
Use similar reasoning to show that ðððŽðŽðŽðŽï¿œâᅵᅵᅵâ (ð¶ð¶) = ðµðµ.
Again, we use a reflection in our reasoning. ðšðš is on the line of reflection, ðšðšðšðšï¿œâᅵᅵᅵâ , so ðððšðšðšðšï¿œâᅵᅵᅵâ (ðšðš) = ðšðš.
Since reflections preserve angle measures, the measure of the reflected angle ðððšðšðšðšï¿œâᅵᅵᅵâ (â ðªðªðšðšðšðš) equals the measure of â ð©ð©ðšðšðšðš, implying that ðððšðšðšðšï¿œâᅵᅵᅵâ ï¿œðšðšðªðªï¿œï¿œï¿œï¿œï¿œâ ï¿œ = ðšðšð©ð©ï¿œï¿œï¿œï¿œï¿œï¿œâ .
Reflections also preserve lengths of segments. This means that the reflection of ðšðšðªðªï¿œï¿œï¿œï¿œ, or the image of ðšðšðªðªï¿œï¿œï¿œï¿œ (ðšðšð©ð©ï¿œï¿œï¿œï¿œ), has the same length as ðšðšðªðªï¿œï¿œï¿œï¿œ. By hypothesis, ðšðšð©ð© = ðšðšðªðª, so the length of the reflection is also equal to ðšðšð©ð©. This implies ðððšðšðšðšï¿œâᅵᅵᅵâ (ðªðª) = ð©ð©. We conclude then that ððâ ð©ð© = ððâ ðªðª.
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 ð¶ð¶ to ðµðµ.
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Lesson 23: Base Angles of Isosceles Triangles
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GEOMETRY
2. Given: ððâ 1 = ððâ 2; ððâ 3 = ððâ 4
Prove: ððâ 5 = ððâ 6
ððâ ðð = ððâ ðð
ððâ ðð = ððâ ðð
Given
ðšðšðšðš = ð©ð©ðšðš
ðšðšðšðš = ðªðªðšðš
If two angles of a triangle are equal in measure, then the sides opposite the angles are equal in length.
ð©ð©ðšðš = ðªðªðšðš Transitive property of equality
ððâ ðð = ððâ ðð If two sides of a triangle are equal in length, then the angles opposite them are equal in measure.
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Lesson 23: Base Angles of Isosceles Triangles
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GEOMETRY
3. Given: Rectangle ðŽðŽðµðµð¶ð¶ðŽðŽ; ðœðœ is the midpoint of ðŽðŽðµðµï¿œï¿œï¿œï¿œ
Prove: â³ ðœðœð¶ð¶ðŽðŽ is isosceles
ðšðšð©ð©ðªðªðšðš is a rectangle. Given
ððâ ðšðš = ððâ ð©ð© All angles of a rectangle are right angles.
ðšðšðšðš = ð©ð©ðªðª Opposite sides of a rectangle are equal in length.
ð±ð± is the midpoint of ðšðšð©ð©ï¿œï¿œï¿œï¿œ. Given
ðšðšð±ð± = ð©ð©ð±ð± Definition of midpoint
â³ ðšðšð±ð±ðšðš â â³ ð©ð©ð±ð±ðªðª SAS
ð±ð±ðšðš = ð±ð±ðªðª Corresponding sides of congruent triangles are equal in length.
â³ ð±ð±ðªðªðšðš is isosceles. Definition of isosceles triangle
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Lesson 23: Base Angles of Isosceles Triangles
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GEOMETRY
4. Given: ð ð ð ð = ð ð ð ð ; ððâ 2 = ððâ 7
Prove: â³ ð ð ð ð ð ð is isosceles
ð¹ð¹ð¹ð¹ = ð¹ð¹ð¹ð¹ Given
â³ ð¹ð¹ð¹ð¹ð¹ð¹ is isosceles If at least two sides of a triangle are equal in length, the triangle isosceles.
ððâ ðð = ððâ ðð Base angles of an isosceles triangle are equal in measure.
ððâ ðð = ððâ ðð Given
ððâ ðð +ððâ ðð + ððâ ðð = ðððððð°
ððâ ðð +ððâ ðð + ððâ ðð = ðððððð°
The sum of angle measures in a triangle is ððððððË.
ððâ ðð +ððâ ðð + ððâ ðð = ððâ ðð + ððâ ðð+ ððâ ðð Substitution property of equality
ððâ ðð +ððâ ðð + ððâ ðð = ððâ ðð + ððâ ðð+ ððâ ðð Substitution property of equality
ððâ ðð = ððâ ðð Subtraction property of equality
ððâ ðð +ððâ ðð = ðððððð°
ððâ ðð +ððâ ðð = ðððððð°
Linear pairs form supplementary angles.
ððâ ðð +ððâ ðð = ððâ ðð+ ððâ ðð Substitution property of equality
ððâ ðð +ððâ ðð = ððâ ðð+ ððâ ðð Substitution property of equality
ððâ ðð = ððâ ðð Subtraction property of equality
ð¹ð¹ð¹ð¹ = ð¹ð¹ð¹ð¹ If two angles of a triangle are equal in measure, then the sides opposite the angles are equal in length.
â³ ð¹ð¹ð¹ð¹ð¹ð¹ is isosceles. Definition of isosceles triangle
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Lesson 24: Congruence Criteria for TrianglesâASA and SSS
M1
GEOMETRY
Lesson 24: Congruence Criteria for TrianglesâASA 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 ânone.â 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.
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Lesson 24: Congruence Criteria for TrianglesâASA and SSS
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GEOMETRY
2. ðŽðŽðŽðŽðŽðŽðŽðŽ 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: â³ ðšðšðšðšðšðš â â³ ðšðšðšðšðšðš, â³ ðšðšðšðšðšðš â â³ ð©ð©ðšðšðšðš, and â³ ðšðšðšðšð©ð© â â³ ðšðšðšðšð©ð©. All three pairs can be supported by SAS/SSS/ASA.
3. Given: ðð = ð ð and ðððð = ðððð Prove: ðð = ðð
ðð = ðð Given
ð·ð·ð·ð· = ðšðšð·ð· Given
ðð = ðð Vertical angles are equal in measure.
â³ ð·ð·ð·ð·ð·ð· â â³ ðšðšð·ð·ðžðž ASA
ð·ð·ð·ð· = ð·ð·ðžðž Corresponding sides of congruent triangles are equal in length.
â³ ð·ð·ð·ð·ðžðž is isosceles. Definition of isosceles triangle
ðð = ðð Base angles of an isosceles triangle are equal in measure.
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Lesson 24: Congruence Criteria for TrianglesâASA and SSS
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GEOMETRY
4. Given: Isoscelesâ³ ðŽðŽðŽðŽðŽðŽ; ðŽðŽðð = ðŽðŽðŽðŽ Prove: â ðŽðŽðððŽðŽ â â ðŽðŽðŽðŽðŽðŽ
Isosceles â³ ðšðšðšðšðšðš Given
ðšðšðšðš = ðšðšðšðš Definition of isosceles triangle
ðšðšð·ð· = ðšðšðšðš Given
ððâ ðšðš = ððâ ðšðš Reflexive property
â³ ðšðšðšðšðšðš â â³ ðšðšð·ð·ðšðš SAS
ð·ð·ðšðš = ðšðšðšðš Corresponding sides of congruent triangles are equal in length.
ðšðšð·ð· + ð·ð·ðšðš = ðšðšðšðš
ðšðšðšðš +ðšðšðšðš = ðšðšðšðš
Partition property
ðšðšð·ð· + ð·ð·ðšðš = ðšðšðšðš+ ðšðšðšðš Substitution property of equality
ðšðšð·ð· + ð·ð·ðšðš = ðšðšð·ð· + ðšðšðšðš Substitution property of equality
ð·ð·ðšðš = ðšðšðšðš Subtraction property of equality
ðšðšðšðš = ðšðšðšðš Reflexive property
â³ ð·ð·ðšðšðšðš â â³ðšðšðšðšðšðš SSS
â ðšðšð·ð·ðšðš â â ðšðšðšðšðšðš Corresponding angles of congruent triangles are congruent.
I must prove two sets of triangles are congruent in order to prove â ðŽðŽðððŽðŽ â â ðŽðŽðŽðŽðŽðŽ.
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GEOMETRY
Lesson 25: Congruence Criteria for TrianglesâAAS and HL
Lesson 25: Congruence Criteria for TrianglesâAAS 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.
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GEOMETRY
Lesson 25: Congruence Criteria for TrianglesâAAS and HL
4. Given: ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ â ð¶ð¶ðŽðŽï¿œï¿œï¿œï¿œ, ð¶ð¶ð¶ð¶ï¿œï¿œï¿œï¿œ ⥠ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ,ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ ⥠ð¶ð¶ðŽðŽï¿œï¿œï¿œï¿œ,
ð¶ð¶ is the midpoint of ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ.
ðŽðŽ is the midpoint of ð¶ð¶ðŽðŽï¿œï¿œï¿œï¿œ
Prove: â³ ðŽðŽðŽðŽð¶ð¶ â â³ ð¶ð¶ðŽðŽðŽðŽ
ðšðšðšðšï¿œï¿œï¿œï¿œ â ðªðªðšðšï¿œï¿œï¿œï¿œ Given
ðŒðŒ is the midpoint of ðšðšðšðšï¿œï¿œï¿œï¿œ.
ðœðœ is the midpoint of ðªðªðšðšï¿œï¿œï¿œï¿œ.
Given
ðšðšðšðš = ðððšðšðŒðŒ
ðªðªðšðš = ðððªðªðœðœ
Definition of midpoint
ðððšðšðŒðŒ = ðððªðªðœðœ Substitution property of equality
ðšðšðŒðŒ = ðªðªðœðœ Division property of equality
ðªðªðŒðŒï¿œï¿œï¿œï¿œ ⥠ðšðšðšðšï¿œï¿œï¿œï¿œ
ðšðšðœðœï¿œï¿œï¿œï¿œ ⥠ðªðªðšðšï¿œï¿œï¿œï¿œ
Given
ððâ ðšðšðŒðŒðšðš = ððâ ðªðªðœðœðšðš = ðððð° Definition of perpendicular
ððâ ðšðšðšðšðŒðŒ = ððâ ðªðªðšðšðœðœ Vertical angles are equal in measure.
â³ ðšðšðšðšðŒðŒ â â³ ðªðªðšðšðœðœ AAS
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GEOMETRY
Lesson 25: Congruence Criteria for TrianglesâAAS and HL
5. Given: ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ ⥠ðŽðŽðµðµï¿œï¿œï¿œï¿œ, ð¶ð¶ð¶ð¶ï¿œï¿œï¿œï¿œ ⥠ðŽðŽðµðµï¿œï¿œï¿œï¿œ,
ðŽðŽðŽðŽ = ðµðµð¶ð¶, ðŽðŽð¶ð¶ = ðµðµðŽðŽ
Prove: â³ ðŽðŽðŽðŽðŽðŽ â â³ ð¶ð¶ðµðµð¶ð¶
ðšðšðšðš = ð«ð«ðªðª Given
ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ðšðšð«ð«ï¿œï¿œï¿œï¿œï¿œ
ðªðªðªðªï¿œï¿œï¿œï¿œ ⥠ðšðšð«ð«ï¿œï¿œï¿œï¿œï¿œ
Given
ððâ ðšðšðšðšðšðš = ððâ ðªðªðªðªð«ð« = ðððð° Definition of perpendicular
ðšðšðªðª = ð«ð«ðšðš Given
ðšðšðªðª = ðšðšðšðš + ðšðšðªðª
ð«ð«ðšðš = ð«ð«ðªðª + ðªðªðšðš
Partition property
ðšðšðšðš + ðšðšðªðª = ð«ð«ðªðª+ ðªðªðšðš Substitution property of equality
ðšðšðšðš = ð«ð«ðªðª Subtraction property of equality
â³ ðšðšðšðšðšðš â â³ ðªðªð«ð«ðªðª HL
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Lesson 26: Triangle Congruency Proofs
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GEOMETRY
Lesson 26: Triangle Congruency Proofs
1. Given: ðŽðŽðŽðŽ = ðŽðŽðŽðŽ, ððâ ðŽðŽðŽðŽðŽðŽ = ððâ ðŽðŽðŽðŽðŽðŽ = 90°
Prove: ðŽðŽðŽðŽ = ðŽðŽðŽðŽ
ðšðšðšðš = ðšðšðšðš Given
ððâ ðšðšðšðšðšðš = ððâ ðšðšðšðšðšðš = ðððð° Given
ððâ ðšðš = ððâ ðšðš Reflexive property
â³ ðšðšðšðšðšðš â â³ ðšðšðšðšðšðš AAS
ðšðšðšðš = ðšðšðšðš Corresponding sides of congruent triangles are equal in length.
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Lesson 26: Triangle Congruency Proofs
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GEOMETRY
2. Given: ððððᅵᅵᅵᅵ bisects â ðððððð. ð·ð·ððᅵᅵᅵᅵ ⥠ððððᅵᅵᅵᅵ, ð·ð·ððᅵᅵᅵᅵ ⥠ððððᅵᅵᅵᅵ.
Prove: ð·ð·ðððððð is a rhombus.
ð»ð»ð»ð»ï¿œï¿œï¿œï¿œ bisects â ðºðºð»ð»ðºðº Given
ððâ ð«ð«ð»ð»ð»ð» = ððâ ððð»ð»ð»ð» Definition of bisect
ððâ ð«ð«ð»ð»ð»ð» = ððâ ððð»ð»ð»ð»
ððâ ððð»ð»ð»ð» = ððâ ð«ð«ð»ð»ð»ð»
If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
ð»ð»ð»ð» = ð»ð»ð»ð» Reflexive property
â³ ð«ð«ð»ð»ð»ð» â â³ ððð»ð»ð»ð» ASA
ððâ ð«ð«ð»ð»ð»ð» = ððâ ð«ð«ð»ð»ð»ð»
ððâ ððð»ð»ð»ð» = ððâ ððð»ð»ð»ð»
Substitution property of equality
â³ ð«ð«ð»ð»ð»ð» is isosceles; â³ ððð»ð»ð»ð» is isosceles
When the base angles of a triangle are equal in measure, the triangle is isosceles
ð«ð«ð»ð» = ð«ð«ð»ð»
ððð»ð» = ððð»ð»
Definition of isosceles
ð«ð«ð»ð» = ððð»ð»
ð«ð«ð»ð» = ððð»ð»
Corresponding sides of congruent triangles are equal in length.
ð«ð«ð»ð» = ððð»ð» = ð«ð«ð»ð» = ððð»ð» Transitive property
ð«ð«ð»ð»ððð»ð» is a rhombus. Definition of rhombus
I must remember that in addition to showing that each half of ð·ð·ðððððð is an isosceles triangle, I must also show that the lengths of the sides of both isosceles triangles are equal to each other, making ð·ð·ðððððð a rhombus.
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3. Given: ððâ 1 = ððâ 2
ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ ⥠ððððᅵᅵᅵᅵ, ðŽðŽð·ð·ï¿œï¿œï¿œï¿œ ⥠ððððᅵᅵᅵᅵ
ððð·ð· = ðððŽðŽ
Prove: â³ ðððððð is isosceles.
ððâ ðð = ððâ ðð Given
ððâ ðžðžðšðšðšðš = ððâ ð¹ð¹ðšðšð«ð« Supplements of angles of equal measure are equal in measure.
ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ðžðžð¹ð¹ï¿œï¿œï¿œï¿œ, ðšðšð«ð«ï¿œï¿œï¿œï¿œ ⥠ðžðžð¹ð¹ï¿œï¿œï¿œï¿œ Given
ððâ ðšðšðšðšðžðž = ððâ ðšðšð«ð«ð¹ð¹ = ðððð° Definition of perpendicular
ðžðžð«ð« = ð¹ð¹ðšðš Given
ðžðžð«ð« = ðžðžðšðš + ðšðšð«ð«
ð¹ð¹ðšðš = ð¹ð¹ð«ð« +ð«ð«ðšðš
Partition property
ðžðžðšðš + ðšðšð«ð« = ð¹ð¹ð«ð« + ð«ð«ðšðš Substitution property of equality
ðžðžðšðš = ð¹ð¹ð«ð« Subtraction property of equality
â³ ðšðšðžðžðšðš â â³ ðšðšð¹ð¹ð«ð« AAS
ððâ ðšðšðžðžðšðš = ððâ ðšðšð¹ð¹ð«ð« Corresponding angles of congruent triangles are equal in measure.
â³ ð·ð·ðžðžð¹ð¹ is isosceles. When the base angles of a triangle are equal in measure, then the triangle is isosceles.
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Lesson 27: Triangle Congruency Proofs
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GEOMETRY
Lesson 27: Triangle Congruency Proofs
1. Given: â³ ð·ð·ð·ð·ð·ð· and â³ ðºðºð·ð·ðºðº are equilateral triangles
Prove: â³ ð·ð·ðºðºð·ð· â â³ ð·ð·ðºðºð·ð·
â³ ð«ð«ð«ð«ð«ð« and â³ ð®ð®ð«ð«ð®ð® are equilateral triangles.
Given
ððâ ð«ð«ð«ð«ð«ð« = ððâ ð®ð®ð«ð«ð®ð® = ðððð° All angles of an equilateral triangle are equal in measure
ððâ ð«ð«ð«ð«ð®ð® = ððâ ð«ð«ð«ð«ð«ð«+ ððâ ð«ð«ð«ð«ð®ð®
ððâ ð«ð«ð«ð«ð®ð® = ððâ ð®ð®ð«ð«ð®ð®+ ððâ ð«ð«ð«ð«ð®ð®
Partition property
ððâ ð«ð«ð«ð«ð®ð® = ððâ ð®ð®ð«ð«ð®ð®+ ððâ ð«ð«ð«ð«ð®ð® Substitution property of equality
ððâ ð«ð«ð«ð«ð®ð® = ððâ ð«ð«ð«ð«ð®ð® Substitution property of equality
ð«ð«ð«ð« = ð«ð«ð«ð«
ð®ð®ð«ð« = ð®ð®ð«ð«
Property of an equilateral triangle
â³ ð«ð«ð®ð®ð«ð« â â³ð«ð«ð®ð®ð«ð« SAS
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GEOMETRY
2. Given: ð ð ð ð ð ð ð ð is a square. ðð is a point on ð ð ð ð ᅵᅵᅵᅵ, and ðð is on ð ð ð ð ï¿œâᅵᅵᅵâ such that ððð ð ᅵᅵᅵᅵᅵ ⥠ð ð ððᅵᅵᅵᅵ.
Prove: â³ ð ð ð ð ðð â â³ð ð ð ð ðð
ð¹ð¹ð¹ð¹ð¹ð¹ð¹ð¹ is a square. Given
ð¹ð¹ð¹ð¹ = ð¹ð¹ð¹ð¹ Property of a square
ððâ ð¹ð¹ð¹ð¹ð¹ð¹ = ððâ ð¹ð¹ð¹ð¹ð¹ð¹ = ðððð° Property of a square
ððâ ð¹ð¹ð¹ð¹ð¹ð¹+ ððâ ð¹ð¹ð¹ð¹ð¹ð¹ = ðððððð° Angles on a line sum to ðððððð°.
ððâ ð¹ð¹ð¹ð¹ð¹ð¹ = ðððð° Subtraction property of equality
ððâ ð¹ð¹ð¹ð¹ð¹ð¹ = ððâ ð¹ð¹ð¹ð¹ð¹ð¹ If parallel lines are cut by a transversal, then alternate interior angles are equal in measure.
ððâ ð¹ð¹ð¹ð¹ð¹ð¹ = ððâ ð¹ð¹ð¹ð¹ð¹ð¹ Complements of angles of equal measures are equal.
â³ ð¹ð¹ð¹ð¹ð¹ð¹ â â³ ð¹ð¹ð¹ð¹ð¹ð¹ ASA
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Lesson 27: Triangle Congruency Proofs
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GEOMETRY
3. Given: ðŽðŽðŽðŽðŽðŽð·ð· and ð·ð·ð·ð·ðºðºðºðº are congruent rectangles.
Prove: ððâ ð·ð·ð·ð·ð ð = ððâ ð·ð·ð·ð·ð ð
ðšðšðšðšðšðšð«ð« and ð«ð«ð«ð«ð®ð®ð®ð® are congruent rectangles.
Given
ð®ð®ð«ð« = ðšðšð«ð«
ð«ð«ðšðš = ð«ð«ð«ð«
Corresponding sides of congruent figures are equal in length.
ððâ ð®ð®ð«ð«ð«ð« = ððâ ðšðšð«ð«ðšðš Corresponding angles of congruent figures are equal in measure.
ððâ ð®ð®ð«ð«ðšðš = ððâ ð®ð®ð«ð«ð«ð« + ððâ ð·ð·ð«ð«ðšðš
ððâ ðšðšð«ð«ð«ð« = ððâ ðšðšð«ð«ðšðš+ ððâ ðšðšð«ð«ð·ð·
Partition property
ððâ ð®ð®ð«ð«ðšðš = ððâ ðšðšð«ð«ðšðš+ ððâ ð·ð·ð«ð«ðšðš
Substitution property of equality
ððâ ð®ð®ð«ð«ðšðš = ððâ ðšðšð«ð«ð«ð« Substitution property of equality
â³ ð®ð®ð«ð«ðšðš â â³ ðšðšð«ð«ð«ð« SAS
ððâ ð®ð®ðšðšð«ð« = ððâ ðšðšð«ð«ð«ð« Corresponding angles of congruent triangles are equal in measure.
ððâ ð·ð·ð«ð«ð·ð· = ððâ ð·ð·ð«ð«ð·ð· Reflexive property
â³ ð«ð«ð·ð·ðšðš â â³ ð«ð«ð·ð·ð«ð« ASA
ððâ ð«ð«ð·ð·ð¹ð¹ = ððâ ð«ð«ð·ð·ð¹ð¹ Corresponding angles of congruent triangles are equal in measure.
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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: â³ ðŽðŽðŽðŽðŽðŽ â â³ ð¶ð¶ðŽðŽðŽðŽ Prove: Quadrilateral ðŽðŽðŽðŽð¶ð¶ðŽðŽ is a parallelogram
Proof:
â³ ðšðšðšðšðšðš â â³ ðªðªðšðšðšðš Given
ððâ ðšðšðšðšðšðš = ððâ ðªðªðšðšðšðš; ððâ ðšðšðšðšðšðš = ððâ ðªðªðšðšðšðš
Corresponding angles of congruent triangles are equal in measure.
ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ðªðªðšðšï¿œï¿œï¿œï¿œ,ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ðšðšðªðªï¿œï¿œï¿œï¿œ If two lines are cut by a transversal such that alternate interior angles are equal in measure, then the lines are parallel.
Quadrilateral ðšðšðšðšðªðªðšðš is a parallelogram
Definition of parallelogram (A quadrilateral in which both pairs of opposite sides are parallel.)
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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: ðŽðŽðŽðŽ â ðŽðŽð¶ð¶;ðŽðŽðŽðŽ â ðŽðŽðŽðŽ Prove: Quadrilateral ðŽðŽðŽðŽð¶ð¶ðŽðŽ is a parallelogram
Proof:
ðšðšðšðš â ðªðªðšðš;ðšðšðšðš â ðšðšðšðš Given
ððâ ðšðšðšðšðšðš = ððâ ðªðªðšðšðšðš; ððâ ðšðšðšðšðšðš = ððâ ðªðªðšðšðšðš
Vertical angles are equal in measure.
â³ ðšðšðšðšðšðš â â³ ðªðªðšðšðšðš; â³ ðšðšðšðšðšðš â â³ ðªðªðšðšðšðš
SAS
ððâ ðšðšðšðšðšðš = ððâ ðªðªðšðšðšðš; ððâ ðšðšðšðšðšðš = ððâ ðªðªðšðšðšðš
Corresponding angles of congruent triangles are equal in measure
ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ðªðªðšðšï¿œï¿œï¿œï¿œ,ðšðšðšðšï¿œï¿œï¿œï¿œ ⥠ðšðšðªðªï¿œï¿œï¿œï¿œ If two lines are cut by a transversal such that alternate interior angles are equal in measure, then the lines are parallel
Quadrilateral ðšðšðšðšðªðªðšðš is a parallelogram
Definition of parallelogram (A quadrilateral in which both pairs of opposite sides are parallel)
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Lesson 28: Properties of Parallelograms
3. Given: Diagonals ðŽðŽð¶ð¶ï¿œï¿œï¿œï¿œ and ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ bisect each other;
â ðŽðŽðŽðŽðŽðŽ â â ðŽðŽðŽðŽðŽðŽ Prove: Quadrilateral ðŽðŽðŽðŽð¶ð¶ðŽðŽ is a rhombus
Proof:
Diagonals ðšðšðªðªï¿œï¿œï¿œï¿œ and ðšðšðšðšï¿œï¿œï¿œï¿œï¿œ bisect each other
Given
ðšðšðšðš = ðšðšðªðª;ðšðšðšðš = ðšðšðšðš Definition of a segment bisector
â ðšðšðšðšðšðš â â ðšðšðšðšðšðš Given
ððâ ðšðšðšðšðšðš = ððâ ðšðšðšðšðšðš = ððððË
ððâ ðšðšðšðšðªðª = ððâ ðšðšðšðšðªðª = ððððË
Angles on a line sum to ððððððË and since both angles are congruent, each angle measures ððððË
â³ ðšðšðšðšðšðš â â³ ðšðšðšðšðšðš â â³ ðªðªðšðšðšðš â â³ðªðªðšðšðšðš
SAS
ðšðšðšðš = ðšðšðªðª = ðªðªðšðš = ðšðšðšðš Corresponding sides of congruent triangles are equal in length
Quadrilateral ðšðšðšðšðªðªðšðš is a rhombus
Definition of rhombus (A quadrilateral with all sides of equal length)
In order to prove that ðŽðŽðŽðŽð¶ð¶ðŽðŽ 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.
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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 ðŽðŽðŽðŽð¶ð¶ðŽðŽ is a parallelogram.
4. Given: Parallelogram ðŽðŽðŽðŽð¶ð¶ðŽðŽ, â ðŽðŽðŽðŽðŽðŽ â â ð¶ð¶ð¶ð¶ðŽðŽ Prove: Quadrilateral ðŽðŽðŽðŽðŽðŽð¶ð¶ is a parallelogram
Proof:
Parallelogram ðšðšðšðšðªðªðšðš, â ðšðšðšðšðšðš â â ðªðªðªðªðšðš Given
ðšðšðšðš = ðšðšðªðª;ðšðšðšðš = ðªðªðšðš Opposite sides of parallelograms are equal in length.
ððâ ðšðš = ððâ ðªðª Opposite angles of parallelograms are equal in measure.
â³ ðšðšðšðšðšðš â â³ ðªðªðªðªðšðš AAS
ðšðšðšðš = ðªðªðªðª;ðšðšðšðš = ðªðªðšðš Corresponding sides of congruent triangles are equal in length.
ðšðšðšðš + ðšðšðšðš = ðšðšðšðš; Partition property
ðªðªðªðª+ ðªðªðšðš = ðªðªðšðš
ðšðšðšðš + ðšðšðšðš = ðªðªðªðª+ ðªðªðšðš Substitution property of equality
ðšðšðšðš + ðšðšðšðš = ðšðšðšðš + ðªðªðšðš Substitution property of equality
ðšðšðšðš = ðªðªðšðš Subtraction property of equality
Quadrilateral ðšðšðšðšðšðšðªðª is a parallelogram If both pairs of opposite sides of a quadrilateral are equal in length, the quadrilateral is a parallelogram
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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 1â4, all the segments within the triangles are midsegments.
1. ðð = ðððð ðð = ðððð ðð = ðððð.ðð
2.
ðð = ðððððð ðð = ðððð
The ðððððð° angle and the angle marked ðð° are corresponding angles; ðð = ðððððð. This means the angle measures of the large triangle are ðððð° (corresponding angles), ðððððð°, and ðð°; this makes ðð = ðððð because the sum of the measures of angles of a triangle is ðððððð°.
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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., â 1,â 2,â 3) to help identify the correspondences.
3. Find the perimeter, ðð, of the triangle.
ð·ð· = ðð(ðð) + ðð(ðððð) + ðð(ðððð) = ðððð
4. State the appropriate correspondences among the four congruent triangles within â³ ðŽðŽðŽðŽðŽðŽ.
â³ ðšðšð·ð·ðšðš â â³ð·ð·ð·ð·ð·ð· â â³ðšðšð·ð·ðžðž â â³ð·ð·ðšðšð·ð·
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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. ð¹ð¹ is the centroid of triangle ðŽðŽðŽðŽðŽðŽ. If the length of ðŽðŽð¹ð¹ï¿œï¿œï¿œï¿œ is 14, what is the length of median ðŽðŽðµðµï¿œï¿œï¿œï¿œ?
ðððð
(ð©ð©ð©ð©) = ð©ð©ð©ð©
ðððð
(ð©ð©ð©ð©) = ðððð
ð©ð©ð©ð© = ðððð
2. ðŽðŽ is the centroid of triangle ð ð ð ð ð ð . If ðŽðŽð¶ð¶ = 9 and ðŽðŽð ð = 13, what are the lengths of ð ð ð¶ð¶ï¿œï¿œï¿œï¿œ and ð ð ðµðµï¿œï¿œï¿œï¿œ?
ðððð
(ð¹ð¹ð¹ð¹) = ðªðªð¹ð¹
ðððð
(ð¹ð¹ð¹ð¹) = ðð
ð¹ð¹ð¹ð¹ = ðððð ðððð
(ðºðºð©ð©) = ðªðªðºðº
ðððð
(ðºðºð©ð©) = ðððð
ðºðºð©ð© =ðððððð
= ðððð.ðð
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Lesson 30: Special Lines in Triangles
ðµðµðŽðŽï¿œï¿œï¿œï¿œ and ðŸðŸðŽðŽï¿œï¿œï¿œï¿œ are the shorter and longer segments, respectively, along each of the medians they belong to.
3. ð ð ð¶ð¶ï¿œï¿œï¿œï¿œ,ðððµðµï¿œï¿œï¿œï¿œ, and ððððᅵᅵᅵᅵ are medians. If ð ð ð¶ð¶ = 18, ðððð = 12 and ð ð ðð = 17, what is the perimeter of â³ ðŽðŽð ð ðð?
ðªðªðªðª =ðððð
(ðªðªð¹ð¹) = ðððð
ðªðªðªðª =ðððð
(ðœðœðªðª) = ðð
ðªðªðªðª =ðððð
(ðªðªð»ð») = ðð.ðð
Perimeter (â³ ðªðªðªðªðªðª) = ðððð + ðð + ðð.ðð = ðððð.ðð
4. In the following figure, â³ ðµðµðŽðŽðŸðŸ is equilateral. If the perimeter of â³ ðµðµðŽðŽðŸðŸ is 18 and ðµðµ and ð¶ð¶ are midpoints of ðœðœðŸðŸï¿œï¿œï¿œ and ðœðœðœðœï¿œ respectively, what are the lengths of ðµðµðœðœï¿œï¿œï¿œï¿œ and ðŸðŸð¶ð¶ï¿œï¿œï¿œï¿œ?
If the perimeter of â³ ð©ð©ðªðªðð is ðððð, then ð©ð©ðªðª = ðððªðª = ðð. ðððð
(ð©ð©ðð) = ð©ð©ðªðª
ð©ð©ðð = ðð(ðð)
ð©ð©ðð = ðððð
ðððð
(ððð¹ð¹) = ðððªðª
ððð¹ð¹ =ðððð
(ðð)
ððð¹ð¹ = ðð
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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 ð¶ð¶ð¶ð¶ï¿œâᅵᅵᅵâ , I need it as a guide to find the intersection with ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ.
Lesson 31: Construct a Square and a Nine-Point Circle
1. Construct the midpoint of segment ðŽðŽðŽðŽ and write the steps to the construction.
1. Draw circle ðšðš: center ðšðš, radius ðšðšðšðš.
2. Draw circle ðšðš: center ðšðš, radius ðšðšðšðš.
3. Label the two intersections of the circles as ðªðª and ð«ð«.
4. Label the intersection of ðªðªð«ð«ï¿œâᅵᅵᅵâ with ðšðšðšðšï¿œï¿œï¿œï¿œ as midpoint ðŽðŽ.
2. Create a copy of ðŽðŽðŽðŽï¿œï¿œï¿œï¿œ and label it as ð¶ð¶ð¶ð¶ï¿œï¿œï¿œï¿œ and write the steps to the construction. 1. Draw a segment and label one endpoint ðªðª. 2. Mark off the length of ðšðšðšðšï¿œï¿œï¿œï¿œ along the drawn segment; label the marked point as ð«ð«.
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Lesson 31: Construct a Square and a Nine-Point Circle
I must remember that the intersections ðð or ðð may lie outside the triangle, as shown in the example.
3. Construct the three altitudes of â³ ðŽðŽðŽðŽð¶ð¶ and write the steps to the construction. Label the orthocenter
as ðð.
1. Draw circle ðšðš: center ðšðš, with radius so that circle ðšðš intersects ðšðšðªðªï¿œâᅵᅵᅵâ in two points; label these points as ð¿ð¿ and ðð.
2. Draw circle ð¿ð¿: center ð¿ð¿, radius ð¿ð¿ðð.
3. Draw circle ðð: center ðð, radius ððð¿ð¿.
4. Label either intersection of circles ð¿ð¿ and ðð as ðð.
5. Label the intersection of ðšðšððï¿œâᅵᅵᅵâ with ðšðšðªðªï¿œâᅵᅵᅵâ as ð¹ð¹ (this is altitude ðšðšð¹ð¹ï¿œï¿œï¿œï¿œ)
6. Repeat steps 1-5 from vertices ðšðš and ðªðª.
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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 ðŽðŽðŽðŽ.
2. Construct the circle that circumscribes â³ ðŽðŽðŽðŽðŽðŽ. Label the center of the circle as ðð.
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Lesson 32: Construct a Nine-Point Circle
3. Construct a square ðŽðŽðŽðŽðŽðŽðŽðŽ based on the provided segment ðŽðŽðŽðŽ.
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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 ðŽðŽ, ðµðµ, and ð¶ð¶ are collinear. ðŽðŽðµðµ = 1.5 and ðµðµð¶ð¶ = 3. What is the length of ðŽðŽð¶ð¶ï¿œï¿œï¿œï¿œ, and what assumptions do we make in answering this question?
ðšðšðªðª = ðð.ðð. The Distance and Ruler Axioms.
2. Find the angle measures marked ð¥ð¥ and ðŠðŠ and justify the answer with the facts that support your reasoning.
ðð = ðððð°, ðð = ðððð°
The angle marked ðð and ðððððð° are a linear pair and are supplementary. The angle vertical to ðð has the same measure as ðð, and the sum of angle measures of a triangle is ðððððð°.
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.
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Lesson 33: Review of the Assumptions
I must remember that this is referred to as âangles at a pointâ.
When parallel lines are intersected by a transversal, I must look for special angle pair relationships.
4. Find the measures of angle ð¥ð¥.
The sum of the measures of all adjacent angles formed by three or more rays with the same vertex is ðððððð°.
ðð(ðððð°) + ðð(ðððð°) + ðð(ðððð°) + ðð(ðð) = ðððððð°
ðððð° + ðððððð° + ðððððð° + ðððð = ðððððð°
ðð = ðð°
5. Find the measures of angles ð¥ð¥ and ðŠðŠ.
ðð = ðððð°, ðð = ðððð°
â ð¹ð¹ð¹ð¹ð¹ð¹ and â ð¹ð¹ð¹ð¹ðžðž are same side interior angles and are therefore supplementary. â ð¹ð¹ð¹ð¹ð¹ð¹ is vertical to â ðŽðŽð¹ð¹ðŽðŽ and therefore the angles are equal in measure. Finally, the angle sum of a triangle is ðððððð°.
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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, â³ ðšðšðšðšðšðš and â³ ðšðšâ²ðšðšâ²ðšðšâ²:
If ðšðšðšðš = ðšðšâ²ðšðšâ² (Side), ððâ ðšðš = ððâ ðšðšâ² (Angle), ðšðšðšðš = ðšðšâ²ðšðšâ²(Side), then the triangles are congruent. (SAS)
If ððâ ðšðš = ððâ ðšðšâ² (Angle), ðšðšðšðš = ðšðšâ²ðšðšâ² (Side), and ððâ ðšðš = ððâ ðšðšâ² (Angle), then the triangles are congruent. (ASA)
If ðšðšðšðš = ðšðšâ²ðšðšâ² (Side), ðšðšðšðš = ðšðšâ²ðšðšâ² (Side), and ðšðšðšðš = ðšðšâ²ðšðšâ² (Side), then the triangles are congruent. (SSS)
If ðšðšðšðš = ðšðšâ²ðšðšâ² (Side), ððâ ðšðš = ððâ ðšðšâ² (Angle), and â ðšðš = â ðšðšâ² (Angle), then the triangles are congruent. (AAS)
Given two right triangles, â³ ðšðšðšðšðšðš and â³ ðšðšâ²ðšðšâ²ðšðšâ², with right angles â ðšðš and â ðšðšâ², if ðšðšðšðš = ðšðšâ²ðšðšâ² (Leg) and ðšðšðšðš = ðšðšâ²ðšðšâ² (Hypotenuse), then the triangles are congruent. (HL)
© 2015 Great Minds eureka-math.orgGEO-M1-HWH-1.1.0-07.2015
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Homework Helper A Story of Functions
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, ðºðºðºðºï¿œï¿œï¿œï¿œ is a midsegment. Find ð¥ð¥ and ðŠðŠ. Determine the perimeter of â³ ðŽðŽðºðºðºðº.
ðð = ðððð°, ðð = ðððð°
Perimeter of â³ ðšðšðšðšðšðš is:
ðððð
(ðððð) +ðððð
(ðððð) + ðððð = ðððð.ðð
3. In the following figure, ðºðºðºðºðºðºðºðº and ðºðºðœðœðœðœðœðœ are squares and ðºðºðºðº = ðºðºðœðœ. Prove that ð ð ðºðºï¿œï¿œï¿œï¿œâ 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 ðð:ðð.
Proof: ðšðšðšðšð®ð®ð®ð® and ð®ð®ð±ð±ð±ð±ð±ð± are squares and ðšðšð®ð® = ð®ð®ð±ð±
Given
â ðšðš and â ð±ð± are right angles All angles of a square are right angles.
â³ ðšðšð®ð®ð®ð® and â³ð±ð±ð®ð®ð®ð® are right triangles
Definition of right triangle.
ð®ð®ð®ð® = ð®ð®ð®ð® Reflexive Property
â³ ðšðšð®ð®ð®ð® â â³ð±ð±ð®ð®ð®ð® HL
ððâ ðšðšð®ð®ð®ð® = ððâ ð±ð±ð®ð®ð®ð® Corresponding angles of congruent triangles are equal in measure.
ð®ð®ð®ð®ï¿œï¿œï¿œï¿œâ is an angle bisector Definition of angle bisector.
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Homework Helper A Story of Functions