Geometrymrboydsmathsite.weebly.com/uploads/8/6/7/8/8678489/properties_of...supplementary to an interior angle. And, that the sum of the measures of the interior angle of a ... Multiplication

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Geometry

Quadrilaterals

2016-03-23

www.njctl.org

http://www.njctl.org

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Table of ContentsPolygonsProperties of Parallelograms

Proving Quadrilaterals are Parallelograms

Constructing Quadrilaterals

Rhombi, Rectangles and Squares

Trapezoids

Kites

Proofs

Click on a topic to go to that section.

Families of Quadrilaterals

PARCC Released Questions & Standards

Coordinate Proofs

4

Polygons

Return to the Table of Contents

5

A polygon is a closed figure made of line segments connected end to end at vertices.

Since it is made of line segments, none its sides are curved.

The sides of simple polygons intersect each other at the common endpoints and do not extend beyond the endpoints.

The sides of complex polygons intersect each other and extend to intersect with other sides.

However, we will only be considering simple polygons in this course and will refer to them as "polygons."

Click to Reveal

Polygons

Would a circle be considered a polygon? Why or why not?

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6

Simple Polygons Complex Polygons

Click to Reveal

This course will only consider simple polygons, which we will refer to as "polygons."

Polygons

7

Types of PolygonsPolygons are named by their number of sides, angles or

vertices; which are all equal in number.

Number of Sides, Angles or Vertices Type of Polygon

3 triangle4 quadrilateral5 pentagon6 hexagon7 heptagon8 octagon9 nonagon10 decagon11 11-gon12 dodecagonn n-gon

8

interiorConvex Polygons

All interior angles are less than 180º.

No side, if extended, will cross into the interior.

interior

Convex or Concave Polygons

9

interior

Concave Polygons

At least one interior angle is at least 180º.

At least two sides, if extended, will cross into

the interior.

A way to remember the difference is that concave

shapes "cave in".

Convex or Concave Polygons

interior

10

1 The figure below is a concave polygon.

True

False

Ans

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11

2 Indentify the polygon.

A Pentagon

B Octagon

C Quadrilateral

D Hexagon

E Decagon

F Triangle

Ans

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12

3 Is the polygon convex or concave?

A Convex

B Concave

Ans

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4 Is the polygon convex or concave?

A ConvexB Concave

Ans

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14

A polygon is equilateral if all its sides are congruent.

Equilateral Polygons

15

A polygon is equiangular if all its angles are congruent.

Equiangular Polygons

16

A polygon is regular if it is equilateral and equiangular.

Regular Polygons

17

5 Describe the polygon. (Choose all that apply)

A PentagonB OctagonC QuadrilateralD HexagonE TriangleF ConvexG ConcaveH EquilateralI EquiangularJ Regular 4

60o

60o

60o

44

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E, F, H, I, J

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A PentagonB OctagonC QuadrilateralD HexagonE TriangleF ConvexG ConcaveH EquilateralI EquiangularJ Regular

Ans

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A PentagonB OctagonC QuadrilateralD HexagonE TriangleF ConvexG ConcaveH EquilateralI EquiangularJ Regular

Ans

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C, F, I

20

Angle Measures of Polygons

In earlier units, we spent a lot of time studying one specific polygon: the triangle.

We won't have to spend that same amount of time on each new polygon we study.

Instead, we will be able to use what we learned about triangles to save a lot of time and effort.

21


Every polygon is comprised of a number of triangles.

The sum of the interior angles of each triangle is 180°.

A triangle is (obviously) comprised of one triangle, and

therefore the sum of its interior angles is 180°.

What is the sum of the interior angles of this polygon?

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22


Every polygon can be thought of as consisting of some number of triangles.

A quadrilateral is comprised of two triangles.

180°

What is the sum of the interior angles of a quadrilateral?180°

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A pentagon is comprised of three triangles.

What is the sum of the interior angles

of a pentagon?180º

180º

180º

Ans

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An octagon is comprised of six triangles.

180º180º180º

180º

180º

180º What is the sum of the interior angles

of an octagon?

Ans

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25


Fill in the final cell of this table with an expression for the number of triangular regions of a polygon based on the number of its sides.

Number of sides Number of triangular regions

3 1

4 2

5 3

6 4

n

Mat

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This slide addresses MP7 & MP8.

Additional Q's to address MP standards:How are the number of triangles contained in a polygon related to the number of sides in the polygon? (MP7)What generalization/formula can you make? (MP8)

26

Angle Measures of PolygonsThe sum of the interior angles of a triangle is 180 degrees.

This will also be true of each triangular region of a polygon.

Using that information, complete this table.

Number of sides Number of triangular regions Sum of Interior Angles

3 1 180º

4 2 360º

5 3 540º

6 4

n

Mat

h Pr

actic

e

This slide addresses MP7 & MP8.

Additional Q's to address MP standards:How is the sum of the interior angles related to the number of sides in the polygon? (MP7)What generalization/formula can you make? (MP8)

27

8 What is the sum of the interior angles of this polygon?

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9 How many triangular regions can be drawn in this polygon?

Ans

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11 How many triangular regions can be drawn in this polygon?

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31


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32

13 Which of the following road signs has interior angles that add up to 360°?

A Yield Sign

B No Passing Sign

C National Forest Sign

D Stop Sign

E No Crossing Sign

Ans

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33

The sum of the measures of the interior angles of a convex polygon with n sides is 180(n-2).

Complete the table.

Polygon Number of Sides

Sum of the measures of the interior angles.

hexagon 6 180(6-2) = 720°

heptagon 7

octagon 8

nonagon 9

decagon 10

11-gon 11

dodecagon 12

Polygon Interior Angles Theorem Q1

34

14 What is the measurement of each angle in the diagram below?

L M

O

x°

(3x)°

146°

(2x+3)°

(3x+4)°

P

N

Ans

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(3x+4) + 146 + x + (3x) + (2x+3) = 5409x + 153 = 540

9x = 387 x = 43

35

Interior Angles of Equiangular Polygons

Regular polygons are equiangular: their angles are equal.

So, it must be true that each angle must be equal to the sum of the interior angles divided by the number of angles.

Sum of the AnglesNumber of Anglesm∠ =

(n - 2)180º n

m∠ =

36

The measure of each interior angle of an regular polygon is given by:

regular polygon number of sides

sum of interior angles

measure of each angle

triangle 3 180°

quadrilateral 4 360°

pentagon 5 540°

hexagon 6 720°

octagon 8 1080°

decagon 10 1440°

(n-2)(180°)

Interior Angles of Regular Polygons

(n-2)180º nm∠ =

37

15 If the stop sign is a regular octagon, what is the measure of each interior angle?

Ans

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135°

38

16 What is the sum of the measures of the interior angles of a convex 20-gon?

A 2880°

B 3060°

C 3240°

D 3420°

Ans

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C

39

17 What is the measure of each interior angle of a regular 20-gon?

A 162°

B 3240°

C 180°

D 60° Ans

wer

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18 What is the measure of each interior angle of a regular 16-gon?

A 2520°

B 2880°

C 3240°

D 157.5° Ans

wer

41

19 What is the value of x?

(5x+15)°

(9x-6)°

(8x)°

(11x+16)°

(10x+8)°

Ans

wer

42

1

23

4

5

The sum of the measures of the exterior angles of a convex polygon, one at each vertex, is 360°.

In this case, m∠1 + m∠2 + m∠3 + m∠4 + m∠5 = 360°

Polygon Exterior Angle Theorem

43

Proof of Exterior Angles TheoremGiven: Any polygon has n exterior angles, where n is the number of its sides, interior angles, or vertices

Prove: The sum of those exterior angles is 360°

We will prove this for any polygon, not for a specific polygon.

To do that, it will help if we introduce one piece of notation: the Greek letter Sigma, Ʃ.

Ʃ represents the phrase "the sum of." So, the phrase,

"the sum of the interior angles" becomes "Ʃ interior ∠s."

Mat

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44

Proof of Exterior Angles TheoremGiven: Any polygon has n interior angles, where n is the number of its sides, interior angles, or vertices

Prove: Ʃ external ∠s = 360°

This proof is based on the fact that each exterior angle is supplementary to an interior angle.And, that the sum of the measures of the interior angle of a polygon with n sides is (n-2) x 180°,

Or, with our Sigma notation: Ʃ internal ∠s = (n-2) x 180°

45

1external

23

4

5

1internal

External angles are formed by extending one side of a polygon.

Take a look at ∠1 in the diagram.

The external and internal angles at a vertex form a linear pair.

What is the sum of their measures?


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Any polygon with n sides has n vertices and n internal ∠s.

That also means if we include one external ∠ at each vertex, that there are n linear pairs.Since 1 linear pair has a measure of 180°, then n of them must have a measure of n x 180°.The sum of the measures of a polygon's linear pairs of internal and external ∠s must be n x 180°.


47

Each linear pair includes an external ∠ and an internal ∠. Adding up all of those means that:

The sum of all the external and internal ∠s of a polygon yields a measure of n x 180°.

Symbolically:

Ʃ external ∠s + Ʃ internal ∠s = n x 180°


48

Ʃ external∠s + Ʃ internal ∠s = n x 180°

But recall that the internal angle theorem tell us that:

Ʃ internal ∠s = (n-2) x 180°

Substituting that in yields

Ʃ external∠s + [(n-2) x 180°] = n x 180°

Multiplying out the second term yields:


49

Ʃ external∠s + [n x 180° - 2 x 180°] = n x 180°

Subtracting n x 180° - 2 x 180° from both sides yields

Ʃ external∠s = 2 x 180°Ʃ external∠s = 360°

Notice that n is not in this formula, so this result holds for all polygons with any number of sides.This is an important, and surprising, result.


50

Statement Reason

1 Any polygon has n interior ∠s Given

2 Each interior ∠ and exterior ∠ are a linear pair

Given in diagram

3 Each interior ∠ is supplementary to an exterior ∠

Linear pair angles are supplementary

4 The sum of the measures of each linear pair is 180°

Definition of supplementary

5 There are n linear pairs Substitution

6 The sum of the linear pairs = n x 180° = sum of exterior ∠s + interior ∠s

Multiplication Property

Proof of Exterior Angles TheoremGiven: Any polygon has n interior∠s, one at each vertex, where n is the number of sides, vertices or angles

Prove: The sum of those n exterior ∠s = 360°

51

Statement Reason

7 The sum of interior ∠s is (n - 2) x 180° Interior Angle Theorem

8 n x 180° = sum of exterior s + (n - 2) x 180° Substitution Property

9 The sum of the exterior ∠s = n x 180° - [(n - 2) x 180°]

Subtraction Property

10 The sum of the exterior ∠s = 180°[n - (n - 2)]

Subtraction Property

11 The sum of the exterior ∠s = 180°(n - n + 2)

Distributive Property

12 The sum of the exterior ∠s = 180°(2) = 360°

Simplifying

Proof of Exterior Angles TheoremGiven: Any polygon has n interior∠s, one at each vertex, where n is the number of sides, vertices or angles

Prove: The sum of those n exterior ∠s = 360°

52

The measure of each exterior angle of a regular polygon with n sides is 360º/n

Polygon Exterior Angle Theorem Corollary

The sum of the external angles of any polygon is 360º.

The number of external angles of a polygon equals n.

All angles are equal in a regular polygon, so their external angles must be as well.

So, each external angle of a regular polygon equals 360º/n.

53

The measure of each exterior angle of a regular polygon with n sides is 360º/n

a

This is a regular hexagon.

n = 6

a = 360 6

a = 60°

Polygon Exterior Angle Theorem Corollary

54

20 What is the sum of the measures of the exterior angles of a heptagon?

A 180°

B 360°

C 540°

D 720°

Ans

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21 If a heptagon is regular, what is the measure of each exterior angle? A 72°

B 60°

C 51.43°

D 45° Ans

wer

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22 What is the sum of the measures of the exterior angles of a pentagon?

Ans

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57

23 What is the measure of each exterior angle of a regular pentagon?

Ans

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58

24 The measure of each interior angle of a regular convex polygon is 172°. How many sides does the polygon have?

Answ

er

59

25The measure of each interior angle of a regular convex polygon is 174°. Find the number of sides of the polygon. A 64

B 62

C 58

D 60

Ans

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60

26 The measure of each interior angle of a regular convex polygon is 162°. Find the number of sides of the polygon.

Ans

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Properties of Parallelograms


62

Lab - Quadrilaterals Party

Lab - Investigating Parallelograms

Click on the link below and complete the two labs before the Parallelogram lesson.

Teac

her N

otes

http://www.alschappell.net/QuadrilateralsPartyWebQuest/QuadrilateralsPartyWebQuest/Introduction.html

http://www.alschappell.net/QuadrilateralsPartyWebQuest/QuadrilateralsPartyWebQuest/Introduction.html

http://njctl.org/courses/math/geometry-2015-16/quadrilaterals/investigating-parallelograms-lab/

http://njctl.org/courses/math/geometry-2015-16/quadrilaterals/investigating-parallelograms-lab/

63

By definition, the opposite sides of a parallelogram are parallel.

D E

G FIn parallelogram DEFG,

DG ?

Parallelograms

?DE

Ans

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64

Theorem

A B

CD

If ABCD is a parallelogram,

then AB ≅ DC and DA ≅ CB

The opposite sides a parallelogram are congruent.

Mat

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Proof of Theorem

A B

CDConstruct diagonal AC as shown and consider it a transversal to sides AD and BC, which we know are ║.

What does that tell about the pairs of angles:∠ACD & ∠CAB and ∠ACB & ∠CAD

Why?

The opposite sides a parallelogram are congruent.Since we are proving that this theorem is true, we are not

using it directly in the proof. Instead, we need to use other properties/theorems that we know.

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66

Given that ∠ACD ≅ ∠CAB and ∠ACB ≅ ∠CADName a pair of congruent ΔsFor what reason are they congruent?


A B

CD

Proof of Theorem

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67

Given that ΔACD ≅ ΔCABWhich pairs of sides are must be congruent?Why?


A B

CD

Proof of Theorem

Ans

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68

27 The opposite sides of a parallelogram are

A Congruent

B Parallel

C Both a and b Ans

wer

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A

CD

B

The opposite angles of a parallelogram are congruent.

Theorem

If ABCD is a parallelogram, then

m∠A = m∠Cand

m∠B = m∠D

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70

A B

CD

The opposite angles of a parallelogram are congruent.

Proof of Theorem

As we did earlier, construct diagonal AC.

We earlier proved ΔACD ≅ ΔCAB

What does this tell you about ∠B and ∠D?

Why?

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28 ΔDAB ≅ ΔBCD. Prove your answer.

True

False

A B

CD

≅

Ans

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72

29 Based on the previous question, m∠A = ?

Prove your answer. A m∠B

B m∠C

C m∠D

D None of the above

∠

A B

CD

Ans

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B

73

The consecutive angles of a parallelogram are supplementary.

y°

x°

x°

y°A

CD

B

If ABCD is a parallelogram, then xo + yo = 180o

Theorem

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y°

x°

x°

y°A

CD

B

Consider AD and BC to be transversals of AB and DC, which are ║.

What are these pairs of angles called: ∠A & ∠D and ∠B & ∠C?

What is the relationship of each pair of angles like that?

Proof of Theorem

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y°

x°

x°

y°A

CD

B

Repeat the same process, but this time consider AB and DC to be transversals of AD and BC, which are ║.

What are these pairs of angles called: ∠A & ∠B and ∠D & ∠C?

What is the relationship of each pair of angles like that?

Proof of Theorem

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A B

CD

12

2y

x-5

9

65°

(5z)°

w°

30 ABCD is a parallelogram. Find the value of x.

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A B

CD

12

2y

x-5

9

65°

(5z)°

w°

31 ABCD is a parallelogram. Find the value of y.

Ans

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78

A B

CD

12

2y

x-5

9

65°

(5z)°

w°

32 ABCD is a parallelogram. Find the value of z.

Ans

wer

79

33 ABCD is a parallelogram. Find the value of w.

A B

CD

12

2y

x-5

9

65°

(5z)°

w°

Ans

wer

80

34 DEFG is a parallelogram. Find the value of w.

D E

FG

70°

15

3x - 3(2w)° (z + 12)°

21

y2

Ans

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81

D E

FG

70°

15

3x - 3(2w)° (z + 12)°

21

y2

35 DEFG is a parallelogram. Find the value of x.

Ans

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8

82

D E

FG

70°

15

3x - 3(2w)° (z + 12)°

21

y2

36 DEFG is a parallelogram. Find the value of y.

Ans

wer

30

83

D E

FG

70°

15

3x - 3(2w)° (z + 12)°

21

y2

37 DEFG is a parallelogram. Find the value of z.

Ans

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84

The diagonals of a parallelogram bisect each other.

A B

CD

E

If ABCD is a parallelogram,

then AE ≅ EC and BE ≅ ED

Theorem

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85

The diagonals of a parallelogram bisect each other.

A B

CD

E

Proof of Theorem

Steps to follow:

1) Identify pairs of triangles in the above diagram that you need to prove congruent to prove this theorem.

2) Then, using what you know about the relationships between pairs of angles and pairs of sides in a parallelogram and the properties of the angles formed by parallel lines & transversals, prove that the triangles are congruent.

3) Then prove the theorem by stating that corresponding parts of congruent triangles are congruent (CPCTC).

Since we are proving that this theorem is true, we are not using it directly in the proof. Instead, we need to use other

properties/theorems that we know.

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Statements; ReasonsABCD is a ; GivenAB || DC, AD || BC; Def. ofAB ≅ CD, AD ≅ CB; Opp. sides of a are ≅∠AEB ≅ ∠CED, ∠AED ≅ ∠CEB; Vertical ∠s are ≅∠BAE ≅ ∠DCE, ∠DAE ≅ ∠BCE; Alt. Int ∠s are ≅ΔBAE ≅ ΔDCE, ΔDAE ≅ ΔBCE; AAS Δ ≅AE ≅ CE, DE ≅ BE; CPCTC

86

38 LMNP is a parallelogram. The diagonals bisect each other. Find QN

LM

NPQ

4

6 Ans

wer

87

LM

NPQ

4

6

39 LMNP is a parallelogram. Find MP

Ans

wer

88

40 BEAR is a parallelogram. Find the value of x.

A

B E

R

S

x 4y

8 10

Ans

wer

89

41 BEAR is a parallelogram. Find the value of y.

A

B E

R

S

x 4y

8 10

Ans

wer

4y = RS = 8y = 2

90

42 BEAR is a parallelogram. Find ER.

A

B E

R

S

x 4y

8 10

Ans

wer

ER = RS + SE = 16

91

43 In a parallelogram, the opposite sides are ________ parallel.

A sometimes

B always

C never

Ans

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92

44 The opposite angles of a parallelogram are __________.

A bisect

B congruent

C parallel

D supplementary Ans

wer

93

45 The consecutive angles of a parallelogram are __________.

A bisect

B congruent

C parallel

D supplementary

Ans

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94

46 The diagonals of a parallelogram ______ each other.

A bisect

B congruent

C parallel

D supplementary Ans

wer

95

47 The opposite sides of a parallelogram are __________.

A bisect

B congruent

C parallel

D supplementary Ans

wer

96

12

M A

TH

R

7

48 MATH is a parallelogram. AH = 12 and MR = 7. Find RT.

A 6

B 7

C 8

D 9

Ans

wer

B

97

49 MATH is a parallelogram. Find AR.

A 6

B 7

C 8

D 9 12

M A

TH

R

7

Ans

wer

98

50 MATH is a parallelogram. Find m∠H.

M A

TH98°

2x - 4

14(3y + 8)°

Ans

wer

99

M A

TH98°

2x

x2 - 24(3y + 8)°

51 MATH is a parallelogram. Find the value of x.

Ans

wer

x2 - 24 = 2xx2 - 2x - 24 = 0

(x - 6)(x + 4) = 0x = 6 or x = -4

since HT = 2x, and 2(-4) = -8, which is not a positive length, the solution x = -4 can be ruled out

(crossed off above).

100

M A

TH98°

2x - 4

14(3y + 8)°

52 MATH is a parallelogram. Find the value of y.

Ans

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101

Proving Quadrilaterals are

Parallelograms


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102

In quadrilateral ABCD,

AB ≅ DC and AD ≅ BC,

so ABCD is a parallelogram.

A B

CD

TheoremIf both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

103

In quadrilateral ABCD,

∠A ≅ ∠D and ∠B ≅ ∠C,


A

CD

B

If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

Theorem

104

53 PQRS is a Parallelogram. Explain your reasoning.

True

False

P

Q

R

S6

6

4

4

Ans

wer True, PQRS is a parallelogram.

The opposite sides are congruent.

105

54 Is PQRS a parallelogram? Explain your reasoning.

Yes

No P Q

RS

Ans

wer

Because PQRS is a quadrilateral, m∠Q + m∠R =

180°. But, we can't assume that ∠Q and ∠R are right angles. We

can't prove PQRS is a parallelogram.

106

55 Is this quadrilateral definitely a parallelogram?

Yes

No

78°

136° 2

Ans

wer

107


Yes

No

3 3

5

4.99

Ans

wer

108


Yes

No

Ans

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109

If an angle of a quadrilateral is supplementary to both of its consecutive angles, then the quadrilateral is a parallelogram.

C

75°

75°

105°

D

A B

In quadrilateral ABCD, m∠A + m∠B = 180°

and m∠B + m∠C = 180°, so ABCD is a parallelogram.

Theorem

110

If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

In quadrilateral ABCD,AE ≅ CE and DE ≅ BE,

so ABCD is a quadrilateral.

A B

CD

E

Theorem

111

If one pair of sides of a quadrilateral is parallel andcongruent, then the quadrilateral is a parallelogram.

In quadrilateral ABCD,AD ≅ CB and AD || BC,


A B

CD

Theorem

112


Yes

No

Ans

wer

No

It could be a trapezoid

113


Yes

No 141°

39°

49°

Ans

wer

114


Yes

No

8 9.5

8

19

Ans

wer

115


Yes

No

Ans

wer

Yes

116

62

Yes

No

Three interior angles of a quadrilateral measure 67°, 67° and 113°. Is this enough information to tell whether the quadrilateral is a parallelogram? Explain.

Ans

wer

NO, the question did not state the position of the measurements in the quadrilateral. We cannot assume their position.

117

In a parallelogram...

the opposite sides are _________________ and ____________,

the opposite angles are _____________, the consecutive angles are _____________,

and the diagonals ____________ each other.

parallel perpendicular

bisect congruent

supplementary

Fill in the blank

Ans

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118

To prove a quadrilateral is a parallelogram...

both pairs of opposite sides of a quadrilateral must be _____________,

both pairs of opposite angles of a quadrilateral must be ____________,

an angle of the quadrilateral must be _____________ to its consecutive

angles, the diagonals of the quadrilateral __________ each other, or one

pair of opposite sides of a quadrilateral are ___________ and

_________.

bisect congruent

parallel perpendicular supplementary

Fill in the blank

Ans

wer

119

63 Which theorem proves the quadrilateral is a parallelogram?

A The opposite angles are congruent.

B The opposite sides are congruent. C An angle in the quadrilateral is supplementary to its consecutive angles. D The diagonals bisect each other. E One pair of opposite sides are congruent and parallel. F Not enough information.

3(2)3

6(7-3)

Ans

wer

120

64 Which theorem proves this quadrilateral is a parallelogram?


B The opposite sides are congruent. C An angle in the quadrilateral is supplementary to its consecutive angles.

D The diagonals bisect each other. E One pair of opposite sides are congruent and parallel. F Not enough information.

Ans

wer

121

65 Which theorem proves the quadrilateral is a parallelogram?


B The opposite sides are congruent. C An angle in the quadrilateral is supplementary to its consecutive angles. D The diagonals bisect each other. E One pair of opposite sides are congruent and parallel. F Not enough information.

6

63(6-4)

Ans

wer

122

Rhombi, Rectanglesand Squares


Mat

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123

Three Special Parallelograms

Rhombus Rectangle

Square

All the same properties of a parallelogram apply to the rhombus, rectangle, and square.

124

A rhombus is a quadrilateral with four congruent sides.

A B

CD

If AB ≅ BC ≅ CD ≅ DA,then ABCD is a rhombus.

Rhombus

125

66 What is the value of y that will make the quadrilateral a rhombus?

A 7.25

B 12

C 20

D 25

35

y

12

Ans

wer

C

126

67 What is the value of y that will make this parallelogram a rhombus?A 7.25

B 12

C 20

D 25 2y+29

6y

Ans

wer

127

A B

CD

The diagonals of a rhombus are perpendicular.

If ABCD is a rhombus, then AC ⟂ BD.

Theorem

128

A B

CD

O


Since ABCD is a parallelogram, its diagonals bisect each other.

AO ≅ CO and BO ≅ DO.

Theorem

129

A B

CD

O


Then by SSS, ΔAOB ≅ ΔBOC ≅ ΔCOD ≅ ΔDOA

And due to corresponding parts of ≅ Δs being ≅ (or CPCTC)

∠AOB ≅ ∠COB ≅ ∠COD ≅ ∠AOD

If the adjacent angles formed by intersecting lines are congruent, then the lines are perpendicular.

Theorem

By definition, the sides of a rhombus are of equal measure, which means that AB = BC = CD = DA.

130

If a parallelogram is a rhombus, then each diagonal bisects a pair of opposite angles.

If ABCD is a rhombus, then

∠DAC ≅ ∠BAC ≅∠BCA ≅ ∠DCAand

∠ADB ≅ ∠CDB ≅ ∠ABD ≅ ∠CBD

Theorem

A B

CD

131

If a parallelogram is a rhombus, then each diagonal bisects a pair of opposite angles.

Theorem

We proved earlier by SSS, that

ΔAOB ≅ ΔBOC ≅ ΔCOD ≅ ΔAOD

Corresponding parts of ≅ Δs are ≅; so

∠ABO ≅ ∠CBO ≅ ∠ADO ≅ ∠CDO

AND

∠DAO ≅ ∠BAO ≅ ∠BCO ≅ ∠DCO

So each diagonal bisects a pair of opposite angles

A B

CD

O

132

68 EFGH is a rhombus. Find the value of x.

E F

G H

72°z°

2x - 6

5y

10

Ans

wer

133

E F

G H

72°z°

2x - 6

5y

10

69 EFGH is a rhombus. Find the value of y.

Ans

wer

134

E F

G H

72°z°

2x - 6

5y

10

70 EFGH is a rhombus. Find the value of z.

F

Ans

wer

135

71 This quadrilateral is a rhombus. Find the value of x.

8

86°

3x + 2

z°

12

y2

Ans

wer

136

72 This quadrilateral is a rhombus. Find the value of y.

8

86°

3x + 2

z°

12

y2

Ans

wer

137

73 This quadrilateral is a rhombus. Find the value of z.

8

86°

3x + 2

z°

12

y2

Ans

wer

138

74 This is a rhombus. Find the value of x.

x°

Ans

wer

90

139


13

x - 3

9

Ans

wer

140


126°x°

Ans

wer

141

77 HJKL is a rhombus. Find the length of HJ.

H J

KL

6 M16

Ans

wer

142

A B

CD

A rectangle is a quadrilateral with only right angles.

If ABCD is a rectangle, then ∠A, ∠B, ∠C and ∠ D are right angles.

If ∠A, ∠B, ∠C and ∠D are right angles, then ABCD is a rectangle.

Rectangles

143

78 What value of y will make the quadrilateral a rectangle?

(6y)°

12

Ans

wer

15

144

The diagonals of a rectangle are congruent.

If ABCD is a rectangle, then AC ≅ BD.

Also, since a rectangle is a parallelogram,

AM ≅ BM ≅ CM ≅ DM

A B

CD

M

Theorem

145

The diagonals of a rectangle are congruent.

Since ABCD is a rectangle (and therefore a parallelogram)AB ≅ DC

Because of the reflexive property, BC ≅ BC .

Since ABCD is a rectanglem∠B = m∠C = 90º

Then ΔABC ≅ ΔDCB by SAS

And, AC ≅ DB since they are corresponding parts of ≅ Δs

A B

CD

Theorem

146

79 RECT is a rectangle. Find the value of x.

2x - 5 13

63°(9y)°

R E

CT

Ans

wer Option A

2x - 5 = 132x = 18

x = 9

147

80 RECT is a rectangle. Find the value of y.

2x - 5 13

63°(9y)°

R E

CT

Ans

wer

The measure of each angle in a rectangle is 90°.

148

81 RSTU is a rectangle. Find the value of z.

R S

TU(8z)°

Ans

wer

11.25

149

82 RSTU is a rectangle. Find the value of z.

R S

TU

4z -

9

7

Ans

wer

150

A square is a quadrilateral with equal sides and equal angles. It is, therefore both a rhombus and a

rectangle.

A square has all the properties of a rectangle

and rhombus.

Square

151

83 This quadrilateral is a square. Find the value of x.

z - 4

(5x)°

6

(3y)° Ans

wer

In a rhombus the diagonals are perpendicular.

152

84 This quadrilateral is a square. Find the value of y.

z - 4

(5x)°

6

(3y)° Ans

wer

In a rhombus the diagonals are perpendicular.

153

z - 4

(5x)°

6

(3y)°

85 This quadrilateral is a square. Find the value of z.

Ans

wer

154

86 This quadrilateral is a square. Find the value of x.

y 2

(12z)°

10 - 3

y

(x2 + 9)°

Ans

wer

155


y 2

(12z)°

10 - 3

y

(x2 + 9)° Ans

wer

156

88 This quadrilateral is a square. Find the value of z.

y 2

(12z)°

10 - 3

y

(x2 + 9)°

Ans

wer

157


(18y)°

Ans

wer

158

90 This quadrilateral is a rhombus. Find the value of x.

2x + 6

4x

Ans

wer

159

112°

(4x)°

91 The quadrilateral is parallelogram. Find the value of x.

Ans

wer

160

92 The quadrilateral is a rectangle. Find the value of x.

10x

3x + 7

Ans

wer

161

Opposite sidesare ||

Diagonals bisectopposite ∠'s Has 4 ≅ sides

Has 4 right ∠'sDiagonals are

Fill in the Table with the Correct Definition

rhombus rectangle square

Ans

wer

Opposite sidesare ||

Diagonals are ≅

162

Lab - Quadrilaterals in the Coordinate Plane

Mat

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http://njctl.org/courses/math/geometry-2015-16/quadrilaterals/quadrilaterals-in-the-coordinate-plane-lab/

163

Trapezoids

Return to theTable of Contents

164

A trapezoid is a quadrilateral with one pair of parallel sides.

base

legbase angles

base

leg

The parallel sides are called bases.

The nonparallel sides are called legs.

A trapezoid also has two pairs of base angles.

Trapezoid

165

An isosceles trapezoid is a trapezoid with congruent legs.

Isosceles Trapezoid

166

If a trapezoid is isosceles, then each pair of base angles are congruent.

If ABCD is an isosceles trapezoid, then ∠A ≅ ∠B and ∠C ≅ ∠D.

A B

CD

Theorem

167


Given: ABCD is an isosceles trapezoidProve: ∠A ≅ ∠B and ∠C ≅ ∠D

A B

CD E F

Theorem

168


Given: ABCD is an isosceles trapezoidProve: ∠A ≅ ∠B and ∠C ≅ ∠D.

A B

CD E F

Theorem

continued...

Mat

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169

Given: ABCD is an isosceles trapezoidProve: ∠A ≅ ∠B and ∠C ≅ ∠D.

A B

CD E F

Theorem

Statement Reason1 ABCD is an isosceles trapezoid

2 AB || DC, AD ≅ BC

3Draw segments from A and B perpendicular to AB and DC. These segments are AE and BF.

Geometric Construction

4 ∠EAB, ∠AEF, ∠BFE, ∠FBA, ∠BFC and ∠AED are right angles

5 All right angles are ≅

6 Definition of congruent and right angles

continued...

170

Theorem7 ABFE is a rectangle

8 AE ≅ BF

9 ΔAED ≅ ΔBFC

10 ∠C ≅ ∠D∠DAE ≅ ∠CBF

11 m∠DAE = m∠CBF

12 m∠DAE + 90 = m∠CBF + 90

13 m∠DAE + m∠EAB = m∠CBF + m∠FBA

14 m∠DAE + m∠EAB = m∠DABm∠CBF + m∠FBA = m∠CBA

15 m∠DAB = m∠CBA

16 ∠DAB ≅ ∠CBA, or ∠A ≅ ∠B

171

If a trapezoid has at least one pair of congruent base angles, then the trapezoid is isosceles.

A B

CD

If in trapezoid ABCD, ∠A ≅ ∠B. Then, ABCD is an isosceles trapezoid.

Theorem

172

A B

CD E F


Theorem

Given: In trapezoid ABCD, ∠C ≅ ∠DProve: ABCD is an isosceles trapezoid

Case 1 - Longer Base ∠s are ≅

Mat

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continued...

173

TheoremStatement Reason

1 In trapezoid ABCD, ∠C ≅ ∠D

2 AB || DC

3Draw segments from A and B perpendicular to AB and DC. These segments are AE and BF.

Geometric Construction

4 ∠EAB, ∠AEF, ∠BFE, ∠FBA, ∠BFC and ∠AED are right angles

5 All right angles are ≅

6 ABFE is a rectangle

7 AE ≅ BF

8 ΔAED ≅ ΔBFC9 AD ≅ BC

10 ABCD is an isosceles trapezoid

174

A B

CD


Case 2 - ∠s of shorter base are ≅

In this case, extend the shorter base and draw a perpendicular line from each vertex of the longer base

to the extended base.

Theorem

175

A B

CD

EF



In this case, extend the shorter base and draw a perpendicular

line from each vertex of the longer base to the extended

base, as shown.

Can you now see how to extend the approach we used for the first

case to this second case?

Theorem

176


Given: In trapezoid ABCD, ∠A ≅ ∠BProve: ABCD is an isosceles trapezoid

Steps to follow: 1) Include the constructions of our new perpendicular segments CE & DF in your proof.2) Identify pairs of triangles in the above diagram that you need to prove congruent to prove this theorem.3) Then, using what you know about the relationships between pairs of angles and sides in isosceles trapezoids and rectangles, prove that the triangles are congruent.4) Then prove the theorem by stating that corresponding parts of congruent triangles are congruent (CPCTC).

A B

CD

EF


Theorem

Ans

wer

177

93 The quadrilateral is an isosceles trapezoid. Find the value of x.

6

11 7x - 10

14

A 3

B 5

C 7

D 9 Ans

wer

178

94 The quadrilateral is an isosceles trapezoid. Find the value of x.

A 3

B 5

C 7

D 964° (9x + 1)° A

nsw

er

179

A trapezoid is isosceles if and only if its diagonals are congruent.

In trapezoid ABCD,

If AC ≅ BD. Then, ABCD is isosceles.

If ABCD is isoscelesThen, AC ≅ BD.

A B

CD

Theorem

180

A trapezoid is isosceles if and only if its diagonals are congruent.A B

CD

Theorem

Statement Reason

1 Trapezoid ABCD is isosceles

2 AB || DC, AD ≅ BC, ∠A ≅∠B, ∠C ≅∠D

4 CD ≅ CD5 SAS

6 AC ≅ BD

A

CD

B

CD

Given: Trapezoid ABCD is isoscelesProve: AC ≅ BD

Teac

her's

Not

e

This proof addresses MP3 & MP6.

Additional Q's to address MP standards:What reason justifies statement #_? (MP3)

181

95 PQRS is a trapezoid. Find the m∠S.

112° 147°

(6w + 2)° (3w)°

P

R

Q

S

Ans

wer

The sum of the interior angles of a quadrilateral is 360°.

The parallel lines in a trapezoid create pairs of consecutive interior angles.

m∠P + m∠S = 180° and m∠Q + m∠R = 180°

(6w+2) + 112 = 1806w + 114 = 180

w = 11

182

112° 147°

(6w + 2)° (3w)°

P

R

Q

S

96 PQRS is a trapezoid. Find the m∠R.

Ans

wer

The sum of the interior angles of a quadrilateral is 360°.

The parallel lines in a trapezoid create pairs of consecutive interior angles.

m∠P + m∠S = 180° and m∠Q + m∠R = 180°

183

97 PQRS is an isosceles trapezoid. Find the m∠Q.

123°

(4w + 1)°

(9w - 3)°P Q

RS

Ans

wer

184

123°

(4w + 1)°

(9w - 3)°P Q

RS

98 PQRS is an isosceles trapezoid. Find the m∠R.

Ans

wer

185

123°

(4w + 1)°

(9w - 3)°P Q

RS

99 PQRS is an isosceles trapezoid. Find the m∠S.

Ans

wer

186

100 The trapezoid is isosceles. Find the value of x.

9

4

6x + 3

2x + 2

Ans

wer

187

101 The trapezoid is isosceles. Find the value of x.

137°

x° Ans

wer

143

188

102 In trapezoid HIJK, can HI and KJ have slopes that are opposite reciprocals?

H I

JK

Ans

wer

YesNo

189

The midsegment of a trapezoid is a segment that joins the midpoints of the legs

Midsegment of a Trapezoid

Lab - Midsegments of a Trapezoid

Click on the link below and complete the lab.

Teac

her N

otes

This lab addresses MP3, MP7 & MP8.

http://njctl.org/courses/math/geometry-2015-16/quadrilaterals/midsegments-of-a-trapezoid-lab/

http://info.mheducation.com/sketchpad.trial.html

190

The midsegment of a trapezoid is parallel to both the bases.

AB ║ EF ║ DC

A B

CD

E F

Theorem

191

The midsegment of a trapezoid is parallel to both the bases.

It's given that E is the midpoint of AD; F is the midpoint of BC; and AB ║ DC.

Since AB ║ DC, they have the same slope and therefore the distance between them, along any perpendicular line, will be the same.

Proof of Theorem

A B

CD

E F

Any line that connects the midpoints of the legs will also be the same distance from each base if measured along a perpendicular line.

Which means that the lines that connect the midpoints will have the same slope as the bases.

If the slopes are the same, the lines are parallel.

That means that AB ║ EF ║ DC, where EF is the midsegment of the trapezoid.

192

A B

CD

E F

The length of the midsegment is half the sum of the bases.

Theorem

EF = 0.5(AB + DC)

193

The length of the midsegment is half the sum of the bases.

Proof of Theorem

Prove: EF = 0.5(AB + DC)

Given: E is the midpoint of AD F is the midpoint of BC AB ║ DC ║ EF

A B

CD

E Fx G

H

yI

J2x 2y

Mat

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continued...

194

Proof of TheoremStatement Reason

1E is the midpoint of AD F is the midpoint of BC AB ║ DC ║ EF

2 ∠AEG ≅ ∠ADH, ∠BFI ≅ ∠BCJ3 ∠EAG ≅ ∠DAH, ∠FBI ≅ ∠CBJ4 ∆AEG ~ ∆ADH, ∆BFI ~ ∆BCJ

5 AD AH DH BC BJ CJAE AG EG , BF BI IF

6 AD = 2AE, BC = 2BF

The midpoints E and F divide AD and BC into 2 congruent segments, so AD and BC are twice the size of AE and BF respectfully.

7 AD = 2 BC = 2AE , BF

8 The scale factor in both of our similar triangles is 2.

====

continued...

195

Proof of Theorem

Statement Reason

9 If EG = x, then DH = 2x.If IF = y, then CH = 2y.

10 EF = AB + x + y11 2EF = 2AB + 2x + 2y12 DC = AB + 2x + 2y13 2x + 2y = DC - AB

14 2EF = 2AB + DC - AB15 2EF = AB + DC

16 EF = 0.5(AB + DC)

196

103 PQRS is a trapezoid. Find LM.

P

Q R

S

L M

15

7

Ans

wer

197

104

P

Q R

S

L M

20

14.5

PQRS is a trapezoid. Find PS.

Ans

wer

198

105

P

QR

S

LM

y

5

10

14

x z

7

PQRS is an trapezoid. ML is the midsegment. Find the value of x.

Ans

wer

x = 18

199

106

P

QR

S

LMy

5

10

14

x z

7

PQRS is an trapezoid. ML is the midsegment. Find the value of y.

Ans

wer

200

107

P

QR

S

LMy

5

10

14

x z

7

PQRS is an trapezoid. ML is the midsegment. Find the value of z.

Ans

wer

201

108 EF is the midsegment of trapezoid HIJK. Find the value of x.

H I

JK

E F

6

x

15

Ans

wer

202

109 EF is the midsegment of trapezoid HIJK. Find the value of x.

HI

J K

EF

x

19

10

Ans

wer

1

203

110 Which of the following is true of every trapezoid? Choose all that apply.

A Exactly 2 sides are congruent.

B Exactly one pair of sides are parallel.

C The diagonals are perpendicular.

D There are 2 pairs of base angles.

Ans

wer

B and D

204

Kites


205

A kite is a quadrilateral with two pairs of adjacent congruent sides whose opposite sides are not congruent.

Kites

Lab - Properties of Kites

Mat

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http://njctl.org/courses/math/geometry-2015-16/quadrilaterals/properties-of-kites-lab/

206

In kite ABCD, ∠B ≅ ∠D

and ∠A ≅ ∠C

If a quadrilateral is a kite, then it has one pair of congruent opposite angles.

Theorem

A

B

C

D

207

If a quadrilateral is a kite, then it has one pair of congruent opposite angles.

Proof of Theorem

Draw line AC and use the Reflexive Property to say that AC ≅ AC.

ΔABC ≅ ΔADC by SSS.

∠B ≅ ∠D since corresponding parts of ≅ Δs are ≅.m∠A ≠ m∠C since AD ≠ CD.

In kite ABCD, m∠B = m∠D

and m∠A ≠ m∠C

A

B

C

D

208

111 LMNP is a kite. Find the value of x.

72°

(x2 - 1)°

48°

M

N

P

L

Ans

wer

m∠L + m∠M + m∠N + m∠P = 360° (Remember ∠M ≅ ∠P)

72 + (x2 - 1) + (x2 - 1) + 48 = 3602x2 + 118 = 360

2x2 = 242x2 = 121x = ±11

209

112 READ is a kite. RE ≅ ____.

A EA

B AD

C DR R

E

A

D

Ans

wer

210

113 READ is a kite. ∠A ≅ ____.

A ∠EB ∠DC ∠R R

E

A

D

Ans

wer

211

114 Find the value of z in the kite.

z 5z - 8

Ans

wer

212

115 Find the value of x in the kite.

68°

(8x + 4)°

44°

Ans

wer

213

116 Find the value of x.

36°

(3x2 - 2)°

24°

Ans

wer

214

Theorem

In kite ABCDAC ⟂ BD

A

BD

C

The diagonals of a kite are perpendicular.

215

Proof of Theorem

In kite ABCDAC ⟂ BD

A

B

C

DE

Given that AD ≅ AB and DC ≅ BC

AC ≅ AC and AE ≅ AE by reflexive property ΔADC ≅ ΔABC by SSS

∠DAE ≅ ∠BAE by CPCTC ΔADE ≅ ΔABE by SAS

∠AED ≅ ∠AEB by CPCTC

AE is perpendicular to BD since intersecting lines that create ≅ adjacent angles are perpendicular

The diagonals of a kite are perpendicular.

216

117 Find the value of x in the kite.

x°

Ans

wer

217

118 Find the value of y in the kite.

(12y)°

Ans

wer

218

Constructing Quadrilaterals


Mat

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219

1. Draw two intersecting lines.

Construction of a Parallelogram

220

2. Place a compass at the point of intersection and draw an arc where it intersects one line in two places.

0°190


221

3. Keeping the compass at the point of intersection, either enlarge, or shorten the radius of the compass and draw an arc where it intersects the other line in 2 places.

0°

140


222

4. Connect the four points where the arcs intersect the lines.


223

Which property of parallelograms does this construction technique depend upon?

Ans

wer


224

Try this!Construct a parallelogram using the lines below.

1)


225

Try this!Construct a parallelogram using the lines below.

2)


226

Videos Demonstrating the Constructions of a Parallelogram using Dynamic Geometric

Software

Click here to see parallelogram video: compass & straightedge

Click here to see parallelogram video: Menu options

http://youtu.be/bT-9d5PloDM

http://youtu.be/_xgG9E_HPqs

227


Construction of a Rectangle

228

2. Place a compass at the point of intersection and draw an arc where it intersects each line, keeping the radius of the compass the same all the way around.

0°190


229



230

Which properties of rectangles does this construction technique depend upon?


Ans

wer

231

Construction of a Rectangle Try this!Construct a rectangle using the lines below.

3)

232

Construction of a Rectangle Try this!Construct a rectangle using the lines below.

4)

233

Videos Demonstrating the Constructions of a Rectangle using Dynamic Geometric Software

Click here to see rectangle video: compass & straightedge

Click here to see rectangle video: Menu options

http://youtu.be/h03mO4Zbp1c

http://youtu.be/Xd6RliRpZx4

234

Construction of a Rhombus

1. Construct a segment of any length.

235

Construction of a Rhombus 2. Find the perpendicular bisector and midpoint of a line segment by constructing the locus between two points.

a. Place the point of the compass on A and open it so the pencil is more than halfway, it does not matter how far, to point B

0°

280

236

Construction of a Rhombus2. Find the perpendicular bisector and midpoint of a line segment by constructing the locus between two points.

b) Keeping the compass locked, place the point of the compass on B and draw an arc so that it intersects the one you just drew

0°

280

237

Construction of a Rhombus2. Find the perpendicular bisector and midpoint of a line segment by constructing the locus between two points.

C

0°

280

c. Draw a line through the arc intersections. The intersection point created with the two perpendicular lines is the center of your rhombus.

238

Construction of a Rhombus3. Place the tip of your compass on your center point (midpoint) C. Extend your compass to any length. Draw 2 arcs where the compass intersects the perpendicular bisector.

CNote: If you wanted to, you could use the intersection points of the 2 arcs from step #2 as your extra points.

0° 135

239

Construction of a Rhombus4. Connect the intersection points of the 2 arcs and the perpendicular bisector from step #3 with the endpoints of your original segment.

C

240

Construction of a Rhombus

C

Which properties of rhombii (pl. for rhombus) does this construction technique depend upon?

Ans

wer

241

Videos Demonstrating the Constructions of a Rhombus using Dynamic Geometric Software

Click here to see rhombus video: compass & straightedge

Click here to see rhombus video: Menu options

http://youtu.be/v8IlTDE4pvk

http://youtu.be/4hVbCnax8oo

242

Construction of a Rhombus Try this!Construct a rhombus using the segment below.

5)

243

Construction of a Rhombus Try this!Construct a rhombus using the segment below.

6)

244

Construction of a Square1. Construct a segment of any length.

245

Construction of a Square2. Find the perpendicular bisector and midpoint of a line segment by constructing the locus between two points.

a. Place the point of the compass on A and open it so the pencil is more than halfway, it does not matter how far, to point B. Drawan arc.

0°

280

246


b) Keeping the compass locked, place the point of the compass on B and draw an arc so that it intersects the one you just drew

0°

280

247


0°

280

c. Draw a line through the arc intersections. The intersection point created with the two perpendicular lines is the center of your circle (and future square).

C

248

Construction of a Square3. Place your compass tip on point C and the pencil point on either A or B and construct your circle.

0°

244

C

249

Construction of a Square4. Connect the intersection points of your perpendicular bisector and the circle with the initial vertices of the line segment to create your square.

C

250

Which properties of squares does this construction technique depend upon?

Construction of a Square

C

Ans

wer

251

Try this!Construct a square using the segment below.

7)


252

Try this!Construct a square using the segment below.

8)


253

Videos Demonstrating the Constructions of a Square using Dynamic Geometric Software

Click here to see square video: compass & straightedge

Click here to see square video: Menu options

http://youtu.be/GdHFD_LIAaw

http://youtu.be/Qt4Tb2adye8

254


Construction of an Isosceles Trapezoid

255

2. Place a compass at the point of intersection and draw an arc where it intersects two of the lines one time each.

0°

190


256

3. Keeping the compass at the point of intersection, either enlarge, or shorten the radius of the compass and draw an arc where it intersects the two lines on the other side (e.g. Since I drew my 1st 2 arcs in the upper half, these 2 arcs will be in the lower half).


0°

144

257



258

Construction of an Isosceles Trapezoid Which properties of isosceles trapezoids does this construction technique depend upon?

Ans

wer

259

Try this!Construct an isosceles trapezoid using the lines below.

9)


260

Try this!Construct an isosceles trapezoid using the lines below.

10)


261

Videos Demonstrating the Constructions of an Isosceles Trapezoid using Dynamic

Geometric Software

Click here to see isosceles trapezoid video: compass &

straightedge

Click here to see isosceles trapezoid video: Menu options

http://youtu.be/Ql0gSz56qxs

http://youtu.be/naPZpGxI62c

262

Construction of a Kite

1. Construct a segment of any length.

263


2. Place your compass tip at any point and draw 2 arcs that intersect your segment.

0°

72

264


3. Find the perpendicular bisector between the 2 red arcs.

0°91

a) Place your compass tip at one of the intersection points. Extend your pencil out more than half way between the 2 arcs. Draw an arc.

265



b) Keeping the radius of your compass the same, place your compass tip at the other intersection point and draw an arc.

0°91

266



c) Connect the intersections of the 2 arcs to construct your perpendicular line.

267

Construction of a Kite4. Place the tip of your compass at the intersection point of our perpendicular lines. Extend your compass to any length. Draw 2 arcs where the compass intersects the perpendicular line.

0°109

Note: If you wanted to, you could use the intersection points of the 2 arcs from step #3 as your extra points.

268

Construction of a Kite5. Connect the intersection points of your perpendicular line and the arcs from step #4 with the initial vertices of the line segment to create your kite.

269


Which properties of a kite does this construction technique depend upon?

Ans

wer

270

Try this!Construct a kite using the segment below.

11)


271

Try this!Construct a kite using the segment below.

12)


272

Videos Demonstrating the Constructions of a Kite using Dynamic Geometric Software

Click here to see kite video: compass & straightedge

Click here to see kite video: Menu options

http://youtu.be/h5kW8mittqw

http://youtu.be/PSBvB63s_tg

273



274

Complete the chart(There can be more than one answer).

squarerectanglerhombusparallelogram

kitetrapezoid isosceles trapezoid

Description Answer(s)

An equilateral quadrilateral

An equiangular quadrilateral

The diagonals are perpendicular

The diagonals are congruent

Has at least 1 pair of parallel sides

rectangle & square

rhombus & square

rhombus, square & isosceles trapezoid

rectangle, square & kite

All except kite

Special Quadrilateral(s)

click to reveal

click to reveal

click to reveal

click to reveal

click to reveal

275

119 A rhombus is a square.

A always

B sometimes

C never

Ans

wer

276

120 A square is a rhombus.

A always

B sometimes

C never

Ans

wer

277

121 A rectangle is a rhombus.

A always

B sometimes

C never

Ans

wer

278

122 A trapezoid is isosceles.

A always

B sometimes

C never

Ans

wer

279

123 A kite is a quadrilateral.

A always

B sometimes

C never

Ans

wer

A

280

124 A parallelogram is a kite.

A always

B sometimes

C never

Ans

wer

281

Sometimes, you will be given a type of quadrilateral and using its qualities find the value of a variable. After substituting the variable back into the algebraic expression(s), the quadrilateral might become a more specific type.

Using what we know about the different types of quadrilaterals, let's try these next few problems.


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282

Quadrilateral ABCD is a parallelogram.

Determine the value of x.

x2 - 75 2x + 5

25

A B

CD

Since we know that ABCD is a parallelogram, we know that the opposite sides are parallel and congruent.

Using the congruence property, we can write an equation & solve it.click

click

Quadrilaterals

283

Quadrilateral ABCD is a parallelogram.

x2 - 75 2x + 5

25

A B

CD

x2 - 75 = 2x + 5

x2 - 2x - 80 = 0

(x - 10)(x + 8) = 0

x = 10 and x = -8click

click

click

click

Do any of these answers not make sense?

Yes, since the value of x cannot be negative, x = -8 can be ruled out. BC = 2(-8) + 5 = -11 which is not possible.

Since ABCD is a parallelogram, AB = 25 because opposite sides are congruent. When substituting x = 10 back into the expressions, then AD = CD = BC = AB = 25. Therefore, ABCD is a rhombus.

click

click

Quadrilaterals

Is parallelogram ABCD a more specific type of of quadrilateral? Explain.

284

125 Quadrilateral EFGH is a parallelogram. Find the value of x.

x2 - 32 3x + 8

32

E F

GH

J

2x + 6.63 3x - 1.37

Ans

wer

285

126 Based on your answer to the previous slide, what type of quadrilateral is EFGH? Explain.

When you are finished, choose the classification of quadrilateral EFGH from the choices below.

A Rhombus

B Rectangle

C Squarex2 - 32 3x + 8

32

E F

GH

J

2x + 6.63 3x - 1.37

Ans

wer

286

127 Quadrilateral KLMN is a trapezoid. Find the value of x.

(x2 - 20)°

(10x)°(7x + 30)°K L

MN

Ans

wer

10x + x2 - 20 = 180x2 + 10x - 200 = 0(x + 20)(x - 10) = 0

x = 10Note: x = -20 does not work

287

128 Based on your answer to the previous slide, what type of quadrilateral is KLMN? Explain.

When you are finished, determine whether trapezoid KLMN is isosceles.

A Isosceles Trapezoid

B Not an Isosceles Trapezoid

(x2 - 20)°

(10x)°(7x + 30)°K L

MN

Ans

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288

Extension: Inscribed QuadrilateralsA quadrilateral is inscribed if all its vertices lie on a circle.

.

.

.

inscribed quadrilateral

289

If a quadrilateral is inscribed in a circle, then the opposite angles are supplementary.

m∠D = 1/2 (mEFG) [green arc]m∠F = 1/2 (mEDG) [red arc]

Since mEFG + mEDG = 360°, m∠D + m∠F = 1/2 (mEFG) + 1/2 (mEDG)m∠D + m∠F = 1/2 (mEFG + mEDG)m∠D + m∠F = 1/2(360°)m∠D + m∠F = 180°

Inscribed Quadrilaterals

.

E

D

F

G

m∠D + m∠F = 180°m∠E + m∠G = 180°

290

Example: Find the m∠G. Then, find the value of a.

Inscribed Quadrilaterals

.

E

D

F

G

121°(3a + 8)°

(2a - 3)°

Answ

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291

129 What is m∠L?

25°

65°

115°

155°

A

B

C

D

.

M

J

L

K

115°

63°

Ans

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292

130 What is m∠M?

23°

63°

117°

153°

A

B

C

D

.

M

J

L

K

115°

63°

Ans

wer

293

131 What is the value of x?

A 8

B 10

C 17

D 19.

M

J

L

K

(7x - 6)°

(5y - 4)°

(3x - 4)°

(8y + 15)°

Ans

wer

294

132 What is the value of y?

A 13

B 14.69

C 7

D 6.08.

M

J

L

K

(7x - 6)°

(5y - 5)°

(3x - 4)°

(8y + 16)°

Ans

wer

295

Can special quadrilaterals be inscribed into a circle?

Throughout this unit, we have been talking about the different quadrilaterals, their properties, and how to construct them.

- Parallelogram- Rectangle- Rhombus- Square- Trapezoid- Kite

Can any of these quadrilaterals be inscribed into a circle?

Find out by completing the lab below.

Lab: Inscribed Quadrilaterals

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http://njctl.org/courses/math/geometry-2015-16/quadrilaterals/inscribed-quadrilaterals-lab/

296

133 Complete the sentence below.A trapezoid can __________ be inscribed into a circle.

A Always

B Sometimes

C Never

Ans

wer

297

134 Complete the sentence below.A rectangle can __________ be inscribed into a circle.

A Always

B Sometimes

C Never

Ans

wer

298

135 Complete the sentence below.A rhombus can __________ be inscribed into a circle.

A Always

B Sometimes

C Never

Ans

wer

299

136 Complete the sentence below.A kite can __________ be inscribed into a circle.

A Always

B Sometimes

C Never

Ans

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300

Proofs


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301

Given: TE ≅ MA, ∠1 ≅ ∠2Prove: TEAM is a parallelogram.

T E

AM

1

2

Proofs

302

Option A

statements reasons

1) TE ≅ MA, ∠1 ≅ ∠2

2) EM ≅ EM

3) ∆MTE ≅ ∆EAM

4) TM ≅ AE

5) TEAM is a parallelogram

Proofs T E

AM

1

2

303

Option B

We are given that TE ≅ MA and ∠2 ≅ ∠3. TE || AM, by the alternate interior angles converse.

So, TEAM is a parallelogram because one pair of opposite sides is parallel and congruent.click to reveal

click to reveal

ProofsT E

AM

1

2

304

Given: FGHJ is a parallelogram, ∠F is a right angleProve: FGHJ is a rectangle

F G

HJ

Proofs

305

F G

HJstatements reasons

1) FGHJ is a parallelogramand ∠F is a right angle 1) Given

2) ∠J and ∠G are right angles

3) ∠H is a right angle

4) TEAM is a rectangle

ProofsGiven: FGHJ is a parallelogram, ∠F is a right angleProve: FGHJ is a rectangle

306

Given: COLD is a quadrilateral, m∠O = 140°, m∠D = 40°, m∠L = 60°Prove: COLD is a trapezoid

C O

LD

140°

40°60°

Proofs

307

C O

LD

140o

40o60o

statements reasons1) COLD is a quadrilateral,m∠O = 140°, m∠L = 40°, m∠D = 60° 1) Given

2) m∠O + m∠L = 180°m∠L + m∠D = 100°

3) Definition of Supplementary Angles

4) ∠L and ∠D are not supplementary

5) Consecutive Interior Angles Converse

6) CL || OD

7) COLD is a trapezoid

ProofsGiven: COLD is a quadrilateral, m∠O = 140°, m∠D = 40°, m∠L = 60°Prove: COLD is a trapezoid

308

Try this ...

Given: ΔFCD ≅ ΔFEDProve: FD ⊥ CE

F

C

D

E

Hin

t

Proofs

309

Coordinate Proofs


310

Recall that Coordinate Proofs place figures on the Cartesian Plane to make use of the coordinates of key features of the figure to help prove something. Usually the distance formula, midpoint formula and slope formula are used in a coordinate proof.

We'll provide a few examples and then have you do some proofs.

Quadrilateral Coordinate Proofs

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This lesson addresses MP3, MP5 & MP6.

Additional Q's to address MP standards:What information are you given? (MP1)What do you need to prove? (MP1)What do you know that is not stated in the problem? (MP3)Which formula would be the best one to use in this proof? Explain. (MP5)Why is this statement true? (MP6)What else is there to do? (MP6)

311

Given: PQRS is a quadrilateralProve: PQRS is a kite

y

x

2

4

6

8

10

-2

-4

-6

-8

-10

2 4 6 8 10-2-4-6-8-10 0

P

Q

R

(-1,6)

(-4,3) (2,3)

(-1,-2)

S

Coordinate Proofs

312

A kite has one unique property.The adjacent sides are congruent.

PS = (6 - 3)2 + (-1 - (-4))2 PQ = (3 - 6)2 + (2 - (-1))2

= 32 + 32 = (-3)2 + 32

= 9 + 9 = 9 + 9 = 18 = 18 = 4.24 = 4.24

√√√√

√√√√

P

Q

R

(-1,6)

(-4,3) (2,3)

(-1,-2)

S

click click

Coordinate Proofs

313

RS = (3 - (-2))2 +(-4 - (-1))2 RQ = (-2 - 3)2 + (-1 - 2)2

= 52 + (-3)2 = (-5)2 + (-3)2

= 25 + 9 = 25 + 9 = 34 = 34 = 5.83 = 5.83

√

√√√

√

√

√√

Because PS = PQ and RS = RQ, PQRS is a kite.

P

Q

R

(-1,6)

(-4,3) (2,3)

(-1,-2)

S

click click

click

Coordinate Proofs

314

y

x

2

4

6

8

10

-2

-4

-6

-8

-10

2 4 6 8 10-2-4-6-8-10 0

J(1, 3)

K(4, -1)L(0, -4)

M(-3, 0)

Given: JKLM is a parallelogramProve: JKLM is a square

Coordinate Proofs

315

Since JKLM is a parallelogram, we know the opposite sides are parallel and congruent.

We also know that a square is a rectangle and a rhombus.We need to prove the adjacent sides are congruent and

perpendicular.

Let's start by proving the adjacent sides are congruent.

JM = (3 - 0)2 + (1 - (-3))2 JK = (-1 - 3)2 + (4 - 1)2

= 32 + 42 = (-4)2 + 32

= 9 + 16 = 9 + 16 = 25 = 25 = 5 = 5

√√√√

√√√√

J(1, 3)

K(4, -1)L(0, -4)

M(-3, 0)

click click

Coordinate Proofs

316

mJM = = 3 - 0 31- (-3) 4

-1 - 3 -4 4 - 1 3

mJK = =

MJ ⊥ JK and JM ≅ JKWhat else do you know?

MJ ≅ LK and JK ≅ LM (Opposite sides are congruent)MJ ⊥ LM and JK ⊥ LK (Perpendicular Transversal Theorem)

Therefore, JKLM is a square

Now, let's prove that the adjacent sides are perpendicular.

J(1, 3)

K(4, -1)L(0, -4)

M(-3, 0)

click click

click

click

Coordinate Proofs

317

y

x

2

4

6

8

10

-2

-4

-6

-8

-10

2 4 6 8 10-2-4-6-8-10 0

P(2, 2)L(1, 0) Q(5, 1)

M(7, -2)R(9, -5)

S(0, -2)

Try this ...

Given: PQRS is a trapezoidProve: LM is the midsegment

Hin

t

Remember, the midsegment is the segment that joins the

midpoints of the legs and is parallel to the bases.

You need to show that:1. SL = LP2. QM = MR

3. slope of LM = slope of SR

Coordinate Proofs

318

Sometimes, shapes might be placed in a coordinate plane, but without numbered values. Instead, algebraic expressions will be used for the coordinates. You can still use the distance formula, midpoint formula and slope formula in this type of coordinate proof.

We'll provide a few examples and then have you do some proofs.


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319

E (q, r) F(p, r)

G(p, u)H(q, u)

x

yGiven: EFGH is a parallelogramProve: EFGH is a rectangle


320

E (q, r) F(p, r)

G(p, u)H(q, u)

x

y

Since EFGH is a parallelogram, we know the opposite sides are parallel and congruent. Since we need to prove that EFGH is a rectangle, we can either prove that adjacent sides are perpendicular or diagonals are congruent.

Let's choose the 1st option: proving the adjacent sides are perpendicular.


continued...

321

E (q, r) F(p, r)

G(p, u)H(q, u)

x

y

mEF = = = 0 r - r 0 p - q p - q

r - u r - u p - p 0

mFG = = = undef.

EF ⊥ FG and both pairs of opposite sides are ||What else do you know?

EF ≅ GH and FG ≅ EH (Opposite sides are congruent)EF ⊥ EH and FG ⊥ GH (Perpendicular Transversal Theorem)

Therefore, EFGH is a rectangleclick

click click

click


322

E (q, r) F(p, r)

G(p, u)H(q, u)

x

y

Since EFGH is a parallelogram, we know the opposite sides are parallel and congruent. Since we need to prove that EFGH is a rectangle, we can either prove that adjacent sides are perpendicular or diagonals are congruent.

This problem can also be done using the 2nd option: proving that the diagonals are congruent


323

E (q, r) F(p, r)

G(p, u)H(q, u)

x

y

We proved that EG ≅ HF (Diagonals are congruent)We also know that EF ≅ GH and FG ≅ EH (Opp. sides are congruent)

Therefore, EFGH is a rectangle

EG = (p - q)2 + (u - r)2 HF = (p - q)2 + (r - u)2

= (p - q)2 + (-(u - r))2

= (p - q)2 + (u - r)2

√ √

click

√√

click

click


324

Given: JKLM is a parallelogramProve: JL and KM bisect each other y

x

J (a, 2b)K (0, 2b)

M(0, 0)L(-a, 0)

U


325

y

x

J (a, 2b)K (0, 2b)

M(0, 0)L(-a, 0)

U

Since JKLM is a parallelogram, we know the opposite sides are parallel and congruent and the diagonals bisect each other.

If we can prove the last part, that the diagonals bisect each other using one of our formulas, we will achieve our goal.

Which formula would prove that the diagonals bisect each other?

Ans

wer

Midpoint formula: If the coordinates of U are the same

when using both the coordinates of JL and KM, then the diagonals

will bisect each other.

Note: If you said distance formula, you could use that too, but you would need to find the

coordinates of U w/ the midpoint formula 1st.


326

y

x

J (a, 2b)K (0, 2b)

M(0, 0)L(-a, 0)

U

Midpoint of KM:

U

U(0, b)

0 + 0 2b + 0 2 , 2

Midpoint of JL:

U

U(0, b)

-a + a 0 + 2b 2 , 2

click click

Because the midpoint U is the same for both of the diagonals, it proves that KM & JL bisect each other.click


327

137 If Quadrilateral STUV is a rectangle, what would be the coordinates for point U?

A U (e, d)

B U (d + e, e)

C U (d, e)

D U (d, e - d)

x

y

V(0, e)

S(0, 0) T(d, 0)

U(?, ?)

Ans

wer

C

328

138 If Quadrilateral OPQR is a parallelogram, what would be the coordinates for point Q?

A Q (b, a)

B Q (b + c, a)

C Q (b + c, a + c)

D Q (b, a + c)

x

y

P(0, a)

O(0, 0) R(b, c)

Q(?, ?)

Ans

wer D

329

139 If Quadrilateral ABCD is an isosceles trapezoid, what would be the coordinates for point B?

A B (-2f, 2h)

B B (-2g, 2h)

C B (-2g, -2h)

D B (2f, 2h)

x

y

B(?, ?)

A(-2f, 0) D(2f, 0)

C(2g, 2h)

Ans

wer B

330

140 Since ABCD is an isosceles trapezoid, BD and AC are congruent. Use the coordinates to verify that BD and AC are congruent.

When you are finished, type in the number "1".

x

y

B(?, ?)

A(-2f, 0) D(2f, 0)

C(2g, 2h)

Ans

wer

BD: d = √(2f - (-2g))2 + (0 - 2h)2

d = √(2f + 2g)2 + (-2h)2

d = √(2f + 2g)2 + 4h2

AC: d = √(2g - (-2f))2 + (2h - 0)2

d = √(2g + 2f)2 + (2h)2

d = √(2f + 2g)2 + 4h2

AC ≅ BD

331

PARCC Released Questions


The remaining slides in this presentation contain questions from the PARCC Sample Test. After finishing unit 10, you should be able to answer these questions.

Good Luck!

332

141 Let BE = x2 - 48 and let DE = 2x. What are the lengths of BE and DE? Justify your answer.

A

B C

D

E16

not drawn to scale

The figure shows parallelogram ABCD with AE = 16.

Part A

PARCC Released Question - PBA - Calculator Section - #7

Answ

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333

142 What conclusion can be made regarding the specific classification of parallelogram ABCD? Justify your answer.

Once you are finished with the question, type in the word to describe the classification of parallelogram ABCD.

A

B C

D

E16

not drawn to scale

The figure shows parallelogram ABCD with AE = 16.

Part B

Answ

er

Parallelogram ABCD is a rectangle becauseAE = BE = DE = 16, which means that CE = 16. Therefore, the diagonals are congruent, making ABCD a rectangle


334


143 Find the coordinates of point Q in terms of a, b, and c.Enter only your answer.

The figure shows parallelogram PQRS on a coordinate plane. Diagonals SQ and PR intersect at point T.

P(2b, 2c) Q(?, ?)

R(2a, 0)S(0, 0)

T

y

x

Answ

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Part A

335

144 Since PQRS is a parallelogram, SQ and PR bisect each other. Use the coordinates to verify that SQ and PR bisect each other.

Which formula(s) could be used to answer this question?A Distance Formula

B Slope

C Midpoint Formula

D A and C only

E A, B and C


Part B

Answ

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336

145 Since PQRS is a parallelogram, SQ and PR bisect each other. Use the coordinates to verify that SQ and PR bisect each other.When you are finished, type in the word "Done".


Part B Continued

Answ

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337

?

One method that can be used to prove that the diagonals of a parallelogram bisect each other is shown in the given partial proof.

P Q

RS

T

Given: Quadrilateral PQRS is a parallelogram.Prove: PT = RT, ST = QT

338

146 Which reason justifies the statement for step 3 in the proof?A When two parallel lines are intersected by a transversal,

same side interior angles are congruent.B When two parallel lines are intersected by a transversal,

alternate interior angles are congruent.C When two parallel lines are intersected by a transversal,

same side interior angles are supplementary.D When two parallel lines are intersected by a transversal,

alternate interior angles are supplementary.

PARCC Released Question - EOY - Calculator Section - #19

Part A

Answ

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339

147 Which statement is justified by the reason for step 4 in the proof?A PQ ≅ RSB PQ ≅ SP

C PT ≅ TR

D SQ ≅ PR


Part B

Answ

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340

148 Which reason justifies the statement for step 5 in the proof?A side-side-side triangle congruence

B side-angle-side triangle congruence

C angle-side-angle triangle congruence

D angle-angle-side triangle congruence


Part C

Answ

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341

Another method of proving diagonals of a parallelogram bisect each other uses a coordinate grid.

Part D

342

149 What could be shown about the diagonals of parallelogram PQRS to complete the proof?A PR and SQ have the same length.B PR is a perpendicular bisector of SQ.

C PR and SQ have the same midpoint.D Angles formed by the intersection of PR and SQ each

measure 90°.


Answ

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343

Throughout this unit, the Standards for Mathematical Practice are used.

MP1: Making sense of problems & persevere in solving them.MP2: Reason abstractly & quantitatively.MP3: Construct viable arguments and critique the reasoning of others. MP4: Model with mathematics.MP5: Use appropriate tools strategically.MP6: Attend to precision.MP7: Look for & make use of structure.MP8: Look for & express regularity in repeated reasoning.

Additional questions are included on the slides using the "Math Practice" Pull-tabs (e.g. a blank one is shown to the right on this slide) with a reference to the standards used.

If questions already exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab.

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Geometrymrboydsmathsite.weebly.com/uploads/8/6/7/8/8678489/properties_of...supplementary to an interior angle. And, that the sum of the measures of the interior angle of a ... Multiplication