Practice chapter 4iii answers

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Chapter 4 Review

Multiple Choice

Identify the choice that best completes the statement or answers the question.

____ 1. Apply the transformation M to the polygon with the given vertices. Identify and describe the transformation.

M: (x, y)  (–x, –y)

A(–3, 6), B(–3, 1), C(1, 1), D(1, 6)

a.

This is a rotation of 180° about the origin.

c.

This is a reflection over the x-axis.

b.

This is a rotation of 180° about the origin.

d.

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____ 2. Determine whether triangles EFG and PQR are congruent.

a. The triangles are congruent because EFG can be mapped to PQR by a reflection: (x,y)(x,y).

b. The triangles are congruent because EFG can be mapped to PQR by a rotation: (x,y)(y,x).

c. The triangles are congruent because EFG can be mapped to PQR by a reflection: (x,y)(x,y).

d. The triangles are congruent because EFG can be mapped to PQR by a rotation: (x,y)(y,x).

____ 3. Classify DBC by its angle measures, given mDAB60, mABD75, and mBDC25.

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____ 4. ABC is an isosceles triangle. AB is the longest side with length 8x5. BC = 4x4 and CA = 3x9. Find AB.

a. AB = 93 c. AB = 24

b. AB = 45 d. AB = 5

____ 5. Daphne folded a triangular sheet of paper into the shape shown. Find mECD, given mCAB61, mABC22, and mBCD42.

a. mECD = 41 c. mECD = 22

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____ 6. Find mK.

a. mK = 63 c. mK = 79

b. mK = 55 d. mK = 39

____ 7. Given that ABC DEC and mE = 23°, find mACB.

a. mACB = 77° c. mACB = 23°

b. mACB = 67° d. mACB = 113°

____ 8. Using the information about John, Jason, and Julie, can you uniquely determine the distances from John to Julie and from Julie to Jason? Explain your answer.

Statement 1: John and Jason are standing 12 feet apart.

Statement 2: The angle from Julie to John to Jason measures 31°. Statement 3: The angle from John to Jason to Julie measures 49°. a. No. There is no unique configuration.

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____ 9. Use AAS to prove the triangles congruent. Given: AB GH, AC FH, AC  FH Prove: ABC HGF

Complete the flowchart proof. Proof:

AB GH B G

Given 1.

AC



FH ACB HFG ABC HGF

Given 2. AAS

AC FH Given

a. 1. Alternate Exterior Angles Theorem 2. Alternate Interior Angles Theorem b. 1. Alternate Interior Angles Theorem 2. Alternate Exterior Angles Theorem c. 1. Alternate Exterior Angles Theorem 2. Alternate Exterior Angles Theorem d. 1. Alternate Interior Angles Theorem

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____ 10. For these triangles, select the triangle congruence statement and the postulate or theorem that supports it.

a. ABC JLK, HL c. ABC JLK, SAS

b. ABC JKL, HL d. ABC JKL, SAS

____ 11. A pilot uses triangles to find the angle of elevation A from the ground to her plane. How can she find mA?

a. ABO CDO by SAS and A C by CPCTC, so mA40 by substitution.

b. ABO CDO by CPCTC and A C by SAS, so mA40 by substitution.

c. ABO CDO by ASA and A C by CPCTC, so mA40 by substitution.

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____ 12. Given: MLN PLO, MNL POL, MO NP Prove:MLP is isosceles.

Complete the proof. Proof:

Statements Reasons

1. MLN PLO, MNL POL 1. Given

2. MO NP 2. Given

3. MONP 3. Definition of congruent line segments 4. NONO 4. Reflexive Property of Equality 5. MONONPNO 5. Subtraction Property of Equality 6. MONOMN and NPNOOP 6. Segment Addition Postulate 7. MNOP 7. Substitution Property of Equality

8. MLN PLO 8. [1]

9. MLPL 9. [2]

10. MLP is isosceles. 10. Definition of isosceles triangle

a. [1] CPCTC [2] ASA

c. [1] CPCTC [2] AAS b. [1] ASA

[2] CPCTC

d. [1] AAS [2] CPCTC ____ 13. Find the value of x.

a. x = 6 c. x = 2

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____ 14. Determine whether or not the lines are perpendicular. Explain your answer.

a. The slope of LM  1

2 and the slope of MN 2. LM is perpendicular to MN because

1

2(2)1.

b. The slope of LM  1

2 and the slope of MN 6

5. LM is not perpendicular to MN

because 1 2 (

6

5) 1.

c. The slope of LM 2 and the slope of MN  1

2. LM is perpendicular to MN because

2(1

2) 1.

d. The slope of LM  1

2 and the slope of MN2. LM is perpendicular to MN because

1

2(2) 1.

____ 15. Write an equation for the line parallel to the line shown that passes through the point (–2, 3).

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____ 16. Position a right triangle with leg lengths r and 2s4 in the coordinate plane and give the coordinates of each vertex.

a. c.

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____ 17. Given: Qis a right angle in PQR. S is the midpoint of PQ and T is the midpoint of RP. Prove: The area of PST is one fourth the area of PQR.

Complete the paragraph proof.

Proof: PQR is a right triangle with height 2d units and base 2c units. The area of PQR[1] square units. By the [2] Formula, the coordinates of S are (0, d) and the coordinates of T are (c, d). Thus PST is a right triangle with height d units and base c units. So the area of PST[3] square units. Since [3] 1

4(2cd), the

area of PST is one fourth the area of PQR. a. [1] 4cd

[2] Midpoint [3] cd

c. [1] 4cd [2]Distance [3] cd b. [1] 2cd

[2] Distance [3] 1₂cd

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____ 18. Given: Qis a right angle in the isosceles PQR. X is the midpoint of PR. Y is the midpoint of QR. Prove: QXY is isosceles.

Complete the paragraph proof.

Proof: Draw a diagram and place the coordinates of PQR and QXY as shown.

By [2],the coordinates of X are 0 1 2 ,

2a0 2       

   3 ,aand

the coordinates of Y are 0 1 2 ,

00 2       

  4 , 0

By [5], XY 0a2_ 4  3 _2  6 andQY 002_ 4 0_2   7 .

Since XYQY, XY  QY by definition. So QXY is isosceles.

a. [1] a

[2] the Distance Formula [3] a₂, [4] a₂

[5] the Midpoint Formula [6] a

2, [7] a 2

c. [1] a

[2] the Midpoint Formula [3] a₂, [4] a₂

[5] the Distance Formula [6] a

2, [7] a 2 b. [1] 2a

[2] the Distance Formula [3] a, [4] a

[5] the Midpoint Formula [6] a, [7] a

d. [1] 2a

[2] the Midpoint Formula [3] a, [4] a

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____ 19. Find the measure of each numbered angle.

a. m1 = 54, m2 = 117, m3 = 63 b. m1 = 117, m2 = 63, m3 = 63 c. m1 = 54, m2 = 63, m3 = 63 d. m1 = 54, m2 = 63, m3 = 117

____ 20. Which list shows all the segments on AC that contain the point B?

a. AC

b. AB, BC, BD

c. AB, AC, AD, BC, BD d. AB, AC, AD, BC, BD, CD

____ 21. M is between R and S. If RM 21, RS15x3, and MS 9x12, what is RS?

a. 12 c. 87

b. 66 d. 147

____ 22. K is the midpoint of VW. If KV3x and KW5x10, what is VW?

a. 7.5 c. 22.5

b. 15 d. 30

____ 23. Which is an obtuse angle?

a. PQR c. R

b. PSQ d. P

____ 24. The midpoint of a segment is _8, 5_. If one endpoint is _0, 1_, what is the other endpoint?

a. 16, 9

  c. 4, 2

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____ 25. What is the contrapositive of the statement?

If a triangle has at least two congruent angles, then it is an isosceles triangle. a. If a triangle has no congruent angles, then it is not an isosceles triangle. b. If a triangle is an isosceles triangle, then it has at least two congruent angles.

c. If a triangle does not have at least two congruent angles, then it is an isosceles triangle. d. If a triangle is not an isosceles triangle, then it does not have at least two congruent

angles.

____ 26. Which is a counterexample of the statement? If an animal has wings, then it can fly.

a. penguin c. duck

b. robin d. rabbit

____ 27. Which completes the proof? Given: mTKVmUKW

Prove: mTKUmVKW Proof:

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____ 28. Which angle is the alternate interior angle with 4?

a. 1 c. 7

b. 5 d. 8

____ 29. If rs, what is the value of y?

a. 18 c. 120

b. 60 d. 162

____ 30. What is the slope of a line that passes through the points _3, 2_ and 7,12

 

a. 7₂ c. 2₇

b. 1 d. 1

____ 31. What is the equation of the line that passes through _8, 8_ and has a slope of 3? a. y8x3 c. y83x8

b. y  8x3 d. y83x8

____ 32. The graph of which line is perpendicular to the graph of y 2x1?

a. y2x1 c. y 2x6

b. y 1₂x1 d. y 1₂x6

____ 33. What is mA?

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____ 34. Which information would you need for the shortest proof that MNP UVW by ASA?

a. P W c. MP UW

b. MN UV d. NPVW

____ 35. What would allow you to prove QRS XYZ by HL?

a. QS XZ

b. R Y c. Q X

d. Q and X are right angles.

____ 36. For a coordinate proof concerning an isosceles triangle, which coordinates might be easiest to use? a. (0, 0), (2a, 0), (a, b)

b. (0, 0), (a, b), (2a, 2b) c. (a, a), (b, b), (c, c) d. (a, b), (c, d), (e, f)

Numeric Response

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Chapter 4 Review

Answer Section

MULTIPLE CHOICE 1. ANS: B

Feedback

A Check that you found the correct coordinates for the transformed vertices.

B Correct!

C This transformation is not a reflection. Check that you found the correct coordinates for the transformed vertices.

D Check that you found the correct coordinates for the transformed vertices.

PTS: 1 DIF: Average REF: 9141f35b-6ab2-11e0-9c90-001185f0d2ea

OBJ: 4-1.1 Drawing and Identifying Transformations NAT: NT.CCSS.MTH.10.9-12.G.CO.5 TOP: 4-1 Congruence and Transformations

KEY: transformation | coordinate geometry MSC: DOK 2 2. ANS: C

Feedback

A Check the rule for the transformation. B This transformation is not a rotation.

C Correct!

D This transformation is not a rotation.

PTS: 1 DIF: Average REF: 91442ea6-6ab2-11e0-9c90-001185f0d2ea

OBJ: 4-1.2 Determining Whether Figures are Congruent NAT: NT.CCSS.MTH.10.9-12.G.CO.6 TOP: 4-1 Congruence and Transformations

KEY: reflection | rotation | transformation | translation | coordinate geometry MSC: DOK 2

3. ANS: A

ABD and DBC form a linear pair, so they are supplementary. Therefore mABDmDBC180. By

substitution, 75 mDBC180. So mDBC105. DBC is an obtuse triangle by definition.

Feedback A Correct!

B An acute triangle has three acute angles.

C A right triangle has one right angle.

D An equiangular triangle has three congruent acute angles.

PTS: 1 DIF: Average REF: 1a624596-4683-11df-9c7d-001185f0d2ea

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4. ANS: B

Step 1 Find the value of x.

BC  CA

4x4  3x9

x  5

Step 2 Find AB. AB  8x5

 8(5)5

 45

Feedback

A First, find the value of x. Then, use substitution to find AB. B Correct!

C This is equals BC and CA. Now find AB. D This is the value of x. Now find AB.

PTS: 1 DIF: Average REF: 1a64a7f2-4683-11df-9c7d-001185f0d2ea OBJ: 4-2.3 Using Triangle Classification TOP: 4-2 Classifying Triangles KEY: equilateral triangle | side length MSC: DOK 2

5. ANS: A

Step 1 Find mACB.

mCABmABCmACB180 Triangle Sum Theorem

61 22 mACB180 Substitute 61° for mCAB and 22° for mABC. 83 mACB180 Simplify.

mACB97 Subtract 83° from both sides. Step 2 Find mECD.

mACBmBCDmECD180 Linear Pair Theorem and Angle Addition Postulate 97 42 mECD180 Substitute 97° for mACB and 42° for mBCD. 139 mECD180 Simplify.

mECD41 Subtract 139° from both sides.

Feedback A Correct!

B You found the measure of angle ACB. Use this to find the measure of angle ECD. C First, use the Triangle Sum Theorem to find the measure of angle ACB. Then, use the

Linear Pair Theorem and the Addition Postulate to find the measure of angle ECD.

D First, use the Triangle Sum Theorem to find the measure of angle ACB. Then, use the Linear Pair Theorem and the Addition Postulate to find the measure of angle ECD.

PTS: 1 DIF: Average REF: 1a696caa-4683-11df-9c7d-001185f0d2ea OBJ: 4-3.1 Application

STA: MI.MIGLC.MTH.06.9-12.G1.2.1 | MI.MIGLC.MTH.06.9-12.G1.2.2 LOC: MTH.C.11.02.01.01.005 | MTH.C.11.03.02.04.002

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6. ANS: A

mKmLmLMN Exterior Angle Theorem 6x9

  4x7 118 _mLMNSubstitute 6x_. 9 for mK, 4x7 for mL, and 118 for

10x2118 Simplify.

10x120 Add 2 to both sides. x12 Divide both sides by 10. mK6x96 12 963

Feedback A Correct!

B You found the measure of angle L. Find the measure of angle K instead.

C First, use the Exterior Angle Theorem to find the value of x. Then substitute the value

for x to find the measure of angle K.

D First, use the Exterior Angle Theorem to find the value of x. Then substitute the value

for x to find the measure of angle K.

PTS: 1 DIF: Average REF: 1a6bcf06-4683-11df-9c7d-001185f0d2ea OBJ: 4-3.3 Applying the Exterior Angle Theorem

STA: MI.MIGLC.MTH.06.9-12.G1.2.1 | MI.MIGLC.MTH.06.9-12.G1.2.2

LOC: MTH.C.11.03.02.04.004 TOP: 4-3 Angle Relationships in Triangles KEY: exterior angle theorem MSC: DOK 2

7. ANS: B

mDCEmCEDmEDC180 Triangle Sum Theorem mDCE23 90 180 Substitution.

mDCE113 180 Simplify.

mDCE67 Subtract 113 from both sides.

DCE BCA Corresponding parts of congruent triangles are _congruent.

mDCEmBCA Definition of congruent angles

mACB67 Corresponding parts of congruent triangles are _congruent.

Feedback

A The sum of all angle measures in a triangle is equal to 180 degrees.

B Correct!

C Check which angles are corresponding angles.

D Check your calculations.

PTS: 1 DIF: Average REF: 1a72f61a-4683-11df-9c7d-001185f0d2ea

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8. ANS: C

Statements 2 and 3 determine the measures of two angles of the triangle. Statement 1 determines the length of the included side.

By ASA, the triangle must be unique.

Feedback

A Draw a diagram. There is enough information to determine a unique triangle.

B There is not enough information for SAS. Draw a diagram to help you. C Correct!

D There is not enough information for SSS. Draw a diagram to help you.

PTS: 1 DIF: Average REF: 1a814442-4683-11df-9c7d-001185f0d2ea

OBJ: 4-6.1 Problem-Solving Application STA: MI.MIGLC.MTH.06.9-12.G1.3.1 LOC: MTH.C.14.06.01.005 TOP: 4-6 Triangle Congruence: ASA, AAS, and HL

KEY: application | triangle congruence MSC: DOK 2 9. ANS: B

1. B and G are alternate interior angles and AB GH. Thus by the Alternate Interior Angles Theorem,

B G.

2. ACB and HFG are alternate exterior angles and AC FH. Thus by the Alternate Exterior Angles Theorem, ACB HFG.

Feedback

A You switched the definitions of alternate interior and alternate exterior angles. B Correct!

C If line segment AB is parallel to line segment GH, are angle B and angle G alternate exterior angles or alternate interior angles?

D If line AC is parallel to line FG, are angle ACB and angle HFG alternate interior angles or alternate exterior angles?

PTS: 1 DIF: Average REF: 1a83a69e-4683-11df-9c7d-001185f0d2ea OBJ: 4-6.3 Using AAS to Prove Triangles Congruent

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10. ANS: B

Because BAC and KJL are right angles, ABCandJKL are right triangles.

You are given a pair of congruent legs ACJL and a pair of congruent hypotenuses CB LK. So a hypotenuse and a leg of ABC are congruent to the corresponding hypotenuse and leg of JKL.

ABC JKL by HL.

Feedback

A Segment AC is congruent to segment JL. Make sure the triangle vertices correspond

accordingly.

B Correct!

C Segment AC is congruent to segment JL. Make sure the triangle vertices correspond accordingly. For SAS, the angle is included between the sides.

D For SAS, the angle is included between the sides.

PTS: 1 DIF: Advanced REF: 1a86300a-4683-11df-9c7d-001185f0d2ea STA: MI.MIGLC.MTH.06.9-12.G1.3.1 | MI.MIGLC.MTH.06.9-12.G2.3.1 |

MI.MIGLC.MTH.06.9-12.G2.3.2 LOC: MTH.C.11.08.02.02.02.010 | MTH.C.11.08.02.02.02.011 TOP: 4-6 Triangle Congruence: ASA, AAS, and HL KEY: proof | congruent triangles | HL MSC: DOK 3

11. ANS: A

From the figure, CO AO, andDO BO. AOB COD by the Vertical Angles Theorem. Therefore,

ABO CDO by SAS and A C by CPCTC. mA40 by substitution.

Feedback A Correct!

B First, show that the triangles are congruent. Then, show that their corresponding parts

are congruent.

C First, show that the triangles are congruent. Then, show that their corresponding parts

are congruent.

D First, show that the triangles are congruent. Then, show that their corresponding parts

are congruent.

PTS: 1 DIF: Average REF: 1a886b56-4683-11df-9c7d-001185f0d2ea OBJ: 4-7.1 Application

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12. ANS: D

[1] Steps 1 and 7 state that two angles and a nonincluded side of MLN andPLO are congruent. By AAS, MLN PLO.

[2] Since MLN PLO, by CPCTC, ML PL.

Feedback

A Before using CPCTC, you must prove that triangle MLN and triangle PLO are

congruent. Since steps 1 and 7 state that two angles and a nonincluded side are congruent, which triangle congruence theorem states that the triangles are congruent?

B Steps 1 and 7 state that two angles and a nonincluded side of triangle MLN and triangle PLO are congruent. Which triangle congruence theorem states that the triangles are congruent?

C Before using CPCTC, you must prove that triangle MLN and triangle PLO are

congruent.

D Correct!

PTS: 1 DIF: Average REF: 1a8af4c2-4683-11df-9c7d-001185f0d2ea OBJ: 4-7.3 Using CPCTC in a Proof LOC: MTH.P.08.02.03.017 | MTH.C.11.08.02.01.003 TOP: 4-7 Triangle Congruence: CPCTC KEY: congruent triangles | corresponding parts | CPCTC MSC: DOK 2

13. ANS: A

The triangles can be proved congruent by the SAS Postulate. By CPCTC, 3x52x1.

Solve the equation for x. 3x52x1

3x2x6 x6

Feedback A Correct!

B When solving, you can either add 5 or subtract 1 from each side. C Remember to combine the like terms when solving.

D These two triangles have SAS congruence, so the two expressions are equal by CPCTC.

PTS: 1 DIF: Advanced REF: 1a8f926a-4683-11df-9c7d-001185f0d2ea NAT: NT.CCSS.MTH.10.9-12.G.SRT.5

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14. ANS: D

Feedback

A The slopes of perpendicular lines are negative reciprocals. B Check that you found the slopes correctly.

C Check that you found the slopes correctly. D Correct!

PTS: 1 DIF: Average REF: 914b55b7-6ab2-11e0-9c90-001185f0d2ea NAT: NT.CCSS.MTH.10.9-12.G.GPE.5 TOP: 4-7-Ext. Lines and Slopes

KEY: slope | perpendicular MSC: DOK 2 15. ANS: A

Feedback A Correct!

B Does this line contain the given point?

C Slopes of perpendicular lines are negative reciprocals. Slopes of parallel lines are equal.

D Slopes of perpendicular lines are negative reciprocals. Slopes of parallel lines are equal. PTS: 1 DIF: Average REF: 914b7cc7-6ab2-11e0-9c90-001185f0d2ea NAT: NT.CCSS.MTH.10.9-12.G.GPE.5 TOP: 4-7-Ext. Lines and Slopes

KEY: slope | parallel MSC: DOK 2 16. ANS: D

Since the triangle has a right angle, place the vertex of the right angle at the origin and position a leg along each axis. This can be done in two ways: with the leg of length r along the x-axis or along the y-axis. In the first case, the three vertices are (0, 0), (r, 0), and (0, 2s4). In the second case, the three vertices are (0, 0), (0, r), and (2s4, 0).

Feedback

A Can you also position the leg of length r along the x-axis? B Can you also position the leg of length r along the y-axis?

C Coordinates lying on the x-axis are of the form (x, 0). Coordinates lying on the y-axis

are of the form (0, y).

D Correct!

PTS: 1 DIF: Average REF: 1a921bd6-4683-11df-9c7d-001185f0d2ea OBJ: 4-8.3 Assigning Coordinates to Vertices

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17. ANS: D

PQR is right triangle with height 2d units and base 2c units. The area of PQR 1

2bh  1

2  2c 2d2cd

square units. By the Midpoint Formula, the coordinates of S are 00 2 ,

02d 2       

  0,d and the

coordinates of T are 2c₂0, 0₂2d

     

 c,d. Thus PST is a right triangle with height 2ddd units

and base c units. So the area of PST 1 2bh 

1

2  c  d  1

2cd square units. Since 1 2cd

1

4 2cd, the area

of PST is one fourth the area of PQR.

Feedback

A The area of a right triangle is one half the base times the height.

B The Distance Formula calculates the distance between two points, not the coordinates of the midpoint of a line segment.

C The area of a right triangle is one half the base times the height. The Distance Formula calculates the distance between two points, not the coordinates of the midpoint of a line segment.

D Correct!

PTS: 1 DIF: Average REF: 1a945722-4683-11df-9c7d-001185f0d2ea OBJ: 4-8.4 Writing a Coordinate Proof NAT: NT.CCSS.MTH.10.9-12.G.GPE.7 STA: MI.MIGLC.MTH.06.9-12.G1.4.2 | MI.MIGLC.MTH.06.9-12.G2.3.2

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18. ANS: D

Draw a diagram and place the coordinates of PQR and QXY as shown.

By the Midpoint Formula, the coordinates of X are 02a 2 ,

2a0 2       

  a,a and

the coordinates of Y are 0₂2a , 0₂0

     

  a, 0.

By the Distance Formula, XY 0a2aa2 a QY 002a02 a.

and

Since XYQY, XY  QY by definition. So QXY is isosceles.

Feedback

A The Midpoint Formula tells how to find the midpoint of a line segment, and the

Distance Formula tells how to calculate the distance between two points. If triangle PQR is isosceles, what are the coordinates of point R?

B The Midpoint Formula tells how to find the midpoint of a line segment, and the Distance Formula tells how to calculate the distance between two points.

C If triangle PQR is isosceles, what are the coordinates of point R? D Correct!

PTS: 1 DIF: Average REF: 1a9ba546-4683-11df-9c7d-001185f0d2ea OBJ: 4-9.4 Using Coordinate Proof

STA: MI.MIGLC.MTH.06.9-12.G1.4.2 | MI.MIGLC.MTH.06.9-12.G2.3.2 LOC: MTH.P.08.02.03.01.008 | MTH.C.11.03.02.03.02.001

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19. ANS: C

Step 1: 2 is supplementary to the angle that is 117. 117 m2180. So m263.

Step 2: By the Alternate Interior Angles Theorem, 2 3. So m2m363.

Step 3: By the Isosceles Triangle Theorem, 2 and the angle opposite the other side of the isosceles triangle are congruent. Let 4 be that unknown angle.

Then, 2 4 and m2m463.

m1m2m4180 by the Triangle Sum Theorem. m163 63 180. So m154.

Feedback

A Angle 2 is supplementary to the angle that measures 117 degrees. B To find the measure of angle 1, use the Isosceles Triangle Theorem.

C Correct!

D By the Alternate Interior Angles Theorem, angle 2 is congruent to angle 3.

PTS: 1 DIF: Advanced REF: 1a9de092-4683-11df-9c7d-001185f0d2ea STA: MI.MIGLC.MTH.06.9-12.G1.2.1 | MI.MIGLC.MTH.06.9-12.G1.2.2

LOC: MTH.C.11.03.02.04.002 TOP: 4-9 Isosceles and Equilateral Triangles KEY: multi-step | isosceles triangle theorem MSC: DOK 2

20. ANS: C PTS: 1 DIF: 2 TOP: Cumulative Test, Chapter 4 MSC: DOK 1

22. ANS: D PTS: 1 DIF: 2 TOP: Cumulative Test, Chapter 4 MSC: DOK 2

23. ANS: B PTS: 1 DIF: 2 TOP: Cumulative Test, Chapter 4 MSC: DOK 1

24. ANS: A PTS: 1 DIF: 2 TOP: Cumulative Test, Chapter 4 MSC: DOK 2

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32. ANS: B PTS: 1 DIF: 2 NAT: NT.CCSS.MTH.10.9-12.G.GPE.5 TOP: Cumulative Test, Chapter 4

33. ANS: B PTS: 1 DIF: 2 TOP: Cumulative Test, Chapter 4 MSC: DOK 2

34. ANS: B PTS: 1 DIF: 2

NAT: NT.CCSS.MTH.10.9-12.G.CO.10 | NT.CCSS.MTH.10.9-12.G.SRT.5 TOP: Cumulative Test, Chapter 4 MSC: DOK 2

35. ANS: D PTS: 1 DIF: 2

NAT: NT.CCSS.MTH.10.9-12.G.CO.10 | NT.CCSS.MTH.10.9-12.G.SRT.5 TOP: Cumulative Test, Chapter 4 MSC: DOK 2

36. ANS: A PTS: 1 DIF: 2 NAT: NT.CCSS.MTH.10.9-12.G.GPE.4 TOP: Cumulative Test, Chapter 4 MSC: DOK 2

NUMERIC RESPONSE 1. ANS: 21.6

PTS: 1 DIF: Average REF: 1aa507a6-4683-11df-9c7d-001185f0d2ea LOC: MTH.C.11.03.02.03.03.002 | MTH.C.11.03.02.03.03.004

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_____ 1.B

_____ 2.C

_____ 3.A

_____ 4.B

_____ 5.A

_____ 6.A

_____ 7.B

_____ 8.C

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_____ 10.B

_____ 11.A

_____ 12.D

_____ 13.A

_____ 14.D

_____ 15.A

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_____ 18.D _____ 19.C

_____ 20.C

_____ 21.C

_____ 22.D

_____ 23.B

_____ 24.A

_____ 25.D

_____ 26.A

_____ 27.D

_____ 28.D

_____ 29.C

_____ 30.A

_____ 31.C

_____ 32.B

_____ 33.B

_____ 34.B

_____ 35.D