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107Geometry

Chapter 8 Resource Book

LESSON

8.3 Practice CFor use with pages 542–549

For what value of x is the quadrilateral a parallelogram?

1. 3x 1 1

5x 2 9

2.

10x 2 33

4x 1 15 3. 6x 2 11

9x 2 53

4.

(17x 2 13)8

(7x 1 25)8 5.

(4x 1 11)8

(10x 1 1)8 6.

(21x 2 13)8

(8x 1 19)8

Decide whether you are given enough information to determine that the quadrilateral is a parallelogram.

7. Opposite sides are parallel. 8. Opposite sides are congruent.

9. Two pairs of consecutive sides 10. Two pairs of consecutive angles are congruent. are congruent.

11. Diagonals are congruent. 12. Diagonals bisect each other.

13. All four sides are congruent. 14. Consecutive angles are supplementary.

Prove that the points represent the vertices of a parallelogram. Use the method indicated.

15. A(24, 7), B(3, 0), C(2, 25), D(25, 2); Both pairs of opposite sides are parallel.

16. A(22, 8), B(2, 7), C(5, 1), D(1, 2); Both pairs of opposite sides are congruent.

Find all the possible coordinates for the fourth vertex of a parallelogram with the given vertices. Then draw the parallelogram on the graph.

17. (4, 21), (24, 1), (0, 8) 18. (3, 24), (22, 21), (1, 2)

x

y

2

2

x

y

2

2

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108GeometryChapter 8 Resource Book

Complete the proof.

19. GIVEN: Regular hexagon JKLMNO J K

N

O L

M

PROVE: OKLN is a parallelogram.

Statements Reasons

1. ? 1. Given

2. } JO > } NM 2. ? } JK > } ML ∠ J > ∠ M

3. ? 3. SAS Congruence Postulate

4. } OK > } NL 4. ?

5. ? 5. Defi nition of regular polygon

6. OKLN is a ~. 6. ?

20. Write a two-column proof.

GIVEN: VWKJ and SJRU are parallelograms. J RK

U

WV

ST

PROVE: ∠ W > ∠ U

21. Write a paragraph proof.

GIVEN: ABCD is a ~. A B

D C

E F

E is the midpoint of }

AD .F is the midpoint of

} BC .

PROVE: Quadrilateral ABFE is a parallelogram.

LESSON

8.3 Practice C continuedFor use with pages 542–549

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A15Geometry

Chapter 8 Resource Book

m } EF 5 8k

} 2 2

3k } 2 }

5h

} 2 2

h } 2 5

5k

} 2 }

2h 5

5k }

4h ;

m }

HG 5 5k

} 2 2 0

}

7h

} 2 2

3h } 2 5

5k

} 2 }

2h 5

5k }

4h

Because the slopes are the same, } EH i } FG and } EF i } HG .

Step 4: Prove EFGH is a parallelogram. Because both pairs of opposite sides are parallel, EFGH is a parallelogram. Because EFGH is a parallelogram and E, F, G, and H are midpoints of the respective sides of A, B, C, and D, a parallelogram is formed when the midpoints of adjacent sides are connected by segments.

Lesson 8.3Practice Level A

1. Theorem 8.9 2. Theorem 8.10

3. Theorem 8.7 4. Sample answer: Theorem 8.7

5. Theorem 8.8 6. Theorem 8.8 7. 5 8. 3

9. 6 10. 69 11. 25 12. 12

13.

x

y

1

1

A B

D C

Sample answer: AB 5 DC 5 5, BC 5 DA 5 2 Ï

}

5

14.

x

y

1

4

AB

CD

Sample answer: AB 5 DC 5 Ï

}

26 , BC 5 DA 5 3

15. Sample answer: }

AD i } BC or }

AB > }

DC

16. Sample answer: }

AD > }

BC or }

AB i } DC

17. Sample answer: ∠ ADC > ∠ ABC

18. Sample answer: }

AE > }

CE

19. Sample answer: m∠ DAB 1 m∠ ABC 5 1808

20. Sample answer: }

AB > }

CD

21. n MJK > n KLM; Corresponding parts of > n’s are >.; Theorem 8.7

22. n MJK > n KLM; Corresponding parts of > n’s are >.; } JK i } LM ; Theorem 8.9

Practice Level B

1. Theorem 8.8 2. Theorem 8.7

3. Theorem 8.10 4. Theorem 8.9 5. 6 6. 8

7. 1 8. 79 9. 20 10. 31

11.

x

y

1

1

B C

DA

The slope of }

BC and }

AD is 0, so }

BC i } AD . Also, BC 5 AD 5 6. By Theorem 8.9, ABCD is a parallelogram.

12.

x

y

6

2

A

BC

D

AB 5 CD 5 2 Ï}

10 and BC 5 AD 5 2 Ï}

17 . So ABCD is a parallelogram by Theorem 8.7.

13. Use Corresponding Angles Converse to show }

AB i } CD , then apply Theorem 8.9.

14. Use Right Angle Congruence Theorem, then apply Theorem 8.8. 15. D(6, 23)

16. a. }

GF > } HJ and }

GH > } FJ , so FGHJ is a parallelogram by Theorem 8.7. b. FGHJ is always a ~, so

} GH i } FJ . Because

} GH is parallel to

the ground, then } FJ is also parallel to the ground by the Transitive Property of Parallel Lines. So, the moving log is always parallel to the ground.

17.

Statements Reasons

1. n ABC > nCDA 1. Given2. } AB >

} CD , 2. Corresp. sides

}

CB > }

AD of > triangles are >.

3. ABCD is a g. 3. Theorem 8.7

Practice Level C

1. 5 2. 8 3. 14 4. 7 5. 12 6. 6 7. yes

8. yes 9. no 10. no 11. no 12. yes 13. yes

14. yes 15. slope of }

AB 5 slope of }

CD 5 21 and slope of

} BC 5 slope of

} DA 5 5, so ABCD is

a ~ by defi nition. 16. AB 5 CD 5 Ï}

17 and BC 5 DA 5 3 Ï

}

5 , so ABCD is a ~ by Theorem 8.7.17. (8, 6), (0, 28), 18. (6, 21), (0, 27),

Lesson 8.2, continuedA

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A16GeometryChapter 8 Resource Book

and (28, 10) and (24, 5)

x

y

2

6

x

y

2

4

19. Regular hexagon JKLMNO; Defi nition of regular polygon; n OJK > n NML; Corresponding parts of > n’s are >.;

} ON > } KL ; Theorem 8.7

20.

Statements Reasons

1. VWKJ and SJRU 1. Givenare ~.

2. ∠ W > ∠ J 2. Opposite ?∠J > ∠U of a ~ are >.

3. ∠ W > ∠ U 3. Transitive Prop. of >

21. Because ABCD is a ~, opposite sides are i. So

} AD i } BC and segments contained within i

segments are also i so }

AE i } BF . We know opposite sides of ~ are > so

} AD >

} BC . Since E and F are

given as midpoints we can show }

AE > } ED and }

CF > } FB , so through Segment Addition Postulate and Division Property of Equality we can show }

AE > } FB . So quadrilateral ABFE is a ~ since one pair of opposite sides is i and >.

Review for Mastery

1. Theorem 8.10 2. Theorem 8.7

3. Theorem 8.8 4. Theorem 8.9 5. 1 }

2

6. LM 5 Ï}}}

(25 2 (24))2 1 (22 2 2)2 5 Ï}

17 ;

NO 5 Ï}}}

(0 2 (21))2 1 (0 2 (24))2 5 Ï}

17 ;

because LM 5 NO 5 Ï}

17 , } LM > }

NO .

Slope of } LM 5 22 2 2

} 25 2 (24)

5 4;

slope of }

NO 5 0 2 (24)

} 0 2 (21)

5 4; because } LM and }

NO

have the same slope, they are parallel.

} LM and }

NO are congruent and parallel. So, LMNO is a parallelogram by Theorem 8.9.

Problem Solving Workshop:Mixed Problem Solving

1. a. 6 b. 7208 c. 3608

2. a–c. check drawings; 9 d. Every angle in the fi gure measures 1408. The sum of the measures of the interior angles for a 9 sided fi gure is (9 2 2) p 1808 5 12608. Because each angle is

1408, then there are 1260

} 140

5 9 angles, which means there are 9 sides.

3. Sample answer:

4. 109 5. a. } JK and } ML have the same slope,

3 } 7 , so they are parallel. } JM and } KL have the same

slope, 4, so they are parallel. By defi nition, JKLM is a parallelogram. b. The length of } JK and }

ML is Ï}

58 . The length of } JM and } KL is 2 Ï}

17 . By Theorem 8.3, JKLM is a parallelogram.

6. The adjacent angles described in the problem are supplementary. You can fi nd their measures by solving the equation x 1 (5x 2 18) 5 180. By Theorem 8.4, opposite angles are congruent. So, the measures of the angles are 338, 1478, 338, and 1478. 7. a. 608 b. increases c. decreases d. It increases because AD increases.

Challenge Practice

1. Sample answer: First, we know

} AF i } CD ,

} AB i } ED , and

} BC i } FE .

Next, draw } BE parallel to }

CD , }

CF parallel to }

AB and

} AD parallel to

} BC and } FE . Label the

intersection point G.

Make sure }

BC > }

GD , }

GD > } FE , and }

AB > }

FG . Because you now have one pair of opposite sides of each quadrilateral congruent and parallel, the quadrilaterals are parallelograms.

2. PTRU is a parallelogram. Sample answer: It is given that PQRS and QTSU are parallelograms. Because the diagonals of a parallelogram bisect each other, PX 5 RX and TX 5 UX. Because } PR and } TU are diagonals of PTRU that bisect each other, PTRU is a parallelogram.

3. Sample answer: Use the compass to measure the length of

} AB . With that measure, draw a circle

with center C. Use the compass to measure the length of

} BC . With that measure, draw a circle

with center A. Now label the bottom point of

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