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102GeometryChapter 8 Resource Book

LESSON

8.2 Challenge PracticeFor use with pages 535–541

The given point coordinates represent three vertices of a parallelogram. Write the coordinates of each other point that could be the fourth vertex. Justify your answers.

1. A(2, 0), B(3, 5), C(6, 0) 2. F(2, 2), G(21, 4), H(21, 5)

3. J(a, b), K(a 1 2, b), L(a 1 4, b 1 3) 4. Q(a, b), R(a2, b), S(a2, b2)

5. In general, given three of the vertices of a parallelogram, how many different possible points are there for the fourth vertex?

6. Given two vertices of a parallelogram, describe the set of points that the third and fourth vertices of the parallelogram cannot come from.

7. Describe a way to draw one segment that divides any parallelogram into two congruent triangles. In how many ways can this be done for a given parallelogram?

8. Describe a way to divide a parallelogram into two pairs of congruent triangles using the least number of segments.

9. Describe two different ways to divide a parallelogram into two pairs of congruent triangles using three segments.

10. Is there a way to divide a parallelogram into exactly three congruent triangles?

11. Use the diagram to write a two-column proof. S

T

R

U

W Y

Z

X

GIVEN: RSTU and WXYZ are parallelograms.

PROVE: n RWX > n TYZ

12. Write a two-column proof to show that if a segment has endpoints on opposite sides of a parallelogram and the segment passes through the point of intersection D of the diagonals of the parallelogram, then D is the midpoint of the segment.

13. Write a coordinate proof to show that if segments are drawn between the midpoints of adjacent sides of any quadrilateral, then a parallelogram is formed.

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A13Geometry

Chapter 8 Resource Book

27. 558 28. a. 1358 b. As the lift is raised, vertices F and H of the parallelogram become farther apart causing the measure of ∠F to decrease and the measure of ∠E to increase. 29. 20

30.

Statements Reasons

1. MNOP and PQRO 1. Givenare g’s.

2. } MN > }

OP ; 2. Opp. sides of }

OP > }

QR a ~ are > .3. } MN >

} QR 3. Transitive prop. of

congruence.

Practice Level C

1. a 5 11, b 5 12 2. c 5 6, d 5 9

3. e 5 8, t 5 3 4. g 5 21, h 5 8

5. j 5 14, k 5 2 6. m 5 7, n 5 3

7. p 5 4, q 5 8 8. r 5 5, s 5 7 9. t 5 9, v 5 4

10.

x

y4

1

AB

C

E

D

; 1 1 } 2 , 1 2

11. 3; Diagonals of ~ bisect each other.

12. 5; Opposite sides of ~ are >.

13. 4; Pythagorean Theorem 14. 8; Diagonals of ~ bisect each other, so DB 5 2 p EB.

15. 5; Pythagorean Theorem or SAS > Theorem

16. 12; AE 5 3, EB 5 4, and AB 5 5

17. 378; Alternate Interior Angles Theorem

18. 908; Defi nition of a right triangle

19. 538; Triangle Sum Theorem

20. 538; Corresponding ? of similar n’s are >.

21. 20; All 4 n’s are > with hypotenuse 5 5.

22. 508 and 1308 23. 61.18 and 118.98

24. MP 5 8 Ï}

2 and NO 5 8 Ï}

2 so } MP > }

NO .

25. MN 5 4 and PO 5 4 so } MN > }

PO .

26. slope } MP 5 slope }

NO 5 1

27. Because i lines have equal slopes

28. MQ 5 2 Ï}

5 , QO 5 2 Ï}

5 , }

MQ > }

QO , NQ 5 2 Ï

}

13 , QP 5 2 Ï}

13 , }

NQ > }

QP

29. Given; } MN > }

AT ; Opposite sides of ~ are >.; } MN > } MH ; Base Angles Theorem

30. } PT > } IP and ~ATRO are given. ∠ I > ∠ T by the Base Angles Theorem. ∠ T > ∠ AOR because opposite ? of a ~ are >. ∠ I > ∠ AOR by the Transitive Property of >.

Review for Mastery

1. x 5 5, y 5 14 2. a 5 10, b 5 5 3. p 5 30.75

4. m 5 13 5. 30 6. 8 7. 508 8. 1308

9. 1 2 3 } 2 ,

1 }

2 2 10. 1 2

1 } 2 , 3 2

Challenge Practice

1. (21, 5), (7, 5), (5, 25)

2. (2, 1), (2, 3), (24, 7)

3. (a 1 2, b 1 3), (a 1 6, b 1 3), (a 2 2, b 2 3)

4. (a, b2), (a 1 a2, b2), (a, 2b2) 5. 3

6. All four vertices of a parallelogram can only have at most two equal x-values and two equal y-values. So the third and fourth vertices of the parallelogram cannot have the same x-values or y-values as the other vertices if there are already two equal x-values or two equal y-values.

7. Sample answer: Draw the diagonal of one vertex to the opposite vertex to create two congruent triangles. This can be done in two ways.

8. Sample answer: Draw a diagonal connecting two opposite vertices and then draw another diagonal connecting the other two opposite vertices.

9. Sample answer: First Way: Connect one vertex to the midpoint of the opposite side. Repeat for the opposite vertex. Next draw the diagonal connecting the remaining opposite vertices.Second Way: Repeat the fi rst method but connect the other opposite vertices with a diagonal.

10. No.

11. Sample answer:

Statements Reasons

1. RSTU and WXYZ are 1. Givenparallelograms.

2. } RS i } UT , } RU i } ST , 2. Defi nition of a}

WX i } ZY , and parallelogram}

WZ i } XY 3. ∠ RWX > ∠ YXW 3. Alternate Interior

Angles Theorem 4. ∠ YXW > ∠ WZY 4. Opposite angles

of a parallelogram are congruent.

Lesson 8.2, continuedA

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A14GeometryChapter 8 Resource Book

Statements Reasons

5. ∠ WZY > ∠ TYZ 5. Alternate Interior Angles Theorem

6. ∠ RWX > ∠ TYZ 6. Substitution Prop. of Equality

7. ∠ XRW > ∠ UZR 7. Alternate Interior Angles Theorem

8. ∠ SXT > ∠ YTZ 8. Alternate Interior Angles Theorem

9. ∠ XRW > ∠ SXT 9. Corresponding Angles Postulate

10. ∠ XRW > ∠ YTZ 10. Substitution Prop. of Equality

11. n RWX , n TYZ 11. Angle-Angle Similarity Post.

12. W XA

BZ Y

D

Given: WXYZ is a parallelogram and } WY and } XZ are diagonals of WXYZ.

Prove: D is the midpoint of the segment with endpoints on opposite sides passing through the point of intersection D.

Statements Reasons

1. WXYZ is a 1. Givenparallelogram, } WX i } YZ , and } WZ i

} XY .

2. } DW > } DY and 2. Theorem 8.10 } DX > } DZ

3. } WX > } YZ , and 3. Theorem 8.7 } WZ > } YX

4. n WDX > n YDZ, 4. SSS Cong. Post.n WDZ > n YDZ

5. Altitude of n WDX 5 5. If two triangles alt. of n YDZ, and alt. are congruent,of n WDZ 5 alt. of their altitudesn YDX. are equal.

6. Let }

AB pass through 6. AssumeD and have its endpoints on }

WX and } YZ . 7. The angle formed by 7. Vertical Angles

}

AD and the altitude Cong. Theoremof n WDX > the angle formed by } BD and the alt. of n YDZ.

8. ∠ 1 and ∠ 2 are 8. Defi nition of an right angles. altitude

Statements Reasons

9. ∠ 1 > ∠ 2 9. Right Angle Congruence Thm.

10. The triangles formed 10. ASA Congruenceby the altitudes and Postulate }

AD and } BD are cong.11. } AD >

} BD 11. Corresponding

parts of cong. triangles are congruent.

12. D is the midpoint 12. Def. of a of

} AB . midpoint

13. Sample answer:

Given: ABCD is a quadrilateral and E, F, G, and H are midpoints of their respective segments.

Prove: When the midpoints of adjacent sides are connected by segments, a parallelogram is formed.

Step 1: Place ABCD and assign coordinates. Let E, F, G, and H be midpoints of their respective segments. Find the coordinates of the midpoints.

x

y

A(0, 0)

F

E

G

H

B(h, 3k)

C(4h, 5k)

D(3h, 0)

E: 1 0 1 h }

2 ,

0 1 3k }

2 2 5 1 h } 2 ,

3k }

2 2 ;

F: 1 h 1 4h }

2 ,

3k 1 5k }

2 2 5 1 5h

} 2 , 4k 2

G: 1 4h 1 3h }

2 ,

5k 1 0 }

2 2 5 1 7h

} 2 ,

5k }

2 2 ;

H: 1 3h 1 0 }

2 ,

0 1 0 }

2 2 5 1 3h

} 2 , 0 2

Step 2: Connect the midpoints of adjacent sides with segments.

Step 3: Prove } EH i } FG and } EF i } HG by showing their slopes are congruent.

m } EH 5 0 2

3k } 2 }

3h

} 2 2

h } 2 5

23k

} 2 }

h 5 2

3k }

2h

m }

FG 5 5k

} 2 2

8k } 2 }

7h

} 2 2

5h } 2 5

2 3k

} 2 }

h 5 2

3k }

2h

Lesson 8.2, continuedA

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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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