LESSON Challenge Practice 8 - RJS · PDF fileDescribe two different ways to divide a parallelogram into two pairs of congruent triangles using three ... Challenge Practice 1. ... Theorem
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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.
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.
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}
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),