Download Advanced Geometry Semester Exam Review

Advanced Geometry Final Exam Review
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Lesson
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1-1
1-2
1-5
1-6
2-3
803
2-5
804
2-7
2-8
3-1
805
806
807
808
809
811
812
813
3-2
3-3
3-4
3-5
4-1
4-2
4-3
4-4
4-5
4-6
4-7
5-1
6-1
6-2
6-3
6-4
6-5
6-6
7-2
Problem #
4. DE
12. c = 8, BC = 25
7. 8
2. hexagon; concave; not regular
5. P = 10.5 units; A = 4 units²
9. If three points are collinear, then they lie
of the same line. True
3. Sometimes, the perpendicular line may or
may not be coplanar with points B and C.
6. Subtraction Property
3. m∠14 = 35, m∠15 = 90, m∠16 = 55
3. BM , AL, EP, OP, PL, LM , MN
4. consective interior (CIA)
8. x = 12, y = 65
9. neither
11. y = x – 7
7. -2
9. b=12; TA=AC=37; TC=70
9. 52
3. ∆RTS ≅ ∆UVW
3. PROOF – Answer at the end.
2. PROOF – Answer at the end.
2. ∠FBG ≅ ∠FGB
5. A(0, b), B(-a, 0)
4.n = 5; ZR = 54
10. 156, 24
14. 4
6. x = 3, y = 6
4. 21
7. 10.5
5. 18
2. Yes; corresponding angles are congruent
and 4 6 10
8
7-3
814
7-4
7-5

3
4

20
3
1. Yes; ∆LNM ~ ∆YXZ; SAS Similarity
4. MNL ~ PNO; x  2.5;
PN  7.5; MN  10.5
4. 5
2
4. 41
3
Advanced Geometry Final Exam Review
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Lesson
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8-1
815
8-2
8-3
816
817
818
819
8-4
8-5
8-7
9-1
9-2
9-4
9-5
10-1
10-2
10-4
820
10-5
821
822
823
824
825
826
10-6
10-8
11-1
11-2
11-3
11-4
12-2
12-3
12-4
12-5
12-6
13-1
13-2
13-3
Problem #
6
2
10. 8 6
10. yes; no
3. x =15, y = 15 2
3
4
1. 3  0.60 ,  0.80 ,  0.75 ;
4
5
5
4
3
4
 0.80 ,  0.60 ,  1.33
5
5
3
11. 77.2
2. 153.8 ft
5. c = 29.1; mA = 80; mB = 45
2. GRAPH - Answer at the end.
8. R’(-1, 1), S’(1, -5), T’(5, -3)
2. no
5. OM = 6
3. r = 6 m; d = 12 m
7. 10π in.
12. 245
5. m∠1 =50, m∠2 =40,
m∠3 =90, m∠4 =90,
m∠5 =40, m∠6 =50
8.
4. 5 10
5. 25
6. (x – 4)² + y² = 64
2. 74 ft, 178.2 ft²
1. 432 units²
3. 173.8 ft²
1. 187.2 units²
7. 7.5 in.
6. 1737.3 ft²
1. 175 cm²
5. 81.7 cm²
6. 805.4 in.²
2. 5102.4 ft³
3. 240 in³
5. 2031.9 m³
Advanced Geometry Final Exam Review
Exam Review Part 2
2 15
1.
15
71
2.
90
3. -68.75°
3
4.
3
3
5.
2
1
2
6. y   x  1  5
12
7.
8.
 x  2
2
16
2
 x  1
9
 y  4


2
5
2
 y  3
27
1
1
Lesson 4-4 #3
Given: GWN is equilateral; WS  WI ; SWG  IWN
Prove: SWG  IWN
Proof:
Statements
Reasons
a) GWN is equilateral
a) Given
WS  WI
SWG  IWN
b) WG  WN
b) def. of equilateral
c) SWG  IWN
c) SAS Congruence
Lesson 4-5 #2
Given: S  W ; SY  YW
Prove: ST  WV
Proof:
Statements
Reasons
a) S  W ; SY  YW
a) Given
b) SYT  WYV
b) Vertical Angles Theorem
c) SWG  IWN
c) ASA Congruence
Lesson 9-1 #2
Advanced Geometry Final Exam Review