Advanced Geometry Selected Solutions: Chapter 1
[copy diagrams in your notebooks, do NOT draw in the book]
1.1 Introduction to Geometry
1. AB, BA, l 2. CED, DEC, E, 7 3. No 4. RS, ST, and RT 5. a)B b) AC or CEA c) E d)
 e) EC f) ABC g) BEC 6. a) RPO, RPS, and SPR b) O c) 3 d) TSO, PST, OST e) 8 8. a) Points
b) rays, endpoint 12. A_C_B_D
1.2 Measurement of Segments and Angles
p. 620: 1. a) x=6,y=4 b) x=2,y=5 c) x=-3, y=-7 d) x=3, y=2 2 a) x=4,y=3 or x=4,y=-3
1.2: 1. a) 61˚40’ b) 71˚42’ 2. a) 132.5˚ b) 19 3/4˚ 3. 1 and 2 4.a) T b) VW c) PR d) QSR e) 6
6. a) 30˚ b) 20˚ c) 130˚ d) Obtuse 7. a) 5 b) neither, they are same size c) 4 8. 55 9. a) 90 b) 45 c) 100 d) 142.5
10. a) 5 b) PQ  QR c) 8 15. y=x+17 16. 67.5 17. a) 0<mP<90 b) 20<x<50
1.3 Collinear, Between, and Diagram Tricks
1. 134 3. a) B and D b) No; yes c) AB and BC d) yes e) Not necessarily f) B g) G h) AF i) EB, ED j) E
and B 4. a) No since right angles can’t be assumed unless are marked b) Yes since collinearity of points can be
assumed from a diagram. 6. 54 7. a) ex: 33˚ and 40˚ b) ex: 60˚ and 70˚ c) 45˚ and 45˚ 9. 135 10. O_K_A_Y
11. two segments cross 13. a) 15 b) 3
1.4 Introduction to Proofs
1. Given, Given, Right Angle Congruence Theorem
2. Given diagram, given diagram, Straight Angle Congruence Theorem.
3. 1) A is a right angle (given) 2) B is a right angle (given) 3) A  B (Right Angle Congr Thm).
4. 1) CDE=110˚ (given) 2) FGH=110˚ (given) 3) CDE  FGH (definition of congruence).
5. 1) JK=2.5 (given) 2) NO=2.5 (given) 3) JK  NO (def. of congruence)
7. 1) 1=20˚, 2=40˚, and 3=30˚ (given) 2) XYZ=90˚(Measurement Addition Axiom for all three
angles) 3) XYZ is a right angle (def. of right angle)
11. 1) list all given as in #7 (given) 2) DBC= 90˚ (Measurement Subtraction Axiom for ABC - ABD)
3) XFE=90˚ (Measurement Addition Axiom for EFY+XFY) 4) DBC  XFE (def. of congruence)
HW1: 1) ABC and CBD are both right angles (given)
2) ABC and CBD are both 90˚ (def. of right angle)
3) ABD is 180˚ (Measurement Addition Axiom)
4) ABD is straight (def. of straight angle)
HW2: 1) ABC is a right angle and ABD is a straight angle (given)
2) ABC = 90˚(def. of right angle).
3) ABD = 180˚ (def. of straight angle)
4) CBD = 90˚ (Meas. Subtraction Axiom for ABD - ABC)
5) CBD is a right angle (def. of right angle)
1.5 Division of Segments and Angles
1. a) CO  DO b) WX  WV 2. a) NRO  PRO b) SXT  TXV  WXV
3. a) JG b) OK 4a) 30˚, b) 24˚25’ 5) a) 2;9 b) 14 6. Yes 7. 43 8. a) 4 b) 8 c) 1 d) 9 e) BD f) No
13. 1) what’s given (given) 2) conclusion (def. of midpoint). This is what I call a 2-step proof, where the first step is
what’s given and the second step is immediately what you’re trying to prove. These problems are all about finding a
reason for the 2nd step. From now on, my answers to these problems will look like “2-Step (def. of midpoint).”
Also, I may skip the first step in future answers, which you should assume is just listing what’s given.
18. 80 19. 16
1.6 Paragraph Proofs
1. See problem1 on p.36. 2. x+3x have to add up to 180 since DEH is a straight angle. So 4x=180˚. So x=45˚.
Then FEH= 3*45=135˚. Since FEH and JKM have the same measure, they are congruent.
3. Like above 4. Like above 5. Cannot be proven (may or may not be true). 6. A different kind of trick. Not only
can’t you prove this, but the conclusion can’t possibly be true. Ever. Do you know why?
7. From given diagram, we know that 1 and 2 add up to a straight angle by the Measurement Add. Axiom. This
means they have sum 180˚ by definition of straight angle. Since they are congruent, m1= m2. Now solve the two
equations 1= 2 and 1+2=180 to find that 1=2=90˚. By def. of right angle, they are right angles.
9. Given everything we know, DCE is 45˚ since CE is a bisector and the other angle is 45˚. By Measurement
Addition Axiom, BCD is 90˚. So is A since it’s a right angle. Thus, A  BCD
1.7 Deductive Structure
1. undefined terms, postulates, definitions, and theorems 2. a 3. a) Yes b) no 4. a) Theorem b) Definition
5a. i) If B, then A ii) Wet=>Rain iii) If an angle is acute, then it is a 45˚ angle. iv) If a point divides a segment into
two congruent segments, it is the midpoint of the segment. 5b i) May be either true or false since we don’t know
what A and B are ii) False since could be wet due to snow and not rain iii) False iv) True 8. Correct 9. Not correct,
since C could be obtuse 10. Not correct, are reasoning from the converse of first statement which may not be true
11. Not correct for same reason as #10. 12. Correct 14. Wendy is a monkey, Katie is the crocodile, and Jody is the
lizard. This problem gives more info than is necessary, I think. What piece of info is not necessary?
1.8 Statements of Logic
1. a) If a person is 18 years old, then he or she may vote in a federal election b) If two angles are opposite angles of a
parallelogram, then they are congruent. 2. a) Conv: If a triangle has a perimeter of 30, then each side has a length of
10; false. Inverse: If each side length of a triangle is not 10, then the perimeter is not 30; false. Contrapositive: If the
perimeter of a triangle is not 30, then each side length is not 10; true. 3. a 4. a) true b) false c) false d) false. 5. a)
d=>f b) s=> ~p c) If bobcats begin to browse, then horses head for home. 9. p=> ~g
1.9 Probability
1. 3/5 2. 1/5 3. 1/5 4. 0 5.1/3 6. 3/10 7. 2/5 8. 1/10 9. 9/25 10. 2/5 11. 5/12 12. 1/10 15. a) 1/5 b) 4/5
Advanced Geometry Selected Solutions: Chapter 2
2.1 Perpendicularity
1.a) A, B, C, D b) s EHF, GHF, and EFG 2. a) MJK b) NOR, POR c) None 3. b) 45 4. a) 30 b) 1.2
7. 2) D is a right angle (def. of perpendicular) 3) C  D (Right  Congr. Thm) 8. ROP 10. a) (-3,2) b) 117
11. 15; 30; 45 13. 2) HGK is 90˚ (Measurement Addition Axiom) 3) HGK is a right angle (def. or right ) 4)
HJK is a right angle (def. of perpendicular) 5) conclusion (right angle congruence thm).
2.2 Complementary and Supplementary Angles
1. A and C 2. 110 3. (90-y)˚ 5. 30 and 60 9. 2-Step (def. of complementary angles) 10. 60 11. 125 13. a) each
is 90˚ b) each is 45˚ 16. 110 and 70 19. 27 21. 30 22. 163 1/3 23. 12 24. a) 1/5 b) 1/10
2.3 Conclusions
1. ABC is a right angle 2. HED is a right angle 3. XS is an angle bisector 4. PQT  TQS  SQR 5. AE
 EC 6. CA  AR  RD 7. PQR is a straight angle or PQS and SQR are supplementary angles 8.
VZ  ZY, Z is the midpoint of VY 9. PAR is a right angle, PAR = 90˚ 10. BF  FD, F is the
midpoint of BD 11. EN and EG trisect AEL, EN bisects AEG, EG bisects NEL 12. PQ 
RS, PQS is a right angle
2.4 Congruency and Complementary/Supplementary Angles
1. a) 49 b) 131 c) 49 d) 41 e) 139 f) 41 g) 139 4. 144˚ 5. 35 and 55 6. Conclusion: 6  7, 2-Step (Thm 5) 7. 2Step (Thm 7) 10. 2) AOC is a right angle and DOB is a right angle (def. of perpendicular) 3) 1 and 2 are
complementary, so are 2 and 3 (def. of complementary s). 4) 1  3 (Thm 6: both complem. to 2).
11. Kind of like above, use def. of bisect, Thm 7. 12. 40 16. Is like #11 but extra step at the end to prove it bisects
17. 165˚ 18. Algebraic proof: 1+2+3= 180; 1+2=2+3; => a1=a3. 19. 98
2.5 Addition and Subtraction Properties
1. a) AD  AC b) JFG  JHG 2. a) RS  TV b) YXZ  CBD 3. 2-Step (Addition Property)
4. 2-Step (Addition Property). 5. 2-Step (Subtraction Property) 6. 32 7. 3 and 4 are congruent, use 2-Step
(Subtr. Prop) 9. 12;21 10. show have 2 right angles, they are congruent, subtraction property 11. two of the smaller
ones are congruent, then Subtr. Property 12. show HJ  KM, then use Add Property. 14. 57 17. a) Yes b)
ABC=180˚; BF  AC
2.6 Multiplication and Division Properties
1. 2-Step (Mult. Prop) 2. a) 20 b) 16 3. 2-Step (Div. Property) 4. 2-Step (Div. Property) 5. 2-Step (Div. Property)
7. a) x=6 b) y=8 11. show ZX  TW, then use Add Property or Mult. Property to prove conclusion. 12. will use
Thm 4 13. show ABF  DCF, then use Div Property for conclusion 14. 165
2.7 Transitive and Substitution Properties
1. 2-Step (Trans. Prop) 2. Same as #1 3. have to use transitive property twice 7. use def. of bisect, then transitive
property, then Mult. prop 8. 68 9. 70 10. Trans Prop, then Add Prop 11. 2 and 3 are suppl., substitution prop.
12. One way is to do def. of bisector, then Thm 7 14. 50 15. Right Angle Congr Thm, then Transitive Prop. 16.
Substitution Prop doesn’t apply to statements of perpendicularity.
2.8 Vertical Angles
1. a) FE and FC; FD and FA; BA and BC b) EFA and CFD; EFD and CFA
2. a) 119˚28’ b) 60˚32’ c) 119˚28’ 3. 43 4. 2) ACB  DCE (Vertical Angle Congruence Thm) 3) conclusion
(Transitive Property using steps 1 and 2) 5. Use Vertical   Thm twice, Transitive property twice 7. No since if
we let 3x-3=4x, x=-3 which results in negative angle measurements 8. like #4 9. first say s 1 and 2 are
supplementary (def. of supp), then use Substitution Property. 10. first Vert.   Thm, then some Trans. Prop
11. show 1  3, show 2  3, and then conclusion 12. 36;90;54 13. They’re right angles.
Advanced Geometry Selected Solutions: Chapter 3
3.1 Congruent Figures
1. TCR, Reflexive Property 2. RTE 3. YTW, Vertical Angle Congruence Thm
4. a) P’=(-5,0) Q’=(-2,-6) R’=(7,0) b) P’=(5,0) Q’=(2,6) R’=(-7,0) c) P’=(-8,0) Q’=(-5,6) R’=(4,0)
5. a) P’=(5,0) Q’=(2,-6) R’=(-7,0) b) P’=(0,0) Q’=(3,6) R’=(12,0) c) P’=(5,0) Q’=(2,6) R’=(-7,0)
3.2 Proving Triangle Congruence
1. a) GH  KO; J  M b) PS  TR;PVS TVR c) BZ  AX; BWZ  AYX
2. a) SAS b) None c) None d) ASA 3. Reflexive Prop, SAS 4. Vertical   Thm, ASA 5. def. or , Reflexive
Prop, SAS 6. 2) TSV  BSV (def. of bisect for SV) 3)TVS  BVS (def. of bisect for VS) 4) SVSV
(Reflexive Property) 5) TSV  BSV (ASA) 7. 2-Step (SSS) 8. BD  FD (Addition Property), Reflexive
Prop, SAS 9. reflexive, SSS 11. ASA at the end 12. SAS at the end 13. Division Prop, SSS 14. Thm5, SAS
15. Add Prop, SSS 16. Subtraction Prop, SAS 17. Thm5, Vertical   Thm, ASA 18. A long one: Division Prop,
the def. of midpoint (a few times), and SAS 19. Mult Prop, Reflexive, SSS 23. Thm5, def. of trisect, ASA
24. Solve for x, show PZ  PV, reflexive, SAS 25. Reflexive Prop, Transitive Prop, SAS 26. Add Prop, Right
Angle Cong Thm, ASA
3.3 CPCTC and Circles
At this point, I will start giving hints about main parts of the solution instead of writing out all the parts.
1. 2) ABC  DEF (SSS) 3) A  D (CPCTC) 2. Reflexive, ASA, CPCTC 3. Thm 19, Reflexive, def. of ,
SAS, CPCTC 4. end w/ SAS 5. end w/ ASA 6. end w/ SSS 7. A=490.9 cm2, C=78.5 cm 8. 5/3, 100, 8100
9. use FGH  FJH (ASA) 10. use Subtraction Prop, ASA 11. Area of Square- Area of circle 12. 67, 25, 13.5
13. Thm7, ASA 14. Thm5, ASA, and CPCTC (should be used at end of all of these) 15. Subtraction, Reflex, ASA
16. Transitive Prop, SAS 17. Subt Prop, ASA 18. Subtr Prop, Thm 7, Refl, ASA 19. a) (18,0) b) 100 or about 314
20. show PQO  RQO first w/ SSS
3.4 Beyond CPCTC
1. a) Median b) Altitude c) Altitude d) Both 2. SAS, CPCTC, def. of angle bisector 4. ASA, CPCTC, def. of median
5. Draw JO (2-Point Postulate), SSS, CPCTC 6. 54;54;52 7. SSS, KRP  KMP (CPCTC), Supp to Congruent
Thm 8. ASA, CPCTC, Add Prop. 9. Supp to  Thm, Thm 19, SAS 10. SAS, Subtr. Prop 11. Draw FG (2-Point
Postulate), SSS, CPCTC, Transitive Prop, Add Prop.
3.5 Overlapping Triangles
1. SSS 2. SAS 3. ASA 4. Add Prop, ASA 5. Add Prop (adding same angle to 2  s) 6. Prove ABC  DCB,
ASA 7. prove OPR  SRP with SAS, then use Subt Prop 8. GFJ  HJF 9. prove ZWX  YWA
10. prove BCD  EDC, Transitive, Add Prop 11. Prove QPS  QPR
3.6 Types of Triangles
1. Scalene 2. Scalene, Isosceles, Equilateral, Scalene, Scalene, Isosceles 3. Right, Obtuse, Right, Acute, Right,
Acute 4. Vertical   Thm 5. Radii Congruence Thm(Thm19) 6. 7; 63 7. AD  CD (def. of isosceles ),
Draw DB (2-Point Postulate), Reflexive Prop, SSS, CPCTC 8. RID  BDI (SSS), 3  1 (CPCTC), 1
and 2 are complementary (Substitution Prop), and so on 9. EFJ  HGJ (SAS) 12. Find each side length
3.7 Angle-Side Theorems
1. Thm 20, 1 and ABC also 2 and ACB are supplementary angles (def. of supplementary s), Supp to
Congruent Thm 2. Thm 21, ASA, CPCTC 3. SSS, CPCTC, Thm 21 4. Complementary to Congruent Thm, Thm
21, def. of isosceles  5. Thm 20, SAS 6. def. of altitude, Right   Thm, Reflexive, ASA, CPCTC, def. of
isosceles 7. B, A, C 8. 6<x<18 9. Division Prop, Thm 20, SAS, CPCTC 10. Radii  Thm, Trans. Prop, def.
of equilateral  11. No (try setting AC=BC and see) 12. a) 36˚ b) 54˚ 13. Thm 20, Add Prop 14. Thm 20, def. of
median, SAS, CPCTC, def. of bisector 15. Reflex, Subtr, Trans, …SAS, CPCTC, Thm 21 16. Refl, Add Prop,
SAS, CPCTC, Thm 21 20. 77 21. F  E (Thm 20 for Isos ECF), 3  E (Trans. Prop), 3 is suppl to 5
(Substitution Prop), 3  4 (Suppl to Same Thm), conclusion (Thm 21)
3.8 The HL Postulate
2. Reflexive, HL Postulate 3. Draw OX (2-Point Postulate), Reflexive, HL 4. Add Prop, HL, CPCTC 5. def. of
altitude, Reflexive, HL 8. Subtraction Prop, HL, CPCTC, Thm21 9. Subtr Prop, Thm20, ASA 11. Draw and label
2 s, CPCTC (a few times), Division Prop, SAS 12. Reflexive, Add Prop, def. of , HL, CPCTC 13. ABC 
ABD (HL), then CPCTC, def. of isosceles 
Advanced Geometry Selected Solutions: Chapter 4
4.1 Detours and Midpoints
3. prove AEB  AED
4. (1,4); (6,2); (1,1) 5. (7,2)
4.2 Now You Set Up the Proof
I am not going to draw the diagrams, but you should have some idea from what’s here.
1. Given: ABC is isosceles, BC is the base, AX is the median to base BC
Conclusions: AX is the bisector of BC, AX is  to BC
2. Draw polygon ABCD with 2 segments joining opposite vertices. Given: AB  BC  CD  DA.
Conclusion: AC  BD
3. Given: AC bisects BAD, FE  AB, GE  AD. Conc: FE  EG
5. Given: scalene POQ, PR is an altitude, Prove: PRO  PRQ. Easy proof that uses def. of altitude and
Right   Thm.
7. Given: ABC is isosceles, BC is the base. Prove: 3  4 (s 1 and 2 are inside the triangle on my diagram)
Use Thm 20, def. of supplementary, Suppl. to  s Thm
9. the vertex angle is the one opposite the base (learned in 3.6), SAS. 11. the two medians are part of what’s given,
SAS. 12. Assume the angle bisectors are segments from the vertex to the opposite side (though could also assume
they’re rays) 13. Will need to prove 3 pairs of s are  (talk about a detour!)
4.3 Right-Angle Theorem
1. Draw in PQ, PR 5. (1,-8) 6. 8 sq units
9. Thm 21 (Conv Isos  Thm) 13. small algebraic proof
4.4 Equidistance Theorems
1. Thm 24 9. (15,3) 12. after all the given info is listed, BE  AC, AD  DC 13. Trans Prop
4.5 Transversals
1. ans in back of book. 2. b) Corresponding, CL and AO,LR c) Alternate interior; CA and LR, AO
3. back of book 4. c) 3 5.
4.6 Slope
1. back of book 2. -3/5 3. -4 4. ½ 5. a) ¼ b) -4 c) -7/2 d) ¼ (I wonder why they say it’s through A, is it
necessary?) 6. 32 7. show slopes are = 8. a) yes b) yes c) show slopes are opp reciprocals 9. a) ½ b) 1/2
c) Collinear 10. a) No b) yes 12. is it the midpoint? 13. AC
Review
2. a) C and ABE b) EBD and CDB
Advanced Geometry Selected Solutions: Chapter 5
4.5 Transversals
1. ans in back of book. 2. b) Corresponding, CL and AO,LR c) Alternate interior; CA and LR, AO
3. back of book 4. c) 3 5.
5.1 Proof by Contradiction (Indirect Proof)
9. a) Corresponding b) Alternate interior
5.2 Proving that Lines Are Parallel
6. use proof by contradiction 10. 1/3 11. 0<x<110
18. Yes by SSSIA Thm 23. No since if x=25, the lines must be ||
5.3 The Converse Thms (Parallel Lines ⇒  Angles)
5. 41 6. Use 2-Point Postulate 8. Make sure you don’t use the fact that the sum of the measures of a triangle’s
interior angles is 180˚ (we don’t know that yet). 8. 70; 70; Isosceles  10. 150 11. only 2  5 17. 60 22. 70
5.4 Quadrilaterals
7. STR  VRT, TSV  RVS 8. A,S,S,A,A,N,N 9. Circle is not made of segments
13. SSSIT used 14. ASA 20. 10 21. see back of book
5.5 Properties of Parallelograms
5. 44 6. 134;46 7. 8;5;5 11. 240 12. 373.2 13. m; p; q and r 19. 28
20. 16 21. KI  KE, IT  ET 25. 1 27. 22.7
5.6
1. prop 4 of ▱; 1-Pair ▱ Prop; prop2 of ▱; can’t prove is a ▱. 11 see back of book
5.7
3. 21 6. Rhombus; Square; Rectangle; Isos. Trap. 13. ▱; Trap; Isos. Trap; Rect; Rect; Rhombus; Kite; Quad
Review
1.
4. 3 14. 6; 19 30. 60;4;102
Advanced Geometry Selected Solutions: Unit 7 (Chs 7 and 6)
I noticed that the book for some reason sometimes incorrectly says “Substitution” instead of correctly saying
“Transitive Property.”
7.1 Triangle Thms
1. 70 2. 50;60;120;70;110 3. 40;140 4. 115 5. 48;60;72 6. 48 7. 9. A, A, N, A, N (since 3rd angle stays same)
10. 90 12. 100 15. 110
7.2 More  Thms
1. No Choice Thm 4. AAS 5. Subtraction Prop, AAS, 7. Trans Prop, AIAT, CAT, def of ▱
10. Proof by contradiction that uses AAS 14. Use one of the || lines => something thms twice and then Transitive
Prop. 16. Use the Midline Thm 3 times to see that each side of the inner triangle is ½ the corresponding side of the
big triangle.
7.3 Polygon Formulas
1. 360, 900, 1080, 1800, 16380 2. 40 3. 5,9,2,0 6. same ans for all= 360 8. 1800, 360
11. Quadrilateral, hexagon 12. 5 14. a)A, b)N, c)S ,d)A 19. ¾
7.4 Regular Polygons
1. 120, 90, 45, 24, 15 15/23 2. 108, 120, 140, 150, 162 6/7 3. 6, 9, 10, 180, 48 4. 10, 6, 15, 20, 50
5. SAS 7. Equiangular Decagon 9. Use Isos  Thm, but key is F  CED by No Choice Thm
11. Pentagon 12. Equiangular Nonagon 13. a)A b)S (depends on whether it’s regular)
c) A d) S e)
6.1 Relating Lines to Planes
2. a) AB b) AB c) A,D,B or C, E, T d) one possible ans is A,C,D,E e) s f) s g) E h) C i) Yes j) CDB and
CDA 5. 3 points are guaranteed to lie in same plan but 4 may not 6. If 2 planes intersect, their intersection is a
straight line. 11. yes, no 13. not necessarily 14. Use HL Postulate 15. 7
6.2 Perpendicularity of a Line and Plane
1. 12 2. ASA 3. SAS 4. SAS 5. HL twice (have 2 dif conclusions!) 7. Thm 48, def of line  plane
8. no 10. ASA 11. SAS 12. I would use an equidistance thm at the end.
6.3 Facts about Parallel Planes
1. F, T, F, T, T 3. a) Prop #3 of || Planes, b) Thm 47 c) Thm 49 d) ABEF is a ▱ (def of ▱), AB  EF (Prop 2
of ▱) 6. AC ||  BD (Thm 49), ABCD is a ▱ (def of ▱), AB  CD (Prop 2 of ▱)
7. RS || VT (Thm
49), RSTV is a ▱ (Method 3 of proving ▱), def. of ▱ 9. SV || TW (Thm 49), T  W (Isos  Thm),
CAT Converse, Trans. Prop, def. of isos  10. Prove that it’s an isos trap and then use its prop
of base angles 
Unit 8: Similar Polygons (Ch 8)
8.1 Ratio and Proportion
2. yes 3. 4, 54/7, 47/3 5. -1/2, -1/2, yes 6. 3/2, 3/1, 1/3 7. 5/36 8. 5/7, 25/49 9. Rectangle 10. 1, ½
13 8,12,20,28 17. 150 18. yes and you don’t need prob 6 for this 19. (c+d)/(a+b) 22. 756in, 7.25in,15.75in
8.2 Similarity
1. b,c,d 2. b and c 3. 90, 30, 7.5 4. 5 5. 6,9 9. 5.6,7.5 10. 3/2, 8, 32, 3/2 11. 205 14. 38.4
15. 11 ft by 14ft
8.3 Methods of Proving s are Similar
1. Supp to  Thm, AA~ 2. CAT Converse, AA~ 3. def. of ▱, AIAT Conv, Vertical  Thm 4. Isos  Thm,
AA~ 5. def. of trapezoid, use AIAT Conv twice 8. Yes by SSS~ 9. Typo in conclusion (~ not ), Ans is Reflexive,
AA~ 11. Proof by contradiction 12. Reflexive Prop, Thm 1, AA~ 15. GHM ~ GJK (SAS~), def of ~s, CAT
16. S, A, N, S, A, S (not a  so can’t use it’s methods!) 17. Isos  Thm, AA~ 18. SAS~ or SSS~ 19. Yes by
SAS~, Yes by CAT 20. -1/2, decrease by 1.5
8.4 If 2 s are Similar, Then …
1. B  E, AA~, def. of ~ s 2. AA~ 3. AA~ 4. AA~ 5. def of ~ s, Cross-Product 6. CAT Conv twice,
AA~, def. of  ~, Cross-Product 9. 25m 10. ¾ 11. explain why s are ~, use def. of ~, CAT
12. Prop 3 of ▱, AA~, def. of ~, Cross-Product, Algebraic Substitution 13. Prove half s are ~
15. Thm 49, CAT Conv, AA~, def. of ~ 17. 6 6/7 20. 2 7/13 21. SAS~, BDC, 18
8.5 Three Theorems Involving Proportions
1. 8, 35/3 2. 6, 72/5 4. 10/3, 5, 20/3 5. 1 6. 4 7. 9 9. 5/2,10 11. 51 12. 28/5 13. 27/11,72/11 14. (0,18)
15. AIAT, Side-Splitter Thm 16. Vertical  Thm, Trans Prop, def. of bis,  Bis Thm 17. def. of trap, trans prop
of || lines (34), Thm 66 twice, Trans Prop. of Equality 18. 32x/5, 25x/4 19. 24, 4x+8 21. 10.5, 17.5
22. 20/3 23. Comp. to  Thm,…, Bis Thm 24.
Unit 9: The Pythagorean Thm (Ch 9)
Remember that odd answers are usually in the back of the book, in case I don’t provide them here.
9.1 Review of Radicals and Solving Quad Eqs
2. 15sqrt(2), sqrt(13), 5, 13, 2sqrt(3)/3, 7sqrt(3) 4. 11sqrt(3), 8sqrt(2)+7sqrt(3) , 5sqrt(3), 6sqrt(2)+sqrt(3)
6. +-3, +-8, +-5, +-3, +-4, +-4sqrt(5) 8. {0,4}, {0,10}, {0,13}, {0,8}
9.3 Altitude-to-Hypotenuse Theorem
Odd numbers are in the book. 14. 12m
9.4 Pythagorean Theorem
2. 3sqrt(2) 4. 46 10. 13 14. 6sqrt(2) 16. use Pythagorean, then thm68 to find AD, then 68 again. ans=4.8 18.
4sqrt(3)
9.5 Distance Formula
4. find distances b/w all 3 pairs of points and make sure they are equal 6. sqrt(85), 3*sqrt(5) 14. make sure that all
points are equidistant from the center 23 and 24 set up a diagram as in p.393
9.7 Special Right Triangles
2. a) short leg is 2*sqrt(3), hyp is 4*sqrt(3) b) short is 15/2, long-leg is 15*sqrt(3)/2 c) short-leg is 5*sqrt(3), hyp
is 10*sqrt(3) d) short is 7*sqrt(3)/3, hyp is 14*sqrt(3)/3
4. 8sqrt(2), 4*sqrt(2), 17*sqrt(2), 19*sqrt(2) 12. a) (1, sqrt(3)) , b) sqrt(3) , c) sqrt(3) 14. Use the 30-60-90
property twice
9.8 The Pythagorean Thm in 3D
4. 17; 17 14. 12, 8, 6, 2sqrt(34)
Unit 10: Circles
9.2 Intro to Circles
1. 19.6, 96.04 2. 90,2.5 3. 180,12.6 4. (0,4), 4 5. 12.5 6. 45 7. 40,40 8. 130,25 9. (3,3) 10. (4,-1)
11. (-3,14), -1 12. 120,9,27 14. Vertical s are , B  C since they intercept same arc!, AA~ 16. 90,3
10.1 The Circle
2. thm74, SAS 3 . thm75 5. 8mm 6. 16cm 9. Method 3 of para 11. 8 13. AEAT Conv, ASA 15. thm 19, isos
△ thm, CAT Conv 18. show that AZ is ⊥ bis, then Thm 76 at the end
10.2 Congruent Chords
2. 8 3. thm 77, isos △ thm
11. 8; 5
4. thm 78 5. show AD and BC are ≅ and || 8. use thm 74 to get RS = ST 9. AAS
10.3 Arcs of a Circle
2. a) arcQRP or arcQPR b) arcBC or arcAb c) 180 d) measure of arcPQ e) No 4. 130 5. 2) arcWX  arcYZ
(Subtr Prop) 3) conclusion (Arc To Chord Thm) 6. 2) thm 21 3) Chord to Arc Thm 8. Chords to Arcs Thm, Add
Prop, Arcs to Chords Thm 10. 216; 200; 252 12. central angle is 60, so other 2 angles must be 120; those angles are
 cause sides are , so is an equiangular (and equilateral) triangle: all sides are 12 13. you want arc length, not
measure! 14. Arc to Angle Thm, SAS, CPCTC, Thm 24 15. arcAB  arcBC, Arc to Chord Thm, thm 18, SSS,
CPCTC 16. Arc to Angle Thm, Thm 19, SAS, CPCTC, Method 2 of para. Can also be done w/ method 3 if you use
AIAT 17. Thm 77, Subt Prop 18. divide by 6,5,8 to get 60, 72, 45 19. use thm 19 and add prop to show that
diameters are , then arcs AB and DE are  by Chord to Arc Thm, subtr. Prop gives arcs AC and CE , Arcs to
Chords Thm, SSS, CPCTC
10.4 Secants and Tangents
2. use pyth thm 3. SSS or HL 4. Tangent Postulate twice, Thm 36 5. (16,0), (38,0); 3 6. Two-Tangent Thm to
find x; ans= 2.5 8. thm 24 9. Two Point Postulate twice and Subt Prop (4 step proof) 10. 23 16. 3;5;8 17. 9;
17 18. Two-Tangent Thm 3 times, Trans Prop 19.
10.5 Angles Related to a Circle
2. 65 4. 42; 50; 40 10. 76 12. 40 14. 110 24. 107; 75 26. 52
10.6 More Angle-Arc Theorems
1. def. Of midpoint, Inscribed s Thm, def. Of bis 2. Thm 91, HL 3. 10 4. 17, 81.5 5. 97 7. 137 9. 42 16. 86
17. a)29 b)81 18. Inscribed s Thm twice, AA~
10.7 Inscribed and Circumscribed Polygons
2. 105, 75 4. Supp to Same  Thm 10. 24 13. given: is a trap inside circle, prove: is isos trap. Use thm 93, def.
Of trap, SSSIT conv, Supp to Same Thm, Method 2 of isos trap. 14. 18
10.8 The Power Theorems
2. 50; 58
6. 18; 4 8. y=9 15. x^2 =8100 * 100
10.9 Circumference and Arc Length
2. 28.00, 49.97, 8.50,14.01 4. 94.2,942.5, 94,247.8
8. 20π; 72 degrees 10. 1257
Unit 11: Area
11.1 Intro to Area
1. 207,78,80 2. 75 sq cm 3. 120,84 4. (height = altitude) 8mm,10mm. 5. 144,50,7,36,81 6. 104, 102, 116
9. 256 sq cm [assume symmetry!] 10. 184 sq cm 11. 9m by 15m 12. A=32, P=16sqrt(2) [use pyth thm on halves of
ABCD] 13. (18,8) [find area of AECT, then find EC]
11.2 Areas of the Parallelogram and Triangle
1. 198 sq mm, 102 sq cm, 35 2. 120 (they give you extra info) 3. 35, 144
4. 6 5. 48 (draw an altitude, what does it do?) 6. 187 7. 14 8. 28(find the height of dashed , it is also the height
of one you care about), 60 9. 18sqrt(3), 72, 36sqr(3) 10. 173.2; 84.9 11. 84, 128 13. 12 14. all have same area
of 80 sq mm since have same base and height! 15. 85 (subtraction method) 16. 12; 8 17. 33 (subtraction) 18. 90;
18
11.3 Area of the Trapezoid
1. 104, 13 2. 75; 72; 78; 312 4. 9 5. 11 6. 22 7. 396 [use divide and conquer on these]; 78
8. 30*sqrt(3); 115 9. 24 10. 10; 5; 5; 70 11. 89 [other base is 13, use pyth to find height of traps]
12. 50 13. 72
11.4 Area of a Kite
1. 60 2. 56; 85 [remember the ]; 160 3. 5 4. 168 5. 336; 500 6. 85 7. 78 [blast from the past, how to find the
height of the ?] 8. 50*sqrt(3) 9.
11.5 Areas of Regular Polygons
1. 36 2. 9*sqrt(3); 49*sqrt(3)/4; 16*sqrt(3); 3*sqrt(3) 3. apothem is NOT the height! Use 30-60-90 to find side.
5. 12; 6*sqrt(3); 216*sqrt(3) 6. [notice 45-45-90] 100; 576; 49; 50; 72; 9 7. 3mm 8. 6km 9. 324 10.
27sqrt(3) 11. 288*sqrt(3) 14. 32 15. 108*sqrt(3); 288; 216*sqrt(3)
11.6 Areas of Circles, Sectors, and Segments
2. 4; 13 3. 20π cm 4. 81π sq dm 5. π; 32π; 2π; 27π; 75π; 9π 6. 57sq m; 72sq m; 15sq m
7. 150+25π 10. 144 11. 16π-32; 32π/3 - 16sqrt(3); 12π – 9 sqrt(3) 12. 16π 14. c) 100-25π d) 25*sqrt(3)25π/2 e) 50π – 100 f) 150sqrt(3) – 75π
15. 12π [assume little circles have same radii]
11.7 Ratios of Areas
1. 1:1; 1:2; 15:8; 5:6 2. 25:4; 4:9; 4:1; 1:9 3. 1:1; 1:2; 2:1 4. 16:81 5. 3:4 6. Why, of course she does! Not.
new garden is 4 times as big. 7. 1:16 8. 49:16 9. 64:225 [watch out for unnecessary info]; 1:2; 1:1; 4:9
10. 4:81 [don’t forget to get the whole big side]; 2:9; 1:3; 4:9 11. 5:4 (use Pythag) 12. 2:3; 2:3 13. 250
15. 16:81
11.8 Hero’s and Brahmagupta’s Formulas
2. 16*sqrt(3) 4. a) not a real triangle, fails the  inequality b) same thing 5. 32.50 8. 168/13; 12; 56/5
9. 2*sqrt(6) 10. 76.5 [use the Distance formula!]
Unit 12: Surface Area and Volume
12.1 Surface Areas of Prisms
1. 550 sq cm; 282 sq mm; 810 sq in 2. 150; 126 3. 550; 120; 790 4. 90+18sqrt(3); 360+72*sqrt(3)
5. 150; 294 6. 236; 144+66*sqrt(3) 7. a) 480;552 b) 120;132 c) 2500;2620 d) 360; 360+108*sqrt(3)
8. 226 [22=top, 22=bottom, 63=2 faces facing you, 28=two bases, 91=2 back faces]
12.2 Surface Areas of Pyramids
1. 60; 240; 340 2. 120; 64*sqrt(3); 360+64*sqrt(3) 4. no; 9; 740 5. [use similar s SAS~ to find AB=6]192; 48;
36; 144; 1:4; 36 6. 105 7. 72; 432; 324; 756 8. a)700;896 b) 544; 800
12.3 Surface Areas of Circular Solids
1. 196π; 36π; 36π; 25π 2. LA=24π,TA=33π; LA=140π;TA=238π; LA=40π,TA=72π; LA=3π,TA=4π
3. 6 4. 34π; 108π [So that’s what hemisphere means!] 6. 35π+20 7. 90π; 66π; 480π 8. 24 [notice that height
of cone must be 3 and l=5] 10. 93π
12.4 Volumes of Prisms and Cylinders
1. 300π; 720 2. (6+5π) cubic m 3. 357 4. V=300; TA=280 6. 22 [l=height in a right prism!] 8. (2,0,8)
9. V=600, TA=620 10. 243π, 148.5π; 360π, 156π+240 11. 6 14. 6.75 [Volume/Base=height] 15. 189
16. Yes by 3.6 cubic cm 17. 621 cubic in; 439 sq. in 18. [Do Vol of outer cyl – Vol of inner, notice height of
outer cyl=20, inner h=19] ans=770 cubic ft
12.5 Volumes of Pyramids and Cones
1. 490sqrt(3) 2. Find the height first, ans=314.16
3. 400;360 4. 9 5. 1080π; 369π; 450π 6. 533 cubic m
7. 1451 cubic ft 8. 391.8 [use 30-60-90] 9. 200 11. 72π (not a frustrum question!) 12. 4423 13. 720 cubic m
14. 15;9 and 15; 1018; 4712; 3695
12.6 Volumes of Spheres
1. 36π; 972π; 500/3 π 2. V=288π, TA=144π 3. 481 cubic m 5. 523 cubic ft 6. [subtract big sphere from
small] 19 cubic cm 7. 240π; 132π 8. height=radius here 56,549 cubic m; 2827 sq m; twice as much; 42m
9. 71 cubic mm 10. 8:125; 4:25 11. 138π cubic ft; 105π sq ft
Chapters 14 and 15
14.3 The Concurrence Theorems
4. Midpoint of the hypotenuse. 6. 6, 15, 4, sqrt(2)
8. Right 10. None
12. TN=18
14. (5,4)
15.1 Inequality Properties
6. a) PX < PQ 8. Comp of X > Comp of Y 10. angle is between 60 and 90, comp is between 0 and 30 12. a)
equal b) exterior > interior
15.2 Inequalities in a Triangle
10. a) AD b) XY 12. a) 55<x<110, b) 0 < x < 90 c) 0 < x < 20 d) x=7.5 sqrt(3) 14) a) Acute b) None c)
Obtuse d) Right 16. look at it as two triangles (draw 1 diagonal) 18. think about exterior angle
15.3 Hinge Theorems
2. R 4. XZ > AB