Preparing for the Regents Examination Geometry, AK
Preparing for the Regents Examination Geometry, AK Preparing for the Regents Examination Geometry, AK
6 1 3 because they are vertical angles. 4 2 because they are vertical angles. By the substitution postulate, 2 3. By the transitive postulate, 4 3, or 3 4 by the symmetric property. 7 Since −− PQ bisects ABC, then 2 CBQ. Since CBQ and 1 are vertical pairs, CBQ 1. By the transitive postulate, 2 1, or 1 2 by the symmetric property. 8 Since CDE is a right triangle, the two angles that are not the right angle are complementary because there sum is 180 (whole triangle) 90 (right angle) 90. Therefore, FED is complementary to FCD. By substitution, EDF is complementary to FCD. 9 mAED mBED 180 4x 20 9x 4 8 180 x 16 mCEB mAED 4x 20 4(16) 20 84 10 mAED mCEB 5 x 7x 26 x 13 mAED 5(13) 65 mAED mDEB 180 65 mDEB 180 mDEB 115 11 mAED mCEB 1 2x 6x 10y 6x 10y mAED mDEB 180 12x 6x 180 18 x 180 x 10 6x 10y 6 0 10y y 6 mAEC mDEB 10y 10(6) 60 12 mCEB mDEB 180 9y 12y 1 2 180 y 8 mAEC mDEB 12(8) 12 108 Note: Since there are many variations of proofs, the following is simply one set of acceptable statements to complete each proof. Depending on the textbook used, the wording and format of reasons may differ, so they have not been supplied for the method of congruence applied in each problem. (These solutions are intended to be used as a guide—other possible solutions may vary.) 13 1. −− AC −− DF 2. A D 3. C F 4. ABC DEF (ASA ASA) 14 1. −− HE −− FE 2. −−− HG −− FE 3. −− HF −− HF 4. EFH GHF (SSS SSS) 15 1. A, B, C, D lie on circle O. 2. −−−− AOC ; −−−− BOD 3. −−− AO , −− BO , −−− CO , −−− DO are radii. 4. −−− AD −−− CO 5. −− BO −−− DO 6. AOB COD (If two lines intersect, the vertical angles are congruent.) 7. ABO CDO (SAS SAS) 16 1. −− QT −− RT 2. QT RT 3. −− TV −− TU 4. TV TU 5. QT TV RT TV 6. QT TV RT TU 7. −−− QTV ; −−− RTU 8. QU RU 9. −−− QV −−− RU 10. −− PU −− VS 11. −− PU −−− UV −− VS −−− UV 12. −− PV −− SU 13. −− PQ −− RS 14. PQU SRU (SSS SSS) 17 1. C is the midpoint of −− AE . 2. −− AC −− CE 3. 1 2 4. 3 BCD 5. 4 BCD 6. 3 4 7. ABC EDC (ASA ASA) Chapter Review 17
18 1. I is the midpoint of −− EG . 2. −− EI −− IG 3. EI IG 4. −− EG −− EI −− IG 5. EG EI IG 6. EG EI EI or EG 2EI 7. EH 2EI 8. EG EH 9. −− EG −− EH 10. FGE IEH 11. FEG HIE 12. FGE IEH (ASA ASA) 5-1 Line Segments Associated With Triangles (page 71) 1 One line 2 C Median CHAPTER 5 A E F D B Angle Bisector 3 −− AE and −− BE 4 −− AB and −−− CD 5 ACF and BCF 6 ADC and BDC Altitude 18 Chapter 5: Congruence Based on Triangles 19 1. AEB, CED are right angles. 2. AEB CED 3. A C 4. −− AE −− CE 5. ABC CDE (ASA ASA) 20 1. −−− DC −−− AD 2. −− AB −−− AD 3. −−− DC −− AB 4. −− DF −− BE 5. −− EF −− EF 6. DF EF BE EF 7. −− DE −− BF 8. 1 2 9. ABF DCE (SAS SAS) Congruence Based on Triangles 5-2 Using Congruent Triangles to Prove Line Segments Congruent and Angles Congruent (pages 72–73) Note: Since there are many variations of proofs, the following is simply one set of acceptable statements to complete each proof. Depending on the textbook used, the wording and format of reasons may differ, so they have not been supplied for the method of congruence applied in each problem. (These solutions are intended to be used as a guide—other possible solutions may vary.)
- Page 1 and 2: ANSWER KEY Preparing for the REGENT
- Page 3 and 4: Contents Chapter 1: Essentials of G
- Page 5 and 6: 1-4 Angles (pages 9-10) 1 (4) It is
- Page 7 and 8: 6 True 7 True 8 True 9 True 10 Answ
- Page 9 and 10: 3 1. e ∨ ~f 2. ~f → g 3. ~e 4.
- Page 11 and 12: CHAPTER 3 3-1 Inductive Reasoning (
- Page 13 and 14: 6. BAC DAE 6. Transitive property.
- Page 15 and 16: CHAPTER 4-1 Setting Up a Valid Proo
- Page 17 and 18: 6 Given: BA bisects CBE 1 2 2 4
- Page 19: 1 1. −−−− ABCD 2. 1 2 3. _
- Page 23 and 24: 5-3 Isosceles and Equilateral Trian
- Page 25 and 26: 12. −−− HG −−− DC 13.
- Page 27 and 28: 5 1. −− FG is the perpendicular
- Page 29 and 30: Chapter Review (pages 84-85) Note:
- Page 31 and 32: 20 Use constructing congruent angle
- Page 33 and 34: 6-3 Line Reflections and Symmetry (
- Page 35 and 36: 11 A(0, 8), B(2, 2), C(6, 4) (8) 10
- Page 37 and 38: 5 (3) (x, y) → (x, 2y) 6 (1) tran
- Page 39 and 40: 11. mABC mADC 12. 2mABD 2mADB 13.
- Page 41 and 42: 6. mDAB mCAD mDCB mACD 7. mCAB
- Page 43 and 44: g e 17 a _ f d b Undefined c a _
- Page 45 and 46: 5 BIG is isosceles because it has t
- Page 47 and 48: ___ 27 a M KA (5, 1), M ___ AT (4
- Page 49 and 50: 8 106 9 mA 75, mC 67 10 57 11 60
- Page 51 and 52: 9-5 The Sum of the Measures of the
- Page 53 and 54: 9-7 The Converse of the Isosceles T
- Page 55 and 56: 12 1. ___ CE ___ BA , ___ BD ___
- Page 57 and 58: 12 a mx 45, my 45 b mx 98, my 8
- Page 59 and 60: Quadrilaterals 10-2 The Parallelogr
- Page 61 and 62: 7. MAD RCB 8. MAD RCB (SAS SAS)
- Page 63 and 64: 5. RSQ TSV (Vertical angles) 6. QR
- Page 65 and 66: 3 4x 2 3x 3 x 5 RS 18 4 Perime
- Page 67 and 68: Note: Since there are many variatio
- Page 69 and 70: 28 Enclose PAT in a large rectangle
6 1 3 because <strong>the</strong>y are vertical angles.<br />
4 2 because <strong>the</strong>y are vertical angles. By<br />
<strong>the</strong> substitution postulate, 2 3. By <strong>the</strong><br />
transitive postulate, 4 3, or 3 4<br />
by <strong>the</strong> symmetric property.<br />
7 Since −−<br />
PQ bisects ABC, <strong>the</strong>n 2 CBQ.<br />
Since CBQ and 1 are vertical pairs,<br />
CBQ 1. By <strong>the</strong> transitive postulate,<br />
2 1, or 1 2 by <strong>the</strong> symmetric<br />
property.<br />
8 Since CDE is a right triangle, <strong>the</strong> two angles<br />
that are not <strong>the</strong> right angle are complementary<br />
because <strong>the</strong>re sum is 180 (whole<br />
triangle) 90 (right angle) 90. There<strong>for</strong>e,<br />
FED is complementary to FCD. By substitution,<br />
EDF is complementary to FCD.<br />
9 mAED mBED 180<br />
4x 20 9x 4 8 180<br />
x 16<br />
mCEB mAED<br />
4x 20<br />
4(16) 20<br />
84<br />
10 mAED mCEB<br />
5 x 7x 26<br />
x 13<br />
mAED 5(13) 65<br />
mAED mDEB 180<br />
65 mDEB 180<br />
mDEB 115<br />
11 mAED mCEB<br />
1 2x 6x 10y<br />
6x 10y<br />
mAED mDEB 180<br />
12x 6x 180<br />
18 x 180<br />
x 10<br />
6x 10y<br />
6 0 10y<br />
y 6<br />
mAEC mDEB<br />
10y<br />
10(6) 60<br />
12 mCEB mDEB 180<br />
9y 12y 1 2 180<br />
y 8<br />
mAEC mDEB<br />
12(8) 12 108<br />
Note: Since <strong>the</strong>re are many variations of proofs,<br />
<strong>the</strong> following is simply one set of acceptable<br />
statements to complete each proof. Depending<br />
on <strong>the</strong> textbook used, <strong>the</strong> wording and <strong>for</strong>mat<br />
of reasons may differ, so <strong>the</strong>y have not been<br />
supplied <strong>for</strong> <strong>the</strong> method of congruence applied<br />
in each problem. (These solutions are intended<br />
to be used as a guide—o<strong>the</strong>r possible solutions<br />
may vary.)<br />
13 1. −−<br />
AC −−<br />
DF<br />
2. A D<br />
3. C F<br />
4. ABC DEF (ASA ASA)<br />
14 1. −−<br />
HE −−<br />
FE<br />
2. −−−<br />
HG −−<br />
FE<br />
3. −−<br />
HF −−<br />
HF<br />
4. EFH GHF (SSS SSS)<br />
15 1. A, B, C, D lie on circle O.<br />
2. −−−−<br />
AOC ; −−−−<br />
BOD<br />
3. −−−<br />
AO , −−<br />
BO , −−−<br />
CO , −−−<br />
DO are radii.<br />
4. −−−<br />
AD −−−<br />
CO<br />
5. −−<br />
BO −−−<br />
DO<br />
6. AOB COD (If two lines<br />
intersect, <strong>the</strong><br />
vertical angles are<br />
congruent.)<br />
7. ABO CDO (SAS SAS)<br />
16 1. −−<br />
QT −−<br />
RT<br />
2. QT RT<br />
3. −−<br />
TV −−<br />
TU<br />
4. TV TU<br />
5. QT TV RT TV<br />
6. QT TV RT TU<br />
7. −−−<br />
QTV ; −−−<br />
RTU<br />
8. QU RU<br />
9. −−−<br />
QV −−−<br />
RU<br />
10. −−<br />
PU −−<br />
VS<br />
11. −−<br />
PU −−−<br />
UV −−<br />
VS −−−<br />
UV<br />
12. −−<br />
PV −−<br />
SU<br />
13. −−<br />
PQ −−<br />
RS<br />
14. PQU SRU (SSS SSS)<br />
17 1. C is <strong>the</strong> midpoint of −−<br />
AE .<br />
2. −−<br />
AC −−<br />
CE<br />
3. 1 2<br />
4. 3 BCD<br />
5. 4 BCD<br />
6. 3 4<br />
7. ABC EDC (ASA ASA)<br />
Chapter Review 17
Change language