chapter 3 • parallel lines and planes page 75 ... to chapter 3, pages 87-88 2 7 19. a. false b....

• CLASSROOM EXERCISES

1. a. L i , L5; L2, L6; L3, L7; L4, L8 b . L 2 , L8; £3, £5 c . £ 2 , £5; £3, L8d. L i , £7; £4, £6 e . L i , £6; £4, L7

2. s-s. in A 3 . corr. A 4 . none 5 . alt . int. A 6 . none7. corr. A 8 . s-s. in A 9 . alt . in A

10. a. I I b . I I c . skew d . i n e . skew f . skew11. A B , EJ , FK, HM , I N , DC12. H I , ID, FE, EA, JK, JB, MN, NC, BC, AD13. E J , FK, GL , HM , I N 1 4 . Answers may vary; for example, EF, HI; BJ, LM15. never 1 6 . always 1 7 . sometimes 1 8 . never19. a . sometimes b . sometimes c . sometimes

Pages 76-77 • WRI TTEN EXERCISES

A 1 . alt . int. A 2 . corr. A: 3 . s-s. int. A 4 . alt . int. A 5 . corr. A

6. corr. A 7 . PQ, SR; SQ 8 . SP , QR; SQ 9 . PQ, SR; PS10. PS, QR; SR 1 1 . PQ, SR; QR 1 2 . corr. A 1 3 . corr. A14. alt . in L s 1 5 . s-s. in A 1 6 . s-s. in A 1 7 . corr. A 1 8 . Corr. A are19. Al t . in A are 2 0 . S-s. int. A are supp.

B 21 . a. Answers may vary. b . Same as mL1 + mL 2 c . Same as mL1 + mL 2d. When 2 nonparallel lines are cut by transversals, the sum of the measures ofs-s. in A is a constant.

22. Check students' drawings. 2 3 . B H, CI , DJ, EK, FL 2 4 . GH, ED, KJ25. Answers may vary. FL, EK, DJ, CI , GL, LK, J I ,I N 2 6 . CDJ I ; GHIJKL27. ABHG, BCIH, CDJI, DEKJ 2 8 . 429. I f the top and bottom lie in 11 planes, then CD and LI are the lines of intersection of

DCIJ with 2 11 planes, and are therefore 11 •30. always 3 1 . sometimes 3 2 . never 3 3 . always 3 4 . sometimes35. sometimes 3 6 . sometimes 3 7 . always 3 8 . sometimes 3 9 . sometimes

C 40-42. Sketches may vary.40.

CHAPTER 3 • Parallel Lines and Planes

23

41.

24 K e y to Chapter 3, pages 78-82

42. /

/

Page 78 • EXPLORATIONS

A: corr. A , alt. in A , vert. A, alt. ext. IS_supp. s - s . in , s-s. ext. A , adj. A

Page 80 • CLASSROOM EXERCISES

1. / II p 2 . I f 2 11 lines are cut by a trans., then corr. A are =.3. I f 2 11 lines are cut by a trans., then alt. in A are4. I f 2 11 lines are cut by a trans., then s-s. int. A are supp.5. I f 2 11 lines are cut by a trans., then corr. A are6. I f 2 11 lines are cut by a trans., then alt. in A are 7 . Vert . A are8. I f a trans. is I to one of two 11 lines, then it is I to the other one also.9. I f 2 11 lines are cut by a trans., then s-s. int. A are supp.

10. m L 4 = mL 5 = mL8 1 3 0 ; mL2 = mL 3 = mL 6 = mL 7 -= 5011. mL 4 = mL5 = mL 8 = x; mL2 = mL 3 = mL 6 = mL 7 = 180 — x12. mL 3 + mL4 = 180; 3mL3 = 180; mL3 = 60, so mL6 = 6013. m L 5 + mL 6 = 180; 2mL6 + 20 = 180; mL6 = 80 = mL2; mL1 = 10014. I n Step 2 he used Thm. 3-2, which relies on Post. 10.


A 1 . L 3 , L6, L8 2 . L 6 , L9, L14 3 . L 4 , L5, L7, L10, L12, L13, L154. L i , L3, £6, £8, £9, £11, £14, L16 5 . 110, 70 6 . x , 180 - x7. x = 60, y = 61 8 . 4x + 14x = 180, x = 10; 2y = 90; y = 459. 120 + x = 180, x = 60; 60 -= 3y + 6, y = 18

10. x 7 0 ; 50 + 70 + y = 180, y 6 011. 3x = 42, x = 14; 3(14) + 6y — 6 = 90, y = 912. x = 55; y + 55 + 50 = 180, y = 7513. 1. Given 2 . Def . of I lines 3 . / II n 4 . I f 2 11 lines are cut by a trans., then

corr. A are 5 . m L 2 = 90 6 . Def . of I linesB 14 . 5 6 ; 56 + 24 + y 1 8 0 , y 1 0 0 ; 56 + 24 + 4z = 180, z = 25

15. x 7 0 ; 5y + 10 = 70, y = 12; z + 32 = 5(12) + 10, z = 38

Key to Chapter 3, pages 80-82 2 5

19. 4x - 2y = 110, 4x + 2y = 130; 8x = 240, x = 30; y = 5

20. Statements R e a s o n s1. k I 1. Given2. L2 L4 2. If 2 11 lines are cut by a trans., then

corr. A are3. L4 L7 3. Vert. A are4. L2 L7 4. Trans. Prop.

16. 3x = 90, 3 0 ; 8y + 4 = 68, y = 8; 2z + 8(8) + 4 = 90, z = 1117. a . mL DA B + 116 = 180, mLDAB = 64; mLKAB = 32; mLDKA = 32

b. More information is needed.

18. 2x + y 6 0 , 2x - y = 40; 4x = 100, x = 25; y = 10

21. Statements R e a s o n s

1. k II 12. L i L 8 , or mL1 = mL 8

3. m L 8 + mL 7 = 1804. m L 1 + mL 7 = 180

5. L i is supp. to L7.


1. k II n2. L i L 2 , or mL1 = mL 2

3. m L 2 + mL 4 = 1804. mL 1 + mL 4 = 180

5. L i is supp. to L4.

23. a .Statements

1. A B II DC; AD 11 B C2. L A is supp. to LB;

LC is supp. to LB.3. L A L C

b. Yes, by the same reasoning as in part (a).

1. Given

2. I f 2 II lines are cut by a trans., thenalt. int. A are

3. L Add. Post.

4. Substitut ion Prop.5. Def . of supp. A

1. Given

2. I f 2 11 lines are cut by a trans., thenalt. int. A, a r e3. L Add. Post.

4. Substitut ion Prop.5. Def . of supp.

Reasons

1. Given

2. I f 2 II lines are cut by a trans., thens-s. int. A are supp.

3. I f 2 A are supps. of the same L,then the 2 A are


C 24 . Statements R e a s o n s

1. A S 11 B T2. m L 1 = mL 4

3. m L 2 = mL 5

4. m L 4 = mL 55. m L 1 = mL 26. S A bisects L. BSR.

1. Given

2. I f 2 11 lines are cut by a trans., thencorr. A are

3. I f 2 II lines are cut by a trans., thenalt. in A are

4. Given5. Substitut ion Prop.6. Def . of L bis.

25. Steps 1-5 of the proof in Ex. 24 prove that InL1 = mL2. SB bisects LAST,so mL2 = mL3. Since mL1 + mL2 + mL 3 1 8 0 , 3 m L 1 = 180 bySubstitution, and mL1 = 60.

Page 82 • MI XED REVI EW EXERCISES

1. a. True b . I f 2 lines form a d j . A, then the lines are I . c . True2. a . True b . I f 2 lines are not skew, then they are 1 - c . False3. a . True b . I f 2 A are supp., then the sum of their measures is 180. c . True4. a . True b . I f 2 planes do not intersect, then they are - c . True


1. KG 11 DE. I f 2 lines are cut by a trans. and s-s. in A are supp.,then the lines are II

2. O X / Z . I f 2 lines are cut by a trans. and corr. A are t h e n the lines are3- LA 11 TS. If 2 lines are cut by a trans. and s-s. in A_ are supp.,

then the lines are 114. GA 1 EM. I f 2 lines are cut by a trans. and alt. in A are t h e n the lines are5. P L 11 A R 6 . P A 11 L R 7 . no segs. 11 8- P L I I A R 9 . n o s e g s .

10. P L 11 A R 1 1. P A 1 1 L i

12. Through a pt. outside a line, there is a line II t o the given line. Through a pt. outsidea line, there is no more than one line 11 t o the given line.

13. Through a point outside a line, there is a line I to the given line. Through a pointoutside a line, there is no more than one line I to the given line.

14. one 1 5 . one 1 6 . one 1 7 . one; Protractor Post. 1 8 . Inf initely many


19. a. False b . True c . True d . True

20. I f k I , then L l •-•-• L 2 . I f k I I n , t h e n L i L 3. T h e re f o re , L2 L3 and / It.

(Substitution Prop.; if 2 lines are cut by a trans. and corr. A are t h e n the lines areIL)


A 1 . A B II P C 2 . A E 11 B D 3 . A B 11 P C 4 . F B 11 E C 5 . none6. A E II B D 7 . none 8 . none 9 . A E 11 B D 1 0 . A E 11 B D 1 1 . A E 11 B D

12. F B 11 E C 1 3 . A E 11 B D 1 4 . none 1 5 . F T E C ; AE 11 B D16. A B II FC; AE 11 B D17. 1. Given 2 . Vert . A are 3 . Trans. Prop. 4 . I f 2 lines are cut by a trans.

and corr. A are t h e n the lines are

B 18 . ( x 4 0 ) + (x + 40) = 180,2x = 180, x = 90; (x - 40) + y = 180,(90 - 40) + y = 180, y = 130

19. 3x = 105, x = 35; 105 = 180 - (2y + x), 105 = 180 - (2y + 35), 2y -= 40, y 2 020. PQ R S . L l =•-•-- L 2 , L 2 L 5 ( V e r t. A a re and L5 L4, so Li L4. Since

alt. int. A are P Q R S •

21. L l '-- - -- L 4 ; L2 L5. If L3 L6, then PQ 11 RS because alt. int. A are

If PQ II RS, then L i L 4 because they are alt. int. A. L 2 L 5 becausevert. A are


1. L i is supp. to L2.2. m L 2 + mL3 = 180

3. L 3 is supp. to L2.4. L 1 L 3

5. k n


1. k t ; n t2. mL 1 = 90; mL2 = 903. m L 1 = mL2, or L i L 24. k II n

1. Given2. L Add. Post.3. Def . of supp. A4. I f 2 A are supps. of the same L,

then the 2 A are

5. I f 2 lines are cut by a trans. and alt.int. A are t h e n the lines are II

1. Given2. Def . of I lines3. Substitut ion Prop.4. I f 2 lines are cut by a trans. and corr.

A are t h e n the lines are 11




1. BE bisects LDBA. 1. Given

2. L2 '----- L3 2. Def. of L bis.

3. L3 L l 3. Given4. L2 L l 4. Trans. Prop.5. CD II B E 5. If 2 lines are cut by a trans. and alt.

int. A are' ----,, t h e n t h e l i ne s are II

25. Statements Reasons

1. BE D A ; CD DA 1. Given

2. CD 11 B E 2. In a plane, 2 lines I to the same lineare II

3. L l L 2 3. I f 2 II lines are cut by a trans., then

29. 2x = 5y, x - y 30; x 3 0 + y, 2(30 + y) = 5y; 60 + 2y = 5y, 3y = 60,y 2 0 ; x - y 30, x - 20 = 30, x = 500

1. L C '--- L3

2. CD II B E

3. B E I DA4. CD I DA

27. mL RS T = 40 + 70 = 110

alt. in A are

1. Given2. I f 2 lines are cut by a trans. and corr.

A are t h e n the lines are II3. Given4. I f a trans. is I to one of 2 II lines,

then it is I to the other one also.

28. m L 1 = 70, mL2 = 60, mLRST = 130

30. The bisectors appear to be 11Given: BD 11 CP' BG bisects LABD;

CH bisects L B CF .Prove: BG I I CH

• APPLICATION

Key to

31.

Chapter 3, pages 89-92

Statements

29

Reasons

1. B D I I CI2. mL A B D = mL B CT

13. ' f ri' / LABD = i mL B CF

4. BG bisects LABD;CH bisects LBCF.

15. mL A B G = -mLABD;2

1mL B CH = - mL B CF2

6. mL A B G m L B C H7. BG II C H

1. Given

2. I f 2 lines are cut by a trans., thencorr. A are

3. Mult . Prop. of =

4. Given

5. L Bis. Thm.

6. Substitut ion Prop.7. I f 2 lines are cut by a trans. and corr.

A are m, then the lines are II

X2 + 3x = 180; x2 + 3x - 180

= 0; (x + 15)(x - 12) - 0; x + 15 - 0 orx 1 2 = 0; x = - 15 (reject) or x = 12; x = 12

Page 89 • SELF-TEST 1

1. sometimes 2 . never 3 . always 4 . sometimes 5 . always6. L 3 , L6; L4, L5 7 . Answers may vary; L i , L5; L2, L6; L3, L7; L4, L88. L 3 , L5 or L4, £6 9 . £ 4 ; £3 1 0 . £ 2 , £8; £4, £7 1 1 . £ 2 , £8 1 2 . 65; 115

13. E B 11 DC 1 4 . none 1 5 . A E II B D 1 6 . one, one


The sum of the measures of the A inside the A is 180. The sum of the measures ofthe A outside the A is 360.

5. c6. a


7.


1. sometimes 2 . always 3 . never 4 . sometimes5. The sums of the meas. of the 2 A in each A are = T h e meas. of the third L in each

A must = 180 — sum.6. Le t meas. of each L = x; 3x = 180, x = 607. I n AABC, if mLA 9 0 and mLB 9 0 , then m.LA + mL B + mL C > 180 since

mL C > O.8. I n AABC, if mLC = 90, then InLA + mL B = 180 — 90 = 90.9. x = 90 1 0 . x = 105 1 1 . x = 35 + (180 — 140) = 75

12. The bis. of LJ may not contain the midpt. of PE.13. The line through P I to JE may not contain the midpt. of JE.14. P X may not be 11 t o JE15. B y the Substitution Prop., mL1 + mL 2 + mL 3 = mL 3 + mL4. B y the

Subtraction Prop. of = , mL1 + mL2 = mL4, which proves Thm. 3-12.16. mL 1 + mL 2 = 90, so this illustrates Cor. 4,

the acute A of a rt. A are comp.

17. mL 1 + mL 2 + mL 3 = 180, so this illustrates Thm. 3-11, the sum of the meas. ofthe A of a A is 180.


A 1 . a.

Ex. 16

b.

8.

C.

X


2. a.

3. not possible 4 .

5. 180 6 . 30 7 . 95 8 . x + (x - 20) = 80; 5 09. 4x + 30 = 6x 2 0 ; x = 25

10. m L 9 + mL10 + mL11 = (mL7 + mL8) + (mL6 + mL8) + (mL6 + mL7) =2(mL6 + mL 7 + mL8) = 2 • 180 = 360

11. 3 0 ; y = 50 + 30 = 80 1 2 . x = 110; y = 110 - 40 = 7013. x = 40; y 9 0 - 40 = 50

B 14 . x = 65 + 25 = 90; y 9 0 - 65 = 25_15. y = 90 - 40 = 50; x = 90 - 50 = 4016. y = 90 - (40 + 20) = 30; x + 30 = 90 - 20, x = 4017. Yes; 4n = 2n + 10; n = 5; the sides are 4(5) = 20, 2(5) + 10 = 20, 7(5) - 15 = 20.18. a . 3 t 5 t 1 2 , t = 6; 3t = t + 20, t = 10; 5t 1 2 = t + 20, t = 8

b. No; there is no value of t such that 3t = 5t - 12 = t + 20.

19. Le t x be the measure of the smallest angle; x + 2x + 3x = 180; 6x = 180; 3 0 ;the meas. of the angles are 30, 60, 90.

20. x + (x + 28) + 2x = 180; 4x + 28 = 180; 4x = 152; x = 38; 38, 66, 7621. m L A + mL B + mL C = 180; mL A + mL B < 120, so mLC > 60.22. m L R + mLS + m L T 1 8 0 ; mLR + mLS > 110, so mL T < 70.23. a. 22 b . 23 c . L A B D and LC are comps. of LCBD.24. a . 130 b . 130 c . I f mLE = 80, then mLFI G will always be 130.

25. Statements R e a s o n s1. L A B D L A E D2. L A L A3. L C L F

26. m L M T R = 180 - 85 = 95; mLSTR = 180 - (30 + 95) = 55;mL1 = 90 - 55 = 35; mL NRT = 55; mL2 = 180 - 55 = 125

1. Given

2. Ref l. Prop.3. I f 2 A of one A are t o 2 A of

another A, then the third A are

32

27.

Key to Chapter 3, pages 97-99

Given: AABCProve: mL1 + mL 2 + mL 3 = 180Statements R e a s o n s

1. Dr a w CD through C 11 t o AB.

2. L 2 £ 5 , or mL2 = mL 5

3. L l L 4 , or mL1 = mL4

1.

2.

3.

Through a pt. outside a line, there isexactly 1 line 11 t o the given line.I f 2 11 lines are cut by a trans., thenalt. in A are

I f 2 II lines are cut by a trans., thencorr. A are

4. n t L A CD + mL 4 1 8 0 ;ntLACD = mL 3 + mL 5

4. L Add. Post.

5. m L 3 + mL 4 + mL 5 = 180 5. Substitution Prop.6. m L 1 + mL 2 + mL 3 = 180 6. Substitution Prop.


1. n i LJ GI = m L H + 1. The meas. of an ext. L of a A = thesum of the meas. of the 2 remote int.

2. m L I I = mL I 2. Given

3. I nLJ G1 = 2 mL H 3. Substitution Prop.1

4. m U J G I = mL11 4. Div. Prop. of =2

5. G K bisects LJGI. 5. Given1

6. m L 1 = 6. L Bis. Thm.

7. m L 1 = m/.11 7. Substitution Prop.

8. G K II H I 8. If 2 lines are cut by a trans. and corr.A are m, then the lines are II

29. 2x + y + 125 = 180, 2x + y = 55, y = 55 — 2x; (x + 2y) + (2x + y) = 90,(x + 2y) + 55 = 90, x + 2y = 35; x + 2(55 — 2x) = 35, x + 110 — 4x = 35,3x = 75, x = 25; 2x + y = 55, 50 + y = 55, y = 5

30. (5x + y) + (5x — y) + 100 = 180, 10x = 80, x = 8; 2x + y = 5x — y,2y = 3x, 2y = 24, y = 12

31. L I £ 2 £ 5 ; L3 £ 4 £ 6C 32 . L 7 L 8 , L11 L 1 2

33. a—b. Check students' drawings. See figure at the right.c. The angle measures 90, so the bisectors ared. Given: AB 11 CD; AE bisects LBAC;

CP' bisects LACD.Prove: AE I CP'


Statements R e a s o n s

1. A B 11 CD2. I nLBAC + mZ_ACD = 180

1 13. - mL B A C + - mL A CD = 902 24. A E bisects LBAC;

CT bisects LACD.1

5. m L 2 = -mLBAC;1mL3 = •/11LACD

6. m L 2 + mL 3 = 907. m L A X F = mL2 + mL 3

8. m L A X F = 909. A R I C.E

1. Given

2. I f 2 11 lines are cut by a trans., thens-s. in A are supp.; clef, of supp. A

3. Div . Prop. of =

4. Given

5. L Bis. Thm.

6. Substitut ion Prop.7. The meas. of an ext. L of a A = the

sum of the meas. of the 2 remotein A

8. Substitut ion Prop.9. Def . of I lines

34. Since 3x and 3y are meas. of s-s. int. A,3x + 3y = 180, and x + y = 60. Then mLEDF =-mLCDA = 180 - (x + y) = 120. L E B F isthe third L of a A with A of meas. 2x and 2y, somLCBA = 180 - (2x + 2y) = 180 - 120 = 60.Then, in ABCD, rnLCDA + mL CB A = 120 + 60 =180. Also, LBCD is an ext. A of AECF with remote int.A of meas. 2x and y, so TrtLBCD = 2x + y. Similarly,nI LBAD = 2y + x. So, mLBCD + mZ_BAD =3x + 32j = 180. Therefore, in ABCD opp. a r e supp.


1-4. Sketches and angle measures will vary. 1 . False; true for acute A2. False; true for acute A 3 . True 4 . False; true for rt. A


1. convex polygon 2 . nonconvex polygon 3 . not a polygon 4 . nonconvex polygon5. not a polygon 6 . nonconvex polygon 7 . I t has the same shape.8. (102 - 2)180 = 18,000; 360


No. of sides 6 10 20 36 18 360 4

Meas. of each ext. L 60 36 18 10 20 1 90

Meas. of each int. L 120 144 162 170 160 179 90

No. of sides 9 15 30 60 45 24 180

Meas. of each ext. L 40 24 12 6 8 15 2

Meas. of each int. L 140 156 168 174 172 165 178

9.


A 1 . (4 — 2)180 = 360; 360 2 . (5 — 2)180 = 540; 360 3 . (6 — 2)180 = 720; 3604. (8 — 2)180 = 1080; 360 5 . (10 — 2)180 = 1440; 360 6 . ( n — 2)180; 3607. 360; yes

8.

9. Le t x me a s . of each of the 2: ----f i . 2 x + 3 ( 9 0 ) = ( 5 — 2 ) 1 8 0 , 2 x + 2 7 0 = 5 4 0,

2x = 270, x = 13510. Le t x me a s . of the fifth L. x + 40 + 80 + 115 + 165 = (5 — 2)180,

x + 400 = 540, x = 14011. 120 1 2 - 1 5 . Sketches may vary. 1 2 .

13. 14. not possible

/-715. not possible (An ext. L would have meas. 50, and 360 is not a multiple of 50.)

B 16 . Le t n = number of sides; (n 2 ) 1 8 0 = 5(360), n = 12.

17. Le t n = number of sides; ( n —n2 ) 1 8 0 1 1 (3-6 0) n = 2 4 .

n18. a. 108 b . No (360 is not a multiple of 108.) 1 9 .

20. The sum of the meas. of the in A of 2 hexagonsand 1 pentagon at any common vertex is120 + 120 + 108 = 348. A sum of 360 isnecessary to tile a plane.

21. x + 2x + 3x + 4x = (4 — 2)180, 10x = 360,x = 36; mLA = 36, mLB = 72, mLC = 108,mL D = 144; mLA + mL D = 180 (also, mLB + mL_C = 180), so AB

Ex. 19

11 CD.


mL W = 180 - (mLWBC + mLWCB) = 180 - 2 ( ) 360-- n2 2

2100 < (n - 2)180 < 2200; 13T3- < n < 1 4 5 ; n = 1 4

- 180n - 720)?

a. [ (n + 1) - 21180 = [ (n 2 ) + l]180 = (n - 2)180 + 180 S + 180b. (2n - 2)180 = [2(n - 1)180] = 2[(n 2) + 1]180 = 2[(n 2 ) 1 8 0 + 180] =2(S + 180)a. Sketches may vary.b. Yes; 90 + 90 + 50 + 260 + 50 = 540 50' 5 0 '

260'

22. a . Le t mLiT = x, then mLS = m LT = 3x; 60 + 130 + x + 3x + 3x =(5 - 2)180, 7x + 190 = 540, 7x = 350, 5 0 ; mLR = 50, mLS = m L T = 150b. m L Q + mL R = 180, so PQ II RS.

36023. L K B C and LKCB are ext. A of a reg. decagon, so mLKBC = mLKCB = -1 0 =36. mL K = 180 - (mLKBC + mLKCB), so mL K = 180 - (36 + 36) = 108.

36024. L WB C and LWCB are each ext. A of the n-gon, so mLWBC = mLWCB = it

25.

C 26 .

27.

(n 2 ) 1 8 0 3 6 0 n - 228. a . = x • x =n 2b. Even values 4


1. inductive 2 . induct ive 3 . deductive 4 . induct ive5. deductive 6 . inductive

Pages 107-109 • WRI TTEN EXERCISES1 1

A 1 . 256, 1024 2 . 6, 3 3 . -8 1 ' 2 4 3 4 . 2 5 , 3 6 5 . 1 7 , 2 3 6 . 4 0 , 5 2 7 . 1 5, 4

1 18. -8- 9. 500, 250 10. Chan is older than Sarah. 11. none

12. Polygon G has 7 sides. 1 3 . none 1 4 . No; deductively15. 1234 x 9 + 5 = 11111 1 6 . 9876 x 9 + 4 =- 88888 1 7 . 99992 = 9 9 9 8 0 0 0 1

B 18 . TrueGiven: BA B CProve: L A L C

19. TrueGiven: L A L CProve: BA B C


20. False

21. TrueGiven: ABCDE is a reg. pentagon.Prove: AC'-- -- A D '-- -- BE '--- - BDr z--- CE

AAttYfrif

Ex. 21

22. True

Given: AB 11 DC; AD II B CProve: AE E C ; DE E B

23. False

Prove: AC I BD

25. TrueGiven: ABCD is equilateral.

24. False

26. a. 16 b . Guess: 32, Actual Count: 31

27. a. I f both pairs of opp. sides of a quad. are II , then opp. A areb. I f both pairs of opp. A of a quad. are t h e n opp. sides areGiven: ABCD is a quad.; mL A = mLC; mLB =- n i L DProve: AD 11 BC; AB 11 CD

Key to Chapter 3, page 109 3 7

No. of sides 3 4 5 6 7 8 n

No. of diagonals 0 2 5 9 14 20 n(n — 3)2

42; 4422; 444222; 6666 x 6667 = 4444222264; 9604; 996004; 9998 x 9998 = 9996000463; 7623; 776223; 7777 x 9999 = 77762223

29.

Statements R e a s o n s1. m L A + mL B + mL C + r i t LD =

360

2. m L A = mLC; mLB = mL D3. 2 mL A + 2mLB = 360;

2mLC + 2mLB = 3604. m L A + mL B = 180;

mL C + mL B = 180

5. L A and LB are supp.;LB and LC are supp.

6. A D 11 BC; AB 11 CD

Page 109 • CALCULATOR KEY-IN

1. 1; 121; 12321; 1111 x 1111 = 12343212.3.4.

1. The sum of the meas. of the in A ofa quad. is 360.

2. Given

3. Substitut ion Prop.

4. Div . Prop. of =-

5. Def . of supp. A

6. I f 2 lines are cut by a trans. ands-s. in A are supp., then the linesare 11

c. Both pairs of opp. A of a quad. are i f and only if Opp. sides are 11C 28 . a . 13, 17, 23, 31, 41, 53, 67, 83, 101 b . Guess: a prime number c . 121, 143,

neither of which is prime

30. a. There are 5 small A each with one of the points A, B, C, D, E as one vertex. Theother two A of each of the A are ext. A of a pentagon. There are two complete setsof ext. A of the pentagon, with each set having total meas. 360. ThenmL A + mL B + mL C + mL D + mL E + 360 + 360 = 5(180) = 900 andmL A + mL B + mL C + mL D + mL E = 180. b . Us ing the same reasoningas in part (a), mL A + mL B + mL C + mL D + mL E + m L F + 360 + 360 =6(180) = 1080 and mLA + t nLB + mL C + mL D + mL E + mL F = 360.c. Fo r each additional point of a star, the sum of the meas. of the A increases by180. The sum of the L meas. for an n-pointed star is 180(n — 4). d . I f a starhas n points, mL A + mL B + mL C + ••• + m L N + 360 + 360 = n(180)and mLA + mL B + mL C + • • • + m L N = n(180) — 720 = 180(n — 4).


Page 110 • SELF-TEST 2

1. acute 2 . scalene 3 . 60 4 . 105, 35 5 . (2x + 4) + (3x - 9) = 90; x = 196. y -= 50; 110 z + 50, z 6 07. 2x + 5 = 3x + 10, x 5 (reject); 2x + 5 = x + 12, x = 7;

3x + 10 = x + 12, x 1

8. 8 9 . equilateral, equiangular 1 0 . 360; (10 - 2)180= 14410360

11. 180 - 174 = 6. = 6 0 1 2 . 3 2 1 3. 32 14. 36 15. 16

6

Pages 111-112 • CHAPTER REVI EW

1. 2 2 . corr. 3 . alt . int. 4 . No; they can be skew. 5 . 105, 1056. 70 = 6x - 2; x = 12 7 . (8y - 40) + (2y + 20) = 180; y 2 08. b 1 c; if a trans. is I to one of 2 11 lines, then it is I to the other one also.9. DE ; L A is supp. to LADE, and if 2 lines are cut by a trans. and s-s. in A are

supp., then the lines are 1110. B E 11 C T ; both are I to D.11. corr. A a l t . in A s - s . in A supp.; in a plane, both lines are I to a third line;

both lines are 11 t o a third line.12. x + (2x 1 5 ) = 90; x = 35 1 3 . 180 1 4 . 10015. L 3 L 6 (I f 2 A are supps. of A , then the 2 A are L 2 L 8 (I f 2 A of one

A are t o 2 Zs_ of another A, then the third A are

16. a .

(18 - 2)180 3 6 0 ( n - 2)18017. = 160 1 8 . - = 15 1 9 . - 150; n = 12fl18 2 41 1

20. 75, 90 2 1 . 100' 1 0 0 0

b. (6 - 2)180 = 720c. 360

Pages 112-113 • CHAPTER TEST1. sometimes 2 . sometimes 3 . never 4 . never 5 . never 6 . always7. (3x - 20) + x 1 8 0 ; 5 0 8 . 2x + 12 = 4(x - 7); x 2 09. mL 1 m L 2 = 60, mL3 = 120

10. mL 1 = 58, mL2 = 90, mL3 = 32, mt_LI = 180 - (32 + 35) = 113, mL5 = 35,mL6 = 55

Key to Chapter 3, page 113 3 9

1. B F bisects LABE;DG bisects LCDB.

12. mL G DB = —mLCDB;2

1.

2.

Given

L Bis. Thm.

1mL FB E = --mLABE2

3. A B 11 CD 3. Given4. mL CDB = mL A B E 4. I f 2 11 lines are cut by a trans., then

corr. A are1 15. —mLCDB = —mLABE2 2

5. Div. Prop. of =

6. mL G DB = mL FB E 6. Substitution Prop.7. B F I I DG 7. I f 2 lines are cut by a trans. and corr.

A are t h e n the lines are 11

14. 15, 17

Page 113 • ALGEBRA REVI EW

1. 3 2 . 2 3 . (0, 0) 4 . Z 5 . (3, 5) 6. (4, 3) 7 . (4, 0) 8 . (0, 4)9. (— 5, 0) 1 0 . 4 , 3) 1 1 . 2 , 2) 12. ( - 4 , —2) 1 3 . ( - 2 , —3)

14. (3, —2) 1 5 . K , 0, S 1 6 . 0 , R , Z 17. 3 18. c , e 1 9 . M, N, P20. T, U 2 1 . V , W 2 2 . J , Q

23-34. YA•

E. A

0F.K.

35. (2, 38. 3, 0) 39. ( — 4, 2 ) 4 0 . (1, —1)1) 3 6 . (2, 5) 3 7 . (0, 3)

(5 — 11. mL 4 = 2 ) 1 8 0 = 108, niL5 = 108 — 72 = 36, mL 1 = 180 — 108 = 72,5

mL2 = 180 — 108 = 72, mL3 = 180 — (72 + 72) = 3612. L E B C L 2 (If 2 lines are cut by a trans. and alt. int. A are t h e n the lines

are 11 .), or £5 L 3 (I f 2 lines are cut by a trans. and corr. A are t h e n thelines are II .)



Pages 114-115 • CUMULATIVE REVIEW: CHAPTERS 1-3

A 1 . sometimes 2 . always 3 . sometimes 4 . always 5 . always6-8. Sketches may vary.6. 7 . A 8 .

•

- 3.5 + 8.5 59. not possible 1 0 . - 2 or 2.52

B 45 . Statements R e a s o n s

1. W X X Y2. L i is comp. to £2.

3. L i is comp. to £3.4. L 2 L 3

11. 5x + 13 = 9x - 39, x = 13; InLPQR = 2m,LPQX = 2[5(13) + 13] = 15612. 180 - x = 2(90 - x) + 35, x = 35; A measure: 35, supp. measure: 145; comp.

measure: 5513. x + 5x + 6x = 180, 1 5 ; L measures: 15, 75, 90 1 4 . 2(60) = 120 1 5 . 9016. 90 - 60 = 30 1 7 . 90 1 8 . 60 1 9 . 180 - 2(60) = 6020. 180 - 60 = 120 2 1 . 180 - 120 = 60 2 2 . 180 - 2(60) = 6023. False. I f 2 lines are 11 , then they do not intersect; true.24. True. I f 2 lines are I , then they intersect to form rt. A; true.25. True. I f an A is not obtuse, then it is acute; false.26. True. I f a A is isos., then it is equilateral; false.27. Vert . A are 2 8 . Seg. Add. Post. 2 9 . A Add. Post.30. I f 2 lines are cut by a trans. and alt. int. A are t h e n the lines are 1131. The meas. of an ext. A of a A = the sum of the meas. of the 2 remote int. A.32. Def . of I lines 3 3 . The sum of the meas. of the A of a A is 180.

(n - 2)18034. Def . of I lines 3 5 . X 3 6 . supp. 3 7 . = 108, n = 5; pentagon

38. 2(12) = 24 3 9 . 4 0 . induct ive 4 1 . biconditional 4 2 . 36043. (8 - 2)180 = 1080 4 4 . acute

1. Given

2. I f the ext. sides of 2 adj. acute A areI , then the A are comp.

3. Given4. I f 2 A are comps. of the same A,

then the 2 A are

int. A are -----', then the lines are 11



1. RU 1 S7' 1. Given2. L i £2 2. If 2 11 lines are cut by a trans., then

alt. int. A are3. LR L T 3. Given4. Z.3 L4 4. If 2 A of one A are t o 2 A of

another A, then the third A are5. RS 11 U T 5. If 2 lines are cut by a trans. and alt.

chapter 3 • parallel lines and planes page 75 ... to chapter 3, pages 87-88 2 7 19. a. false b....

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