resources for geometry a - congruent triangles

Congruent FiguresCongruent FiguresGEOMETRY LESSON 4-1GEOMETRY LESSON 4-1

Pages 182-185 Exercises

1. CAB DAB;C D;ABC ABD;

AC AD; AB ABCB DB

2. GEF JHI;GFE JIH;EGF HJI;

GE JH; EF HIFG IJ

3. BK

4. CM

5. ML

6. B

7. C

8. J

9. KJB

10. CLM

11. JBK

12. MCL

13. E, K, G, N

14. PO SI; OL ID;LY DE; PY SE

15. P S; O I;L D; Y E

16. 33 in.

17. 54 in.

18. 105

19. 77


20.36 in.

21.34 in.

22.75

23.103

24.Yes; RTK UTK, R U (Given) RKT UKT

If two of a are to two of another , the third are . TR TU, RK UK (Given)

TK TK (Reflexive Prop. of ) TRK TUK(Def. of )

25.No; the corr. sides are not .

26.No; corr. sides are not necessarily .

s s

s

s

27. Yes; all corr. sides and are .

28. a. Given

b. If || lines, then alt. int.

are .

c. Given

d. If 2 of one are

to two of another ,

then 3rd are .

e. Reflexive Prop. of

f. Given

g. Def. of

s

s

s

s

s

s


29. A and H; B and G; C and E; D and F

30. x = 15; t = 2

31. 5

32. m A = m D = 20

33. m B = m E = 21

34. BC = EF = 8

35. AC = DF = 19

36. Answers may vary. Sample: It is important that PACH OLDE for the patch to completely fill the hole.

37. Answers may vary. Sample: She could arrange them in a neat pile and pull out the ones of like sizes.

38. JYB XCH

39. BCE ADE

40. TPK TRK

41. JLM NRZ;JLM ZRN

42. Answers may vary. Sample: The die is a mold that is used to make items that are all the same size.

43. Answers may vary. Sample:

TKR MJL: TK MJ; TR ML;KR JL;

TKR MJL; TRK MLJ;KTR JML


44. a. Given

b. If || lines, then alt. int. are .

c. If || lines, then alt. int. are .

d. Vertical are .

e. Given

f. Given

g. Def. of segment bisector

h. Def. of

45. Answers may vary. Sample: Since the sum of the of a is 180, and if 2 of one are the same as 2 of a second , then their sum subtracted from 180 has to be the same.

46. KL = 4; LM = 3; KM = 5

47. 2; either (3, 1) or (3, –7)

s

s

s

s

s

s

s

48. a. 15

b.


49. 1.5

50. 4.25

51. 40

52. 72


54.

55. 100

56. RS = PQ

57. 1

58. 12

59. AB GH

Triangle Congruence by SSS and SASTriangle Congruence by SSS and SASGEOMETRY LESSON 4-2GEOMETRY LESSON 4-2

1. SSS

2. cannot be proved

3. SAS

4. SSS

5. Yes; OB OB by Refl. Prop.; BOP

BOR since all rt. are ; OP OR(Given); the are by SAS.

s

s


6. Yes; AC DB (Given); AE CE and BE DE(Def. of midpt.); AEB

CED (vert. are ) AEB CED by SAS.

7. a. Given

b. Reflexive

c. JKM

d. LMK

8. WV, VU

9. W

s

10. U, V

11. WU

12. X

13. XZ, YZ

14. LG MN

15. T Vor RS WU

16. DC CB


17. additional information not needed

18. Yes; ACB EFDby SAS.

19. Yes; PVQ STRby SSS.

20. No; need YVWZVW or YW ZW.

21. Yes; NMO LOMby SAS.

22. ANG RWT; SAS

23. KLJ MON; SSS

24. Not possible; need H P or DY TK.

25. JEF SVF or JEF SFV; SSS

26. BRT BRS; SSS

27. PQR NMO; SAS

28. No; even though the are , the sides may not be.

s

29. No; you would need H K or GI JL.

30. yes; SAS

31.


32.

33. a. Vertical are .

b. Given

c. Def. of midpt.

d. Given

e. Def. of midpt.

f. SAS

s


35. a–b. Answers may vary. Sample:

a. wallpaper designs; ironwork on a bridge; highway warning signs

35. (continued)

b. produce a well-balanced, symmetric appearance. In construction, enhance designs. Highway warning signs are more easily identified if they are .

36. ISP PSO; ISP OSP

by SAS.

37. IP PO; ISPOSP by SSS.

s

s


38. Yes; ADB CBDby SAS; ADB

DBC because if || lines, then alt. int. are .

39. Yes; ABC CDAby SAS; DAC

ACB because if || lines, then alt. int. are .

40. No; ABCD could be a square with side 5 and EFGH could be a polygon with side 5 but no rt. .

s

s

s

41. 1. FG || KL (Given)

2. GFK FKL (If || lines, then alt. int. are .)

3. FG KL (Given)

4. FK FK (Reflexive Prop. of )

5. FGK KLF(SAS)

s

42. AE and BD bisect each other, so AC CE and BC CD.

ACBDCE because vert. are .

ACB ECD by SAS.

s


43.

44. AM MB because Mis the midpt. of AB.

B AMCbecause all right are

. CM DB is given. AMC MBD by

SAS.

s

51. FG

52. C

53. The product of the slopes of two lines is –1 if and only if the lines are .

54. If x = 2, then 2x = 4. If 2x = 4, then x = 2.

55. If 2x = 6, then x = 3. The statement and the converse are both true.

45. D

46. G

47. C

48. [2] a. AB AB; Reflexive Prop. of

b. No; AB is not a corr. side.

[1] one part correct

49. E

50. AB


56. If x2 = 9, then x = 3. The statement is true but the converse is false.

Triangle Congruence by ASA and AASTriangle Congruence by ASA and AASGEOMETRY LESSON 4-3GEOMETRY LESSON 4-3


1. PQR VXW

2. ACB EFD

3. RS

4. N and O

5. yes

6. not possible

7. yes

8. a. Reflexive

b. ASA

9. AAS

10. ASA

11. not possible

12. FDE GHI;DFE HGI

13. a. UWV

b. UW

c. right

d. Reflexive

14. B D

15. MU UN

16. PQ QS

17. WZV WZY


18. a. Vert. are .

b. Given

c. TQ QR

d. AAS

19. PMO NMO; ASA

20. UTS RST; AAS

21. ZVY WVY; AAS

22. TUX DEO; AAS

23. The are not because no sides are

.

s

s

24. TXU ODE; ASA

25. The are not because the are not included .

26. Yes; if 2 of a are to 2 of another , then the 3rd are . So, an AAS proof can be rewritten as an ASA proof.

s

s

s

s

27. a. SRP

b. PR

c. alt. int.

d. PR

e. Reflexive

28. a. Given

b. Def. of bis.

c. Given

d. Reflexive Prop. of

e. AAS

s

s


29. a. Def. of

b. All right are .

c. QTP STR

d. Def. of midpt.

e. AAS

30.

s

31. Yes; by AAS since MON QOP.

32. Yes; by AAS since FGJ HJG

because when lines are ||, then alt. int. are and GJ GJ by the Reflexive Prop. of .

33. Yes; by ASA, since EAB DBC

because || lines have corr. .

s

s

34. Yes; by ASA since BDH FDH by

def. of bis. and DH DH by the Reflexive Prop. of .


36. a. Check students’ work.

b. Answers may vary; most likely ASA.


37. AEB CED, BEC DEA, ABC CDA, BAD DCB

38. AEB CED, BEC DEA, ABC CDA, ABD DCA, BAD DCB,ABD DCB, CBA DAB, BCD ADC

39. They are bisectors; ASA.

40.

41.

42. D

43. F

1320

44. [2] a. RPQ SPQ,RQP SQP

(Def. of bisector)

b. ASA



45. [4] a. Def. of midpt.

b. Yes; JLM KGM because they are alt. int. of || lines, and LMJ GMK because vertical are . So the are by ASA.

c. Yes; if two of one are to 2 of another , the third are .

[3] incorrect for part b or c, but otherwise correct

[2] correct conclusions but incomplete explanations for parts b and c

[1] at least one part correct

46. ONL MLN; SAS

47. not possible

s

s s

s s

s

s

48. AC and CB

49. If corr. are , then the lines are ||.

50. 56 photos

51. 36 photos

52. 60% more paper

s

Using Congruent Triangles: CPCTCUsing Congruent Triangles: CPCTCGEOMETRY LESSON 4-4GEOMETRY LESSON 4-4


1. PSQ SPR; SQ RP; PQ SR

2. AAS; ABC EBD; A E; CB DB;

DE CA by CPCTC

3. SAS; KLJ OMN; K O; J N;

KJ ON by CPCTC

4. SSS; HUG BUG; H B; HUGBUG; UGHUGB by CPCTC

5. They are ; the are by AAS, so all corr. ext.

are also .

6. a. SSSb. CPCTC

7. ABD CBD by ASA because BD BD by Reflexive Prop. of ; AB CB by CPCTC.

8. MOE REO by SSS because OE OEby Reflexive Prop. of ;

M R by CPCTC.

s

s

9. SPT OPT by SAS because TP TPby Reflexive Prop. of

; S O by CPCTC.

10. PNK MNL by SAS because KNP

LNM by vert. are ; KP LM by

CPCTC.

11. CYT RYP by AAS; CT RP by CPCTC.

s


12. ATM RMT by SAS because ATM

RMT by alt. int. are ; AMT

RTM by CPCTC.

13. Yes; ABD CBDby SSS so A Cby CPCTC.

14. a. Given

b. Given

c. Reflexive Prop. of

d. AAS

s

15. PKL QKL by def. of bisect, and KL KLby Reflexive Prop. of , so the are by SAS.

16. KL KL by Reflexive Prop. of ; PL LQ by Def. of bis.; KLP

KLQ by Def. of ; the are by SAS.

17. KLP KLQbecause all rt are ; KL KL by Reflexive Prop. of ; and PKL

QKL by def. of bisect; the are by ASA.

s

s

s

s

18. The are by SAS so the distance across the sinkhole is 26.5 yd by CPCTC.

19. a. Given

b. Def. of

c. All right are .

d. Given

e. Def. of segment bis.

f. Reflexive Prop. of

g. SAS

h. CPCTC

s

s


20. ABX ACX by SSS, so BAX

CAX by CPCTC. Thus AX bisects

BAC by the def. of bisector.

21. Prove ABE CDFby SAS since AEFC by subtr.

22. Prove KJM QPMby ASA since P

J and K Q by alt. int. are .

23. e or b, e or b, d, c, f, a

s

24. BA BC is given; BD BD by the Reflexive Prop. of and since BD bisects ABC, ABD CBDby def. of an bisector; thus, ABD CBD by SAS; AD DC by CPCTC so BD bisects AC by def. of a bis.; ADB CDB by CPCTC and ADBand CDB are suppl.; thus, ADB and CDB are right and BD AC by def. of .

25. a. AP PB; AC BC

b. The diagram is constructed in such a way that the are by SSS. CPA CPB by CPCTC. Since these are and suppl., they are right . Thus, CP is to .

s

s

s

s


26. 1. PR || MG; MP || GR (Given)

2. Draw PG. (2 pts. determine a line.)

3. RPG PGM and RGP GPM (If || lines, then alt. int. are .)

4. PGM GPR (ASA) A similar proof can be written if diagonal RM is drawn.

27. Since PGM GPR (or PMR GRM), then PR MG and MP GR by CPCTC.

28. C

29. C

s

30. D

31. B

32. C

33. [2] a. KBV KBT; yes; SAS

b.CPCTC


34. ASA

35. AAS


36. 95; 85

37. The slope of line m is the same as the slope of line n.

38. not possible

39. not possible

Isosceles and Equilateral TrianglesIsosceles and Equilateral TrianglesGEOMETRY LESSON 4-5GEOMETRY LESSON 4-5


1. a. RS

b. RS

c. Given

d. Def. of bisector

e. Reflexive Prop.

of

f. AAS

2. a. KM

b. KM

c. By construction

2. (continued)

d. Def. of segment

bisector

e. Reflexive Prop.

of

f. SSS

g. CPCTC

3. VX; Conv. of the Isosc. Thm.

4. UW; Conv. of the Isosc. Thm.

5. VY; VT = VX (Ex. 3) and UT = YX (Ex. 4), so VU = VY by the Subtr. Prop. of =.

6. Answers may vary. Sample: VUY; opp. sides are .

7. x = 80; y = 40

8. x = 40; y = 70

9. x = 38; y = 4

s


10. x = 4 ; y = 60

11. x = 36; y = 36

12. x = 92; y = 7

13. 64

14. 2

15. 42

16. 35

17. 150; 15

18. 24, 48, 72, 96, 120

12

12

19. a.

30, 30, 120

b. 5; 30, 60, 90, 120, 150

c. Check students’ work.

20. 70

21. 50

22. 140

23. 6

24. x = 60; y = 30

25. x = 64; y = 71

26. x = 30; y = 120

27. Two sides of a are if and only if the

opp. those sides are .

28. 80, 80, 20; 80, 50, 50

s


29. a. isosc.

b. 900 ft; 1100 ft

c. The tower is the bis. of the base of each .

30. No; the can be positioned in ways such that the base is not on the bottom.

31. 45; they are = and have sum 90.

s 32. Answers may vary. Sample: Corollary to Thm. 4-3: Since XYYZ, X Z by Thm. 4-3. YZ ZX, so Y

X by Thm. 4-3 also. By the Trans. Prop., Y

Z, so X YZ. Corollary to Thm.

4-4: Since X Z, XY YZ by Thm. 4-4.

Y X, so YZ ZXby Thm. 4-4 also. By the Trans. Prop. XYZX, so XY YZ ZX.

33. a. Given

b. A D

c. Given

d. ABE DCE

34. m = 36; n = 27

35. m = 60; n = 30

36. m = 20; n = 45

37. (0, 0), (4, 4), (–4, 0),(0, –4), (8, 4), (4, 8)

38. (5, 0); (0, 5); (–5, 5);

(5, –5); (0, 10); (10, 0)


39. (5, 3); (2, 6); (2, 9); (8, 3); (–1, 6); (5, 0)

40. a. 25

b. 40; 40; 100

c. Obtuse isosc. ;

2 of the are

and one is

obtuse.

s

41. AC CB and ACDDCB are given. CD

CD by the Refl. Prop. of , so ACD BCD

by SAS. So AD DBby CPCTC, and CDbisects AB. Also ADC

BDC by CPCTC, m ADC + m BDC = 180 by Add. Post., so m ADC = m BDC = 90 by the Subst. Prop. So CD is the bis. of AB.

42. The bis. of the base of an isosc. is the bis. of the vertex ; given isosc. ABCwith bis. CD, ADC

BDC and AD DBby def. of bis. Since CD CD by Refl. Prop., ACD BCDby SAS. So ACD

BCD by CPCTC, and CD bisects ACB.


43. a. 5

b.

44. 0 < measure of base< 45

45. 45 < measure of base< 90

46. C

47. G

48. D

49. [2] a. 60; since m PAB= m PBA, and m PAB + m PBA = 120, m PAB = 60.

b. 120; m APB = 60 so m PAB = 60. Since PABand QAB are compl., m QAB= 30. QABis isosc. so m AQB = 120.



50. RC = GV; RC GVby CPCTC since

RTC GHVby ASA.

51. AAS

52. SSS

53. 24 sides

Congruence in Right TrianglesCongruence in Right TrianglesGEOMETRY LESSON 4-6GEOMETRY LESSON 4-6


1. ABC DEF by HL. Both are rt. , AC DF, and CB FE.

2. SPR QRP by HL. Both are rt. , SPQR (given) and PRPR by the Reflexive Prop. of .

3. LMP OMN by HL. Both are rt. because vert. are ; LP NO, and LMOM.

s s

s s

s s

s

4. AEB DCB by HL. Both are rt. .AB BD and EB CBby the def. of midpt.

5. T and Q are rt. .

6. RX RT or XV TV

7. TY ER or RT YE

8. Right are needed, either A and G or

AQC and GJC.

s s

s

s

9. BC FA

10. RT NQ

11. a. Given

b. Def. of rt.

c. Reflexive Prop. of

d. Given

e. HL


12. a. suppl. are rt.

b. Def. of rt.

c. Given

d. Reflexive Prop. of

e. HL

13. PS PT so S Tby the Isosc. Thm.

PRS PRT. PRS PRT by

AAS.

14. Yes; RS TU andRT TV.

s s 15. Yes; PM PM and PMW is a rt. since

JP || MW.

16. a. Given

b. Def. of

c. MLJ and KJL are rt. .

d. Given

e. LJ LJ

f. HL

s

17. a. Given

b. IGH

c. Def. of rt.

d. I is the midpt. of HV.

e. Def. of midpt.

f. IGH ITV

18. HL; each rt. has ahyp. and side.

19. x = 3; y = 2

20. x = –1; y = 3


21. whether the 7-yd side is the hyp. or a leg

22. ABX ADX; HL ABC ADC; SSS

or SAS BXC DXC; HL

23. a. Answers may vary. Sample: You could show that suppl. AXB and AXD are .

b. ABC ADC by SSS so BAC

DAC by CPCTC.ABX ADX by

SAS so AXBAXD. AXB is

suppl. and to AXD so they are

both rt. .

s

s

24.

25.

26.

27.


28. 1. EB DB; A and C are rt. (Given)

2. BEA BDCare rt. (Def. of rt. )

3. B is the midpt. of AC (Given)

4. AB BC(Def. of midpt.)

5. BEA BDC(HL)

s

s

29. 1. LO bisects MLN,OM LM, ON LN, (Given)

2. M and N are rt. (Def. of )

3. MLO NLO(Def. of bis.)

4. M N(All rt. are .)

5. LO LO (Reflexive Prop. of )

6. LMO LNO(AAS)

s

s

30. Answers may vary. Sample: Measure 2 sides of the formed by the amp. and the platform’s corner. Since the will be by HL or SAS, the are the same.

s

s


31. a.

b. slope of DG = –1; slope of GF = –1; slope of GE = 1

c. EGD and EGFare rt. .

d. DE = 26 ; FE = 26

s

31. (continued)

e. EGD EGF by HL. Both are rt. ,

DE FE, and EG EG.

32. An HA Thm. is the same as AAS with AAS corr. to the rt. , an acute , and the hyp.

s s

33. Since BE EA andBE EC, AEB and

CEB are both rt. .AB BC because

ABC is equilateral, and BE BE. AEB

CEB by HL.

34. No; AB CB because AEB CEB, but

doesn’t have to be to AB or to CB.

35. A

36. H

s


37. D

38. [2] a. TFW TGW

b. RFW and RGW are rt. .


39. isosceles

40. equilateral

s

41. BC || AD because each slope = –1. BT BA, BA AD because product of slopes is –1.

42. If || lines then alt. int. are .

43. If two lines are ||, then same-side int. are suppl.

44. Vert. are .

45. If two lines are ||, then corr. are .

s

s

s

s

46. If two lines are ||, then corr. are .

47. If two lines are ||, then same-side int. are suppl.

s

s

Using Corresponding Parts of Congruent TrianglesUsing Corresponding Parts of Congruent TrianglesGEOMETRY LESSON 4-7GEOMETRY LESSON 4-7


1. M

2. DF

3. XY

4.

5.

6.

7.

8.

9.

10. a. Given

b. Reflexive Prop. of

c. Given

d. AAS

e. CPCTC

11. LQP PML; HL


12. RST UTS; SSS

13. QDA UAD; SAS

14. QPT RUS; AAS

15. TD RO if TDIROE by AAS. TID

REO if TEIRIE. TEI RIE

by SSS.

16. AE DE if AEBDEC by AAS. AB

DC and A D since they are corr. parts of

ABC and DCB, which are by HL.

17. QET QEU by SAS if QT QU. QTand QU are corr. parts of QTB and QUBwhich are by ASA.

18. ADC EDG by ASA if A E. Aand E are corr. parts in ADB and

EDF, which are by SAS.

19–22. Answers may vary. Samples are given.

19.

20.


21. a.

b.

22. a.

b.

23. ACE BCD by ASA; AC BC, A

B (Given) C C(Reflexive Prop. of )

ACE BCD (ASA)

24. WYX ZXY by HL; WY YX, ZX YX, WX

ZY (Given) WYXand ZXY are rt. (Def. of ) XY XY(Reflexive Prop. of )

WYX ZXY (HL)

s

25. m 1 = 56; m 2 = 56; m 3 = 34; m 4 = 90; m 5 = 22; m 6 = 34; m 7 = 34; m 8 = 68; m 9 = 112

26. ABC FCG; ASA

27. a. Given

b. Reflexive Prop. of

c. Given

d. ETI


27. (continued)e. IRE

f. CPCTC

g. Given

h. All rt. are .

i. TDI

j. ROE

k. CPCTC

s

28. a. Given

b. Def. of

c. Def. of rt.

d. Given

e. BC BC

f. Reflexive

g. HL

h. CPCTC

28. (continued)i. DEC

j. Vert. are .

k. AAS

l. AE DE

m. CPCTC

s


29–30. Proofs may vary. Samples are given.

29. It is given that 1 2 and 3 4.

Since QB QB by the Reflexive Prop. of

, QTB QUBby ASA. So QT QUby CPCTC. Since QE

QE by the Reflexive Prop. of , then QET QEUby SAS.

30. 1. AD ED (Given)

2. D is the midpt. of BF. (Given)

3. FD DB (Def. of midpt.)

4. FDE ADB (Vert. are .)

5. FDE BDA(SAS)

6. E A (CPCTC)

s

30. (continued)7. GDE CDA

(Vert. are .)

8. ADC EDG(ASA)

31. a. AD BC;AB DC;AE EC;DE EB

s


31. (continued)b. Use DB DB (refl.)

and alt. int. to show ADB

CBD (ASA). ABDC and AD BC

(CPCTC). AEBCED (ASA) and AED CEB

(ASA). Then AEEC and DE EB(CPCTC).

s

32. 1. AC EC; CB CD (Given)

2. C C (Reflexive Prop. of )

3. ACD ECB(SAS)

4. A E (CPCTC)

33. PQ RQ and PQTRQT by Def. of

bisector. QT QT so PQT RQT by

SAS. P R by CPCTC. QT bisects

VQS so VQTSQT and PQT

and RQT are both rt. . So VQP

SQR since they are compl. of . PQV

RQS by ASA so QV QS by CPCTC.

34. C

35. F

s

s


36. A

37. [2] a. HBC HED

b. HB HE by CPCTC if HBC

HED by ASA. Since BDC

CED by AAS, then DBC

CED by CPCTC and CHB

DHE because vertical are .


s

38. [4] a. HL

b.

c. x = 30. In ADC m A + m ADC + m ACD= 180. Substituting, 90 + x + x + x = 180. Solving, x = 30.

d. 120; it is suppl. to a 60°

e. 6 m; DC = 2(AD)


38. (continued)[3] 4 parts answered

correctly

[2] 3 parts answered correctly

[1] 2 parts answered correctly

39. a. right

b.

c. Reflexive

d. HL

40.

41.

42. y + 6 = (x – 2)

43. y – 5 = 1(x – 0)

44. y – 6 = –2(x + 3)

12

45. y – 0 = – (x – 0)

46–48. Eqs. may vary, depending on pt. chosen.

46. y – 4 = 2(x – 1)

47. y + 5 = (x – 3)

48. y + 3 = – (x + 4)

13

53

56

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