CHAPTER Solutions Key 5 Properties and Attributes of Triangles

CHAPTER

5

Solutions Key Properties and Attributes of Triangles −− b. Since DG = GE and ⊥ DE, is the ⊥ bisector −− of DE by the Conv. of the ⊥ Bisector Thm. 1 DE EF = _ 2 _ EF = 1 (20.8) = 10.4 2

ARE YOU READY? PAGE 297 1. E

2. C

3. A

4. D

5. B 6. acute

7. right

8. acute

9. obtuse

2 10. 8 = 64

11. (-12) 2 = 144

49 = 7 12. √

13. - √ 36 = -6

14. √ 9 + 16 = √ 25 = 5

15. √ 100 - 36 = √ 64 = 8

16.

√ 81 9 81 = _ _ =_ 25

17.

5

√ 25

2a. WX = WZ WX = 3.05 −−− −− −−− −− b. Since XW = ZW, XW ⊥ XY, and ZW ⊥ ZY, YW bisects ∠XYZ by the Conv. of the ∠ Bisector Thm. m∠XYZ = 2m∠WYZ m∠XYZ = 2(63°) = 126°

2

bisects 3. By the Conv. of the ∠ Bisector Thm., QS ∠PQR.

2 =2

4.

18. d + 5 < 1 d < -4

y

P

19.

x

-4 ≤ w - 7 3≤w w≥3

Q

−− Step 1 Graph PQ. −− −− The ⊥ bisector of PQ is ⊥ to PQ at its midpoint −− Step 2 Find the midpoint of PQ. + (-4) −− + 1 , 2_ = (3, -1) midpoint of PQ = 5_ 2 2 Step 3 Find the slope of the perpendicular bisector. −− -4 - 2 -6 = _ 3 =_ slope of PQ = _ 2 1-5 -4 Since the slopes of ⊥ lines are opposite reciprocals, 2. the slope of the ⊥ 1bisector is - _ 3 Step 4 Use point-slope form to write an equation. −− 2 and passes The ⊥ bisector of PQ has slope - _ 3 through (3, -1). y - y 1 = m(x - x 1) 2 (x - 3) y - (-1) = - _ 3 _ y + 1 = - 2 (x - 3) 3

(

m -2 > _

10 -20 > m m < -20

20. -3s ≥ 6 s ≤ -2

21.

22. Let p and q represent the following: p: Lines and m intersect. q: Lines and m are not parallel. Given: p → q, and p. So by the Law of Detachment, q is true: Lines and m are not parallel. 23. Let p, q, and r represent the following: −− p : M is the midpoint of AB q : AM = MB 1 AB and MB = _ 1 AB r : AM = _ 2 2 Given: p → q and q → r. So by the Law of Syllogism, −− 1 AB and p → r : If M is the midpoint of AB, then AM = _ 2 1 AB. MB = _ 2

)

THINK AND DISCUSS 1. Yes; no; since PY = QY = 3, Y is the midpoint −− of PQ, and thus by the def. of bisector, is a −− −− bisector of PQ. If is the ⊥ bisector of PQ, then PX would equal QX by the ⊥ Bisector Thm. However, PX = 8.5 and QX = 8.4, so is not the ⊥ bisector −− of PQ. 2. No; although MJ = ML, to apply the Conv. of the −− −− ∠ Bisector Thm., you must know that MJ ⊥ KJ −− −− and ML ⊥ KL.

5-1 PERPENDICULAR AND ANGLE BISECTORS, PAGES 300–306 CHECK IT OUT! 1a. DG = EG DG = 14.6

93

Holt McDougal Geometry

ʡ"ISECTOR

3.

4HM)FAPTISON THEʡBISECTOROFA SEG THENITIS EQUIDISTANTFROM THEENDPOINTSOF THESEG

Since the slopes of ⊥ lines are opposite reciprocals, the slope of the ⊥ bisector is 1. Step 4 Use point-slope form to write an equation. −−− The ⊥ bisector of MN has slope 1 and passes through (-2, 1). y - y 1 = m(x - x 1) y - 1 = 1[x - (-2)] y-1=x+2

Ȝ"ISECTOR

#ONV)FAPTIS EQUIDISTANTFROM THEENDPOINTSOFA SEG THENTHE PTISONTHE ʡBISECTOROFASEG

4HM)FAPTISON THEBISECTOROFAN Ȝ THENITIS EQUIDISTANTFROM THESIDESOF THEȜ

#ONV)FAPTIN THEINTOFANȜIS EQUIDISTANTFROM THESIDESOFTHEȜ THENTHEPTISON THEBISECTOROFTHEȜ

10.

EXERCISES

y

GUIDED PRACTICE

1. perpendicular bisector

−− 2. Since PS = QS and m ⊥ PQ, m is the ⊥ bisector −− of PQ by the Conv. of the ⊥ Bisector Thm. PQ = 2QT

U

−− Step 1 Graph UV. −− −− The ⊥ bisector of UV is ⊥ to UV at its midpoint −− Step 2 Find the midpoint of UV . −− 2 + 4 -6 + 0 midpoint of UV = _, _ = (3, -3) 2 2 Step 3 Find the slope of the perpendicular bisector. −− 0 - (-6) 6 =3 slope of UV = _ = _ 2 4-2 Since the slopes of ⊥ lines are opposite reciprocals, 1. the slope of the ⊥ bisector is - _ 3 Step 4 Use point-slope form to write an equation. −− 1 and passes The ⊥ bisector of UV has slope - _ 3 through (3, -3). y - y 1 = m(x - x 1) 1 (x - 3) y - (-3) = - _ 3 _ y + 3 = - 1 (x - 3) 3

3. SP = SQ SP = 25.9

(

4. PS = QS 4a = 2a + 26 2a = 26 a = 13 So QS = 2(13) + 26 = 52. 5. AD = CD AD = 21.9 −− −− −− −− 6. Since AD = CD, AC ⊥ AB, and CD ⊥ BC, BD bisects ∠ABC by the Conv. of the ∠ Bisector Thm. 1 m∠ABC m∠CBD = _ 2 1 (48°) = 24° m∠CBD = _ 2 −− −− −− −− 7. Since DA = DC, AD ⊥ AB, and CD ⊥ BC, BD bisects ∠ABC by the Conv. of the ∠ Bisector Thm. m∠DBC = m∠DBA 10y + 3 = 8y + 10 2y + 3 = 10 2y = 7 7 y=_ 2 7 + 3]° = 38° So m∠DBC = [10 _ 2 −− −− 8. The braces can be installed so that PK ⊥ JL, −− −− PM ⊥ NL, and PK = PM. Then by the Conv. of the ∠ Bisector Thm., P will be on the bisector of ∠JLN.

11.

x

K

−− Step 1 Graph JK. −− −− The ⊥ bisector of JK is ⊥ to JK at its midpoint −− Step 2 Find the midpoint of JK. −− -7 + 1 5 + (-1) midpoint of JK = _, _ = (-3, 2) 2 2 Step 3 Find the slope of the perpendicular bisector. −− 3 -1 - 5 = _ -6 = - _ slope of JK = _ 4 8 1 - (-7) Since the slopes of ⊥ lines are opposite reciprocals, 4. the slope of the ⊥ bisector is _ 3 Step 4 Use point-slope form to write an equation. −− 4 and passes The ⊥ bisector of JK has slope _ 3 through (-3, 2).

y

(

x

)

y

J

()

V x

PQ = 2(47.7) = 95.4

9. M

N

−−− Step 1 Graph MN. −−− −−− The ⊥ bisector of MN is ⊥ to MN at its midpoint. −−− Step 2 Find the midpoint of MN.

(

)

−−− -5 + 1 4 + (-2) midpoint of MN = _, _ = (-2, 1) 2 2

)

y - y 1 = m(x - x 1) 4 [x - (-3)] y-2=_ 3 4 (x + 3) y-2=_ 3

Step 3 Find the slope of the perpendicular bisector. −−− -6 = -1 -2 - 4 = _ slope of MN = _ 6 1 - (-5)

94


y_ 2 - y1 x2 - x1 −− -1 - 5 = _ -6 = –1 slope of XY = _ 6 -1 - (-7) Since the slopes of ⊥ lines are opposite reciprocals, the slope of the ⊥ bisector is 1. Step 4 Use point-slope form to write an equation. −− 1 and passes The ⊥ bisector of XY has slope - _ 2 through (-2, -3). y - y 1 = m(x - x 1) y - 2 = 1[x - (-4)] y-2=x+4 −−− 21. Step 1 Graph MN. −−− −−− The ⊥ bisector of MN is ⊥ to MN at its midpoint −−− Step 2 Find the midpoint of MN. x_ 1 + x2 y 1 + y2 ,_ 2 2 −−− -3 + 7 1 + (-5) midpoint of MN = _ , _ = (2, -3) 2 2 Step 3 Find the slope of the ⊥ bisector. y2 - y1 slope = _ x2 - x1 -5 - (-1) −−− _ -4 = - _ 2 slope of MN = =_ 5 10 7 - (-3) Since the slopes of ⊥ lines are opposite reciprocals, _. the slope of the bisector is 5 2 Step 4 Use point-slope form to write an equation. 1 and passes through The bisector of ⊥ has slope - _ 2 (-2, -3). y - y 1 = m(x - x 1) _ (x - 2) y - (-3) = 5 2 _ (x - 2) y+3= 5 2 QS = QT 22. PS = PT 6n - 3 = 4n + 14 3m + 9 = 5m - 13 2n - 3 = 14 9 = 2m - 13 2n = 17 22 = 2m n = 8.5 11 = m slope =

PRACTICE AND PROBLEM SOLVING

12. GJ = GK GJ = 8.25

JG = KG x + 12 = 3x - 17 12 = 2x - 17 29 = 2x 14.5 = x So KG = 3(14.5) - 17 = 26.5. −− −− 14. Since GJ = GK and t ⊥ JK, t is the ⊥ bisector of JK by the Conv. of the ⊥ Bisector Thm. JK = 2JH JK = 2(26.5) = 53 13.

15. RQ = TQ RQ = 1.3 −− −− −− −− 16. Since RQ = TQ, RQ ⊥ RS, and TQ ⊥ TS, SQ bisects ∠RST by the Conv. of the Bisector Thm. m∠RST = 2m∠RSQ m∠RST = 2(58°) = 116°

(

17. m∠QSR = m∠QST 9a + 48 = 6a + 50 3a + 48 = 50 3a = 2 2 + 50 = 54° m∠QST = 6 _ 3 18. They can position Main St. so that the ∠ formed by Elm St. and Main St. is to the ∠ formed by Grove St. and Main St. Then by the ∠ Bisector Thm., every point on Main St. will be equidistant from Elm St. and Grove St. −− 19. Step 1 Graph EF. −− −− The ⊥ bisector of EF is ⊥ to EF at its midpoint −− Step 2 Find the midpoint of EF. x1 + x2 _ y + y2 _ , 1 2 2 −− -4 + 0 -7 + 1 midpoint of EF = _ , _ = (-2, -3) 2 2 Step 3 Find the slope of the perpendicular bisector. y2 - y1 slope = _ x2 - x1 −− 1 - (-7) 8 slope of EF = _ = _ =2 0 - (-4) 4 Since the slopes of ⊥ lines are opposite reciprocals, 1. the slope of the ⊥ bisector is - _ 2 Step 4 Use point-slope form to write an equation. −− 1 and passes The ⊥ bisector of EF has slope - _ 2 through (–2, –3). y - y 1 = m(x - x 1) 1 [x - (-2)] y - (-3) = -_ 2 _ y + 3 = - 1 (x + 2) 2 −− 20. Step 1 Graph XY. −− −− The ⊥ bisector of XY is ⊥ to XY at its midpoint −− Step 2 Find the midpoint of XY. x_ 1 + x2 y 1 + y2 ,_ 2 2 -7 + (-1) 5 + (-1) −− midpoint of XY = _ , _ = (–4, 2) 2 2 Step 3 Find the slope of the perpendicular bisector.

()

(

(

(

(

)

)

)

) (

23. JK = LK JK = 38 25.

MK ML + LK ML ML

)

24. GN = 2GZ GN = 2(36) = 72 = HK = HJ + JK = HJ = 38

27. JL = 2LX JL = 2(12) = 24

26. HY = MY HY = 24

28.

NK NM + ML + LK NM + 38 + 38 NM

= GK = 114 = 114 = 38

26 ; BC =

26 ; 29. Possible answer: C(3, 2); AC = √

so AC = BC, and by the Conv. of the ⊥ Bisector −− Thm., C is on the ⊥ bisector of AB. −− 30. Draw line ⊥ to AB through X. So m∠AYX = 90° and m∠BYX = 90° by the def. of ⊥. It is given that −− −− AX = BX. So AX BX by def. of segs. By the −− −− Reflex. Prop. of , XY XY. So AYX BYX −− −− by HL. Then AY BY by CPCTC. By the def. of −− −− midpoint, Y is the midpoint of AB. Since is ⊥ to AB −− at its midpoint, is the ⊥ bisector of AB. Therefore −− X is on the ⊥ bisector of AB.

)

95


31.

Statements bisects ∠QPR, 1. PS −− −− , SR ⊥ PR SQ ⊥ PQ 2. ∠QPS ∠RPS 3. 4. 5. 6. 7. 8.

∠SQP and ∠SRP are rt. . ∠SQP ∠SRP −− −− PS PS PQS PRS −− −− SQ SR SQ = SR

34. In the construction of the perpendicular bisector −− of AB, the same compass setting is used to draw an arc from each end point of the segment. So in the diagram, AX = BX and AY = BY. By the Converse of the Perpendicular Bisector Theorem, both X and −− Y lie on the perpendicular bisector of AB. So is the −− perpendicular bisector of AB.

Reasons 1. Given 2. Def. of ∠ bisector 3. Def. of ⊥ 4. Rt. ∠ Thm. 5. Reflex. Prop. of 6. AAS 7. CPCTC 8. Def. of segs.

Ű X A

TEST PREP

35. D; −− J is on the perpendicular bisector of XY, so by the Perpendicular Bisector Theorem, JX = JY. 36. G; 37. Possible answer: All locations that are equidistant from Park St. and Washington Ave. lie on the bisector of the ∠ formed by the 2 streets. All locations that are equidistant from the museum and the library lie on the perpendicular bisector of a segment formed by the museum and the library. So the visitor center should be built at point V, where the angle bisector and the perpendicular bisector intersect.

P

C B

−− 33a. Step 1 Graph AC. −− −− The ⊥ bisector of AC is ⊥ to AC at its midpoint −− Step 2 Find the midpoint of AC. x_ y + y2 1 + x2 _ , 1 2 2 −− -3 + 3 -2 + 6 midpoint of AC = _ , _ = (0, 2) 2 2 Step 3 Find the slope of the perpendicular bisector. y2 - y1 slope = _ x2 - x1 - (-2) −− 6 _ = 4_ slope of AC = _ = 8 6 3 3 - (-3) Since the slopes of ⊥ lines are opposite 3. reciprocals, the slope of the ⊥ bisector is - _ 4 Step 4 Use point-slope form to write an equation. −− _ and passes The ⊥ bisector of AC has slope - 3 4 through (0, 2). y - y 1 = m(x - x 1) _ (x – 0) y-2=- 3 4 _ y = - 3x + 2 4 b. There are 2 points on the ⊥ bisector that are 4 mi −− dist. from the midpoint of AC. −− c. Distance of warehouse from midpoint of AC = 4 mi. −− Distance of midpoint of AC from A 2 3 + 4 2 = 5 mi. = 2 Distance of warehouse from A = 4 + 5 2 ≈ 6.4 mi.

(

(

)

B

Y

32. Possible answer: By stating that the point must be in the int. of the ∠, the thm. implies that it must be in the same plane as the ∠. It is possible for a point to be equidistant from the sides of an ∠ but to lie in a different plane. In the diagram, ∠ABC is in plane Z, and P is equidistant from the sides of ∠ABC, but P does not lie in plane Z. Thus P cannot be on the bisector of the ∠, because the bisector must lie in the same plane as the ∠.

A

P

CHALLENGE AND EXTEND

are both and from P to BC 38a. The dist. from P to BA , and and BC 2 √ 5. So P is equidistant from BA therefore by the Converse of the Angle Bisector Theorem, P is on the bisector of ∠ABC.

)

b. Possible answer: y = 3x - 6. 39. The distance of a point (x, y) from x-axis is ⎪y⎥, and its distance from y-axis is ⎪x⎥. So locus is ⎪y⎥ = ⎪x⎥, or the lines y = x and y = -x. 40. 1. 2. 3. 4. 5.

Statements −− −− , VZ ⊥ YZ , VX ⊥ YX VX = VZ ∠VXY and ∠VZY are rt. . −− −− YV YV YXV YZV ∠XYV ∠ZYV

bisects ∠XYZ. 6. YV

96

Reasons 1. Given 2. Def. of ⊥ 3. Reflex. Prop. of 4. HL 5. CPCTC 6. Def. of ∠ bisector


−− −− 41. It is given that KN is the perpendicular bisector of JL −− and LN is the perpendicular bisector of −− KM. By the Perpendicular Bisector Theorem, JK = KL and KL = ML. Thus JK = ML by the Trans. Prop. −− −− of =. By the definition of segs., JK ML. By the Seg. Add. Post., JR + RL = JL and .MT + TK = MK. By the definition of the perpendicular bisector, −− R is the midpoint of JL and T is the midpoint of −− −− −− −− −− MK. Thus JR RL and MT TK. By the definition of cong segs., JR = RL and MT = TK. By Subst., JR + JR = JL and MT + MT = MK. It is given −− −− that JR MT. So JR = MT by definition of segs. By Subst., JR + JR = MK. By the Trans. Prop. of =, −− −− JL = MK, so JL MK by the definition of segs. −− −− By the Reflex. Prop. of , JM JM. Therefore JKM MLJ by SSS, and ∠JKM ∠MLJ by CPCTC.

2.

y x

O

H x

y

G

Step 1 Graph the . Step 2 Find equations for two perpendicular bisectors. Since two sides of lie along the axes, use the graph to find the perpendicular bisectors of these two sides. the perpendicular bisector of GO is y = -4.5, and the perpendicular bisector of OH is x = 4. Step 3 Find the intersection of the two equations. The lines y = -4.5 and x = 4 intersect at (4, -4.5), the circumcenter of GOH.

SPIRAL REVIEW

3a. X is the incenter of PQR. By the Incenter Theorem, X is euqidistant from the sides of PQR. −− The distance from X to PR is 19.2, so −− the distance from X to PQ is also 19.2.

42. C

2 4-2 =_ = _ 43. slope of RS 1+4

5

-5 + 1 _ = _ =2 slope of VT

-7 - 3 5 The slopes are the same, so the lines are parallel. -5 - 2 = _ 7 = _ 44. slope of RV 3 -7 + 4 5 -1 - 4 = -_ = _ slope of ST 2 3-1 The slopes are not the same, so the lines are not parallel. The product of the slopes is not -1, so the lines are not perpendicular.

b. m∠PRQ = 2m∠PRX m∠PRQ = 2(12°) = 24° m∠RQP + m∠PRQ + m∠QPR = 180° 52 + 24 + m∠RQP = 180° m∠RQP = 104° 1 m∠RQP m∠PQX = _ 2 1 (104°) = 52° m∠PQX = _ 2 4. By the Incenter Theorem, the incenter of a is equidistant from the sides of the . Draw the formed by the streets and draw the ∠ bisectors to find the incenter, point M. The city should place the monument at point M.

3 -1 - 2 = -_ = _ 45. slope of RT 7 3+4 -5 2 7 _ _ = slope of VR = 3 -7 + 4 The product of the slopes is -1, so the lines are perpendicular. -9 - (-1) 46. m = _ = -8 47. y - y 1 = m(x - x 1) 2-1 y + 15 = -0.5(x - 10) y - y 1 = m(x - x 1) y + 15 = -0.5x + 5 y + 1 = -8(x - 1) 1 x - 10 y = -8x + 7 y = -_ 2 5 0 5 _ _ 48. m = = 4 0 - (-4) y = mx + b 5x + 5 y=_ 4

M

THINK AND DISCUSS 1. Possible answer:

5-2 BISECTORS OF TRIANGLES, PAGES 307–313 CHECK IT OUT!

2. Q; P. Possible answer: the incenter is always inside the , so Q cannot be the incenter. Therefore P must be the incenter, and Q must be the circumcenter.

1a. GM = MJ = 14.5 b. GK = KH = 18.6 c. Z is circumcenter of GHJ, By the Circumcenter Theorem, Z is equidistant from the vertices of GHJ. JZ = GZ = 19.9

97


3.

#IRCUMCENTER

9. F is the incenter of CDE. By the Incenter Theorem, F is equaldistant from the sides of −− CDE. The distance from F to DE is 42.1, so the −− distance from F to CD is also 42.1.

)NCENTER

$EFINITION

4HEPTOF CONCURRENCYOF THEʡBISECTORS

4HEPTOF CONCURRENCYOF THEȜBISECTORS

$ISTANCE

%QUIDISTANTFROM THEVERTICESOF THE̱

%QUIDISTANTFROM THESIDESOF THE̱

,OCATION)NSIDE /UTSIDE OR/N

#ANBEINSIDE OUTSIDE ORON THE̱

)NSIDETHE̱

10. m∠DCE = 2m∠FCD m∠DCE = 2(17°) = 34° m∠DCE + m∠CDE + m∠CED = 180° 34 + 54 + m∠CED = 180° m∠CED = 92° 1 _ m∠FED = m∠CED 2 1 (92°) = 46° m∠FED = _ 2 11. The largest possible in the int. of the is its inscribed , and the center of the inscribed is the incenter. Draw the and its ∠ bisectors. Center the at E, the point of concurrency of the ∠ bisectors.

EXERCISES GUIDED PRACTICE

1. They do not intersect at a single point. 2. circumscribed about 3. N is the circumcenter of PQR. By the Circumcenter Theorem, N is equidistant from vertices of PQR. NR = NP = 5.64

x

4. RV = PV = 5.47

O

L

x

x

Step 1 Graph . Step 2 Find equations for the two perpendicular bisectors. Since the two sides of the lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of KO is y = 6, and the perpendicular bisector of OL is x = 2. Step 3 Find the intersection of the two equations. The lines y = 6 and x = 2 intersect at (2, 6), the circumcenter of KOL. x

15. AY = YB = 63.9

19. m∠TSR = 2m∠JSR m∠TSR = 2(14°) = 28° m∠TSR + m∠SRT + m∠RTS = 180° 28 + 42 + m∠TSR = 180° m∠TSR = 110° 1 _ m∠RTJ = m∠TSR 2 _ m∠RTJ = 1 (110°) = 55° 2 20. By the Circumcenter Theorem, the circumcenter of the is equidistant from the vertices. Draw the formed by the cities, and draw the perpendicular bisectors of the sides. The main office should be located at M, the circumcenter.

x

O

14. DB = AD = 62.8

18. J is the incenter of RST. By the Incenter Theorem, J is equaldistant from the sides of RST. −− The distance from J to ST is 8.37, so the distance −− from J to RS is also 8.37.

A

13. YC = YB = 63.9

17. Step 1 Write equations of the perpendicular bisectors −− −−− of OV and OW. −− The perpendicular bisector of OV is y = 9.5; the −− perpendicular bisector of WO is x = -1.5. Step 2 Find the circumcenter of the . The circumcenter is at the intersection of the perpendicular bisectors, (-1.5, 9.5).

y

8.

12. CF = FA = 59.7

16. Step 1 Write equations of the perpendicular bisectors −−− −− of MO and NO. −−− The perpendicular bisector of MO is x = -2.5; the −− perpendicular bisector of NO is y = 7. Step 2 Find the circumcenter of the . The circumcenter is at the intersection of the perpendicular bisectors, (-2.5, 7).

ÓäÂ


y K y

xäÂ

6. N is the circumcenter of PQR. By the Circumcenter Theorem, N is equidistant from vertices of PQR. QN = NP = 5.64 7.

*

+

5. TR = QT = 3.95

y

B

Step 1 Graph the . Step 2 Find equations for the two perpendicular bisectors. Since the two sides of the lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of AO is x = -3.5, and the perpendicular bisector of OL is y = -5. Step 3 Find the intersection of the two equations. The lines x = -3.5 and y = -5 intersect at (-3.5, -5), the circumcenter of AOB.

98


21. Possible answer: if ∠JML is a rt. ∠, then m∠MJL + m∠MLJ = 90° because the acute of a rt. −− are comp. Since M is the incenter of JKL, JM −− and LM are ∠ bisectors of JKL. So by the def. of ∠ bisector, m∠KJL = 2m∠MJL and m∠KLJ = 2m∠MLJ. By subst., m∠KJL + m∠KLJ = 2(m∠MJL + m∠MLJ) = 2(90°) = 180°. But by the Sum Theorem, m∠K = 180° - (m∠KJL + m∠KLJ) = 180° - 180° = 0°. This would mean that JKL is not a . Therefore ∠JML cannot be a rt. ∠.

36.

4. 5. 6. 7.

1. Given 2. Def. of ∠ bisector 3. Reflex. Prop. of 4. SAS 5. CPCTC 6. Lin. Pair Thm. 7. supp. → rt. 8. Def. of rt. ∠

PQS RQS ∠PSQ ∠RSQ ∠PSQ and ∠RSQ are supp. ∠PSQ and ∠RSQ are rt. .

8. ∠PSQ = ∠RSQ = 90° −−

⊥ PR 9. QS −− −− 10. PS RS −− 11. S is midpoint of PR.

S H

22. The angle bisector; m∠BAE = m∠EAC −− −− 23. The perpendicular bisector; AD = BD, AD ⊥ DG −− −− and BD ⊥ DG

9. Def. of ⊥ 10. CPCTC 11. Def. of midpoint 12. Def. of the perpendicular bisector

is the perpendicular 12. QS −− bisector of PR.

24. The angle bisector; m∠ABG = m∠GBC −− −− 25. The angle bisector; since AE and BG are ∠ −− bisectors, R is the incenter. CR intersects the incenter, so it is an ∠ the bisector.

37a. The new store is at the circumcenter of ABC. −− The perpendicular bisector of AB is x = 4. −− 3 −− 3 . ; the midpoint of AC is 2, _ The slope of AC is _ 4 2 −− The perpendicular bisector of AC is 3 = -_ 4 (x - 2). y-_ 3 2 3 = -_ 4 (4 - 2) At the intersection, x = 4 and y - _ 3 2 8 9 - 16 = -_ 7. = -_, so y = _ 3 6 6 7 . The new store is located at 4, -_ 6 b. outside, since y > 0 for all int. poins. of the , but 7 <0 -_ 6 c. distance from each store = distance from store C 7 = 4_ 1 ≈ 4.2 mi = 3 - -_ 6 6 38. Possible answers: Similarities: Both are circles. Both intersect the triangle in exactly 3 points. Differences: The inscribed circle is smaller than the circumscribed circle. Except for the points of intersection, the inscribed circle lies inside the triangle, while the circumscribed circle lies outside. The center of the inscribed circle always lies inside the triangle, while the center of the circumscribed circle may be inside, outside, or on the triangle. The center of the inscribed circle is the point of concurrency of the angle bisectors, while the center of the circumscribed circle is the point of concurrency of the perpendicular bisectors.

( )

27. neither

28. never 29. sometimes

Reasons

−− −− 3. QS QS

P

26. neither

Statements −− −− −− 1. QS bisects ∠PQR, PQ RQ. 2. ∠PQS ∠RQS

30. sometimes

(

31. never

( )

32. sometimes

−− −− 33. The slope of OA is 2; the midpoint of OA is (2, 4). −− The perpendicular bisector of OA is 1 (x - 2). y - 4 = -_ 2 −− The perpendicular bisector of OB is x = 4. 1 (4 - 2) = At the intersection, x = 4 and y - 4 = - _ 2 -1, so y = 3. The circumcenter is at (4, 3). −− 34. The perpendicular bisector of OY is y = 6. −− −− The slope of OZ is = 1; the midpoint of OZ is (3, 3). −− The perpendicular bisector of OZ is y - 3 = -(x - 3). At the intersection, y = 6 and 6 - 3 = 3 = -x + 3, so x = 0. The circumcenter is at (0, 6). 35a. ∠ Bisector Theorem

)

39a. Check students’ constructions. b. Check students’ constructions. TEST PREP

40. B; PX = PY by the Incenter Theorem. 41. F; m = 1, y + 2 = x - 5, or y = x - 7.

b. the bisector of ∠B

c. PX = PZ

99


−− 1-4 =_ 1 = opposite reciprocal of 52. slope of MY = _ 2 -4 - 2 2; so Y is on the perpendicular bisector

42. 14.75 KN = MN 5z - 4 = z + 11 4z = 15 z = 3.75 LN = MN = 3.75 + 11 = 14.75

−− -8 - 4 53. slope of MZ = _ = -3 ≠ opposite reciprocal -2 - 2 of 2; so Z is not on the ⊥ bisector


43a. Possible answer: −− Given: M is the midpoint of QR. Prove: PM = QM = RM Proof: The coordinates of M are 0 + 2a _ 2b + 0 _ = (a, b). , 2 2

(

5-3 MEDIANS AND ALTITUDES OF TRIANGLES, PAGES 314–320 CHECK IT OUT!

)

1a.

By the Distance Formula, PM = √ (a - 0) 2 + (b - 0) 2 2 2

2. 3; 4; possible answer: the x-coordinate of the centroid is the average of the x-coordinates of the vertices of the , and the y-coordinate of the centroid is the average of the y-coordinates of the vertices of the . −− 3. Possible answer: An equation of altitude to JK is 1 x + 3. It is true that 4 = -_ 1 (-2) + 3, so y = -_ 2 2 (-2, 4) is a solution of this equation. Therefore this altitude passes through the orthocenter.

and - 2a) 2 + (b - 0) 2 √(a 2 (-a) 2 + b 2 = a + b2. = √

RM =

Therefore, PM = QM = RM. b. Possible answer: The circumcenter of a rt. is the midpoint of the hyp. 44. Let C be the circumcenter of the . Given: AC = 28 1 AC. cm; so by the properties of 30-60-90 , BC = _ 2 So AB = AC + BC 3 AC =_ 2 3 (28) = 42 cm. =_ 2 SPIRAL REVIEW

t =_ 10 45. _ 26 65 65t = 260 t= 4 47.

_2 KW + ZW = KW 3 1 KW ZW = _ 3 1 KW 7=_

2 LX b. LZ = _ 3 2 (8.1) =_ 3 = 5.4

3 21 = KW

a +b , 2 2 QM = √(a - 0) + (b - 2b) 2 a 2 + (-b) 2 = a + b2, = √ =

KZ + ZW = KW

THINK AND DISCUSS 1. Possible answer: is isosc. B

2.5 = _ 6 46. _ x 1.75 2.5x = 10.5 x = 4.2

A

D

C

2. Possible answer: is a rt. . K

420 = _ 7 _

y 2 840 = 7y y = 120

J

3. The ratio of the length of the longer segment to the length of the shorter segment is 2 : 1.

48. m∠AFB + m∠BFE = 180° 55 + m∠BFE = 180° m∠BFE = 125°

4.

49. m∠AFB + m∠BFD + m∠DFE = 180° 55 + 90 + m∠DFE = 180° m∠DFE = 35° So m∠BFC = m∠DFE = 35° 50. m∠BFC + m∠CFE = m∠BFD + m∠DFE 35 + m∠CFE = 90 + 35 m∠CFE = 90° −− _ −− 51. slope of ST = 8 = -2; midpoint of ST = M(2, 4) -4 −− 4 - 3 1 = opposite reciprocal = -_ slope of MX = _ 2 2-0 of 2; so X is on the perpendicular bisector

L

$EFINITION ,OCATION)NSIDE /UTSIDE OR/N

#ENTROID

/RTHOCENTER

4HEPTOF CONCURRENCYOF THEMEDIANS

4HEPTOF CONCURRENCYOF THEALTITUDES

)NSIDETHE̱

#ANBEINSIDE OUTSIDE ORON THE̱


1. centroid

100

2. altitude


2 VX 3. VW = _ 3 2 (204) = 136 =_ 3 5.

1 VX 4. WX = _ 3 1 (204) = 68 =_ 3

2 RY RW = _ 3 2 RY 104 = _ 3 _3 (104) = RY 2 RY = 156

9.

) )

x

K

10.

M

P

y

Q

x

R

yx x

Step 1 Graph the . Step 2 Find an equation of the line containing the −− −− altitude from P to QR. QR is horizontal, the altitude is vertical, so the equation is x = -5. Step 3 Find an equation of the line containing the −− altitude from Q to PR. −− 5 - 8 = -1. Slope of PR = _ -2 + 5 Equation is y - 5 = x - 4, or y = x + 1. Step 4 Solve the system to find the coordinates of the orthocenter. x = -5 and y = -5 + 1 = -4. The coordinates of the orthocenter are (-5, -4).

Step 1 Graph the . Step 2 Find an equation of the line containing the −− −− altitude from L to KM. Since KM is horizontal, the altitude is vertical, so the equation is x = 4. Step 3 Find an equation of the line containing the −− altitude from K to LM. -2 - 6 = -2. Slope of LM = _ 8-4 1 (x - 2). Equation is y + 2 = _ 2 Step 4 Solve the system to find the coordinates of the orthocenter. 1 (4 - 2) = 1, so y = -1. x = 4 and y + 2 = _ 2 The coordinates of the orthocenter are (4, -1).

W

x

Step 1 Graph the . Step 2 Find an equation of the line containing the −− −− altitude from W to UV. Since UV is vertical, the altitude is horizontal, so the equation is y = -3. Step 3 Find an equation of the line containing the −−− altitude from U to VW. −−− -3 - 6 Slope of VW = _ = -1. 5+4 Equation is y + 9 = x + 4, or y = x - 5. Step 4 Solve the system to find the coordinates of the orthocenter. y = -3 and -3 = x - 5, so x = 2. The coordinates of the orthocenter are (2, -3).

U

x

ÞÊ ÊÓÊÊÚÊÝÓ® Ó

yx x

1 RW 6. WY = _ 2 1 (104) = 52 =_ 2

7. 1 Understand the Problem Answer will be the coordinates of the centroid of the . Important information is the location of vertices, A(0, 2), B(7, 4), and C(5. 0). 2 Make a Plan The centroid of the is the point of intersection of the three medians. So write the equations for two medians and find their point of intersection. 3 Solve −− Let M be the midpoint of AB and N be the midpoint −− of BC. 0+7 2+4 M = _ , _ = (3.5, 3) 2 2 5+7 _ 0+4 _ = (6, 2) N= , 2 2 −− AN is horizontal. Its equation is y = 2. −−− 3 - 0 = -2. Its equation is Slope of CM = _ 3.5 - 5 y = -2(x - 5). At the centroid, y = 2 = -2(x - 5), so x = 5 + (-1) = 4. The coordinates of the centroid are D(4, 2). 4 Look Back −− −− Let L be the midpoint of AC. Equation for BL is 2 _ y - 4 = (x - 7), which intersects y = 2 at (4, 2). 3 y 8. L £

( (

y

V

11.

y x

y

D

E

x

C

Step 1 Graph the . Step 2 Find an equation of the line containing the −− −− altitude from E to CD. CD is vertical, the altitude is horizontal, so the equation is y = 2. Step 3 Find an equation of the line containing the −− −− altitude from C to DE. DE is horizontal, altitude is vertical, so the equation is x = -1. Step 4 y = 2 and x = -1. The coordinates of the orthocenter are (-1, 2).

101



20. Step 1 Find an equation of the line containing the −− altitude through A. BC is vertical, the altitude is horizontal, so the equation is y = -3. Step 2 Find an equation of the line containing the altitude through C. −− 5 + 3 Slope of AB is _ = 2. 8-4 1 (x - 8). Equation is y + 8 = -_ 16. Support should be attached at the centroid. Equation 2 of the median through (4, 0) is x = 4. The median Step 3 Find the coordinates of the orthocenter. through (0, 10) also passes through 1 (x - 8), y = -3, so -3 + 8 = 5 = -_ 2 0 + 14 4+8 _ 10 7 1 _ _ _ or x = -10 + 8 = -2. , =- . = (6, 7), and has slope 2 2 2 0-6 The coordinates are (-2, -3). 1x + Equation of the second median is y = -_ 3 GP 1 GP 21. GL = _ 22. PL = _ 2 2 2 1 _ 10. At intersection, x = 4, so y = - (4) + 10 = 8. 3 (8) = 12 1 (8) = 4 2 =_ =_ 2 2 The coordinates of the centroid are (4, 8).

1 HC 12. PC = _ 3 1 (10.8) = 3.6 =_ 3 14. JA = 3PA = 3(2.9) = 8.7

(

2 Hp 13. HC = _ 3 2 (10.8) = 7.2 =_ 3 15. JP = 2PA = 2(2.9) = 5.8

)

17. Step 1 Find an equation of the line containing the −− altitude through X. YZ is vertical, the altitude is horizontal, so the equation is y = -2. Step 2 Find an equation of the line containing the altitude through Z. −− 10 + 2 3. Slope of XY is _ = _ 2 6+2 2 (x - 6). Equation is y + 6 = - _ 3 Step 3 Find the coordinates of the orthocenter. 2 (x - 6), y = -2, so -2 + 6 = 4 = - _ 3 or x = -6 + 6 = 0. The coordinates are (0, -2). 18. Step 1 Find an equation of the line containing the −− altitude through J. GH is horizontal, the altitude is vertical, so the equation is x = 4. Step 2 Find an equation of the line containing the altitude through H. −− -1 - 5 Slope of GJ is _ = -1. 4+2 Equation is y - 5 = x - 6. Step 3 Find the coordinates of the orthocenter. x = 4, so y - 5 = 4 - 6, or y = 5 - 2 = 3. The coordinates are (4, 3). 19. Step 1 Find an equation of the line containing the −− altitude through T. RS is horizontal, the altitude is vertical, so the equation is x = -2. Step 2 Find an equation of the line containing the −− altitude through R. ST is vertical, the altitude is horizontal, so the equation is y = 9. Step 3 Find the coordinates of the orthocenter. x = -2 and y = 9. The coordinates are (-2, 9).

23. HL = LJ = 5 −− −− −− −− 24. GL is the perpendicular bisector of HJ, so GJ GH. GJ = GH = 2GK = 2(6.5) = 13 25. P = GJ + GH + HJ = 2GH + 2LJ = 2(13) + 2(5) = 36 units 1 (HJ)(GL) 26. A = _ 2 1 (10)(12) = 60 square units =_ 2

) ( 1 (8 + 2 + 5), _ 1 (-1 + 7 - 3) = (5, 1) 28. G = ( _ ) 3 3

1 (0 + 14 + 16), _ 1 (-4 + 6 - 8) = (10, -2) 27. G = _ 3 3

29. PZ = 2ZX = 2(27) = 54

30. PX = 3ZX = 3(27) = 81

31. Step 1 Find n. 2n + 17 = 54 2n = 54 - 17 n = 18.5 Step 2 Find QZ. QZ = 4n - 26 = 4(18.5) - 26 = 48 1 (QZ) 32. YZ = _ 2 1 (48) = 24 =_ 2 33. Possible answer: the perpendicular bisector of base; the bisector of vertex ∠; the median to the base; the altitude to the base

102


34. sometimes

TEST PREP

35. always

41. D

42. G; I, III true since incenter, centroid always inside II false since obtuse

43. D CHALLENGE AND EXTEND

36. never

44a. Possible answer: ABC is equil., and is the −−− perpendicular bisector of BC. Since ABC is −− −− equil., AB AC by definition. So AB = AC by the definition of segs. Therefore by the Converse of the Perpendicular Bisector Theorem, A is on line . Similarly, B is on the perpendicular −− bisector of AC, and C is on the perpendicular −− bisector of AB.

37. always

B

38.

Statements −− −− 1. PS and RT are medians −− −− of PQR. PS RT 2. PS = RT 2 RT 2 PS = _ 3. _ 3 3 2 PS, RZ = _ 2 RT 4. PZ = _ 3 3 5. PZ = RZ −− −− 6. PZ RZ 7. ∠SPR ∠TRP −− −− 8. PR PR 9. PTR RSP 10. ∠QPR ∠QRP −− −− 11. PQ RQ 12. PQR is an isosc .

Reasons D

1. Given 2. Def. of segs. 3. Mult. Prop. of =

5. Subst. 6. Def. of segs. 7. Isosc. Thm. 8. Reflex. Prop. of 9. SAS 10. CPCTC 11. Con. of Isosc. Theorem 12. Def. of isosc.

b. DG =

) (

(3) (3)

_8 2 + _8 2

8 √2 ≈ 3.8 mi =_ 3 −− c. Perpendicular from G crosses EF at H(4, 4), 2 2 4 + _ 4 distance = _ 3 3 4 √2 ≈ 1.9 mi =_ 3

() ()

)

A

C

b. Possible answer: By the definition of the −− −− perpendicular bisector, BD CD. So D is the −− −− midpoint of BC by definition, and AD is a median of ABC by the definition of median. Therefore contains the median of ABC through A. Also −− by the definition of the perpendicular bisector, AD −− −− ⊥ BC. So AD is the altitude of ABC by the definition. Therefore contains the altitude of ABC through A. Again by the definition of the −− −− −− −− perpendicular bisector, BD CD. AB AC by the −− −− definition of equil. , and AD AD by the Reflex. Prop. of . So ABD ACD by SSS. Then −− ∠DAB ∠DAC by CPCTC, and AD is the bisector of ∠BAC by the definition of ∠ bisector. Therefore contains the ∠ bisector of ABC through A. The same reason can be applied to the other two ⊥ bisectors.

4. Centroid Thm.

39. Possible answer: The centroid of a is also called its center of gravity because the weight of the shape is evenly distributed in every direction from this point. This means the shape will rest in a horizontal position when supported at this point. 1 (0 + 0 + 8), _ 1 (0 + 8 + 0) = 2_ 2 , 2_ 2 40a. G = _ 3 3 3 3

(

Ű

c. Possible answer: The perpendicular bisectors of a are concurrent at the circumcenter, and the ∠ bisectors are concurrent at the incenter. The medians of a are concurrent at the centroid, and the altitudes of a are concurrent at the orthocenter. But in an equil. , the perpendicular bisector through a given vertex also contains the ∠ bisector, the median, and the altitude through that vertex. So the points of concurrency must all be the same point That is, the circumcenter, the incenter, the centroid, and the orthocenter in an equil. are the same point. c ; slope of ST = _ c ; 45a. slope of RS = _ b b-a slope of RT = 0. −− −− b b. Since ⊥ RS, slope of = -_. Since m ⊥ ST, c −− b-a a - b . Since n ⊥ RT slope of m = -_ = _ , n is c c a vertical line, and its slope is undefined.

103


c. An equation of is b y - 0 = - _ (x - a) c b ab _ y=- x+_ c c

1 AE 3. HF = _ 2 1 (1550) = 775 =_ 2 The distance she measures between H and F is 775 m.

An equation of m is a - b (x - 0) y-0=_ c a b x _ y= c

An equation of n is x = b.

(

ab - b d. b, _ c 2

THINK AND DISCUSS

)

−− 1. The endpoints of XY are not the midpoints of the sides of the .

e. Since the equation of line n is x = b and the x-coordinate of P is b, P lies on n.

2.

f. Lines , m, and n are concurrent at P.

$EFINITIONASEGJOINING THEMDPTSOFSIDES OFA̱

4RIANGLE -IDSEGMENT

SPIRAL REVIEW

46. Let x, y be prices of peanuts and popcorn. x = y + 0.75 or y = x - 0.75 5x + 3y = 21.75 5x + 3(x - 0.75) = 21.75 5x + 3x - 2.25 = 21.75 8x = 24 x=3 Price of peanuts is $3.00. 47. F; Possible answer: a rectangle with width 5 and length 8.

.ONEXAMPLE

%XAMPLE

B D A

B E

-IDSEGMENT

D C

E

A

C


49. KL = 2KP = 2(7.0) = 14.0

48. T

0ROPERTIESȡTOTHE THIRDSIDEHALFTHE LENGTHOFTHETHIRDSIDE

1. midpoints 2. The midpoints are S(-1, 4), T(4, 6); −− 2 −− 2; 4 =_ slope of ST = _ ; slope of PR = _ 10 5 5 −− −− since the slopes are =, ST PR. 2 2 + 5 2 = √ 29 ; ST =

50. QJ = QL = 9.1 51. m∠ JLQ = m∠LJQ = 36° m∠JQL + m∠LJQ + m∠ JLQ = 180° m∠JQL + 36 + 36 = 180° m∠JQL = 108°

2 4 + 10 2 = √ 116 = 2 √ 29 ; PR = 1 _ and ST = PR. 2 1 XY 3. NM = _ 4. XZ = 2LM 2 = 2(5.6) = 11.2 1 (10.2) = 5.1 =_ 2

CONSTRUCTION

1. Check students’ constructions. 2. Possible answer: The orthocenter of an acute is inside the . The orthocenter of an obtuse is outside the . The orthocenter of a rt. is the vertex of the rt. ∠.

1 XZ 5. NZ = _ 2 1 (11.2) = 5.6 =_ 2

5-4 THE TRIANGLE MIDSEGMENT THEOREM, PAGES 322–327

6. m∠LMN = m∠MNZ = 29°

CHECK IT OUT!

7. m∠YXZ = m∠MNZ = 29° 8. m∠XLM + m∠LMN = 180 m∠XLM + 29 = 180 m∠XLM = 151° 1 _ 9. CD < XZ = 15 ft = 5 yd 2 The width of the 2nd floor is less than 5 yd.

1. The midpoints are M(1, 1), N(3, 4); −− 6 _ −−− 3 ; slope of RS = _ = 3; slope of MN = _ 2 4 2 −−− −− since the slopes are =, MN RS. 2 2 + 3 2 = √ 13 ; MN = 2 2

4 + 6 = √ 52 = 2 √ 13 ; 1 _ and MN = RS. 2 1 KL 2a. JL = 2PN b. PM = _ 2 = 2(36) = 72 1 (97) = 48.5 =_ 2 c. m∠MLK = m∠JMP = 102° RS =


10. The midpoints are D(-4, 3), E(0, 4); −− 2 _ −− 1 slope of DE = _ ; slope of CB = _ = 1; 4 4 −− −−8 since the slopes are =, DE CB. 2 1 + 4 2 = √ 17 ; DE = 2 2 + 8 2 = √ 68 = 2 √ 17 ; CB = 1 CB. and DE = _ 2

104


1 GH 12. RQ = _ 2 1 (27) = 13.5 =_ 2 14. m∠PQR = m∠QRJ = 55°

11. GJ = 2PQ = 2(19) = 38

1 GJ 13. RJ = _ 2 1 (38) = 19 =_ 2 15. m∠HGJ = m∠QRJ 16. m∠GPQ + m∠HGJ = 180° = 55° m∠GPQ + 55 = 180° m∠GPQ = 125° −− 17. Yes; DE is a midsegment of ABC, so its length is 1 ft, or 2 _ 1 ft, which is 27 in. This is less than half of 4 _ 2 4 30 in. So the carpenter can use the 30 in. timber to make the crossbar. 19. P = KL + LM + KM 1 GH + _ 1 HJ =7+_ 2 2 1 (12) + _ 1 (8) =7+_ 2 2 = 17

18. P = GH + HJ + GJ = 12 + 2LJ + 2KL = 12 + 2(4) + 2(7) = 34

22. 2(n - 9) = 35 2n - 18 = 35 2n = 53 n = 26.5

23. 2(4n + 5) = 74 8n + 10 = 74 8n = 64 n=8

24. 2n - 23 = 2(9.5) 2n - 23 = 19 2n = 42 n = 21

1 GH 32. FJ = _ 33. m∠DCG = m∠CBA 2 = 57° 1 CG =_ 4 1 (16.5) = 4.125 =_ 4 34. m∠GHE = m∠HCD = m∠ABC = 57° 35. m∠FJG + m∠GHE = 180 m∠FJG + 57 = 180 m∠FJG = 123°

b. XA = CX = 3.5 mi, BC = 2BY = 8 mi trip length = WX + XA + AB + BC + CX + XW = 2.25 + 3.5 + 9 + 8 + 3.5 + 2.25 = 28.5 mi

26. 2(5n) = 8n + 10 10n = 8n + 10 2n = 10 n =5 −− 27. B; possible answer: in ABC, DE is a midsegment −− and BC is the side to it. By the Midsegment Theorem, the length of a midsegment is half the 1 BC. length of the side, so DE = _ 2 28. ∠D ∠FZY ∠YXE ∠ZYX; ∠E ∠ZYF ∠DXZ ∠XZY; ∠F ∠XYE ∠DZX ∠ZXY 25. 6n = 2(n + 8) 6n = 2n + 16 4n = 16 n=4

0 + 2a _ 0 + 2b = (a, b) , (_ 2 2 ) 2a + 2c 2b + 0 b. N = (_, _) = (a + c, b) 2 2

38a. M =

−− 0-0 =0 c. slope of PR = _ 2c - 0 −−− b-b =0 slope of MN = _ a+c-a −− −−− Slopes of MN and PR are =, so MN PR. −− d. PR = 2c; MN = a + c - a = c; the length of PR is −−− 1 PR. twice length of MN, so MN = _ 2

E Y

X

1 DC 31. EH = _ 2 1 CB =_ 2 1 (22) = 11 = _ 2

36. Yes; possible answer: let x be the length of each side of an isosc. . By the Midsegment Theorem, the length of the midsegment to each 1 x. Since these two midsegments of those sides is _ 2 are equal in length, they are . 1 XY 37a. WX = _ 2 1 AB =_ 4 _ = 1 (9) = 2.25 mi 4

20. The perimeter of GHJ is twice the perimeter of KLM. 21. 3n = 2(54) 3n = 108 n = 36

1 AB 30. CG = _ 2 1 (33) = 16.5 =_ 2

TEST PREP D

Z

F

39. D;

29. Possible answer: about 18 parking spaces; the new street is along the midsegment of the triangle plot of land. The length of the street is half of 440 ft, or 220 ft. Estimate the quotient 220 ÷ 23 by rounding 220 to 225 and 23 to 25. Since 225 ÷ 25 = 9, city can put about 9 parking spaces on one side of the street. So the total number of parking spaces is about 2(9), or 18.

105

RT = 2PQ 4x - 27 = 2(x + 9) 4x - 27 = 2x + 18 2x = 45 x = 22.5 RT = 4(22.5) - 27 = 63 m

40. H 41. D


53. NX = 2XS = 2(3) = 6


42. Let the coordinates of the vertices be (u, v), (w, x), (y, z). By the Midpoint Formula, v + x = 2(3) = 6 u + w = 2(-6) = -12 x + z = 2(1) = 2 w + y = 2(2) = 4 v + z = 2(-3) = -6 u + y = 2(0) = 0 2w = -12 + 4 - 0 2x = 6 + 2 - (-6) 2w = -8 2x = 14 w = -4 x =7 So (w, x) = (-4, 7). u + (-4) = -12 v+7=6 u = -8 v = -1 So (u, v) = (-8, -1). -4 + y = 4 7+z=2 y=8 z = -5 So (y, z) = (8, -5).

55. NP = 2NR = 2(4.5) = 9 CONSTRUCTION

1 AC 1. XY = _ 2 2. Possible answer: Find m∠BXY and m∠BAC and confirm that they are =. This means the two segments are by the Converse of the Corr. Post.

READY TO GO ON? PAGE 329 1. PQ = 2PR = 2(4.8) = 9.6

43. The midsegment is equilateral and equiangular 44. n 2 - 3 = 2(39) n 2 = 81 n = ±9

45.

2 n - 6n + 3 = 2(3n - 16) n 2 - 6n + 3 = 6n - 32 n 2 - 12n + 35 = 0 (n - 7)(n - 5) = 0 n = 7 or 5 3(7) - 16 = 5 > 0 3(5) - 16 = -1 < 0 So n = 7 is the only possible solution.

2. JM = ML = 58

32

16

8

b. length of midsegment 8 1 = (length of midsegment 4) _ 2 1 1 _ _ = =4 16 4

()

( )

c. length of midsegment n n 1 = 64 _ 1 = 26 - n n = AB _ 2 2

()

()

SPIRAL REVIEW

2% + 3% 48. concentration = _ = 2.5% 2 3(2%) + 1(3%) __ = 2.25% 49. concentration = 3+1 50. G(-3, -2) → G(-3, 2) = G(-3 + 0, -2 + 4) H(0, 0) → H(0 + 0, 0 + 4) = H(0, 4) J(4, 1) → J(4 + 0, 1 + 4) = J(4, 5) K(1, -2) → K(1 + 0, -2 + 4) = K(1, 2) 51. G(-3, -2) → G(1, -4) = G[-3 + 4, -2 + (-2)] H(0, 0) → H[0 + 4, 0 + (-2)] = H(4, -2) J(4, 1) → J[4 + 4, 1 + (-2)] = J(8, -1) K(1, -2) → K[1 + 4, -2 + (-2)] = K(5, -4) 52. G(-3, -2) → G(3, 0) = G(-3 + 6, -2 + 2) H(0, 0) → H(0 + 6, 0 + 2) = H(6, 2) J(4, 1) → J(4 + 6, 1 + 2) = J(10, 3) K(1, -2) → K(1 + 6, -2 + 2) = K(7, 0)

AB = AC 5z + 16 = 8z - 5 21 = 3z 7=z AC = 8(7) - 5 = 51

equation is y + 1 = -2(x - 3). 5. PS = PT = 83.9 XT = RX = 46.7 6. m∠GJK + m∠KJH + m∠JHK + m∠KHL + m∠LGJ = 180 2m∠GJK + 2(16) + 50 = 180 2m∠GJK = 98 m∠GJK = 49° −− The distance from K to HJ = KL = 21.

4

4

3.

−−− 4 _ 4. Slope of MN = _ = 1 , so the slope of the 8 2 perpendicular bisector = -2;

46. QXY XPZ YZR ZYX; 1 (area of PQR) area of XYZ = _ 4 47a. Number of 1 2 3 4 Midsegment Length of Midsegment

3 MX 54. MR = _ 2 3 (5.5) = 8.25 =_ 2

7. The equations of the two perpendicular bisectors are x = 4.5 and y = -2. So C = (4.5, -2). 1 BD 8. BW = _ 3 1 WE CW = _ 1 _ 2 = (87) = 29 1 (38) = 19 3 =_ CE = 3CW 2 = 3(19) = 57 1 (0 + 8 + 10), _ 1 (4 + 0 + 8) = (6, 4) 9. G = _ 3 3 −− 10. PS is horizontal, the altitude is vertical, so the −− 4 equation is x = 4; the slope of SV = _ = 1, so the 4 slope of the altitude to it is -1; the equation of this altitude is y - 4 = -(x - 2); at he orthocenter O, x = 4, so y = 4 - (4 - 2) = 2, and O = (4, 2).

(

)

1 JM 11. ZV = _ 2 = RM = 45 m∠RZV = m∠PVZ = 36°

PM = 2ZR = 2(53) = 106

12. XY = 2MN = 2(39) = 78 The distance across the pond is 78 m.

106


3.

5-5 INDIRECT PROOF AND INEQUALITIES IN ONE TRIANGLE, PAGES 332–339

B

C

A

CHECK IT OUT! 1. Possible answer: Given: RST Prove: RST cannot have 2 rt . Proof: Assume that RST has 2 rt . Let ∠R and ∠S be the rt. . By the def. of rt. ∠, m∠R = 90° and m∠S = 90°. By the Sum Thm., m∠R + m∠S + m∠T = 180°. But then 90 + 90 + m∠T = 180° by subst., so m∠T = 0°. However, a cannot have an ∠ with a measure of 0°. So there is no RST, which contradicts the given information. This means the assumption is false, and RST cannot have 2 rt. . −− 2a. The shortest side is AC, so the smallest ∠ is ∠B. −− The longest side is AB, so the greatest ∠ is ∠C. from smallest to greatest are ∠B, ∠A, ∠C.

4HEOREM )FBCAB THEN MȜA MȜC )FBCAC THEN MȜA MȜB )FACBC THEN MȜB MȜA )FACAB THEN MȜB MȜC )FABBC THEN MȜC MȜA )FABAC THEN MȜC MȜB

4HEOREM )FMȜA MȜC THENBCAB )FMȜA MȜB THENBCAC )FMȜB MȜA THENACBC )FMȜB MȜC THENACAB )FMȜC MȜA THENABBC )FMȜC MȜB THENABAC

4RIANGLE )NEQUALITY 4HEOREM AB BC A# BC AC A" AB AC BC

EXERCISES

b. m∠F = 90°, m∠E = 180 - (22 + 90) = 68° −− The smallest ∠ is ∠D, so the shortest side is EF. −− The greatest ∠ is ∠F, so the longest side is DE. −− −− −− Sides from shortest to longest are EF, DF, DE.

GUIDED PRACTICE

1. Possible answer: To prove something indirectly, you assume the opposite of what you are trying to prove. Then you use logic to lead to a contradiction of given information, a definition, a postulate, or a previously proven theorem. You can then conclude that the assumption was false and the original statement is true.

3a. 8 + 13 21 21 ≯ 21 No; 8 + 3 = 21, which is not greater than the third side length. b. 6.2 + 7 9 6.2 + 9 7 7 + 9 6.2 13.2 > 9 15.2 > 7 16 > 6.2 Yes; the sum of each pair of the lengths is greater than the third length. c. When t = 4, t - 2 = 2, 4t = 16, t 2 + 1 = 17. 2 + 16 17 2 + 17 16 16 + 17 2 18 > 17 19 > 16 33 > 2 Yes; the sum of each pair of the lengths is greater than the third length. 4. Let s be the length of the 3rd side. Apply the Inequal. Theorem. s + 17 > 22 s + 22 > 17 17 + 22 > s s >5 s > -5 39 > s Combine the inequals. So 5 < s < 39. the length of the 3rd side is > 5 in. and < 39 in.

2. Possible answer: Given: ABC is a scalene triangle. Prove: ABC cannot have 2 . Proof: Assume that ABC does have 2 . Let −− −− ∠A and ∠C be the . Then AB CB by the Converse of the Isosc. Theorem. However, a scalene by definition has no sides. So ABC is not scalene, which contradicts the given information. This means the assumption is false, and therefore ABC can not have 2 . 3. Possible answer: −− Given: PQR is an isosc. with base PR. Prove: PQR cannot have a base ∠ that is a rt. ∠. Proof: Assume that PQR has a base ∠ that is a rt. ∠. Let ∠P be the rt. ∠. By the Isosc. Theorem, ∠P ∠R, so ∠R is also a rt. ∠. By the definition of rt. , m∠P = m∠R = 90°. By the Sum Theorem, m∠P + m∠Q + m∠R = 180°. By Subst. m∠Q = 0°. However, a cannot have an ∠ with a measure of 0°. So there is no PQR, which contradicts the given information. This means the assumption is false, and therefore PQR can not have a base ∠ that is rt. −− 4. The shortest side is PQ, so the smallest ∠ is ∠R. −− The longest side is PR, so the greatest ∠ is ∠Q. from smallest to greatest are ∠R, ∠P, ∠Q.

5. Let d be the distance from Seguin to Johnson City. d + 22 > 50 d + 50 > 22 22 + 50 > d d > 28 d > -28 72 > d 28 < d < 72 The distance from Seguin to Johnson City is > 28 mi and < 72 mi.

THINK AND DISCUSS 1. No; possible answer: the student must consider 2 cases and assume that either the ∠ is acute or the ∠ is rt.

5. m∠Z = 180 - (39 + 46) = 95° −− The smallest ∠ is ∠X, so the shortest side is YZ. −− The greatest ∠ is ∠Z, so the longest side is XY. −− −− −− Sides from shortest to longest are YZ, XZ, XY.

2. Possible answers: 2 cm, 4 cm, 5 cm; 2 cm, 4 cm, 8 cm

107


6. 4 + 7 10 4 + 10 7 7 + 10 4 11 > 10 14 > 7 17 > 4 Yes; the sum of each pair of 2 lengths is greater than the third length.


16. Possible answer: −− −− Given: ABC is scalene. XZ and YZ are midsegments of ABC. Prove: ABC cannot have 2 midsegments. Proof: Assume that ABC does have 2 −− −− midsegments. Let XZ and YZ be the midsegments. By the def. of segs., XZ = YZ. By

7. 2 + 9 12 11 ≯ 12 No; 2 + 9 = 11, which is not greater than the third side length.

1 BC and the Mid segment Thm., XZ = __ 2

8. 3.5 + 3.5 6 3.5 + 6 3.5 7>6 9.5 > 3.5 Yes; the sum of each pair of 2 lengths is greater than the third length.

1 1 1 BA. So __ BC = __ BA by subst. But then YZ = __ 2 2 2 −− −− BC = BA, and by the def. of segs., BC BA. However, a scalene by def. has no sides. So ABC is not scalene, which contradicts the given information. This means the assumption is false, and therefore a scalene cannot have 2 midsegments.

9. 1.1 + 1.7 3 2.8 ≯ 3 No; 1.1 + 1.7 = 2.8, which is not greater than the third side length. 10. When x = 5, 3x = 15, 2x - 1 = 9, x 2 = 25. 9 + 15 25 24 ≯ 25 No; when x = 5, the value of 3x is 15, the value of 2x - 1 is 9 and the value of x 2 is 25. 15 + 9 = 24, which is not greater than the third side length.

B X A

11. When c = 2, 7c + 6 = 20, 10c - 7 = 13, 3c 2 = 12. 12 + 13 20 12 + 20 13 13 + 20 12 25 > 20 32 > 13 33 > 12 Yes; when c = 2, the value of 7c + 6 is 20, the value of 10c - 7 is 13, and the value of 3c 2 is 12. The sum of each pair of 2 lengths is greater than the third length. 12. Let s be the length of the 3rd side. Apply the Inequal. Theorem. s + 8 > 12 s + 12 > 8 8 + 12 > s s>4 s > -4 20 > s Combine the inequals. So 4 < s < 20. The length of the 3rd side is > 4 mm and < 20 mm. 13. Let s be the length of the 3rd side. Apply the Inequal. Theorem. s + 16 > 16 16 + 16 > s s>0 32 > s Combine the inequals. So 0 < s < 32. The length of the 3rd side is > 0 ft and < 32 ft. 14. Let s be the length of the 3rd side. Apply the Inequal. Theorem. s + 11.4 > 12 s + 12 > 11.4 11.4 + 12 > s s > 0.6 s > -0.6 23.4 > s Combine the inequals. So 0.6 < s < 23.4. the length of the 3rd side is > 0.6 cm and < 23.4 cm. 15a. The longest side is opposite the greatest ∠. So the longest side is between the refrigerator and the stove. b. No; 4 + 5 = 9 ≯ 9. By the Inequal. Theorem, a cannot have these side lengths.

Y Z

C

17. Possible answer: Given: ∠J and ∠K are supp. Prove: ∠J and ∠K cannot both be obtuse. Proof: Assume that ∠J and ∠K are both obtuse. Then m∠J > 90° and m∠K > 90° by the definition of obtuse. Add the 2 inequals., m∠J + m∠K > 180°. However, by the definition of supp. , m∠J + m∠K = 180°. So m∠J + m∠K > 180° contradicts the given information. This means the assumption is false, and therefore a pair of supp. cannot both be obtuse. −− 18. The shortest side is KL, so the smallest ∠ is ∠J. −− The longest side is JL, so the greatest ∠ is ∠K.

from smallest to greatest are ∠J, ∠L, ∠K. 19. m∠S = 90, m∠T = 90 - m∠R = 24° −− The smallest ∠ is ∠T, so the shortest side is RS. −− The greatest ∠ is ∠S, so the longest side is RT. −− −− −− Sides from shortest to longest are RS, ST, RT. 20. 6 + 10 15 16 > 15 Yes; the sum of each pair of 2 lengths is greater than the third length. 21. 14 + 18 32 32 ≯ 32 No; 14 + 18 = 32, which is not greater than the third side length. 22. 5.8 + 5.8 11.9 11.6 ≯ 11.9 No; 5.8 + 5.8 = 11.6, which is not greater than the third side length. 23. 41.9 + 62.5 103 104.4 > 103 Yes; the sum of each pair of 2 lengths is greater than the third.

108


24. When z = 6, z + 8 = 14, 3z + 5 = 23, 4z - 11 = 13. 13 + 14 23 27 > 23 Yes; when z = 6, the value of z + 8 is 14, the value of 3z + 5 is 23, and the value of 4z - 11 is 13. The sum of each pair of 2 lengths is greater than the third side length. 25. When m = 3, m + 11 = 14, 8m = 24, m 2 + 1 = 10. 10 + 14 24 24 ≯ 24 No; when m = 3, the value of m + 11 = 14, the value of 8m is 24, and the valu e of m 2 + 1 is 10. the sum of 14 and 10 is 24, which is not greater than the third side length. 26. b + c > s s + 4 > 19 s > 15 15 yd < s < 23 yd 27. s + 23 > 28 s>5 5 km < s < 51 km

4 + 19 > s 23 > s 23 + 28 > s 51 > s

28. s + 3.8 > 9.2 3.8 + 9.2 > s s > 5.4 13.0 > s 5.4 cm < s < 13.0 cm 29. s + 1.89 > 3.07 s > 1.18 1.18 m < s < 4.96 m

1.89 + 3.07 > s 4.96 > s

5 1 > 3_ 30. s + 2_ 8 8 1 s > 1_ 2 3 in. 1 in. < s < 5_ 1_ 2 4

5 >s 1 + 3_ 2_ 8 8 3>s 5_ 4

1 5 > 6_ 31. s + 3_ 6 2 _ s > 22 3 1 ft 2 ft < s < 10_ 2_

5 + 6_ 1 >s 3_ 6 2 1>s 10_ 3

3 3 −− −− −− −− −− 32. AD, BD, AB, BC, CD; possible answer: in ABD, m∠ABD = 50°. In BCD, m∠DBC = 74°. In ABD, the order of the tubes from shortest to −− −− −− longest is AD, BD, AB. In BCD, the order of the −− −− −− tubes from shortest to longest is BD, BC, CD. So AD < BD < AB, and BD < BC < CD. Since AB = 50.8 and BC = 54.1, it is also true that AB < BC. −− −− −− −− −− So AD < BD < AB < BC < CD. 33. a > 7.5, where a is the length of a leg. Possible answer: By the Inequal. Thm., a + a > 15 and a + 15 > a. The solution of the first inequality is a > 7.5. The second inequality simplifies to 15 > 0, which is always a true statement. 34. Step 1 Find x. 2x + 5x - 1 = 90 7x = 91 x = 13 Step 2 Find ∠ measures and order sides. m∠A = 90°, m∠B = 5(13) - 1 = 64°, m∠C = 2(13) = 26° −− −− −− m∠C < m∠B < m∠A, so order is AB, AC, BC.

35. Step 1 Find x. 4.5x - 5 + 10x - 2 + 5x - 8 = 180 19.5x = 195 x = 10 Step 2 Find ∠ measures and order sides. m∠D = 4.5(10) - 5 = 40°, m∠E = 10(10) - 2 = 98°, m∠F = 5(10) - 8 = 42° −− −− −− m∠D < m∠F < m∠E, so order is EF, DE, DF. 36. A rt. ∠ cannot be an acute ∠. So the 1st and the 3rd statements contradict each other. 37. An obtuse ∠ measures > 90°. So the 2nd and the 3rd statements contradict each other. 38. If 1st statement is true, JK = LK. So the 1st and the 3rd statements contradict each other. 39. 2 line segs. cannot be both ⊥ and . So the 1st and the 3rd statements contradict each other. 40. A figure cannot be both a and a quad. So the 2nd and the 3rd statements contradict each other. 41. 4 is not a prime number, so no multiple of 4 is prime. So the 2nd and the 3rd statements contradict each other. 42. m∠P > m∠PQS, so QS > PS. 43. m∠PSQ = 180 - (54 + 75) = 51° m∠PSQ < m∠P, so PQ < QS. 44. m∠R < m∠RSQ, so QS < QR. 45. m∠RQS = 180 - (51 + 78) = 51° = m∠R By Converse of Isosc. Theorem, QS = RS. 46. PQ < QS and QS = RS, so PQ < RS. 47. RS = QS and QS > PS, so RS > PS. 48. AE > BA, so m∠ABE > m∠BEA. 49. CE > BC, so m∠CBE > m∠CEB. 50. CD = DE, so by Isosc. Theorem, m∠DCE = m∠DEC. 51. DE < CE, so m∠DCE < m∠CDE. 52. AE < BE, so m∠ABE < m∠EAB. 53. BE = CE, so by Isosc. Theorem, m∠EBC = m∠ECB. 2 2 54. JK = 6 + 8 2 = 10; KL = 5 + 8 2 ≈ 9.4; JL = ⎪-3 - 8⎥ = 11 KL < JK < JL, so order is ∠J, ∠L, ∠K. 55. JK = ⎪-10 - 2⎥ = 12; KL = 2 12 + 5 2 = 13 JL =

2 2

12 + 7 ≈ 13.9;

JK < JL < KL, so order is ∠L, ∠K, ∠J. 2 1 + 7 ≈ 7.1; KL = 6 + 4 2 ≈ 7.2; 2 2 JL = 7 + 3 ≈ 7.6 JK < KL < JL, so order is ∠L, ∠J, ∠K.

56. JK =

2 2

2 2

10 + 7 ≈ 12.2; KL = 2 2 + 11 2 ≈ 11.2; 2 12 + 4 2 ≈ 12.6 JL = KL < JK < JL, so order is ∠J, ∠L, ∠K.

57. JK =

109


66. Possible answer: Given: P is in the int. of XYZ. Prove: XY + XP + PZ > YZ. Proof: By the Inequal. Theorem, PY + PZ > YZ and XY + XP > YP. Since PZ > 0, the second inequal. is equivalent to XY + XP + PZ > YP + PZ. But then YP + PZ > YZ, so XY + XP + PZ > YZ by the Trans. Prop of Inequal.

58. Possible answer: Assume that the client committed the burglary. A person who commits a burglary must be present at the scene when the crime is committed. However, a witness saw the client in a different city at the time that the crime was committed. This means that the assumption that the client committed the burglary is false. Therefore the client did not commit the burglary.

X

59a. AR + 400 > 600 AR > 200 mi

400 + 600 > AR 1000 mi > AR 200 = 0.4 h. Time to travel Time to travel 200 mi is _ 500 1000 = 2 h. So the range of time is 1000 mi is _ 500 0.4 h < t < 2 h.

P Y

Z

67a. definition of segs.

b. No; AR < 1000, so by the Inequal. Theorem, AM must be less than 1800. 60. Step 1 Write and solve 2 inequals. for n. n+6>8 6+8>n n>2 14 > n Step 2 Combine the inequals. 2 < n < 14

b. Isosc. Theorem

c. definition of

d. m∠1 + m∠3

e. subst.

f. m∠S

g. trans. Prop. of Inequal. 68a. ABC

61. Step 1 Write and solve 2 inequals. for n. 2n + 5 > 7 5 + 7 > 2n 2n > 2 12 > 2n n >1 6 >n Step 2 Combine the inequals. 1
b. AD

c. Isosc. Theorem

d. definition of

e. m∠3

f. subst.

g. in , longer side is opp. larger ∠

h. subst.

i. AC + BC > AB 69. Possible answer: A rt. has a rt. ∠ and 2 acute . By definition, the rt. ∠ has the greatest measure. Since the hyp. is the side opposite the rt. ∠, the hyp. is the longest side by Thm 5-5-2. Similarly, the diagonal of a square forms 2 rt. , with the diagonal being the hyp. of each. Since the diagonal is longer than the leg lengths in both , the diagonal is longer than the side length of the square.

62. Step 1 Write and solve 2 inequals. for n. n+1+3>6 3+6>n+1 n>2 8 >n Step 2 Combine the inequal.s 2n+3 2n + 3 > n + 3 n >0

TEST PREP

70. A; 3+3=6>5

64. Step 1 Write and solve 2 inequals. for n. Use the fact that n + 2 < n + 3. n + 2 + n + 3 > 3n - 2 3n - 2 + n + 2 > n + 3 2n + 5 > 3n - 2 4n > n + 3 7>n 3n > 3 n>1 Step 2 Combine the inequals. 1 n + 2 n + n + 2 > 2n + 1 3n + 1 > n + 2 2 > 1 always true 2n > 1 n > 0.5

71. H; GH + HJ < GJ contradicts the Inequal. Theorem

72. C; −− ∠S must be the largest ∠, so RT is the longest side. CHALLENGE AND EXTEND

()

5 73. The total number of choices is = 10. The 3 choices that form a : 1 + 3 = 4 ≯ 5, 7, or 9 3+5=8>7 1 + 5 = 6 ≯ 7 or 9 3+5=8≯9 1+7=9≯9 3 + 7 = 10 > 9 5 + 7 = 12 > 9 3 or 30%. is possible for 3 choices. So prob. = _ 10 2 p b. _ 2 is rational 74a. √ q2 2 c. 2 q 2 2 d. (2x) = 4x 2 1 p 2 and p 2 is divisible by 4 e. q = _ 2

110


75. 1. 2. 3. 4. 5.

Statements −− PX ⊥ ; Y is any point on other than X. m∠1 = 90° ∠1 is a rt. ∠. XPY is a rt. . ∠2 and ∠P are comp.

6. 90° = m∠2 + m∠P 7. 90° > m∠2 8. m∠1 > m∠2 9. PY > PX

3a.

Reasons

Statements

Reasons

1. C is the midpoint −− of BD; m∠1 = m∠2 m∠3 > m∠4 −− −− 2. BC DC 3. ∠1 ∠2 −− −− 4. AC EC

1. Given 2. Def. of ⊥ 3. Def. of rt. ∠ 4. Def. of rt. 5. Acute of rt. are comp. 6. Def. of comp. 7. Comparison Prop. of Inequal. 8. Subst. 9. In , longer side is opp. larger ∠

1. Given

2. Def. of midpoint 3. Def. of 4. Con. of Isosc. Thm. 5. Hinge Thm.

5. AB > ED b.

Statements

Reasons

1. ∠SRT ∠STR, TU > RU −− −− 2. ST SR

SPIRAL REVIEW

-2 - 2 = -2 77. y - y 1 = m(x - x 1) 76. slope = _ -1 + 3 y - 0 = 2[x - (-3)] y - y 1 = m(x - x 1) y - 0 = 2(x + 3) y - 2 = -2[x - (-3)] y = 2x + 6 y - 2 = -2(x + 3) -2x + y=6 y - 2 = -2x - 6 2x + y = -4

1. Given

−− −− 3. SU SU 4. m∠TSU > m∠RSU

2. Con. of Isosc. Thm. 3. Reflex. Prop. of 4. Con. of the Hinge Thm.

THINK AND DISCUSS 1. Possible answer: kitchen tongs

2

78. QP = 5(-1) - 2 = 3, ST = -1 + 7 = 6, SU = 3(-1) 2 + 1 = 4, so PQR TUS by SSS.

2. No; in this case, 2 sides of the 1st are to 2 sides of the 2nd , but the given ∠ measures are not the measures of included between the sides. Thus you cannot apply the Hinge Theorem.

2

79. BC = 6 - 5(6) + 4 = 10, EF = 2(6) - 1 = 11, m∠ABC = 14(6) + 18 = 102°, so ABC EFD by SAS.

3.

80. Equation of altitude from S is y = 3. Slope of RS is 3 - 5 = -_ 1 , so slope of altitude from T is 2; _ 2 4-0 equation is y - 1 = 2x, or y = 2x + 1. At O, y = 3 and therefore 3 = 2x + 1, so x = 1; thus O = (1, 3).

)NEQUALITIESIN4WO4RIANGLES B A

Y C

X

Z

81. The altitudes from N and P lie along the x- and y-axes, respectively. Therefore O = (0, 0). (INGE4HEOREM )FAB Ɂ XY, AC Ɂ XZ, ANDMȜA MȜX, THEN BCYZ.

5-6 INEQUALITIES IN TWO TRIANGLES, PAGES 340–345

#ONVERSEOF(INGE4HEOREM )FAB Ɂ XY, AC Ɂ XZ, ANDBCYZ, THEN MȜA MȜX.

CHECK IT OUT! 1a. Compare the side lengths in EFG and EHG. EF > EH EG = EG FG = HG By the Converse of the Hinge Theorem, m∠EGF > m∠EGH.


b. Compare the sides and the in ABD and CBD. AD = CD BD = BD m∠CDB > m∠ADB By the Hinge Theorem, BC > AB. 2. The ∠ of swing at full speed is greater than the ∠ of swing at low speed.

1. Compare the sides and the in ABC and XYZ. AB = YZ BC = XY m∠B < m∠Y By the Hinge Theorem, AC < XZ. 2. Compare the side lengths in SRT and QRT. RT = RT RS = RQ ST > QT By the Converse of Hinge Theorem, m∠SRT > m∠QRT. 3. Compare the sides and in KLM and KNM. KM = KM LM = NM m∠KML > m∠KMN By the Hinge Theorem, KL > KN.

111


2z + 7 > 0 13. 2z + 7 < 72 2z > -7 2z < 65 z > -3.5 z < 32.5 Combining, -3.5 < z < 32.5.

4. Step 1 Compare the side lengths in . By the Converse of the Hinge Theorem, 2x + 8 < 25 2x < 17 x < 8.5. Step 2 Since (2x + 8)° is an angle in a , 2x + 8 > 0 2x > -8 x > -4. Step 3 Combine the inequals. The range of values is -4 < x < 8.5.

14. 4z - 6 < z + 11 4z - 6 > 0 3z < 17 4z > 6 3 17 z <_ z>_ 2 3 3
5. Step 1 Compare the sides and the in . By the Hinge Theorem, 5x - 6 < 9 5x < 15 x < 3. Step 2 Since 5x - 6 is a length, 5x - 6 > 0 5x > 6 x > 1.2. Step 3 Combine the inequals. The range of values is 1.2 < x < 3.

16.

6. Step 1 Compare the sides and the in . By the Hinge Theorem, 2x - 5 < x + 7 x < 12. Step 2 Since 2x - 5 is a length, 2x - 5 > 0 2x > 5 x > 2.5. Step 3 Combine the inequals. The range of values is 2.5 < x < 12.

5. JQ + QP = JP, NP + QP = NQ 6. JP > NQ 7. m∠K > m∠M 17. BC = YZ 19. m∠QPR > m∠QRP 21. m∠RSP = m∠RPS

7. The 2nd position; the lengths of the upper and lower arm are the same in both positions, but the distance from the shoulder to the wrist is greater in the 2nd position. So the included ∠ measure is greater by the Converse of the Hinge Theorem. 8.

Statements −− 1. FH is a median of DFG; m∠DHF > m∠GHF −− 2. H is midpoint of DG. −− −− 3. DH GH −− −− 4. FH FH 5. DF > GF

Reasons 1. Given 2. 3. 4. 5.

Def. of median Def. of midpoint Reflex. Prop. of Hinge Thm.


9. BC = CD, CA = CA, AD > AB; by Converse of Hinge Theorem, m∠DCA > m∠BCA. 10. GH = KL, HJ = LM, GJ < KM; by Converse of Hinge Theorem, m∠GHJ < m∠KLM. 11. ST = UV, SU = SU, m∠UST > m∠SUV; by Hinge Theorem, TU > SV.

Statements −− −−− −− −−− 1. JK NM, KP MQ, JQ > NP −− −− 2. QP QP 3. QP = QP 4. JQ + QP > NP + QP

Reasons 1. Given 2. Reflex. Prop. of 3. Def. of segs. 4. Add. Prop. of Inequal. 5. Segment Add. Post. 6. Subst. 7. Con. of the Hinge Thm.

18. m∠QRP < m∠SRP 20.m∠PRS < m∠RSP 22. m∠QPR > m∠RPS

23. m∠PSR < m∠PQR 24. Corr. sides are , and the included are ∠B and ∠E. By the Hinge Theorem, m∠B > m∠E → AC > DF. −− −− −− −− 25. SR ST by definition, and SV SV. So by the Converse of the Hinge Theorem, RV < TV → m∠RSV < m∠TSV. 26. Corr. sides are , and the included are ∠G and ∠K. m∠G = 90° > m∠K, so by the Hinge Theorem, HJ > LM. −− −− −− −− 27. YM MZ by definition, and XM XM. So by the Converse of the Hinge Theorem, YX > ZX → m∠YMX > m∠ZMX. 28. Possible answer: As the angle made by a door hinge gets larger, the width of the door opening increases. As the angle made by the hinge gets smaller, the width of the door opening decreases. This is like the side opposite an angle in a triangle getting larger as the measure of the angle increases or getting smaller as the angle decreases.

4z - 12 > 0 12. 4z - 12 < 16 4z < 28 4z > 12 z<7 z>3 Combining, 3 < z < 7.

112


29. Possible answer: Similarities: Both the SAS Post. and the Hinge Theorem concern the relationship between 2 . Both involve 2 sides and the included ∠ of each . Differences: To apply the SAS Post., you must know that 2 sides and the included ∠ of one are to 2 sides and the included ∠ of the 2nd . To apply the Hinge Theorem, you must know that 2 sides of one are to 2 sides of the 2nd , but the included are ≠ in measure. The SAS Post. allows you to conclude that the 2 are ; then by CPCTC, you can show that the sides opposite the are . The Hinge Theorem involves 2 that are ; in this case, the sides opposite the included are ≠ in length, and the exact relationship between the lengths is determined by the sizes of the included . −− −− −− −− 30a. Newton Springs; NS HS, SJ SJ, and m∠NSJ < m∠HSJ, so NJ < JH by the Hinge Theorem. b. By Inequal. Theorem, NJ + SJ > SN NJ + 182 > 300 NJ > 118 mi Min. distance = SN + NJ > 300 + 118 = 418 mi TEST PREP

31. D; 0 < 3x - 9 < 2x + 1 9 < 3x or 3 < x, and x < 10 3 < x < 10 32. H; −− D lies on AB; AD = DB by the definition of median. 33. Group A is closer to the camp. Possible answer: The 6.5-mi and 4-mi paths together with the distance lines back to the camp form 2 . 2 sides of 1 are to 2 sides of the other . In the for Group A, the measure of the included ∠ is 90° + 35° = 125°. In the for Group B, the measure of the included ∠ is 90° + 45° = 135°. By the Hinge Theorem, the side opposite the 125° ∠ is shorter than the side opposite the 135° ∠. So Group A is closer to the camp. CHALLENGE AND EXTEND

−− −− b. Locate point Q on AC so that BQ bisects ∠PBC. By the definition of ∠ bisector, ∠QBC ∠QBP. It −− −− −− −− is given that BC EF. Since BP EF from part a, −− −− BC BP by the Trans. Prop. of . By the Reflex. −− −−− Prop. of , BQ BQ. So BQP BQC by SAS, −− −− and QP QC by CPCTC. c. AQ + QP > AP by the Inequal. Theorem in AQP. AQ + QC = AC by the Segment Add. −− −− Post. From part b, QP QC, so QP = QC by the definition of segs. Thus AQ + QC > AP by −− subst., and so AC > AP by subst. From part a, AP −− DF. So by the definition of segs., AP = DF. Therefore AC > DF by subst. SPIRAL REVIEW

36. range: 5 - 0.5 = 4.5 mode: 2

37. range: 99 - 85 = 14 mode: none

38. range: 9 - 4 = 5 modes: 4, 5, 7 39. m∠2 = 3(5) + 21 = 36°, m∠6 = 7(5) + 1 = 36° = m∠2; m n by the Converse of the Corr. Post. 40. m∠4 = 2(7) + 34 = 48°, m∠7 = 15(7) + 27 = 132°; so m∠4 + m∠7 = 180°; m n by the Converse of the Same-Side Int. Theorem. 41. By Similar Triangles Theorem: 1 AB DF = _ 2 = AE = 2.5

42. BC = 2DE = 2(2.3) = 4.6

43. m∠BFD = 180 - m∠CFD = 180 - m∠CBA = 180 - 95 = 85°

5-7 THE PYTHAGOREAN THEOREM, PAGES 348–355 CHECK IT OUT! 1a. a 2 + b 2 = c 2 42 + 82 = x 2 80 = x 2 √

80 = x x = √

(16)(5) = 4 √ 5 b.

34. Step 1 Apply Hinge Theorem. −− −− −− By Converse of Isosc. Theorem, VZ VY; VX −− VX; m∠XVZ > m∠XVY. So XZ > XY. Step 2 Write and solve 2 inequals. 5x + 15 > 8x - 6 8x - 6 > 0 21 > 3x 8x > 6 7>x x > 0.75 Step 3 Combine the inequals. 0.75 < x < 7

a 2 + b 2= c 2 x 2 + 12 2 = (x + 4) 2 x 2 + 144 = x 2 + 8x + 16 128 = 8x x = 16

35a. Locate point P outside ABC so that ∠ABP −− −− −− ∠DEF and BP EF. It is given that AB −− −− −− DE, so ABP DEF by SAS. Thus AP DF by CPCTC.

113


2. Let y be the distance in ft from the foot of the ladder to the base of the wall. Then 4y is the distance in ft from the top of the ladder to the base of the wall.

THINK AND DISCUSS 1. The greatest number is substituted for c. The other 2 numbers are substtituted for a and b in any order.

a 2 + b 2= c 2 (4y) 2 + y 2 = 30 2 17y 2 = 900 2 900 y =_ 17 900 y= _ 17 900 ≈ 29 ft 1 in. 4y = 4 _ 17

2. Possible answer: The sum of the areas of the 2 smaller squares equals the area of the largest square. So 3 2 + 4 2 = 5 2, or 9 + 16 = 25. 3. Must be nonzero and whole numbers, and must satisfy the equation a 2 + b 2 = c 2

√ √

3a.

b.

4.

2 2 2 a +b =c 2 8 + 10 = c 2 164 = c 2 c = √ 164 = 2 √41 The side lengths do not form a Pythagorean triple because 2 √ 41 is not a whole number. 2

a2 + b2 = c2 24 2 + b 2 = 26 2 b 2 = 100 b = 10 The side lengths are nonzero whole numbers that satisfy the equation a 2 + b 2 = c 2, so they form a Pythagorean triple.

c.

2 2 2 a +b =c 2 1 + 2.4 = c 2 6.76 = c 2 c = 2.6 The side lengths do not form a Pythagorean triple because 2.4 and 2.6 are not whole numbers.

d.

a2 + b2 = c2 16 2 + 30 2 = c 2 1156 = c 2 c = 34 The side lengths are nonzero whole numbers that satisfy the equation a 2 + b 2 = c 2, so they form a Pythagorean triple.

0YTHAGOREAN 2ELATIONSHIPS 0YTH4HM)NARṮ THESUMOFTHESQUARESOF THELEGLENGTHSEQUALSTHE SQUAREOFTHEHYPOTENUSE

#ONVOFTHE0YTH 4HM)FTHESUMOF THESQUARESOFSIDE LENGTHSOFA̱ EQUALSTHESQUAREOF THETHIRDSIDELENGTH THENTHE̱ISARṮ

0YTH)NEQUAL4HM)NA̱ WITHcASTHELENGTHOFTHE LONGESTSIDE IFc Êa Êb THE ̱ISOBTUSE BUTIFc Êa Êb THE̱ISACUTE


1. No; although it is true that (2.7) 2 + (3.6) 2 = (4.5) 2, the numbers 2.7, 3,6, and 4.5 are not whole numbers. 2. a 2 + b 2 = c 2 32 + 92 = x2 90 = x 2 √ 90 = x x = √ (9)(10) = 3 √ 10

2

4.

4a. Step 1 Determine if the measures form a . By the Inequal. Theorem, 7, 12, and 16 can be the side lengths of a . Step 2 Classify the . c2 a2 + b2 16 2 7 2 + 12 2 256 49 + 144 256 > 193 Since c 2 > a 2 + b 2, is obtuse.

2 2 2 3. a + b = c 2 2 x + 7 = 11 2 x 2 = 72 x = √ 72 x = √ (36)(2) √ =6 2

2 2 2 a +b =c 2 2 2 (x - 2) + 8 = x x 2 - 4x + 4 + 64 = x 2 -4x + 68 = 0 68 = 4x x = 17

5. Let the width and the height of the monitor be w = 5x and h = 4x, respectively. a2 + b2= c2 2 2 2 (4x) + (5x) = 19 2 41x = 361 2 361 x =_ 41 361 x= _ 41 61 ≈ 14.8 in. w = 5x = 5 3_ 41

√ √ 361 ≈ 11.9 in. h = 4x = 4 √_ 41

b. Step 1 Determine if the measures form a . Since 11 + 18 = 29 ≯ 34, these cannot be the side lengths of a . c. Step 1 Determine if the measures form a . By the Inequal. Theorem, 3.8, 4.1, and 5.2 can be the side lengths of a . Step 2 Classify the . c2 a2 + b2 5.2 2 3.8 2 + 4.1 2 27.04 14.44 + 16.81 27.04 < 31.25 Since c 2 < a 2 + b 2, is acute.

2 2 2 6. a + b = c 2 2 4 + 5 = c2 41 = c 2 c = √ 41 The side lengths do not form a Pythagorean triple is not a whole number. because √41

114


7. a 2 + b 2 = c 2 12 2 + b 2 = 20 2 b 2 = 256 b = 16 The side lengths are nonzero whole numbers that satisfy the equation a 2 + b 2 = c 2, so they form a Pythagorean triple. 8.

a2 + b2 = c2 1.5 2 + b 2 = 1.7 2 b 2 = 0.64 b = 0.8 The side lengths do not form a Pythagorean triple because they are not whole numbers.

9. Step 1 Determine if the measures form a . By the Inequal. Theorem, 7, 10, and 12 can be the side lengths of a . Step 2 Classify the . c2 a2 + b2 12 2 7 2 + 10 2 144 49 + 100 144 < 149 Since c 2 < a 2 + b 2, is acute. 10. Step 1 Determine if the measures form a . By the Inequal. Theorem, 9, 11, and 15 can be the side lengths of a . Step 2 Classify the . c2 a2 + b2 15 2 9 2 + 11 2 225 81 + 121 225 > 202 Since c 2 > a 2 + b 2, is obtuse. 11. Step 1 Determine if the measures form a . By the Inequal. Theorem, 9, 40, and 41 can be the side lengths of a . Step 2 Classify the . c2 a2 + b2 41 2 9 2 + 40 2 1681 81 + 1600 1681 = 1681 Since c 2 = a 2 + b 2, is a rt. . 12. Step 1 Determine if measures form a . 3 = 3_ 1 + 1_ 1 ≯ 3_ 1 , these cannot be Since 1_ 2 4 4 4 the side lengths of a . 13. Step 1 Determine if the measures form a . By the Inequal. Theorem, 5.9, 6, and 8.4 can be the side lengths of a . Step 2 Classify the . c2 a2 + b2 8.4 2 5.9 2 + 6 2 70.56 34.81 + 36 70.56 < 70.81 Since c 2 < a 2 + b 2, is acute.

14. Step 1 Determine if the measures form a . can By the Inequal. Theorem, 11, 13, and 7 √6 be the side lengths of a . Step 2 Classify the . c2 a2 + b2 (7 √6 ) 2 11 2 + 13 2 294 121 + 169 294 > 290 Since c 2 > a 2 + b 2, is obtuse. PRACTICE AND PROBLEM SOLVING

15. 6 2 + 8 2 = x 2 100 = x 2 x = 10

16. 9 2 + x 2 = 13 2 81 + x 2 = 169 x 2 = 88 x = √ 88 = 2 √ 22

2 2 2 17. x + 7 = (x + 1) 2 2 x + 49 = x + 2x + 1 48 = 2x x = 24

18. (3x) 2 + (5x) 2 = 8 2 34x 2 = 64 32 x2 = _ 17

32 ≈ 4 ft 1 in. 3x = 3 _ 17 _ 5x = 5 32 ≈ 6 ft 10 in. 17

2 2 2 19. 2.5 + b = 6.5 2 6.25 + b = 42.25 b 2 = 36 b =6 The side lengths cannot form a because 2.5 and 6.5 are not whole numbers.

20. 15 2 + 20 2 = c 2 625 = c 2 c = 25 Yes; the three side lengths are nonzero whole numbers that satisfy a 2 + b 2 = c 2. 2 2 2 21. 2 + b = 7 2 4 + b = 49 b 2 = 45 b = √ 45 = 3 √ 5 is The side lengths cannot form a because 3 √5 not a whole number.

22. 10 + 12 = 22 > 15 23. 8 + 13 = 21 ≯ 23

The side lengths can 15 2 10 2 + 12 2 not form a . 225 < 244 The side lengths form an acute . 1 + 2 = 3_ 1 > 2_ 1 24. 9 + 14 = 23 > 17 25. 1_ 2 2 2 2 2 2 17 9 + 14 2 2 1 1 _ _ 2 1 + 22 289 > 277 2 2 The side lengths form 1 = 6_ 1 6_ an obtuse . 4 4 The side lengths form a rt. .

( ) ( )

115


26. 0.7 + 1.1 = 1.8 > 1.7 27. 7 + 12 = 19 > 6 √ 5 2 1.7 2 0.7 2 + 1.1 2 (6 √5 ) 7 2 + 12 2 2.89 > 1.7 180 < 193 The side lengths form The side lengths form an obtuse . an acute . 28. Possible answer: Shape the rope into a with side lengths of 3, 4, and 5. Because 3 2 + 4 2 = 5 2, the is a rt. with the rt. ∠ opposite the side 5. 2

2

2

29. B; (x + 3) + 4 = (x + 6x + 9) + 16. In the solution shown, the 6x term was omitted. 30. Let a and b be the horizontal leg lengths of the leftand right-hand . 92 + b2 = 152 b2 = 144 b = 12 a + b = 25 a = 25 - 12 = 13

132 + 92 = x 2 250 = x 2 x = √ 250 = 5 √ 10

31. Let a and b be the horizontal leg lengths of the left- and right-hand . 2 2 2 a + 6 = 10 2 a = 64 a=8 x = a + b = 8 + √13

2 2 2 6 +b =7 2 b = 13 b = √ 13

32. Let d be the length of 33. Let h be the height of the shared side. the . 2 22 + d 2 = 72 3 2 + h 2 = ( √ 34 ) 2 2 d = 45 h = 25 x 2 + 52 = d2 x 2 + h 2 = 11 2 x 2 = 20 x 2 = 96 x = √ x = √ 20 = 2 √5 96 = 4 √ 6 34. Let h be the height of the . 5 2 + h 2 = 13 2 h 2 = 144 (x + 5) 2 + h 2 = 20 2 2 x + 10x + 25 + 144 = 400 x 2 + 10x - 231= 0 (x - 11)(x + 21) = 0 Since x > 0, the only possible solution is x = 11. 35. Let b be the base length of each . 18 2 + (2b) 2 = 30 2 4b 2 = 576 b 2 = 144 2 2 18 + b = x 2 468 = x 2 x = √ 468 = 6 √13 2 2 2 36. 3963 + x = (3963 + 250) 2 x = 2,044,000 x = √ 2,044,000 ≈ 1430 mi 37. Possible answer: Outer figure: The length of each side is a + b, so the outer figure has 4 sides. Each ∠ is a rt. ∠ from one of the rt. , so the outer figure has 4 rt. . By definition, it is a square. Inner figure: The length of each side is c, so the inner figure has 4 sides. The 2 acute of a rt. are comp., so the measure of each ∠ in the inner figure is 90°. Therefore the inner figure has 4 rt. . By definition, it is a square.

38. Let b be the base of the . 8 2 + b 2 = 17 2 b 2 = 225 b = 15 P = 8 + 15 + 17 = 40 units 1 (15)(8) A=_ 2 = 60 square units

39. Let 2b be the base of the . 62 + b2 = 82 b 2 = 28 b = 2 √ 7 2b = 4 √ 7 P = 8 + 8 + 4 √ 7 = 16 + 4 √ 7 units 1 (4 √ 7 )(6) A=_ 2 = 12 √ 7 square units

40. Let h be the height of 41. Let h be the height and c be the 3rd side length the . of the . 4 2 + h 2 = 12 2 h 2 = 128 32 + h2 = 52 h = √ 128 h 2 = 16 = 8 √ 2 h =4 P = 12 + 12 + 8 62 + 42 = c2 = 32 units 52 = c 2 1 _ √ A = (8)8 2 c = √ 52 2 = 2 √ 13 2 square units = 32 √ P = 5 + (3 + 6) + 2 √13 = 14 + 2 √ 13 units 1 (3 + 6)(4) A=_ 2 = 18 square units 42. Let a + b = 15 be the 2nd side length and c be the 3rd side length of the . 2 2 2 a + 12 = 15 2 a = 81 a=9 b = 15 - 9 = 6 6 2 + 12 2 = c 2 180 = c 2 c = √ 180 = 6 √5 P = 15 + 15 + 6 √5 = 30 + 6 √ 5 units 1 (15)(12) = 90 square units A=_ 2 43. P = 4 + 5 + 5 + 8 = 22 units 1 (a + b)h A=_ 2 1 (5 + 8)(4) =_ 2 = 26 square units 44. Possible answer: When you use Pythagorean Theorem, you know that the is a rt. . You substitute the known values into a 2 + b 2 = c 2 and solve for the unknown side length. When you use the Converse of Pythagorean Theorem, you are trying to find out whether a given is a rt. . Usually all side lengths are known. You substitute all the values into a 2 + b 2 = c 2 to determine whether the resulting equation is true. If it is true, then you know that the is a rt. .

116


45. Draw PQR with ∠R as the rt. ∠, leg lengths of a and b, and hyp. length of x. In ABC, it is given that a 2 + b 2 = c 2. In PQR, a 2 + b 2 = x 2 by the Pythagorean Theorem. Since a 2 + b 2 = c 2, and 2 2 2 2 2 a + b = x , it follows by subst. that x = c . Take the positive square root of both sides, and x = c. So AB = PQ, BC = QR, and AC = PR. −− −− −− −− By the definition of segs., AB PQ, BC QR , −− −− and AC PR . Then ABC PQR by SSS, and ∠C ∠R by CPCTC. By the definition of rt. ∠, m∠R = 90°. So by the definition of , m∠C = 90°. Therefore ∠C is a rt. ∠ by definition, and ABC is a rt. by definition.


52. Let 3 points be A(-1, 2), B(-10, 5), and C(-4, k). 2 2 2 2 2 2 AB = 9 + 3 = 90 AC = 3 + (k - 2) 2 2 2 2 BC = 6 + (k - 5) = 9 + k - 4k + 4 2 = k 2 - 4k + 13 = 36 + k - 10k + 25 2 = k - 10k + 61 2 2 2 If AB + BC = AC , 90 + k 2 - 10k + 61 = k 2 - 4k + 13 138 = 6k k = 23 If AB 2 + AC 2 = BC 2, 90 + k 2 - 4k + 13 = k 2 - 10k + 61 6k = -42 k = -7 2 2 If AC + BC = AB 2, k 2 - 4k + 13 + k 2 - 10k + 61 = 90 2k 2 - 14k - 16 = 0 k 2 - 7k - 8 = 0 (k - 8)(k + 1) = 0 k = 8 or -1 So k = -7, -1, 8, or 23

b. JL = x 2 - x 1, LK = y 2 - y 1

46a. (x 2, y 1) 2

c. √ n + 1 ; possible answer: the length of the hyp. is the square root of the whole number 1 greater than the number of the , or √ n + 1.

2

2

c. JK = JL + LK = (x 2 - x 1) 2 + (y 2 - y 1) 2 (x - x ) 2 + (y - y ) 2 JK = 2

1

2

1

2 2 2 47a. KR + 500 = 1300 2 KR = 1,440,000 KR = 1200 mi KM 2 + 390 2 = 1200 2 KM 2 = 1,287,900 KM ≈ 1135 mi SK + KM SR + RM 500 + 1135 1300 + 390 1635 < 1690 She should fly first to King City.

b.

SM 2 SR 2 + RM 2 1360 2 1300 2 + 390 2 1,849,600 > 1,842,100 So by the Pythagorean Inequals. Theorem, m∠SRM > 90°.

TEST PREP

53. By the Inequal. Theorem, a + b > c. 2 By the Pythagorean Theorem, c = a + b2. By subst., a + b >

51a. PA 2 = 1 2 + 1 2 = 2 PA = √2 PC 2 = 1 2 + PB 2 = 4 PC = 2 PE 2 = 1 2 + PD 2 = 6 PE = √6

a +b .

c a b

54. c 2 = a 2 + b 2 2 a + b2 c =

1 hc 1 ab = _ A=_ 2 2 ab = hc 2 a + b2 ab = h ab _ =h 2 a + b2

48. GX = HX = 6 GM 2 + MX 2 = GX 2 GM 2 + 4 2 = 6 2 GM 2 = 20 GJ = 2GM = 2 √ 20 ≈ 8.9 49. B; 7 2 + 24 2 25 2 625 = 625

2 2

50. H; 11 2 7 2 + 9 2 121 ≯ 130 2 2 2 PB = 1 + PA = 3 PB = √ 3 PD 2 = 1 2 + PC 2 = 5 PD = √ 5 2 2 PF = 1 + PE 2 = 7 PF = √ 7

55a. No; possible answer: let a = 3, b = 4, and c = 5. So a + 1 = 4, b + 1 = 5, and c + 1 = 6. 3, 4, and 5 form a Pythagorean triple, but 4, 5 and 6 do not because 4 2 + 5 2 ≠ 6 2. b. Yes; possible answer: if a, b and c form a Pythagorean triple, a 2 + b 2 = c 2 is true. Multiply both sides by 4 to get the equation 4a 2 + 4b 2 = 4c 2. This is equivalent to (2a) 2 + (2b) 2 = (2c) 2. So by def., 2a, 2b, and 2c also form a Pythagorean triple.

10 ; possible answer: for each added to the b. √ pattern, the number under the radical symbol increases by 1. So the length of the hyp. of the 7th would be √ 8 , the length of the hyp. of the 8th would be √ 9 , and the length of the hyp. of the 9th would be √ 10 .

c. No; possible answer: let a = 3, b = 4, and c = 5. So a 2 = 9, b 2 = 16, and c 2 = 25. 3, 4, and 5 form a Pythagorean triple, but 9, 16, and 25 do not because 9 2 + 16 2 ≠ 25 2. d. No; possible answer: let a = 3, b = 4, and c = 5. = 2, and √ So √ a = √ 3 , √b c = √ 5 . 3, 4, and 5 do form a Pythagorean triple, but √ 3 , 2 and √5 not because ( √ 3 ) 2 + 2 2 ≠ ( √ 5 ) 2.

117


4. Step 1 Divide the equil. into two 30°-60°-90° . The height of the frame is the length of the longer leg. Step 2 Find the length x of the shorter leg. 30 = x √ 3 30 = x _ √ 3 3 30 √ _ =x 3 =x 10 √3 Step 3 Find the length s of each side of the frame. s = 20 √ 3 ≈ 34.6 cm

SPIRAL REVIEW

56. (4 + x)12 - (4x + 1)6 = 0 48 + 12x - 24x - 6 = 0 42 - 12x = 0 42 = 12x x = 3.5 2x 5 _ 57. =x 58. 4x + 3(x + 2) = -3(x + 3) 3 4x + 3x + 6 = -3x - 9 2x - 5 = 3x 10x = -15 -5 = x 3 x = -1.5 = - _ 2 59. By the Midpoint Formula, the coordinates of M are (a, b). By the Distance Formula, 2 AM = (a - 0) 2 + (b - 0) 2 = a + b 2 and

THINK AND DISCUSS 1. Possible answer: The is a rt. , so the measure of one ∠ is 90°, and the other 2 acute are comp. The is isosc., so its base are . So the measure of each of the base is 45°.

2 - a) 2 + (2b - b) 2 = a + b2. MB = (0 So by subst., AM = MB. −− −− −− −−− 60. JK NP, JL NM, KL < MP m∠J > 0° m∠J < m∠N 4x - 6 > 0 4x - 6 < 68 4x > 6 4x < 74 x > 1.5 x < 18.5 So 1.5 < x < 18.5. −− −− −− −− 61. BA BC, BD BD, m∠ABD < m∠CBD 3x + 1 < 7 3x + 1 > 0 3x < 6 3x > -1 1 x <2 x < -_ 3 1 _ So - < x < 2. 3

2. In figure I, use the relationship x = 2(8). In figure II, first use the relationship 8 = √ 3 (shorter leg), and then use the relationship x = 2(shorter leg). 3.

3PECIAL2IGHT 4RIANGLES ̱ s

е sȖ s

s

5-8 APPLYING SPECIAL RIGHT TRIANGLES, PAGES 356–362 CHECK IT OUT! 1a. The is an isosc. rt. , which is a 45°-45°90° . ) √ x = (10 √2 2 = 20

c. 24 = 2x 12 = x y = x √ 3 y = 12 √ 3

s

е s Ȗ


b. The is an isosc. rt. , which is a 45°-45°90° . 16 = x √2 16 = x _ √2 2 16 √ _ =x 2 x = 8 √2

1. The is an isosc. rt. , which is a 45°-45°-90° . x = 14 √ 2 3. The is an isosc. rt. , which is a 45°-45°-90° . 9 √ 2 = x √ 2 x=9

2. Tessa needs a 45°-45°-90° with hyp. of length [C + 2(8)] cm and leg length of 42 cm. C + 2(8) = 42 √2 C = -16 + 42 √ 2 ≈ 43 cm 3 = 2x 3a. 18 √ =x 9 √3 y = x √ 3 y = (9 √ 3 ) √ 3 y = 27

̱

4. The sign forms a right . Using the Pyth. Thm., we get d = √ 19.5 2+ 19.5 2 d ≈ 27.6 in.

b. x = 5 √3 y = 2(5) = 10

d.

2. The is an isosc. rt. , which is a 45°-45°-90° . 12 = x √ 2 12 = x _ √ 2 √ 2 12 _=x 2 2 x = 6 √

5. 6 = 2x 3=x y = x √ 3 y = 3 √ 3 3 ) √ 3 7. x = (7 √ x = 21 y = 2(7 √ 3) y = 14 √ 3

9 = y √ 3

9 =y _

√ 3

3 9 √ _ =y

6.

15 = x √ 3

15 = x _

√ 3

3 15 √ _ =x 3 5 √ 3=x y = 2x y = 10 √ 3

3 3 √ 3 =y x = 2y x = 6 √3

118


8. Step 1 Divide the equil. into two 30°-60°-90° . The height of the frame is the length of the longer leg. h Step 2 Find the length x of the shorter leg. 5(2.25) = 2x 5.625 = x Step 3 Find the length h of the longer leg. h = 5.625 √ 3 ≈ 9.75 in.

IN

x


9. The is an isosc. rt. , which is a 45°-45°-90° . 15 = x √ 2

15 = x _

√2 15 √2 _ =x 2

10. The is an isosc. rt. , which is a 45°-45°-90° . 2 ) √ 2=8 x = (4 √ 11. The is an isosc. rt. , which is a 45°-45°-90° . 18 √ 2 = x √ 2 18 = x

12. The tabletop is a 45°-45°-90° . 48 = w √2 = 2w 48 √2 ≈ 33.9 in. w = 24 √2 13. x = 2(24) = 48 y = 24 √3

15.

3 = 2x 14. 10 √ 5 √ 3 =x y = x √ 3 √ ( = 5 3 ) √ 3 = 15

3 2 = x √ 2 √ 3 = 3x 3 2 √ _ =x 3 4 √3 y = 2x = _ 3

()

22. Let s be the leg length. 18 = s √ 2 9 √ 2=s 1 = 36 nails Hyp.: 18 ÷ _ 2 1 = 72 √ 2) ÷ _ 2 ≈ 102 nails Legs: 2(9 √ 4 The total is approximately 138 nails. 23. No; possible answer: if the ∠ measures are in ratio 1 : 2 : 3 , then the measures of the angles are 30°-60°-90°, and the is a 30°-60°-90° . Assume the length of the shortest leg is 1. Then the length of the hyp. is 2, and the length of the longer leg is √ 3. So the side lengths would be in the ratio 1 : √ 3 : 2. 24.

y

Q

x

R

−− Let P = (x, y). QR is the hyp. −− From the diagram, QR is a 45° ∠ to the axes. P is in −− quad II → PQ is horizontal → y = y-coordinate of Q −− = 6; PR is vertical → x = x-coordinate of R = -6. So P = (-6, 6).

b. Length of the dog walk = x + 12 + x = 9 + 12 + 9 = 30 ft 18. 28 = 2a 12 = a √2 = 2a 12 √2 14 = a b = a √ 6 √ 2 =a 3 = 14 √ 3 b = a = 6 √ P=a+b+c 2 = 14 + 14 √ P=a+b+c 3 + 28 + 6 √2 + 12 = 6 √2 = (42 + 14 √ 3 ) cm ) in. 1 _ = (12 + 12 √2 A = ab 2 1 ab A=_ _ 2 = 1 (14)(14 √ 3) 2 1 _ ) = (6 √ 2 )(6 √2 2 3 cm = 98 √ 2 2 = 36 in.

P

16a. The ramp forms a 30°-60°-90° . Let the length of the ramp be x. x = 2(4.5) = 9 ft

17.

()

s √ 18 = s √ 20. h = _ 2 3 2 √ 18 2 = 2s = 2 √ 3 9 √ 2=s P = 3s P = 4s = 3(4) = 12 ft = 4(9 √ 2 ) = 36 √ 2m 1 sh 2 A=_ A=s 2 2 = (9 √ 2) _ ) = 1 (4)(2 √3 = 81(2) = 162 m 2 2 2 3 ft = 4 √ s _ √ P = 3s 3 21. h= 2 ) = 60 √3 yd = 3(20 √3 2h = s √ 3 1 _ A = sh 60 = s √ 3 2 60 √ 3 = 3s _ )(30) = 1 (20 √3 20 √ 3=s 2 3 yd 2 = 300 √

19.

25.

T

y

P

x

S

−− Let P = (x, y). PT is the hyp. −− From the diagram, ST is a 45° ∠ to the axes. P is in −− quad I → PT is horizontal → y = y-coordinate of 2= 6 2 + 6 2 √ 2 T = 3; PT = ST √

(

)

= √ 144 = 12; x = (x-coordinate of T ) + 12 = -2 + 12 = 10. So P = (10, 3).

119


26.

y

P

35. Step 1 Identify the pattern.

2 times the length of The length of each hyp. is _ √ 3

x

the previous hyp. The length of the first hyp. is 2. Step 2 Find x. 32 2 4(2) = _ x= _ 9 √ 3

( )

X

W

−− Let P = (x, y). PX is the hyp. −−− From the diagram, P is in quad. II → PW is vertical → x = x-coordinate of W = -1; PX = WX √ 3= ; y = (y-coordinate of W ) + 5 √ 5 √3 3 = -4 + 5 √ 3. √ ) ( So P = -1, -4 + 5 3 . 27.

y

Y

36a. Let f be the length of the face diagonal. 2. Then f = e √ e = 1: f = √ 2 , so

2 e +f2 d =

e +f

2 2

e + 2e

2

= 3e = e √ 3

= √4 +8 = √ 12 = 2 √3 e = 3: f = 3 √ 2 , so

2 e +f2 d =

x

=

2 2

= √ 1 + 2 = √3 e = 2: f = 2 √ 2 , so

2 e +f2 d =

Z

b. d =

= √9 + 18 = √ 27 = 3 √3

0

−− Let P = (x, y). PY is the hyp. −− From the diagram, P is in quad. IV → PZ is vertical → x = x-coordinate of Z = 5; PZ = YZ √ 3 = 12 √ 3; = 10 - 12 √ y = (y-coordinate of Z ) - 12 √3 3. So P = (5, 10 - 12 √ 3 ). 28. Possible answer: Both types of are rt. . In each one, there is a unique relationship among the side lengths. For each type of , if you know 1 side length, you can find the other 2. 29a. NB = 2NL = 2(320) = 640 mi

≈ 453 mi b. IN = NL √2

c. BI = BL - IL = NL √ 3 - NL - 320 ≈ 234 mi = 320 √3

37. Possible answer: Given: ABC is a 30°-60°-90° with m∠A = 30° −− and m∠B = 60°. CD is the altitude to the hyp. Prove: AD = 3DB −− Proof: It is given that CD is the altitude to the hyp. −− −− Thus CD ⊥ AB by the definition of altitude. So ∠ADC and ∠BDC are rt. by the definition of ⊥, and ADC and BDC are rt. by definition. It is given that m∠A = 30° and m∠B = 60°. Since the acute of a rt. are comp., m∠DCA = 60° and m∠DCB = 30° by Subtr. Prop. of =. So ADC and BDC are both 30°-60°-90° . By the 30°-60°-90° (DB). Theorem, AD = √ 3 (DC) and DC = √3 (DB)). This simplifies to By subst., AD = √ 3 ( √3 AD = 3DB.

TEST PREP

30. C

D

31. F; (5, 12, 13) is a Pythagorean triple, and 5 + 13 = 18.

32. B; 33. 32 = 2w 24 = a √ 2 w = 16 a = 12 √ 2 ≈ 17.0 in. = w √ 3 = 16 √3 A = w = 16 √ 3 (16) = 443.4 in. 2 CHALLENGE AND EXTEND

34. Step 1 Identify the pattern. times the length of The length of each hyp. is √2 the previous hyp. Step 2 Write and solve an equation for x. 4 4 = ( √ 2) x 4 = 4x x=1

A

B

C

SPIRAL REVIEW

39. y = x 2 - 10x - 2 38. y = x 2 + 4x + 0 2 2 = (x + 2) + 0 - 2 = (x - 5) 2 - 2 - 5 2 2 = (x + 2) - 4 = (x - 5) 2 - 27 Axis of symmetry: Axis of symmetry: x = 5 x = -2 40. y = x 2 + 7x + 15 = (x + 3.5) 2 + 15 - 3.5 2 = (x + 3.5) 2 + 2.75 Axis of symmetry: x = -3.5 41. m∠ADB - 180 - 70 = 110° is obtuse. So ADB is obtuse. 42. m∠DBC = 180 - (60 + 70) = 50°. All 3 are acute, so BDC is acute.

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−− −− −− −− 7. PQ ST, QR TV, and m∠Q > m∠T. By the Hinge Theorem, PR > SV. −− −− −− −− 8. JK JM, JL JL, and KL < ML. By the Converse of the Hinge Theorem, m∠KJL < m∠MJL. −− −− −− −− 9. AD BC, BD BD, and m∠ADB < m∠DBC. By the Hinge Theorem, AB > 0 AB < CD 4x - 13 > 0 4x - 13 < 15 4x > 13 4x < 28 x > 3.25 x<7 3.25 < x < 7

43. m∠ABC = m∠ABD + m∠DBC = 180 - (30 + 110) + 50 = 90° ∠ABC is a rt. ∠, so ABC is a rt. . 44. ∠PSQ and ∠PQS are comp. By the Converse of the is the bisector of ∠PQR. ∠ Bisector Theorem, QS So m∠PQR = 2m∠PQS = 2(90 - m∠PSQ) = 2(90 - 65) = 50°. 45. ∠QTV and ∠VTS are supp., and ∠TQV and ∠QTV are comp. By the Converse of the ∠ Bisector is the bisector of ∠PQR. So Theorem, QS m∠VTS = 180 - m∠QTV = 180 - (90 - m∠TQV) = 180 - (90 - m∠PQS) = 180 - (90 - 42) = 132°.

10. x 2 = 5 2 + 9 2 x 2 = 106 x = √

106

46. By the ∠ Bisector Theorem, PS = SR and TU = TV. Substitute in the given equation. SR = 3TU PS = 3TV 7.5 = 3TV TV = 2.5

READY TO GO ON? PAGE 365 1. Possible answer: Given: ∠A and ∠B are supplementary. ∠A is an acute angle. Prove: ∠B cannot be an acute angle. Proof: Assume that ∠B is an acute angle. By the def. of acute, m∠A < 90° and m∠B < 90°. When the 2 inequalities are added. m∠A + m∠B < 180°. However, by the def. of supp., m∠A + m∠B = 180°. So m∠A + m∠B < 180° contradicts the given information, and the assumption that ∠B is an acute ∠ is false. Therefore ∠B cannot be acute. −− 2. KM is the shortest side, so ∠L is the least ∠. −− KL is the longest side, so ∠M is the greatest ∠. From smallest to greatest, the order is ∠L, ∠K, ∠M.

12. 10 + 12 = 22 > 16 The side lengths can form a . 16 2 10 2 + 12 2 256 100 + 144 256 > 244 The is obtuse. 13. Length of the walkway = √

50 2 + 80 2 = √

8900 ≈ 94 ft 4 in. 14. Length of the shorter leg of a 30°-60°-90° is 36 ÷ 2 = 18 in. So h = 18 √

3 ≈ 31 in. 2 15. x = 8 √

16.

22 = x √

2

= 2x 22 √2 11 √

2=x

3 = x √

3 17. 5 √

5=x y = 2x = 2(5) = 10

STUDY GUIDE: REVIEW, PAGES 366–369

3. m∠D = 90 - 48 = 42°, m∠E = 90° −− ∠D is the least ∠, so EF is the shortest side. −− ∠E is the greatest ∠, so DF is the longest side. −− −− −− From shortest to longest, the order is EF, DE, DF. 4. No; possible answer: the sum of 8.3 and 10.5 is 18.8, which is not greater than 18.8. By the Inequality Thm., a cannot have these side lengths.

2 2 2 11. a + 9 = 11 2 a + 81 = 121 a 2 = 40 a = √

40 = 2 √

10 The side lengths do not form a Pythagorean triple, because 2 √

10 is not a whole number.

1. equidistant

2. midsegment

3. incenter

4. locus

LESSON 5-1 5. BD = 2CD = 2(3.7) = 7.4 6.

5. Yes; possible answer: when s = 4, the value of 4s is 16, the value of s + 10 is 14, and the value of s 2 is 16. The sum of each pair of 2 lengths is greater than the third length. So a can have sides with these lengths.

XY = YZ 3n + 5 = 8n - 9 14 = 5n n = 2.8 YZ = 8(2.8) - 9 = 13.4

7. HT = FT = 5.8 8. m∠MNV = m∠PNV 2z + 10 = 4z - 6 16 = 2z z=8 m∠MNP = 2m∠MNV = 2[2(8°) + 10°] = 52°

6. Let d be the distance from the theater to the zoo. d + 9 > 16 9 + 16 > d d > 16 - 9 = 7 25 > d Range of the distances: greater than 7 km and less than 25 km.

121


−− 9. The midpoint of AB is (1, 0); −− -10 slope of AB = _ = -1, so the slope of the 10 perpendicular bisector is 1; the equation of the perpendicular bisector is y = x -1. −− 10. The midpoint of XY is (4, 6); −− _ 8 slope of XY = = 4, so the slope of the 2 perpendicular bisector is -0.25; the equation of the perpendicular bisector is y - 6 = -0.25(x - 4). 11. No; to apply the Converse of the Angle Bisector −− −− Theorem, you need to know that AP ⊥ AB and −− −− CP ⊥ CB. −− −− −− −− −− −− 12. Yes; since AP ⊥ AB, CP ⊥ CB, and AP CP, P is on the bisector of ∠ABC by the Converse of the Angle Bisector Theorem.

LESSON 5-2 13. GY = HY = 42.2

14. GP = JP = 46

15. GJ = 2GX 16. PH = JP = 46 = 2(28.8) = 57.6 −− −−− 17. distance from A to UV = distance from A to UW = 18 18. m∠WVU + m∠VUW + m∠UWV = 180 2m∠WVA + 2(20)+ 66 = 180 2m∠WVA = 74 m∠WVA = 37° −−− 19. MO is vertical, so the equation of the horizontal perpendicular bisector is y = 3; −− NO is horizontal, so the equation of the vertical perpendicular bisector is x = 4. The circumcenter is at (4, 3). −− 20. OR is vertical, so the equation of the horizontal perpendicular bisector is y = -3.5; −− OS is horizontal, so the equation of the vertical perpendicular bisector is x = -6. The circumcenter is at (-6, -3.5).

LESSON 5-3

2 DB 21. DZ = _ 22. DB = 3ZB 3 24.6 = 3ZB 2 (24.6) = 16.4 =_ ZB = 8.2 3 23. EZ = 2ZC 24. EC = 3ZC 11.6 = 2ZC = 3(5.8) = 17.4 ZC = 5.8 −− 25. JK is vertical, so the equation of the altitude from L is y = 0; −− KL is horizontal, so the equation of the altitude from J is x = -6. The orthocenter is at (-6, 0). −− 26. AB is horizontal, so the equation of the altitude from C is x = 1; −− AC is vertical, so the equation of the altitude from B is y = 2. The orthocenter is at (1, 2).

−− 27. RT is horizontal, so the equation of the altitude from S is x = 7; −− 5 = 1, so the equation of the altitude RS has slope _ 5 from T is y - 3 = -(x - 8). At the orthocenter, x = 7 and y - 3 = -(7 - 8) = 1 → y = 4, so the orthocenter is at (7, 4). −− 28. XY is horizontal, so the equation of the altitude from Z is x = 3; −− 6 = -1, so the equation of the altitude XZ has slope _ -6 from Y is y - 2 = x - 5 or y = x - 3. At the orthocenter, x = 3 and y = x - 3 = 0, so the orthocenter is at (3, 0).

(

)

1 (0 + 3 + 6), _ 1 (4 + 8 + 0) = (3, 4) 29. G = _ 3 3

LESSON 5-4

1 XY 30. BC = _ 2 1 (70.2) = 35.1 =_ 2 1 XZ 32. XC = _ 2 = AB = 32.4

31. XZ = 2AB = 2(32.4) = 64.8

33. m∠BCZ = m∠ABC = 42°

34. m∠BAX = 180° - m∠ABC = 180° - 42° = 138° 35. m∠YXZ = m∠BCZ = 42°

−−− 2 36. V = (-1, -1); W = (6, 1); slope of VW = _ ; 7 −− _ 4 2 _ slope of GJ = = ; since the slopes are the 14 7 −−− −− same, VW GJ. 2 2 + 7 2 = √

53 ; VW =

GJ =

1 GJ. 4 + 14 = 2 √

53 , so VW = _ 2

2 2

LESSON 5-5 −− 37. ∠A is the smallest ∠, so BC is the shortest side; −− ∠C is the largest ∠, so AB is the longest side; −− −− −− From shortest to longest, the order is BC, AC, AB. −− 38. GH is the shortest side, so ∠F is the smallest ∠; −− FH is the longest side, so ∠G is the largest ∠; From smallest to largest, the order is ∠F, ∠H, ∠G. 39. x + 4.5 > 13.5 4.5 + 13.5 > x x>9 18 > x Range of the values: > 9 cm and < 18 cm 40. 6.2 + 8.1 14.2 14.3 > 14.2 Yes; possible answer: the sum of each pair of 2 lengths is greater than the third length. 41. z + z 3z 2z ≯ 3z No; possible answer: when z = 5, the value of 3z is 15. So the 3 lengths are 5, 5, and 15. the sum of 5 and 5 is 10, which is not greater than 15. By the Inequality Thm., a cannot have these side lengths.

122


42. Possible answer: Given: ABC Prove: ABC cannot have 2 obtuse . Proof: Assume that ABC has 2 obtuse . Let ∠A and ∠B be the obtuse . By the definition of obtuse, m∠A > 90° and m∠B > 90°. If the 2 inequalities are added, m∠A + m∠B > 180°. However, by the Sum Theorem, m∠A + m∠B + m∠C = 180°. So m∠A + m∠B = 180° - m∠C. But then 180° - m∠C > 180° by subst., and thus m∠C < 0°. A cannot have an ∠ with a measure less than 0°. So the assumption that ABC has 2 obtuse is false. Therefore a cannot have 2 obtuse .

LESSON 5-8 55. 45°-45°-90° x = 26 √ 2

56. 45°-45°-90° 12 = x √ 2 = 2x 12 √2 x = 6 √2

57. 45°-45°-90° x = (16 √ 2 ) √ 2 = 32

58. 30°-60°-90° 48 = 2x x = 24 = 24 √3 y = x √3

59. 30°-60°-90° x = 6 √ 3 y = 2(6) = 12

60. 30°-60°-90° 14 = x √ 3 = 3x 14 √3 3 14 √ x=_ 3 y = 2x 3 3 14 √ 28 √ =2 _ =_ 3 3

LESSON 5-6 −− −− −− −− 43. PQ QR, QS QS, and m∠PQS < m∠RQS. By the Hinge Theorem, PS < RS. −− −− −− −− 44. BC DC, AC AC, and AB < AD. By the Converse of the Hinge Theorem, m∠BCA < m∠DCA. 45. m∠GFH < m∠EFH 5n + 7 < 22 5n < 15 n<3 -1.4 < n < 3 46.

XZ < JK 4n - 11 < 39 4n < 50 n < 12.5 2.75 < n < 12.5

m∠GFH > 0 5n + 7 > 0 5n > -7 n > -1.4 XZ > 0 4n - 11 > 0 4n > 11 n > 2.75

(

61. The diagonal forms two 45°-45°-90° . 30 = s √ 2 30 √ 2 = 2s s = 15 √ 2 ≈ 21 ft 3 in. 62. The altitude forms two 30°-60°-90° . The shorter legs measure 9 ft. h = 9 √ 3 ≈ 15 ft 7 in.

CHAPTER TEST, PAGE 370 1. KL = JK = 9.8 2. m∠WXY = 2m∠WXZ = 2(17) = 34° 3.

LESSON 5-7 47. a 2 + b 2 = c 2 22 + 62 = x 2 40 = x 2 x = 2 √ 10

48. a 2 + b 2 = c 2 x 2 + 8 2 = 14 2 x 2 = 132 x = 2 √ 33

49.

2 2 2 a +b =c 2 2 x + (4.5) = (7.5) 2 x 2 = 36 x =6 The side lengths do not form a Pythagorean triple because 4.5 and 7.5 are not whole numbers.

50.

a 2 + b 2= c 2 24 2 + 32 2 = x 2 1600 = x 2 x = 40 The side lengths form a Pythagorean triple because they are nonzero whole numbers that satisfy

AC = BC 2n + 9 = 5n - 9 18 = 3n n =6 BC = 5(6) - 9 = 21

4. RS = 2MS = 2(3.4) = 6.8 RQ = SQ = 4.9

5. m∠DEF + m∠EFD + m∠FDE = 180 2m∠GEF + 2(25) + 42 = 180 2m∠GEF = 88 m∠GEF = 44° −− −− distance from G to DF = distance from G to DE = 3.7 2 _ 6. XW = XC 3 2 (261) = 174 =_ 3 1 _ BW = ZW 2 1 (118) = 59 =_ 2 BZ = 3BW = 3(59) = 177 −− 7. JK is vertical, so the equation of the horizontal altitude is y = 4; −− -6 slope of KL is _ = -1, so the slope of the altitude 6 is 1, and its equation is y - 2 = x + 5, or y - 7 = x. At the orthocenter, y = 4 and x = 4 - 7 = -3. The orthocenter is at (-3, 4).

a 2 + b 2 = c 2. 51. 9 + 12 = 21 > 16

16 2 9 2 + 12 2 256 > 225 The side lengths can form an obtuse .

)

52. 11 + 14 = 25 ≯ 27 The side lengths cannot form a .

53. 1.5 + 3.6 = 5.1 > 3.9 54. 2 + 3.7 = 5.7 > 4.1

3.9 2 1.5 2 + 3.6 2 4.1 2 2 2 + 3.7 2 15.21 = 15.21 16.81 < 17.69 The side lengths can The side lengths can form a rt. . form an acute .

123


1 HJ 8. PR = _ 2 = QJ = 51 GJ = 2PQ = 2(74) = 148 m∠GRP = m∠GJH = 71° 9. Possible answer: Given: ∠1 and ∠2 form a lin. pair. Prove: ∠1 and ∠2 cannot both be obtuse . Proof: Assume ∠1 and ∠2 are both obtuse . By the definition of obtuse, m∠1 > 90° and m∠2 > 90°. If the 2 inequalities are added, m∠1 + m∠2 > 180°. However, by the Lin. Pair Theorem, ∠1 and ∠2 are supp. . By the definition of supp. , m∠1 + m∠2 = 180°. So m∠1 + m∠2 > 180° contradicts the given information. The assumption that ∠1 and ∠2 are both obtuse is false. Therefore ∠1 and ∠2 cannot both be obtuse. −− 10. BH is the shortest side, so ∠E is the smallest ∠. −− BE is the longest side, so ∠H is the largest ∠. From smallest to largest, the order is ∠E, ∠B, ∠H. −− 11. ∠R is the smallest ∠, so TY is the shortest side. −− ∠Y is the largest ∠, so RT is the longest side. −− −− −− From shortest to longest, the order is TY, RY, RT.

−− −− −− −− 14. DH KH, HN HN, and DN < KN. m∠DHN > 0° m∠DHN < m∠KHN 4x - 10 > 0 4x - 10 < 24 4x > 10 4x < 34 x > 2.5 x < 8.5 2.5 < x < 8.5 15. x 2 + 21 2 = 24 2 x 2 = 135 x = 3 √ 15 The side lengths do not form a Pythagorean triple because 3 √ 15 is not a whole number. 2 2 2 16. 18 + 20 = 38 > 27 17. c = 62 + 82 2 2 2 = 10,568 27 18 + 20 729 > 724 c = √ 10,568 The side lengths can ≈ 102 ft 10 in. form an obtuse .

18.

2 20 = x √ 20 √ 2 = 2x x = 10 √ 2

20.

8 = x √ 3 8 √ 3 = 3x 8 √3 x=_ 3 y = 2x 3 3 8 √ 16 √ =2 _ =_ 3 3

12. AC + 114 > 247 114 + 247 > AC AC > 133 361 > AC Range of the distance: > 133 mi and < 361 mi. −− −− −− −− 13. PS PZ, PV PV, and SV < ZV. By the Converse of the Hinge Theorem, m∠SPV < m∠ZPV.

19. 32 = 2x x = 16 y = x √ 3 = 16 √ 3

( )

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CHAPTER Solutions Key 5 Properties and Attributes of Triangles

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