CHAPTER Solutions Key 5 Properties and Attributes … Key 5 Properties and Attributes of Triangles...

Solutions KeyProperties and Attributes of Triangles5

CHAPTER

ARE YOU READY? PAGE 297

1. E 2. C

3. A 4. D

5. B

6. acute 7. right

8. acute 9. obtuse

10. 8 2 = 64 11. (-12 ) 2 = 144

12. √ �� 49 = 7 13. - √ �� 36 = -6

14. √ �� 9 + 16 = √ �� 25 = 5 15. √ �� 100 - 36 = √ �� 64 = 8

16. √ �� 81 _ 25

= √ �� 81

_ √ �� 25

= 9 _ 5 17.

√ � 2 2 = 2

18. d + 5 < 1

d < -4

19. -4 ≤ w - 7 3 ≤ w w ≥ 3

20. -3s ≥ 6 s ≤ -2

21. -2 > m _ 10

-20 > m m < -20

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 = MBr : AM = 1 _

2 AB and MB = 1 _

2 AB

Given: p → q and q → r. So by the Law of Syllogism,

p → r : If M is the midpoint of −−

AB , then AM = 1 _ 2 AB and

MB = 1 _ 2 AB.

5-1 PERPENDICULAR AND ANGLE BISECTORS, PAGES 300–306

CHECK IT OUT!

1a. DG = EG DG = 14.6

b. Since DG = GE and � ⊥ −−

DE , � is the ⊥ bisector of

−− DE by the Conv. of the ⊥ Bisector Thm.

EF = 1 _ 2 DE

EF = 1 _ 2 (20.8) = 10.4

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°

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

∠PQR.

4.

Step 1 Graph −−

PQ .The ⊥ bisector of

−− PQ is ⊥ to

−− PQ at its midpoint

Step 2 Find the midpoint of −−

PQ .

midpoint of −−

PQ = ( 5 + 1 _ 2 ,

2 + (-4) _

2 ) = (3, -1)

Step 3 Find the slope of the perpendicular bisector.

slope of −−

PQ = -4 - 2 _ 1 - 5

= -6 _ -4

= 3 _ 2

Since the slopes of ⊥ lines are opposite reciprocals,

the slope of the ⊥ 1bisector is - 2 _ 3

.

Step 4 Use point-slope form to write an equation.

The ⊥ bisector of −−

PQ has slope - 2 _ 3

and passes through (3, -1). y - y 1 = m(x - x 1 )

y - (-1) = - 2 _ 3 (x - 3)

y + 1 = - 2 _ 3 (x - 3)

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 .

93 Holt McDougal Geometry

ge07_SOLKEY_C05_093-124.indd 93ge07_SOLKEY_C05_093-124.indd 93 12/22/09 4:25:11 PM12/22/09 4:25:11 PM

3.

EXERCISESGUIDED 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

PQ = 2(47.7) = 95.4

3. SP = SQ SP = 25.9

4. PS = QS 4a = 2a + 26 2a = 26 a = 13So QS = 2(13) + 26 = 52.

5. AD = CDAD = 21.9

6. Since AD = CD, −−

AC ⊥ −−

AB , and −−

CD ⊥ −−

BC , �� BD

bisects ∠ABC by the Conv. of the ∠ Bisector Thm.m∠CBD = 1 _

2 m∠ABC

m∠CBD = 1 _ 2 (48°) = 24°

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 y = 7 _

2

So m∠DBC = [10 ( 7 _ 2 ) + 3]° = 38°

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.

9.

Step 1 Graph −−−

MN .The ⊥ bisector of

−−− MN is ⊥ to

−−− MN at its midpoint.

Step 2 Find the midpoint of −−−

MN .

midpoint of −−−

MN = ( -5 + 1 _ 2 ,

4 + (-2) _

2 ) = (-2, 1)

Step 3 Find the slope of the perpendicular bisector.slope of

−−− MN = -2 - 4 _

1 - (-5) = -6 _

6 = -1

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

10.

Step 1 Graph −−

UV .The ⊥ bisector of

−− UV is ⊥ to

−− UV at its midpoint


UV .

midpoint of −−

UV = ( 2 + 4 _ 2 , -6 + 0 _

2 ) = (3, -3)


slope of −−

UV = 0 - (-6)

_ 4 - 2

= 6 _ 2 = 3

Since the slopes of ⊥ lines are opposite reciprocals, the slope of the ⊥ bisector is - 1 _

3 .

Step 4 Use point-slope form to write an equation. The ⊥ bisector of

−− UV has slope - 1 _

3 and passes

through (3, -3). y - y 1 = m(x - x 1 )

y - (-3) = - 1 _ 3 (x - 3)

y + 3 = - 1 _ 3 (x - 3)

11.

Step 1 Graph −−

JK .The ⊥ bisector of

−− JK is ⊥ to

−− JK at its midpoint


JK .

midpoint of −−

JK = ( -7 + 1 _ 2 ,

5 + (-1) _

2 ) = (-3, 2)


slope of −−

JK = -1 - 5 _ 1 - (-7)

= -6 _ 8 = - 3 _

4

Since the slopes of ⊥ lines are opposite reciprocals, the slope of the ⊥ bisector is 4 _

3 .



JK has slope 4 _ 3 and passes

through (-3, 2).

y - y 1 = m(x - x 1 )

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

y - 2 = 4 _ 3 (x + 3)



PRACTICE AND PROBLEM SOLVING

12. GJ = GKGJ = 8.25

13. JG = KG x + 12 = 3x - 17 12 = 2x - 17 29 = 2x 14.5 = xSo 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

15. RQ = TQRQ = 1.3

16. Since RQ = TQ, −−

RQ ⊥ −−

RS , and −−

TQ ⊥ −−

TS , �� SQ

bisects ∠RST by the Conv. of the Bisector Thm.m∠RST = 2m∠RSQm∠RST = 2(58°) = 116°

17. m∠QSR = m∠QST 9a + 48 = 6a + 50 3a + 48 = 50 3a = 2 m∠QST = 6 ( 2 _

3 ) + 50 = 54°

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


EF .

( x 1 + x 2

_ 2 ,

y 1 + y 2 _

2 )

midpoint of −−

EF = ( -4 + 0 _ 2 , -7 + 1 _

2 ) = (-2, -3)


slope = y 2 - y 1

_ x 2 - x 1

slope of −−

EF = 1 - (-7)

_ 0 - (-4)

= 8 _ 4 = 2


the slope of the ⊥ bisector is - 1 _ 2 .



EF has slope - 1 _ 2 and passes

through (–2, –3). y - y 1 = m(x - x 1 )

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

y + 3 = - 1 _ 2 (x + 2)

20. Step 1 Graph −−

XY .The ⊥ bisector of

−− XY is ⊥ to

−− XY at its midpoint


XY .

( x 1 + x 2

_ 2 ,

y 1 + y 2 _

2 )

midpoint of −−

XY = ( -7 + (-1)

_ 2 ,

5 + (-1) _

2 ) = (–4, 2)


slope = y 2 - y 1

_ x 2 - x 1

slope of −−

XY = -1 - 5 _ -1 - (-7)

= -6 _ 6 = –1

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

−− XY has slope - 1 _

2 and passes

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 + x 2

_ 2 ,

y 1 + y 2 _

2 )

midpoint of −−−

MN = ( -3 + 7 _ 2 ,

1 + (-5) _

2 ) = (2, -3)

Step 3 Find the slope of the ⊥ bisector.

slope = y 2 - y 1

_ x 2 - x 1

slope of −−−

MN = -5 - (-1)

_ 7 - (-3)

= -4 _ 10

= - 2 _ 5


the slope of the bisector is 5 _ 2 .

Step 4 Use point-slope form to write an equation.The bisector of ⊥ has slope - 1 _

2 and passes through

(-2, -3). y - y 1 = m(x - x 1 )

y - (-3) = 5 _ 2 (x - 2)

y + 3 = 5 _ 2 (x - 2)

22. PS = PT 3m + 9 = 5m - 13 9 = 2m - 13 22 = 2m 11 = m

QS = QT 6n - 3 = 4n + 14 2n - 3 = 14 2n = 17 n = 8.5

23. JK = LK JK = 38

24. GN = 2GZGN = 2(36) = 72

25. MK = HK ML + LK = HJ + JK ML = HJ ML = 38

26. HY = MY HY = 24

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

28. NK = GK NM + ML + LK = 114 NM + 38 + 38 = 114 NM = 38

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



31. Statements Reasons

1. �� PS bisects ∠QPR, −−

SQ ⊥ �� PQ ,

−− SR ⊥

�� PR

1. Given

2. ∠QPS � ∠RPS 2. Def. of ∠ bisector

3. ∠SQP and ∠SRP are rt. �. 3. Def. of ⊥4. ∠SQP � ∠SRP 4. Rt. ∠ � Thm.

5. −−

PS � −−

PS 5. Reflex. Prop. of �6. PQS � PRS 6. AAS

7. −−

SQ � −−

SR 7. CPCTC8. SQ = SR 8. Def. of � segs.

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 ∠.

33a. Step 1 Graph −−

AC .The ⊥ bisector of

−− AC is ⊥ to

−− AC at its midpoint


AC .

( x 1 + x 2

_ 2 ,

y 1 + y 2 _

2 )

midpoint of −−

AC = ( -3 + 3 _ 2 , -2 + 6 _

2 ) = (0, 2)


slope = y 2 - y 1

_ x 2 - x 1

slope of −−

AC = 6 - (-2)

_ 3 - (-3)

= 8 _ 6 = 4 _

3

Since the slopes of ⊥ lines are opposite reciprocals, the slope of the ⊥ bisector is - 3 _

4 .



AC has slope - 3 _ 4 and passes

through (0, 2). y - y 1 = m(x - x 1 )

y - 2 = - 3 _ 4 (x – 0)

y = - 3 _ 4 x + 2

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

= √

�� 3 2 + 4 2 = 5 mi. Distance of warehouse from A =

√ �� 4 2 + 5 2 ≈ 6.4 mi.

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 .

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.

CHALLENGE AND EXTEND

38a. The dist. from P to �� BA and from P to

�� BC are both

2 √ � 5. So P is equidistant from �� BA and

�� BC , and

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.


1. −−

VX ⊥ �� YX ,

−− VZ ⊥

�� YZ ,

VX = VZ1. Given

2. ∠VXY and ∠VZY are rt. �.

2. Def. of ⊥

3. −−

YV � −−

YV 3. Reflex. Prop. of �4. YXV � YZV 4. HL5. ∠XYV � ∠ZYV 5. CPCTC

6. � �� YV bisects ∠XYZ. 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.

SPIRAL REVIEW

42. C

43. slope of � �� RS = 4 - 2 _ 1 + 4

= 2 _ 5

slope of � �� VT = -5 + 1 _ -7 - 3

= 2 _ 5

The slopes are the same, so the lines are parallel.

44. slope of � �� RV = -5 - 2 _ -7 + 4

= 7 _ 3

slope of � �� ST = -1 - 4 _ 3 - 1

= - 5 _ 2

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.

45. slope of � �� RT = -1 - 2 _ 3 + 4

= - 3 _ 7

slope of � �� VR = -5 - 2 _ -7 + 4

= 7 _ 3

The product of the slopes is -1, so the lines are perpendicular.

46. m = -9 - (-1)

_ 2 - 1

= -8

y - y 1 = m(x - x 1 ) y + 1 = -8(x - 1) y = -8x + 7

47. y - y 1 = m(x - x 1 )y + 15 = -0.5(x - 10)y + 15 = -0.5x + 5

y = - 1 _ 2 x - 10

48. m = 5 - 0 _ 0 - (-4)

= 5 _ 4

y = mx + b

y = 5 _ 4 x + 5

5-2 BISECTORS OF TRIANGLES, PAGES 307–313

CHECK IT OUT!

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

2.

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.

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.

b. m∠PRQ = 2m∠PRXm∠PRQ = 2(12°) = 24°m∠RQP + m∠PRQ + m∠QPR = 180° 52 + 24 + m∠RQP = 180° m∠RQP = 104°

m∠PQX = 1 _ 2 m∠RQP

m∠PQX = 1 _ 2 (104°) = 52°

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.

THINK AND DISCUSS

1. Possible answer:

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.



3.


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

4. RV = PV = 5.47

5. TR = QT = 3.95

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

7.

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.

8.

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.

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.

10. m∠DCE = 2m∠FCDm∠DCE = 2(17°) = 34°m∠DCE + m∠CDE + m∠CED = 180° 34 + 54 + m∠CED = 180° m∠CED = 92°

m∠FED = 1 _ 2 m∠CED

m∠FED = 1 _ 2 (92°) = 46°

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.


12. CF = FA = 59.7 13. YC = YB = 63.9

14. DB = AD = 62.8 15. AY = YB = 63.9

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).

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).

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.

19. m∠TSR = 2m∠JSRm∠TSR = 2(14°) = 28°m∠TSR + m∠SRT + m∠RTS = 180° 28 + 42 + m∠TSR = 180° m∠TSR = 110°

m∠RTJ = 1 _ 2 m∠TSR

m∠RTJ = 1 _ 2 (110°) = 55°

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.



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. ∠.

22. The angle bisector; m∠BAE = m∠EAC

23. The perpendicular bisector; AD = BD, −−

AD ⊥ −−

DG and

−− BD ⊥

−− DG

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.

26. neither 27. neither

28. never

29. sometimes 30. sometimes

31. never

32. sometimes

33. The slope of −−

OA is 2; the midpoint of −−

OA is (2, 4).


OA is y - 4 = - 1 _

2 (x - 2).


OB is x = 4.

At the intersection, x = 4 and y - 4 = - 1 _ 2 (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).


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 b. the bisector of ∠B

c. PX = PZ


1. −−

QS bisects ∠PQR, −−

PQ � −−

RQ . 1. Given 2. ∠PQS � ∠RQS 2. Def. of ∠

bisector

3. −−

QS � −−

QS 3. Reflex. Prop. of �

4. �PQS � �RQS 4. SAS 5. ∠PSQ � ∠RSQ 5. CPCTC 6. ∠PSQ and ∠RSQ are supp. 6. Lin. Pair Thm. 7. ∠PSQ and ∠RSQ are rt. �. 7. � � supp.

→ rt. � 8. ∠PSQ = ∠RSQ = 90° 8. Def. of rt. ∠

9. �� QS ⊥ −−

PR 9. Def. of ⊥10.

−− PS �

−− RS 10. CPCTC

11. S is midpoint of −−

PR . 11. Def. of midpoint

12. �� QS is the perpendicular bisector of

−− PR .

12. Def. of the perpendicular bisector

37a. The new store is at the circumcenter of �ABC.The perpendicular bisector of

−− AB is x = 4.

The slope of −−

AC is 3 _ 4 ; the midpoint of

−− AC is (2, 3 _

2 ) .


AC is

y - 3 _ 2 = - 4 _

3 (x - 2).

At the intersection, x = 4 and y - 3 _ 2

= - 4 _ 3

(4 - 2)

= - 8 _ 3 , so y = 9 - 16 _

6 = - 7 _

6 .

The new store is located at (4, - 7 _ 6

) .

b. outside, since y > 0 for all int. poins. of the �, but

- 7 _ 6 < 0

c. distance from each store = distance from store C

= 3 - (- 7 _ 6 ) = 4 1 _

6 ≈ 4.2 mi

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.

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.



42. 14.75 KN = MN5z - 4 = z + 11 4z = 15 z = 3.75LN = MN = 3.75 + 11 = 14.75


43a. Possible answer: Given: M is the midpoint of

−− QR .

Prove: PM = QM = RMProof: The coordinates of M are

( 0 + 2a _ 2 , 2b + 0 _

2 ) = (a, b).

By the Distance Formula,

PM = √ �� (a - 0 ) 2 + (b - 0 ) 2

= √

�� a 2 + b 2 ,

QM = √ �� (a - 0 ) 2 + (b - 2b ) 2

= √ �� a 2 + (-b ) 2 = √

�� a 2 + b 2 , and

RM = √ �� (a - 2a ) 2 + (b - 0 ) 2

= √ �� (-a ) 2 + b 2 = √

�� a 2 + b 2 . 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 cm; so by the properties of 30-60-90 �, BC = 1 _

2 AC.

So AB = AC + BC

= 3 _ 2

AC

= 3 _ 2

(28) = 42 cm.

SPIRAL REVIEW

45. t _ 26

= 10 _ 65

65t = 260 t = 4

46. 2.5 _ 1.75

= 6 _ x

2.5x = 10.5 x = 4.2

47. 420 _ y = 7 _ 2

840 = 7y y = 120

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

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 _ -4

= -2; midpoint of −−

ST = M(2, 4)

slope of −−

MX = 4 - 3 _ 2 - 0

= - 1 _ 2 = opposite reciprocal

of 2; so X is on the perpendicular bisector

52. slope of −−

MY = 1 - 4 _ -4 - 2

= 1 _ 2 = opposite reciprocal of

2; so Y is on the perpendicular bisector

53. slope of −−

MZ = -8 - 4 _ -2 - 2

= -3 ≠ opposite reciprocal of 2;

so Z is not on the ⊥ bisector

5-3 MEDIANS AND ALTITUDES OF TRIANGLES, PAGES 314–320

CHECK IT OUT!

1a. KZ + ZW = KW

2 _ 3 KW + ZW = KW

ZW = 1 _ 3 KW

7 = 1 _ 3 KW

21 = KW

b. LZ = 2 _ 3 LX

= 2 _ 3 (8.1)

= 5.4

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 y = - 1 _

2 x + 3. It is true that 4 = - 1 _

2 (-2) + 3, so

(-2, 4) is a solution of this equation. Therefore this altitude passes through the orthocenter.

THINK AND DISCUSS

1. Possible answer: � is isosc.

2. Possible answer: � is a rt. �.

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

4.


1. centroid 2. altitude



3. VW = 2 _ 3 VX

= 2 _ 3 (204) = 136

4. WX = 1 _ 3 VX

= 1 _ 3 (204) = 68

5. RW = 2 _ 3 RY

104 = 2 _ 3 RY

3 _ 2 (104) = RY

RY = 156

6. WY = 1 _ 2 RW

= 1 _ 2 (104) = 52

7. 1 Understand the ProblemAnswer 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 PlanThe 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 SolveLet M be the midpoint of

−− AB and N be the midpoint

of −−

BC .

M = ( 0 + 7 _ 2 , 2 + 4 _

2 ) = (3.5, 3)

N = ( 5 + 7 _ 2 , 0 + 4 _

2 ) = (6, 2)

−−

AN is horizontal. Its equation is y = 2. Slope of

−−− CM = 3 - 0 _

3.5 - 5 = -2. Its equation is

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 BackLet L be the midpoint of

−− AC . Equation for

−− BL is

y - 4 = 2 _ 3 (x - 7), which intersects y = 2 at (4, 2).

8.

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 .

Slope of LM = -2 - 6 _ 8 - 4

= -2.

Equation is y + 2 = 1 _ 2 (x - 2).

Step 4 Solve the system to find the coordinates of the orthocenter.x = 4 and y + 2 = 1 _

2 (4 - 2) = 1, so y = -1.

The coordinates of the orthocenter are (4, -1).

9.


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 .

Slope of −−−

VW = -3 - 6 _ 5 + 4

= -1.

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).

10.


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 .

Slope of −−

PR = 5 - 8 _ -2 + 5

= -1.

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).

11.


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).




12. PC = 1 _ 3 HC

= 1 _ 3 (10.8) = 3.6

13. HC = 2 _ 3 Hp

= 2 _ 3 (10.8) = 7.2

14. JA = 3PA= 3(2.9) = 8.7

15. JP = 2PA= 2(2.9) = 5.8

16. Support should be attached at the centroid. Equation of the median through (4, 0) is x = 4. The median through (0, 10) also passes through

( 4 + 8 _ 2 , 0 + 14 _

2 ) = (6, 7), and has slope 10 - 7 _

0 - 6 = - 1 _

2 .

Equation of the second median is y = - 1 _ 2 x +

10. At intersection, x = 4, so y = - 1 _ 2 (4) + 10 = 8.

The coordinates of the centroid are (4, 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.

Slope of −−

XY is 10 + 2 _ 6 + 2

= 3 _ 2 .

Equation is y + 6 = - 2 _ 3 (x - 6).

Step 3 Find the coordinates of the orthocenter. y = -2, so -2 + 6 = 4 = - 2 _

3 (x - 6),

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. Slope of

−− GJ is -1 - 5 _

4 + 2 = -1.

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).

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.

Slope of −−

AB is 5 + 3 _ 8 - 4

= 2.

Equation is y + 8 = - 1 _ 2 (x - 8).

Step 3 Find the coordinates of the orthocenter. y = -3, so -3 + 8 = 5 = - 1 _

2 (x - 8),

or x = -10 + 8 = -2. The coordinates are (-2, -3).

21. GL = 3 _ 2 GP

= 3 _ 2 (8) = 12

22. PL = 1 _ 2 GP

= 1 _ 2 (8) = 4

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

26. A = 1 _ 2 (HJ)(GL)

= 1 _ 2 (10)(12) = 60 square units

27. G = ( 1 _ 3 (0 + 14 + 16), 1 _

3 (-4 + 6 - 8)) = (10, -2)

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

3 (-1 + 7 - 3)) = (5, 1)

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

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

31. Step 1 Find n.2n + 17 = 54 2n = 54 - 17 n = 18.5Step 2 Find QZ.QZ = 4n - 26

= 4(18.5) - 26 = 48

32. YZ = 1 _ 2 (QZ)

= 1 _ 2 (48) = 24

33. Possible answer: the perpendicular bisector of base; the bisector of vertex ∠; the median to the base; the altitude to the base



34. sometimes 35. always

36. never 37. always


1. −−

PS and −−

RT are medians of �PQR.

−− PS �

−− RT

1. Given

2. PS = RT 2. Def. of � segs.

3. 2 _ 3 PS = 2 _

3 RT 3. Mult. Prop. of =

4. PZ = 2 _ 3 PS, RZ = 2 _

3 RT 4. Centroid Thm.

5. PZ = RZ 5. Subst.

6. −−

PZ � −−

RZ 6. Def. of � segs.

7. ∠SPR � ∠TRP 7. Isosc. � Thm.

8. −−

PR � −−

PR 8. Reflex. Prop. of �

9. �PTR � �RSP 9. SAS

10. ∠QPR � ∠QRP 10. CPCTC

11. −−

PQ � −−

RQ 11. Con. of Isosc. � Theorem

12. �PQR is an isosc �. 12. Def. of isosc. �

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.

40a. G = ( 1 _ 3 (0 + 0 + 8), 1 _

3 (0 + 8 + 0)) = (2 2 _

3 , 2 2 _

3 )

b. DG = √ ��

( 8 _ 3 )

2 + ( 8 _

3 )

2

= 8 _ 3 √ � 2 ≈ 3.8 mi

c. Perpendicular from G crosses −−

EF at H(4, 4),

distance = √ ��

( 4 _ 3 )

2 + ( 4 _

3 )

2

= 4 _ 3 √ � 2 ≈ 1.9 mi

TEST PREP

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

43. D


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 .

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.

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.

45a. slope of RS = c _ b ; slope of ST = c _

b - a ;

slope of RT = 0.

b. Since � ⊥ −−

RS , slope of � = - b _ c . Since m ⊥ −−

ST ,

slope of m = - b - a _ c = a - b _ c . Since n ⊥ −−

RT , n is

a vertical line, and its slope is undefined.



c. An equation of � is

y - 0 = - b _ c (x - a)

y = - b _ c x + ab _ c

An equation of m is

y - 0 = a - b _ c (x - 0)

y = a - b _ c x

An equation of n is x = b.

d. (b, ab - b 2 _ c )

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

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

SPIRAL REVIEW

46. Let x, y be prices of peanuts and popcorn.x = y + 0.75 or y = x - 0.75 5x + 3y = 21.755x + 3(x - 0.75) = 21.75 5x + 3x - 2.25 = 21.75 8x = 24 x = 3Price of peanuts is $3.00.

47. F; Possible answer: a rectangle with width 5 and length 8.

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

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°

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. ∠.

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

CHECK IT OUT!

1. The midpoints are M(1, 1), N(3, 4);

slope of −−−

MN = 3 _ 2 ; slope of

−− RS = 6 _

4 = 3 _

2 ;

since the slopes are =, −−−

MN ‖ −−

RS .

MN = √

�� 2 2 + 3 2 = √ �� 13 ;

RS = √

�� 4 2 + 6 2 = √ �� 52 = 2 √ �� 13 ;

and MN = 1 _ 2 RS.

2a. JL = 2PN= 2(36) = 72

b. PM = 1 _ 2 KL

= 1 _ 2 (97) = 48.5

c. m∠MLK = m∠JMP = 102°

3. HF = 1 _ 2 AE

= 1 _ 2 (1550) = 775

The distance she measures between H and F is 775 m.

THINK AND DISCUSS

1. The endpoints of −−

XY are not the midpoints of the sides of the �.

2.


1. midpoints

2. The midpoints are S(-1, 4), T(4, 6);slope of

−− ST = 2 _

5 ; slope of

−− PR = 4 _

10 = 2 _

5 ;

since the slopes are =, −−

ST ‖ −−

PR .

ST = √

�� 2 2 + 5 2 = √ �� 29 ;

PR = √

�� 4 2 + 10 2 = √ �� 116 = 2 √ �� 29 ;and ST = 1 _

2 PR.

3. NM = 1 _ 2 XY

= 1 _ 2 (10.2) = 5.1

4. XZ = 2LM= 2(5.6) = 11.2

5. NZ = 1 _ 2 XZ

= 1 _ 2 (11.2) = 5.6

6. m∠LMN = m∠MNZ = 29°

7. m∠YXZ = m∠MNZ = 29°

8. m∠XLM + m∠LMN = 180 m∠XLM + 29 = 180 m∠XLM = 151°

9. CD < 1 _ 2 XZ = 15 ft = 5 yd

The width of the 2nd floor is less than 5 yd.


10. The midpoints are D(-4, 3), E(0, 4);slope of

−− DE = 1 _

4 ; slope of

−− CB = 2 _

8 = 1 _

4 ;

since the slopes are =, −−

DE ‖ −−

CB .

DE = √

�� 1 2 + 4 2 = √ �� 17 ;

CB = √

�� 2 2 + 8 2 = √ �� 68 = 2 √ �� 17 ;and DE = 1 _

2 CB.



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

12. RQ = 1 _ 2 GH

= 1 _ 2 (27) = 13.5

13. RJ = 1 _ 2 GJ

= 1 _ 2 (38) = 19

14. m∠PQR = m∠QRJ= 55°

15. m∠HGJ = m∠QRJ= 55°

16. m∠GPQ + m∠HGJ = 180° m∠GPQ + 55 = 180° m∠GPQ = 125°

17. Yes; −−

DE is a midsegment of �ABC, so its length is half of 4 1 _

2 ft, or 2 1 _

4 ft, which is 27 in. This is less than

30 in. So the carpenter can use the 30 in. timber to make the crossbar.

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

19. P = KL + LM + KM= 7 + 1 _

2 GH + 1 _

2 HJ

= 7 + 1 _ 2 (12) + 1 _

2 (8)

= 17

20. The perimeter of � GHJ is twice the perimeter of �KLM.

21. 3n = 2(54)3n = 108 n = 36

22. 2(n - 9) = 352n - 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

25. 6n = 2(n + 8)6n = 2n + 164n = 16 n = 4

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 length of the ‖ side, so DE = 1 _

2 BC.

28. ∠D � ∠FZY � ∠YXE � ∠ZYX;∠E � ∠ZYF � ∠DXZ � ∠XZY;∠F � ∠XYE � ∠DZX � ∠ZXY

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.

30. CG = 1 _ 2 AB

= 1 _ 2 (33) = 16.5

31. EH = 1 _ 2

DC

= 1 _ 2

CB

= 1 _ 2

(22) = 11

32. FJ = 1 _ 2 GH

= 1 _ 4 CG

= 1 _ 4 (16.5) = 4.125

33. m∠DCG = m∠CBA = 57°

34. m∠GHE = m∠HCD = m∠ABC = 57°

35. m∠FJG + m∠GHE = 180 m∠FJG + 57 = 180 m∠FJG = 123°

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

of those sides is 1 _ 2 x. Since these two midsegments

are equal in length, they are �.

37a. WX = 1 _ 2 XY

= 1 _ 4 AB

= 1 _ 4 (9) = 2.25 mi

b. XA = CX = 3.5 mi, BC = 2BY = 8 mitrip length = WX + XA + AB + BC + CX + XW

= 2.25 + 3.5 + 9 + 8 + 3.5 + 2.25 = 28.5 mi

38a. M = ( 0 + 2a _ 2 , 0 + 2b _

2 ) = (a, b)

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

2 ) = (a + c, b)

c. slope of −−

PR = 0 - 0 _ 2c - 0

= 0

slope of −−−

MN = b - b _ a + c - a

= 0

Slopes of −−−

MN and −−

PR are =, so MN ‖ PR.

d. PR = 2c; MN = a + c - a = c; the length of −−

PR is

twice length of −−−

MN , so MN = 1 _ 2 PR.

TEST PREP

39. D; RT = 2PQ4x - 27 = 2(x + 9)4x - 27 = 2x + 18 2x = 45 x = 22.5RT = 4(22.5) - 27

= 63 m

40. H

41. D


ge07_SOLKEY_C05_093-124.indd 105ge07_SOLKEY_C05_093-124.indd 105 12/24/09 7:14:17 AM12/24/09 7:14:17 AM


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

v + x = 2(3) = 6x + z = 2(1) = 2v + z = 2(-3) = -6 2x = 6 + 2 - (-6) 2x = 14 x = 7

So (w, x) = (-4, 7).u + (-4) = -12 u = -8

v + 7 = 6 v = -1

So (u, v) = (-8, -1).-4 + y = 4

y = 87 + z = 2

z = -5So (y, z) = (8, -5).

43. The midsegment � is equilateral and equiangular

44. n 2 - 3 = 2(39) n 2 = 81 n = ±9

45. n 2 - 6n + 3 = 2(3n - 16) n 2 - 6n + 3 = 6n - 32 n 2 - 12n + 35 = 0 (n - 7)(n - 5) = 0 n = 7 or 53(7) - 16 = 5 > 03(5) - 16 = -1 < 0So n = 7 is the only possible solution.

46. �QXY � �XPZ � �YZR � �ZYX; area of �XYZ = 1 _

4 (area of �PQR)

47a. Number of Midsegment

1 2 3 4

Length ofMidsegment

32 16 8 4

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

2 )

4

= 4 ( 1 _ 16

) = 1 _ 4

c. length of midsegment

n = AB ( 1 _ 2

) n

= 64 ( 1 _ 2 )

n

= 2 6 - n

SPIRAL REVIEW

48. concentration = 2% + 3% _ 2 = 2.5%

49. concentration = 3(2%) + 1(3%)

__ 3 + 1

= 2.25%

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)

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

54. MR = 3 _ 2 MX

= 3 _ 2 (5.5) = 8.25

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

CONSTRUCTION

1. XY = 1 _ 2 AC

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

2. JM = ML = 58 3. AB = AC5z + 16 = 8z - 5 21 = 3z 7 = zAC = 8(7) - 5 = 51

4. Slope of −−−

MN = 4 _ 8 = 1 _

2 , so the slope of the

perpendicular bisector = -2;

equation is y + 1 = -2(x - 3).

5. PS = PT = 83.9XT = 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.

7. The equations of the two perpendicular bisectors are x = 4.5 and y = -2. So C = (4.5, -2).

8. BW = 1 _ 3 BD

= 1 _ 3 (87) = 29

CE = 3CW = 3(19) = 57

CW = 1 _ 2 WE

= 1 _ 2 (38) = 19

9. G = ( 1 _ 3 (0 + 8 + 10), 1 _

3 (4 + 0 + 8)) = (6, 4)

10. −−

PS is horizontal, the altitude is vertical, so the equation is x = 4; the slope of

−− SV = 4 _

4 = 1, so the

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).

11. ZV = 1 _ 2 JM

= RM = 45

PM = 2ZR= 2(53) = 106

m∠RZV = m∠PVZ = 36°

12. XY = 2MN= 2(39) = 78

The distance across the pond is 78 m.



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

CHECK IT OUT!

1. Possible answer:Given: �RSTProve: �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.

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 .

3a. 8 + 13 � 21 21 ≯ 21No; 8 + 3 = 21, which is not greater than the third side length.

b. 6.2 + 7 � 9 13.2 > 9

6.2 + 9 � 7 15.2 > 7

7 + 9 � 6.2 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 18 > 17

2 + 17 � 16 19 > 16

16 + 17 � 2 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 > 5

s + 22 > 17 s > -5

17 + 22 > s 39 > s

Combine the inequals. So 5 < s < 39. the length of the 3rd side is > 5 in. and < 39 in.

5. Let d be the distance from Seguin to Johnson City.d + 22 > 50 d > 28

d + 50 > 22 d > -28

22 + 50 > d 72 > d

28 < d < 72The 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.

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

3.


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.

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. 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 .



6. 4 + 7 � 10 11 > 10 �

4 + 10 � 7 14 > 7 �

7 + 10 � 4 17 > 4 �

Yes; the sum of each pair of 2 lengths is greater than the third length.

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

8. 3.5 + 3.5 � 6 7 > 6 �

3.5 + 6 � 3.5 9.5 > 3.5 �

Yes; the sum of each pair of 2 lengths is greater than the third length.

9. 1.1 + 1.7 � 3 2.8 ≯ 3No; 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 ≯ 25No; 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.

11. When c = 2, 7c + 6 = 20, 10c - 7 = 13, 3 c 2 = 12.12 + 13 � 20 25 > 20 �

12 + 20 � 13 32 > 13 �

13 + 20 � 12 33 > 12 �

Yes; when c = 2, the value of 7c + 6 is 20, the value of 10c - 7 is 13, and the value of 3 c 2 is 12. The sum of each pair of 2 lengths is greater than the third length.


s + 12 > 8 s > -4

8 + 12 > s 20 > s

Combine the inequals. So 4 < s < 20. The length of the 3rd side is > 4 mm and < 20 mm.


16 + 16 > s 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 > 0.6

s + 12 > 11.4 s > -0.6

11.4 + 12 > s 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.


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

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

YZ = 1 __ 2 BA. So 1 __

2 BC = 1 __

2 BA by subst. But then

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.

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 ≯ 32No; 14 + 18 = 32, which is not greater than the third side length.

22. 5.8 + 5.8 � 11.9 11.6 ≯ 11.9No; 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.



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 ≯ 24No; 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 > ss + 4 > 19 s > 15

4 + 19 > s 23 > s

15 yd < s < 23 yd

27. s + 23 > 28 s > 55 km < s < 51 km

23 + 28 > s51 > s

28. s + 3.8 > 9.2 s > 5.45.4 cm < s < 13.0 cm

3.8 + 9.2 > s 13.0 > s

29. s + 1.89 > 3.07 s > 1.181.18 m < s < 4.96 m

1.89 + 3.07 > s4.96 > s

30. s + 2 1 _ 8 > 3 5 _

8

s > 1 1 _ 2

1 1 _ 2 in. < s < 5 3 _

4 in.

2 1 _ 8 + 3 5 _

8 > s

5 3 _ 4 > s

31. s + 3 5 _ 6 > 6 1 _

2

s > 2 2 _ 3

2 2 _ 3 ft < s < 10 1 _

3 ft

3 5 _ 6 + 6 1 _

2 > s

10 1 _ 3 > s

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 = 13Step 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∠RBy 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.

54. JK = √

�� 6 2 + 8 2 = 10; KL = √

�� 5 2 + 8 2 ≈ 9.4; JL = ⎪-3 - 8⎥ = 11KL < JK < JL, so order is ∠J, ∠L, ∠K.

55. JK = ⎪-10 - 2⎥ = 12; KL = √

�� 12 2 + 7 2 ≈ 13.9;

JL = √

�� 12 2 + 5 2 = 13JK < JL < KL, so order is ∠L, ∠K, ∠J.

56. JK = √

�� 1 2 + 7 2 ≈ 7.1; KL = √

�� 6 2 + 4 2 ≈ 7.2;

JL = √

�� 7 2 + 3 2 ≈ 7.6JK < KL < JL, so order is ∠L, ∠J, ∠K.

57. JK = √

�� 10 2 + 7 2 ≈ 12.2; KL = √

�� 2 2 + 11 2 ≈ 11.2;

JL = √

�� 12 2 + 4 2 ≈ 12.6KL < JK < JL, so order is ∠J, ∠L, ∠K.



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.

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

400 + 600 > AR 1000 mi > AR

Time to travel 200 mi is 200 _ 500

= 0.4 h. Time to travel

1000 mi is 1000 _ 500

= 2 h. So the range of time is

0.4 h < t < 2 h.

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 n > 2

6 + 8 > n 14 > n

Step 2 Combine the inequals.2 < n < 14

61. Step 1 Write and solve 2 inequals. for n.2n + 5 > 7 2n > 2 n > 1

5 + 7 > 2n 12 > 2n 6 > n


62. Step 1 Write and solve 2 inequals. for n.n + 1 + 3 > 6 n > 2

3 + 6 > n + 1 8 > n

Step 2 Combine the inequal.s2 < n < 8

63. Step 1 n + 1 < n + 2 < n + 3, so only need to check 1 inequal. for n.n + 1 + n + 2 > n + 3 2n + 3 > n + 3 n > 0

64. Step 1 Write and solve 2 inequals. for n. Use the fact that n + 2 < n + 3.3n - 2 + n + 2 > n + 3 4n > n + 3 3n > 3 n > 1

n + 2 + n + 3 > 3n - 2 2n + 5 > 3n - 2 7 > n


65. Step 1 Write and solve 2 inequals. for n. Use fact that n < n + 2.2n + 1 + n > n + 2 3n + 1 > n + 2 2n > 1 n > 0.5

n + n + 2 > 2n + 1 2 > 1 always true

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.

67a. definition of � segs.

c. definition of � �

e. subst.

g. trans. Prop. of Inequal.

b. Isosc. � Theorem

d. m∠1 + m∠3

f. m∠S

68a. �ABC

c. Isosc. � Theorem

e. m∠3

g. in �, longer side is opp. larger ∠

i. AC + BC > AB

b. AD

d. definition of � �

f. subst.

h. subst.

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.

TEST PREP

70. A;3 + 3 = 6 > 5 �

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

72. C;∠S must be the largest ∠, so

−− RT is the longest side.


73. The total number of choices is ( 5

3 ) = 10. The

choices that form a �:1 + 3 = 4 ≯ 5, 7, or 9 �1 + 5 = 6 ≯ 7 or 9 �1 + 7 = 9 ≯ 9 �

3 + 5 = 8 > 7 �3 + 5 = 8 ≯ 9 �3 + 7 = 10 > 9 �5 + 7 = 12 > 9 �

� is possible for 3 choices. So prob. = 3 _ 10

or 30%.

74a. √ 2 is rational

c. 2 q 2

b. p 2

_ q 2

d. (2 x) 2 = 4 x 2

e. q 2 = 1 _ 2 p 2 and p 2 is divisible by 4




1. −−

PX ⊥ �; Y is any point on � other than X.

1. Given

2. m∠1 = 90° 2. Def. of ⊥3. ∠1 is a rt. ∠. 3. Def. of rt. ∠4. �XPY is a rt. �. 4. Def. of rt. �5. ∠2 and ∠P are comp. 5. Acute � of rt. �

are comp.6. 90° = m∠2 + m∠P 6. Def. of comp. �7. 90° > m∠2 7. Comparison

Prop. of Inequal.8. m∠1 > m∠2 8. Subst.9. PY > PX 9. In �, longer side

is opp. larger ∠

SPIRAL REVIEW

76. slope = -2 - 2 _ -1 + 3

= -2

y - y 1 = m(x - x 1 ) y - 2 = -2[x - (-3)] y - 2 = -2(x + 3) y - 2 = -2x - 62x + y = -4

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

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

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

80. Equation of altitude from S is y = 3. Slope of RS is 3 - 5 _ 4 - 0

= - 1 _ 2 , so slope of altitude from T is 2;

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).

81. The altitudes from N and P lie along the x- and y-axes, respectively. Therefore O = (0, 0).

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

CHECK IT OUT!

1a. Compare the side lengths in �EFG and �EHG.EF > EH EG = EG FG = HGBy 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∠ADBBy the Hinge Theorem, BC > AB.

2. The ∠ of swing at full speed is greater than the ∠ of swing at low speed.

3a. Statements Reasons

1. C is the midpoint of

−− BD ;

m∠1 = m∠2 m∠3 > m∠4

1. Given

2. −−

BC � −−

DC 2. Def. of midpoint

3. ∠1 � ∠2 3. Def. of � �4.

−− AC �

−− EC 4. Con. of Isosc. �

Thm.5. AB > ED 5. Hinge Thm.

b. Statements Reasons

1. ∠SRT � ∠STR, TU > RU

1. Given

2. −−

ST � −−

SR 2. Con. of Isosc. � Thm.

3. −−

SU � −−

SU 3. Reflex. Prop. of �4. m∠TSU > m∠RSU 4. Con. of the Hinge

Thm.

THINK AND DISCUSS

1. Possible answer: kitchen tongs

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.

3.


1. Compare the sides and the � in �ABC and �XYZ.AB = YZ BC = XY m∠B < m∠YBy the Hinge Theorem, AC < XZ.

2. Compare the side lengths in �SRT and �QRT.RT = RT RS = RQ ST > QTBy 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∠KMNBy the Hinge Theorem, KL > KN.



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.

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.

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.

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.


1. −−

FH is a median of �DFG; m∠DHF > m∠GHF

1. Given

2. H is midpoint of −−

DG . 2. Def. of median

3. −−

DH � −−

GH 3. Def. of midpoint

4. −−

FH � −−

FH 4. Reflex. Prop. of �5. DF > GF 5. 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.

12. 4z - 12 < 16 4z < 28 z < 7

4z - 12 > 0 4z > 12 z > 3

Combining, 3 < z < 7.

13. 2z + 7 < 72 2z < 65 z < 32.5

2z + 7 > 0 2z > -7 z > -3.5

Combining, -3.5 < z < 32.5.

14. 4z - 6 < z + 11 3z < 17

z < 17 _ 3

4z - 6 > 0 4z > 6

z > 3 _ 2

Combining, 3 _ 2 < z < 17 _

3 .

15. The lengths of the arms are the same in both positions, but the included ∠ measure is greater in the 2nd position. Therefore, by the Hinge Theorem, the distance from the cab to the bucket is greater in the 2nd position.


1. −−

JK � −−−

NM , −−

KP � −−−

MQ , JQ > NP

1. Given

2. −−

QP � −−

QP 2. Reflex. Prop. of �3. QP = QP 3. Def. of � segs.4. JQ + QP > NP + QP 4. Add. Prop. of

Inequal.5. JQ + QP = JP,

NP + QP = NQ5. Segment Add.

Post.6. JP > NQ 6. Subst.7. m∠K > m∠M 7. Con. of the Hinge

Thm.

17. BC = YZ 18. m∠QRP < m∠SRP

19. m∠QPR > m∠QRP 20.m∠PRS < m∠RSP

21. m∠RSP = m∠RPS 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.



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 > SNNJ + 182 > 300 NJ > 118 miMin. distance = SN + NJ

> 300 + 118 = 418 mi

TEST PREP

31. D;0 < 3x - 9 < 2x + 19 < 3x or 3 < x, and x < 103 < 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.


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 21 > 3x 7 > x

8x - 6 > 0 8x > 6 x > 0.75

Step 3 Combine the inequals.0.75 < x < 7

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.

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 = 14mode: none

38. range: 9 - 4 = 5modes: 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:DF = 1 _

2 AB

= 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 4 2 + 8 2 = x 2 80 = x 2 √ 80 = x x = √ (16)(5) = 4 √ 5

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



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.

a 2 + b 2 = c 2 (4y ) 2 + y 2 = 30 2 17 y 2 = 900

y 2 = 900 _ 17

y = √ �� 900 _ 17

4y = 4 √ �� 900 _ 17

≈ 29 ft 1 in.

3a. a 2 + b 2 = c 2 8 2 + 10 2 = c 2 164 = c 2 c = √ �� 164 = 2 √ �� 41 The side lengths do not form a Pythagorean triplebecause 2 √ �� 41 is not a whole number.

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

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

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

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 �. c 2 � a 2 + b 2 16 2 � 7 2 + 12 2 256 � 49 + 144256 > 193Since c 2 > a 2 + b 2 , � is obtuse.

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 �. c 2 � a 2 + b 2 5.2 2 � 3.8 2 + 4.1 2 27.04 � 14.44 + 16.8127.04 < 31.25Since c 2 < a 2 + b 2 , � is acute.

THINK AND DISCUSS

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

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

4.


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 3 2 + 9 2 = x 2 90 = x 2 √ �� 90 = x x = √ �� (9)(10)

= 3 √ �� 10

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

= 6 √ � 2

4. a 2 + b 2 = c 2 (x - 2 ) 2 + 8 2 = x 2 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.

a 2 + b 2 = c 2

(4x ) 2 + (5x ) 2 = 19 2

41 x 2 = 361

x 2 = 361 _ 41

x = √ �� 361 _ 41

w = 5x = 5 √ �� 3 61 _ 41

≈ 14.8 in.

h = 4x = 4 √ �� 361 _ 41

≈ 11.9 in.

6. a 2 + b 2 = c 2 4 2 + 5 2 = c 2 41 = c 2 c = √ �� 41 The side lengths do not form a Pythagorean triple

because √ �� 41 is not a whole number.



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

8. a 2 + b 2 = c 2 1.5 2 + b 2 = 1.7 2 b 2 = 0.64 b = 0.8The 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 �. c 2 � a 2 + b 2 12 2 � 7 2 + 10 2 144 � 49 + 100144 < 149Since 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 �. c 2 � a 2 + b 2 15 2 � 9 2 + 11 2 225 � 81 + 121225 > 202Since 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 �. c 2 � a 2 + b 2 41 2 � 9 2 + 40 2 1681 � 81 + 16001681 = 1681Since c 2 = a 2 + b 2 , � is a rt. �.

12. Step 1 Determine if measures form a �.Since 1 1 _

2 + 1 3 _

4 = 3 1 _

4 ≯ 3 1 _

4 , these cannot be

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 �. c 2 � a 2 + b 2 8.4 2 � 5.9 2 + 6 2 70.56 � 34.81 + 3670.56 < 70.81Since c 2 < a 2 + b 2 , � is acute.

14. Step 1 Determine if the measures form a �.By the � Inequal. Theorem, 11, 13, and 7 √ � 6 can be the side lengths of a �.Step 2 Classify the �. c 2 � a 2 + b 2 (7 √ � 6 )

2 � 11 2 + 13 2

294 � 121 + 169 294 > 290Since c 2 > a 2 + b 2 , � is obtuse.


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

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

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

17

3x = 3 √ �� 32 _ 17

≈ 4 ft 1 in.

5x = 5 √ �� 32 _ 17

≈ 6 ft 10 in.

19. 2.5 2 + b 2 = 6.5 2 6.25 + b 2 = 42.25 b 2 = 36 b = 6The 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 = 25Yes; the three side lengths are nonzero whole

numbers that satisfy a 2 + b 2 = c 2 .

21. 2 2 + b 2 = 7 2 4 + b 2 = 49 b 2 = 45 b = √ �� 45 = 3 √ � 5 The side lengths cannot form a � because 3 √ � 5 is not a whole number.

22. 10 + 12 = 22 > 15 15 2 � 10 2 + 12 2 225 < 244The side lengths form an acute �.

23. 8 + 13 = 21 ≯ 23 The side lengths can not form a �.

24. 9 + 14 = 23 > 17 17 2 � 9 2 + 14 2 289 > 277The side lengths form an obtuse �.

25. 1 1 _ 2 + 2 = 3 1 _

2 > 2 1 _

2

(2 1 _ 2 )

2 � (1 1 _

2 )

2 + 2 2

6 1 _ 4

= 6 1 _ 4

The side lengths form a rt. �.



26. 0.7 + 1.1 = 1.8 > 1.7 � 1.7 2 � 0.7 2 + 1.1 2 2.89 > 1.7The side lengths form an obtuse �.

27. 7 + 12 = 19 > 6 √ � 5 �

(6 √ � 5 ) 2 � 7 2 + 12 2

180 < 193The side lengths form 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.

29. B; (x + 3 ) 2 + 4

2 = ( x

2 + 6x + 9) + 16. In the

solution shown, the 6x term was omitted. 30. Let a and b be the horizontal leg lengths of the left-

and right-hand �.

9 2 + b 2 = 15 2 b 2 = 144

b = 12a + b = 25 a = 25 - 12 = 13

13 2 + 9 2 = 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 �.

a 2 + 6 2 = 10 2 a 2 = 64

a = 8

6 2 + b 2 = 7 2 b 2 = 13 b = √ �� 13

x = a + b = 8 + √ �� 13

32. Let d be the length of the shared side. 2 2 + d 2 = 7 2 d 2 = 45 x 2 + 5 2 = d 2 x 2 = 20 x = √ �� 20 = 2 √ � 5

33. Let h be the height of the �. 3 2 + h 2 = ( √ �� 34 )

2

h 2 = 25 x 2 + h 2 = 11 2 x 2 = 96 x = √ �� 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 x 2 + 10x + 25 + 144 = 400 x 2 + 10x - 231 = 0 (x - 11)(x + 21) = 0Since x > 0, the only possible solution is x = 11.

35. Let b be the base length of each �. 18 2 + (2b ) 2 = 30 2 4 b 2 = 576 b 2 = 144 18 2 + b 2 = x 2 468 = x 2 x = √ �� 468 = 6 √ �� 13

36. 3963 2 + x 2 = (3963 + 250 ) 2 x 2 = 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 = 15P = 8 + 15 + 17

= 40 unitsA = 1 _

2 (15)(8)

= 60 square units

39. Let 2b be the base of the �. 6 2 + b 2 = 8 2 b 2 = 28 b = 2 √ � 7 2b = 4 √ � 7 P = 8 + 8 + 4 √ � 7 = 16 + 4 √ � 7 units

A = 1 _ 2 (4 √ � 7 ) (6)

= 12 √ � 7 square units

40. Let h be the height of the �. 4 2 + h 2 = 12 2 h 2 = 128 h = √ �� 128 = 8 √ � 2 P = 12 + 12 + 8

= 32 unitsA = 1 _

2 (8)8 √ � 2

= 32 √ � 2 square units

41. Let h be the height and c be the 3rd side length of the �. 3 2 + h 2 = 5 2 h 2 = 16 h = 4 6 2 + 4 2 = c 2 52 = c 2 c = √ �� 52 = 2 √ �� 13 P = 5 + (3 + 6) + 2 √ �� 13

= 14 + 2 √ �� 13 unitsA = 1 _

2 (3 + 6)(4)

= 18 square units

42. Let a + b = 15 be the 2nd side length and c be the 3rd side length of the �. a 2 + 12 2 = 15 2 a 2 = 81 a = 9b = 15 - 9 = 6 6 2 + 12 2 = c 2 180 = c 2 c = √ �� 180 = 6 √ � 5 P = 15 + 15 + 6 √ � 5

= 30 + 6 √ � 5 unitsA = 1 _

2 (15)(12) = 90 square units

43. P = 4 + 5 + 5 + 8 = 22 units

A = 1 _ 2 (a + b)h

= 1 _ 2 (5 + 8)(4)

= 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. �.



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

a 2 + b 2 = x 2 , it follows by subst. that x 2 = c 2 . 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.

46a. ( x 2 , y 1 ) b. JL = x 2 - x 1 ,LK = y 2 - y 1

c. JK 2 = JL

2 + LK

2

= ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

JK = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

47a. KR 2 + 500 2 = 1300 2 KR 2 = 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. S M 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

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

50. H; 11 2 � 7 2 + 9 2 121 ≯ 130

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

PB 2 = 1 2 + PA 2 = 3 PB = √ 3 PD 2 = 1 2 + PC 2 = 5 PD = √ 5 PF 2 = 1 2 + PE 2 = 7 PF = √ 7

b. √ 10 ; possible answer: for each � added to the 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. √ 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 .


52. Let 3 points be A(-1, 2), B(-10, 5), and C(-4, k).

AB 2 = 9 2 + 3 2 = 90 BC 2 = 6 2 + (k - 5 ) 2

= 36 + k 2 - 10k + 25= k 2 - 10k + 61

AC 2 = 3 2 + (k - 2 ) 2 = 9 + k 2 - 4k + 4= k 2 - 4k + 13

If AB 2 + BC 2 = AC 2 ,90 + k 2 - 10k + 61 = k 2 - 4k + 13 138 = 6k k = 23If AB 2 + AC 2 = BC 2 ,90 + k 2 - 4k + 13 = k 2 - 10k + 61 6k = -42 k = -7If AC 2 + BC 2 = AB 2 , k 2 - 4k + 13 + k 2 - 10k + 61 = 90 2 k 2 - 14k - 16 = 0 k 2 - 7k - 8 = 0 (k - 8)(k + 1) = 0 k = 8 or -1So k = -7, -1, 8, or 23

53. By the � Inequal. Theorem, a + b > c.

By the Pythagorean Theorem, c = √

a 2 + b 2 .

By subst., a + b > √

a 2 + b 2 .

54. c 2 = a 2 + b 2

c = √

a 2 + b 2

A = 1 _ 2 ab = 1 _

2 hc

ab = hc

ab = h √

a 2 + b 2 ab _ √

a 2 + b 2 = h

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 4 a 2 + 4 b 2 =

4 c 2 . This is equivalent to (2a ) 2 + (2b ) 2 = (2 c) 2 . So by def., 2a, 2b, and 2c also form a Pythagorean triple.

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 + 1 6 2 ≠ 2 5 2 .

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



SPIRAL REVIEW

56. (4 + x)12 - (4x + 1)6 = 0 48 + 12x - 24x - 6 = 0 42 - 12x = 0 42 = 12x x = 3.5

57. 2x - 5 _ 3 = x

2x - 5 = 3x -5 = x

58. 4x + 3(x + 2) = -3(x + 3) 4x + 3x + 6 = -3x - 9 10x = -15 x = -1.5 = - 3 _

2

59. By the Midpoint Formula, the coordinates of M are (a, b). By the Distance Formula,

AM = √ �� (a - 0 ) 2 + (b - 0 ) 2 = √

�� a 2 + b 2 and

MB = √ �� (0 - a ) 2 + (2b - b ) 2 = √

�� a 2 + b 2 . So by subst., AM = MB.

60. −−

JK � −−

NP , −−

JL � −−−

NM , KL < MP m∠J < m∠N4x - 6 < 68 4x < 74 x < 18.5

m∠J > 0°4x - 6 > 0 4x > 6 x > 1.5

So 1.5 < x < 18.5.

61. −−

BA � −−

BC , −−

BD � −−

BD , m∠ABD < m∠CBD3x + 1 < 7 3x < 6 x < 2

3x + 1 > 0 3x > -1 x < - 1 _

3

So - 1 _ 3 < x < 2.

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

b. The � is an isosc. rt. �, which is a 45°-45°-90° �. 16 = x

√ � 2

16 _

√ � 2

= x

16 √ � 2 _

2 = x

x = 8 √ � 2

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

3a. 18 √ � 3 = 2x 9 √ � 3 = xy = x √ � 3 y = (9 √ � 3 ) √ � 3 y = 27

b. x = 5 √ � 3 y = 2(5) = 10

c. 24 = 2x12 = xy = x √ � 3 y = 12 √ � 3

d. 9 = y √ � 3 9 _

√ � 3 = y

9 √ � 3

_ 3 = y

3 √ � 3 = yx = 2yx = 6 √ � 3

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 _ √ � 3

= x

30 √ � 3

_ 3 = x

10 √ � 3 = xStep 3 Find the length s of each side of the frame.s = 20 √ � 3 ≈ 34.6 cm

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. 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.


1. The � is an isosc. rt. �, which is a

45°-45°-90° �.x = 14 √ � 2


45°-45°-90° �.9 √ � 2 = x √ � 2 x = 9


45°-45°-90° �. 12 = x √ � 2

12 _ √ � 2

= x

12 √ � 2 _ 2 = x

x = 6 √ � 2

4. The sign forms a right �. Using the Pyth. Thm., we get

d = √ �� 19.5 2 + 19.5 2 d ≈ 27.6 in.

5. 6 = 2x3 = xy = x √ � 3 y = 3 √ � 3

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

6. 15 = x √ � 3

15 _ √ � 3

= x

15 √ � 3

_ 3 = x

5 √ � 3 = xy = 2xy = 10 √ � 3



8. 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.5(2.25) = 2x 5.625 = xStep 3 Find the length h of the longer leg.h = 5.625 √ � 3 ≈ 9.75 in.



45°-45°-90° �.

15 = x √ � 2

15 _ √ � 2

= x

15 √ � 2 _ 2 = x


45°-45°-90° �.

x = (4 √ � 2 ) √ � 2 = 8


45°-45°-90° �.18 √ � 2 = x √ � 2 18 = x

12. The tabletop is a 45°-45°-90° �. 48 = w √ � 2 48 √ � 2 = 2w w = 24 √ � 2 ≈ 33.9 in.

13. x = 2(24) = 48y = 24 √ � 3

14. 10 √ � 3 = 2x 5 √ � 3 = xy = x √ � 3

= (5 √ � 3 ) √ � 3 = 15

15. 2 = x √ � 3 2 √ � 3 = 3x

2 √ � 3

_ 3 = x

y = 2x = 4 √ � 3

_ 3

16a. The ramp forms a 30°-60°-90° �. Let the length of the ramp be x.

x = 2(4.5) = 9 ft

b. Length of the dog walk = x + 12 + x= 9 + 12 + 9 = 30 ft

17. 12 = a √ � 2 12 √ � 2 = 2a 6 √ � 2 = ab = a = 6 √ � 2 P = a + b + c

= 6 √ � 2 + 6 √ � 2 + 12= (12 + 12 √ � 2 ) in.

A = 1 _ 2 ab

= 1 _ 2 (6 √ � 2 ) (6 √ � 2 )

= 36 in. 2

18. 28 = 2a14 = ab = a √ � 3 = 14 √ � 3 P = a + b + c

= 14 + 14 √ � 3 + 28= (42 + 14 √ � 3 ) cm

A = 1 _ 2 ab

= 1 _ 2 (14) (14 √ � 3 )

= 98 √ � 3 cm 2

19. 18 = s √ � 2 18 √ � 2 = 2s 9 √ � 2 = sP = 4s

= 4 (9 √ � 2 ) = 36 √ � 2 m

A = s 2 = (9 √ � 2 )

2

= 81(2) = 162 m 2

20. h = ( s _ 2

) √ � 3

= 2 √ � 3 P = 3s

= 3(4) = 12 ft

A = 1 _ 2

sh

= 1 _ 2

(4) (2 √ � 3 )

= 4 √ � 3 ft 2

21. h = ( s _ 2 ) √ � 3

2h = s √ � 3 60 = s √ � 3 60 √ � 3 = 3s20 √ � 3 = s

P = 3s= 3 (20 √ � 3 ) = 60 √ � 3 yd

A = 1 _ 2 sh

= 1 _ 2 (20 √ � 3 ) (30)

= 300 √ � 3 yd 2

22. Let s be the leg length. 18 = s √ � 2 9 √ � 2 = sHyp.: 18 ÷ 1 _

2 = 36 nails

Legs: 2(9 √ � 2 ) ÷ 1 _ 4 = 72 √ � 2 ≈ 102 nails

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.

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).

25.

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

T = 3; PT = ST √ � 2 = ( √

�� 6 2 + 6 2 ) √ � 2 = √ �� 144 = 12;

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



26.

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 = 5 √ � 3 ; y = (y-coordinate of W ) + 5 √ � 3 = -4 + 5 √ � 3 . So P = (-1, -4 + 5 √ � 3 ) .

27.

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 ; y = (y-coordinate of Z ) - 12 √ � 3 = 10 - 12 √ � 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

b. IN = NL √ � 2 ≈ 453 mi

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

TEST PREP

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

32. B;24 = a √ � 2 a = 12 √ � 2 ≈ 17.0 in.

33. 32 = 2w w = 16 � = w √ � 3 = 16 √ � 3 A = �w

= 16 √ � 3 (16)= 443.4 in. 2


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

4 x

4 = 4xx = 1

35. Step 1 Identify the pattern.

The length of each hyp. is 2 _ √ � 3

times the length of

the previous hyp. The length of the first hyp. is 2.Step 2 Find x.

x = ( 2 _ √ � 3

) 4 (2) = 32 _

9

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

d = √

�� e 2 + f 2 = √ �� 1 + 2 = √ � 3

e = 2: f = 2 √ � 2 , so

d = √

�� e 2 + f 2 = √ �� 4 + 8 = √ �� 12 = 2 √ � 3

e = 3: f = 3 √ � 2 , so

d = √

�� e 2 + f 2 = √ �� 9 + 18 = √ �� 27 = 3 √ � 3

b. d = √

�� e 2 + f 2

= √

�� e 2 + 2 e 2

= √

�� 3 e 2 = e √ � 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 = 3DBProof: 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° � Theorem, AD = √ � 3 (DC) and DC = √ � 3 (DB). By subst., AD = √ � 3 ( √ � 3 (DB)) . This simplifies to AD = 3DB.

SPIRAL REVIEW

38. y = x 2 + 4x + 0= (x + 2 ) 2 + 0 - 2 2 = (x + 2 ) 2 - 4

Axis of symmetry: x = -2

39. y = x 2 - 10x - 2= (x - 5 ) 2 - 2 - 5 2 = (x - 5 ) 2 - 27

Axis of symmetry: x = 5

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.



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

∠ Bisector Theorem, �� QS is the bisector of ∠PQR.

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

Theorem, �� QS is the bisector of ∠PQR. So

m∠VTS = 180 - m∠QTV= 180 - (90 - m∠TQV)= 180 - (90 - m∠PQS)= 180 - (90 - 42) = 132°.

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

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.

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.

6. Let d be the distance from the theater to the zoo.d + 9 > 16 d > 16 - 9 = 7

9 + 16 > d 25 > d

Range of the distances: greater than 7 km and less than 25 km.

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 < CD4x - 13 < 15 4x < 28 x < 7

AB > 04x - 13 > 0 4x > 13 x > 3.25

3.25 < x < 7

10. x 2 = 5 2 + 9 2 x 2 = 106 x = √ 106

11. a 2 + 9 2 = 11 2 a 2 + 81 = 121 a 2 = 40 a = √ 40 = 2 √ 10 The side lengths do not form a Pythagorean triple, because2 √ 10 is not a whole number.

12. 10 + 12 = 22 > 16 �The side lengths can form a �. 16 2 � 10 2 + 12 2 256 � 100 + 144256 > 244The � 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.

15. x = 8 √ 2 16. 22 = x √ 2 22 √ 2 = 2x 11 √ 2 = x

17. 5 √ 3 = x √ 3 5 = xy = 2x

= 2(5) = 10

STUDY GUIDE: REVIEW, PAGES 366–369

1. equidistant 2. midsegment

3. incenter 4. locus

LESSON 5-1

5. BD = 2CD = 2(3.7) = 7.4

6. XY = YZ3n + 5 = 8n - 9 14 = 5n n = 2.8YZ = 8(2.8) - 9 = 13.4

7. HT = FT = 5.8

8. m∠MNV = m∠PNV 2z + 10 = 4z - 6 16 = 2z z = 8m∠MNP = 2m∠MNV

= 2[2(8°) + 10°] = 52°



9. The midpoint of −−

AB is (1, 0);slope of

−− AB = -10 _

10 = -1, so the slope of the

perpendicular bisector is 1;

the equation of the perpendicular bisector is y = x -1.

10. The midpoint of −−

XY is (4, 6);

slope of −−

XY = 8 _ 2 = 4, so the slope of the

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= 2(28.8) = 57.6

16. PH = JP = 46

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

21. DZ = 2 _ 3 DB

= 2 _ 3 (24.6) = 16.4

22. DB = 3ZB24.6 = 3ZB ZB = 8.2

23. EZ = 2ZC11.6 = 2ZC ZC = 5.8

24. EC = 3ZC = 3(5.8) = 17.4

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; −−

RS has slope 5 _ 5 = 1, so the equation of the altitude

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; −−

XZ has slope 6 _ -6

= -1, so the equation of the altitude

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).

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

3 (4 + 8 + 0)) = (3, 4)

LESSON 5-4

30. BC = 1 _ 2 XY

= 1 _ 2 (70.2) = 35.1

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

32. XC = 1 _ 2 XZ

= AB = 32.4

33. m∠BCZ = m∠ABC = 42°

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

35. m∠YXZ = m∠BCZ = 42°

36. V = (-1, -1); W = (6, 1); slope of −−−

VW = 2 _ 7 ;

slope of −−

GJ = 4 _ 14

= 2 _ 7 ; since the slopes are the

same, −−−

VW ‖ −−

GJ .

VW = √

2 2 + 7 2 = √ 53 ;

GJ = √

4 2 + 14 2 = 2 √ 53 , so VW = 1 _ 2 GJ.

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 x > 9

4.5 + 13.5 > x 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 ≯ 3zNo; 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.



42. Possible answer: Given: �ABCProve: �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 5-6

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

m∠GFH > 0 5n + 7 > 0 5n > -7 n > -1.4

46. XZ < JK4n - 11 < 39 4n < 50 n < 12.5

XZ > 04n - 11 > 0 4n > 11 n > 2.75

2.75 < n < 12.5

LESSON 5-7

47. a 2 + b 2 = c 2 2 2 + 6 2 = 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. a 2 + b 2 = c 2 x 2 + (4.5 ) 2 = (7.5 ) 2 x 2 = 36 x = 6The 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 = 40The side lengths form a Pythagorean triple because they are nonzero whole numbers that satisfy

a 2 + b 2 = c 2 .

51. 9 + 12 = 21 > 16 16 2 � 9 2 + 12 2 256 > 225The side lengths can form an obtuse �.

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

53. 1.5 + 3.6 = 5.1 > 3.9 3.9 2 � 1.5 2 + 3.6 2 15.21 = 15.21The side lengths can form a rt. �.

54. 2 + 3.7 = 5.7 > 4.1 4.1 2 � 2 2 + 3.7 2 16.81 < 17.69The side lengths can form an acute �.

LESSON 5-8

55. 45°-45°-90° �x = 26 √ 2

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

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

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

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

60. 30°-60°-90° � 14 = x √ 3 14 √ 3 = 3x

x = 14 √ 3

_ 3

y = 2x

= 2 ( 14 √ 3

_ 3

) = 28 √ 3

_ 3

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. AC = BC2n + 9 = 5n - 9 18 = 3n n = 6BC = 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

6. XW = 2 _ 3 XC

= 2 _ 3 (261) = 174

BW = 1 _ 2 ZW

= 1 _ 2 (118) = 59

BZ = 3BW = 3(59) = 177

7. −−

JK is vertical, so the equation of the horizontal altitude is y = 4;

slope of −−

KL is -6 _ 6 = -1, so the slope of the altitude

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).



8. PR = 1 _ 2 HJ

= QJ = 51GJ = 2PQ

= 2(74) = 148m∠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 .

12. AC + 114 > 247 AC > 133

114 + 247 > AC 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.

14. −−

DH � −−

KH , −−

HN � −−

HN , and DN < KN. m∠DHN < m∠KHN 4x - 10 < 24 4x < 34 x < 8.5

m∠DHN > 0° 4x - 10 > 0 4x > 10 x > 2.5

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.

16. 18 + 20 = 38 > 27 27 2 18 2 + 20 2 729 > 724The side lengths can form an obtuse �.

17. c 2 = 62 2 + 82 2 = 10,568 c = √ �� 10,568

≈ 102 ft 10 in.

18. 20 = x √ � 2 20 √ � 2 = 2x x = 10 √ � 2

19. 32 = 2x x = 16y = x √ � 3 = 16 √ � 3

20. 8 = x √ � 3 8 √ � 3 = 3x

x = 8 √ � 3

_ 3

y = 2x

= 2 ( 8 √ � 3

_ 3 ) =

16 √ � 3 _

3



CHAPTER Solutions Key 5 Properties and Attributes … Key 5 Properties and Attributes of Triangles...

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Transcript of CHAPTER Solutions Key 5 Properties and Attributes … Key 5 Properties and Attributes of Triangles...