CHAPTER
5
Solutions Key
Properties and Attributes of Triangles
−−
b. Since DG = GE and ⊥ DE, is the ⊥ bisector
−−
of DE by the Conv. of the ⊥ Bisector Thm.
1 DE
EF = _
2
_
EF = 1 (20.8) = 10.4
2
ARE YOU READY? PAGE 297
1. E
2. C
3. A
4. D
5. B
6. acute
7. right
8. acute
9. obtuse
2
10. 8 = 64
11. (-12) 2 = 144
49 = 7
12. √
13. - √
36 = -6
14. √
9 + 16 = √
25 = 5
15. √
100 - 36 = √
64 = 8
16.
√
81
9
81 = _
_
=_
25
17.
5
√
25
2a. WX = WZ
WX = 3.05
−−− −−
−−− −−
b. Since XW = ZW, XW ⊥ XY, and ZW ⊥ ZY, YW
bisects ∠XYZ by the Conv. of the ∠ Bisector Thm.
m∠XYZ = 2m∠WYZ
m∠XYZ = 2(63°) = 126°
2
3. By the Conv. of the ∠ Bisector Thm., QS bisects
∠PQR.
2 =2
4.
18. d + 5 < 1
d < -4
y
P
19.
x
-4 ≤ w - 7
3≤w
w≥3
Q
−−
Step 1 Graph PQ.
−−
−−
The ⊥ bisector of PQ is ⊥ to PQ at its midpoint
−−
Step 2 Find the midpoint of PQ.
+ (-4)
−−
+ 1 , 2_
= (3, -1)
midpoint of PQ = 5_
2
2
Step 3 Find the slope of the perpendicular bisector.
−− -4 - 2
-6 = _
3
=_
slope of PQ = _
2
1-5
-4
Since the slopes of ⊥ lines are opposite reciprocals,
2.
the slope of the ⊥ 1bisector is - _
3
Step 4 Use point-slope form to write an equation.
−−
2 and passes
The ⊥ bisector of PQ has slope - _
3
through (3, -1).
y - y 1 = m(x - x 1)
2 (x - 3)
y - (-1) = - _
3
_
y + 1 = - 2 (x - 3)
3
(
m
-2 > _
10
-20 > m
m < -20
20. -3s ≥ 6
s ≤ -2
21.
22. Let p and q represent the following:
p: Lines and m intersect.
q: Lines and m are not parallel.
Given: p → q, and p. So by the Law of Detachment,
q is true: Lines and m are not parallel.
23. Let p, q, and r represent the following:
−−
p : M is the midpoint of AB
q : AM = MB
1 AB and MB = _
1 AB
r : AM = _
2
2
Given: p → q and q → r. So by the Law of Syllogism,
−−
1 AB and
p → r : If M is the midpoint of AB, then AM = _
2
1 AB.
MB = _
2
)
THINK AND DISCUSS
1. Yes; no; since PY = QY = 3, Y is the midpoint
−−
of PQ, and thus by the def. of bisector, is a
−−
−−
bisector of PQ. If is the ⊥ bisector of PQ, then PX
would equal QX by the ⊥ Bisector Thm. However,
PX = 8.5 and QX = 8.4, so is not the ⊥ bisector
−−
of PQ.
2. No; although MJ = ML, to apply the Conv. of the
−− −−
∠ Bisector Thm., you must know that MJ ⊥ KJ
−− −−
and ML ⊥ KL.
5-1 PERPENDICULAR AND ANGLE
BISECTORS, PAGES 300–306
CHECK IT OUT!
1a. DG = EG
DG = 14.6
93
Holt McDougal Geometry
ʡ"ISECTOR
3.
#ONV )F A PT IS
EQUIDISTANT FROM
THE ENDPOINTS OF A
SEG THEN THE
PT IS ON THE
ʡBISECTOR OF A SEG
4HM )F A PT IS ON
THE ʡBISECTOR OF A
SEG THEN IT IS
EQUIDISTANT FROM
THE ENDPOINTS OF
THE SEG
Since the slopes of ⊥ lines are opposite reciprocals,
the slope of the ⊥ bisector is 1.
Step 4 Use point-slope form to write an equation.
−−−
The ⊥ bisector of MN has slope 1 and passes
through (-2, 1).
y - y 1 = m(x - x 1)
y - 1 = 1[x - (-2)]
y-1=x+2
Ȝ"ISECTOR
4HM )F A PT IS ON
THE BISECTOR OF AN
Ȝ THEN IT IS
EQUIDISTANT FROM
THE SIDES OF
THE Ȝ
#ONV )F A PT IN
THE INT OF AN ȜIS
EQUIDISTANT FROM
THE SIDES OF THE Ȝ
THEN THE PT IS ON
THE BISECTOR OF THE Ȝ
10.
EXERCISES
y
GUIDED PRACTICE
1. perpendicular bisector
−−
2. Since PS = QS and m ⊥ PQ, m is the ⊥ bisector
−−
of PQ by the Conv. of the ⊥ Bisector Thm.
PQ = 2QT
U
−−
Step 1 Graph UV.
−−
−−
The ⊥ bisector of UV is ⊥ to UV at its midpoint
−−
Step 2 Find the midpoint of UV .
−−
2 + 4 -6 + 0
midpoint of UV = _, _ = (3, -3)
2
2
Step 3 Find the slope of the perpendicular bisector.
−− 0 - (-6)
6 =3
slope of UV = _ = _
2
4-2
Since the slopes of ⊥ lines are opposite reciprocals,
1.
the slope of the ⊥ bisector is - _
3
Step 4 Use point-slope form to write an equation.
−−
1 and passes
The ⊥ bisector of UV has slope - _
3
through (3, -3).
y - y 1 = m(x - x 1)
1 (x - 3)
y - (-3) = - _
3
_
y + 3 = - 1 (x - 3)
3
3. SP = SQ
SP = 25.9
(
4. PS = QS
4a = 2a + 26
2a = 26
a = 13
So QS = 2(13) + 26 = 52.
5. AD = CD
AD = 21.9
−− −−
−− −−
6. Since AD = CD, AC ⊥ AB, and CD ⊥ BC, BD
bisects ∠ABC by the Conv. of the ∠ Bisector Thm.
1 m∠ABC
m∠CBD = _
2
1 (48°) = 24°
m∠CBD = _
2
−− −−
−− −−
7. Since DA = DC, AD ⊥ AB, and CD ⊥ BC, BD
bisects ∠ABC by the Conv. of the ∠ Bisector Thm.
m∠DBC = m∠DBA
10y + 3 = 8y + 10
2y + 3 = 10
2y = 7
7
y=_
2
7 + 3]° = 38°
So m∠DBC = [10 _
2
−− −−
8. The braces can be installed so that PK ⊥ JL,
−− −−
PM ⊥ NL, and PK = PM. Then by the Conv. of the
∠ Bisector Thm., P will be on the bisector of ∠JLN.
11.
x
K
−−
Step 1 Graph JK.
−−
−−
The ⊥ bisector of JK is ⊥ to JK at its midpoint
−−
Step 2 Find the midpoint of JK.
−−
-7 + 1 5 + (-1)
midpoint of JK = _, _ = (-3, 2)
2
2
Step 3 Find the slope of the perpendicular bisector.
−−
3
-1 - 5 = _
-6 = - _
slope of JK = _
4
8
1 - (-7)
Since the slopes of ⊥ lines are opposite reciprocals,
4.
the slope of the ⊥ bisector is _
3
Step 4 Use point-slope form to write an equation.
−−
4 and passes
The ⊥ bisector of JK has slope _
3
through (-3, 2).
y
(
x
)
y
J
()
V x
PQ = 2(47.7) = 95.4
9. M
N
−−−
Step 1 Graph MN.
−−−
−−−
The ⊥ bisector of MN is ⊥ to MN at its midpoint.
−−−
Step 2 Find the midpoint of MN.
(
)
−−−
-5 + 1 4 + (-2)
midpoint of MN = _, _ = (-2, 1)
2
2
)
y - y 1 = m(x - x 1)
4 [x - (-3)]
y-2=_
3
4 (x + 3)
y-2=_
3
Step 3 Find the slope of the perpendicular bisector.
−−−
-6 = -1
-2 - 4 = _
slope of MN = _
6
1 - (-5)
94
Holt McDougal Geometry
y_
2 - y1
x2 - x1
−−
-1 - 5 = _
-6 = –1
slope of XY = _
6
-1 - (-7)
Since the slopes of ⊥ lines are opposite reciprocals,
the slope of the ⊥ bisector is 1.
Step 4 Use point-slope form to write an equation.
−−
1 and passes
The ⊥ bisector of XY has slope - _
2
through (-2, -3).
y - y 1 = m(x - x 1)
y - 2 = 1[x - (-4)]
y-2=x+4
−−−
21. Step 1 Graph MN.
−−−
−−−
The ⊥ bisector of MN is ⊥ to MN at its midpoint
−−−
Step 2 Find the midpoint of MN.
x_
1 + x2 y
1 + y2
,_
2
2
−−−
-3 + 7 1 + (-5)
midpoint of MN = _ , _ = (2, -3)
2
2
Step 3 Find the slope of the ⊥ bisector.
y2 - y1
slope = _
x2 - x1
-5 - (-1)
−−− _
-4 = - _
2
slope of MN =
=_
5
10
7 - (-3)
Since the slopes of ⊥ lines are opposite reciprocals,
_.
the slope of the bisector is 5
2
Step 4 Use point-slope form to write an equation.
1 and passes through
The bisector of ⊥ has slope - _
2
(-2, -3).
y - y 1 = m(x - x 1)
_ (x - 2)
y - (-3) = 5
2
_ (x - 2)
y+3= 5
2
QS = QT
22.
PS = PT
6n - 3 = 4n + 14
3m + 9 = 5m - 13
2n - 3 = 14
9 = 2m - 13
2n = 17
22 = 2m
n = 8.5
11 = m
slope =
PRACTICE AND PROBLEM SOLVING
12. GJ = GK
GJ = 8.25
JG = KG
x + 12 = 3x - 17
12 = 2x - 17
29 = 2x
14.5 = x
So KG = 3(14.5) - 17 = 26.5.
−−
−−
14. Since GJ = GK and t ⊥ JK, t is the ⊥ bisector of JK
by the Conv. of the ⊥ Bisector Thm.
JK = 2JH
JK = 2(26.5) = 53
13.
15. RQ = TQ
RQ = 1.3
−− −−
−− −−
16. Since RQ = TQ, RQ ⊥ RS, and TQ ⊥ TS, SQ
bisects ∠RST by the Conv. of the Bisector Thm.
m∠RST = 2m∠RSQ
m∠RST = 2(58°) = 116°
(
17. m∠QSR = m∠QST
9a + 48 = 6a + 50
3a + 48 = 50
3a = 2
2 + 50 = 54°
m∠QST = 6 _
3
18. They can position Main St. so that the ∠ formed by
Elm St. and Main St. is to the ∠ formed by Grove
St. and Main St. Then by the ∠ Bisector Thm., every
point on Main St. will be equidistant from Elm St.
and Grove St.
−−
19. Step 1 Graph EF.
−−
−−
The ⊥ bisector of EF is ⊥ to EF at its midpoint
−−
Step 2 Find the midpoint of EF.
x1 + x2 _
y + y2
_
, 1
2
2
−−
-4 + 0 -7 + 1
midpoint of EF = _ , _ = (-2, -3)
2
2
Step 3 Find the slope of the perpendicular bisector.
y2 - y1
slope = _
x2 - x1
−− 1 - (-7) 8
slope of EF = _ = _
=2
0 - (-4) 4
Since the slopes of ⊥ lines are opposite reciprocals,
1.
the slope of the ⊥ bisector is - _
2
Step 4 Use point-slope form to write an equation.
−−
1 and passes
The ⊥ bisector of EF has slope - _
2
through (–2, –3).
y - y 1 = m(x - x 1)
1 [x - (-2)]
y - (-3) = -_
2
_
y + 3 = - 1 (x + 2)
2
−−
20. Step 1 Graph XY.
−−
−−
The ⊥ bisector of XY is ⊥ to XY at its midpoint
−−
Step 2 Find the midpoint of XY.
x_
1 + x2 y
1 + y2
,_
2
2
-7 + (-1) 5 + (-1)
−−
midpoint of XY = _ , _ = (–4, 2)
2
2
Step 3 Find the slope of the perpendicular bisector.
()
(
(
(
(
)
)
)
)
(
23. JK = LK
JK = 38
25.
MK
ML + LK
ML
ML
)
24. GN = 2GZ
GN = 2(36) = 72
= HK
= HJ + JK
= HJ
= 38
27. JL = 2LX
JL = 2(12) = 24
26. HY = MY
HY = 24
28.
NK
NM + ML + LK
NM + 38 + 38
NM
= GK
= 114
= 114
= 38
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.
)
95
Holt McDougal Geometry
31.
Statements
bisects ∠QPR,
1. PS
−−
−−
, SR ⊥ PR
SQ ⊥ PQ
2. ∠QPS ∠RPS
3.
4.
5.
6.
7.
8.
∠SQP and ∠SRP are rt. .
∠SQP ∠SRP
−− −−
PS PS
PQS PRS
−− −−
SQ SR
SQ = SR
34. In the construction of the perpendicular bisector
−−
of AB, the same compass setting is used to draw an
arc from each end point of the segment. So in the
diagram, AX = BX and AY = BY. By the Converse
of the Perpendicular Bisector Theorem, both X and
−−
Y lie on the perpendicular bisector of AB. So is the
−−
perpendicular bisector of AB.
Reasons
1. Given
2. Def. of ∠
bisector
3. Def. of ⊥
4. Rt. ∠ Thm.
5. Reflex. Prop. of 6. AAS
7. CPCTC
8. Def. of segs.
Ű
X
A
TEST PREP
35. D;
−−
J is on the perpendicular bisector of XY, so by the
Perpendicular Bisector Theorem, JX = JY.
36. G;
37. Possible answer: All locations that are equidistant
from Park St. and Washington Ave. lie on the
bisector of the ∠ formed by the 2 streets. All
locations that are equidistant from the museum
and the library lie on the perpendicular bisector of a
segment formed by the museum and the library. So
the visitor center should be built at point V, where
the angle bisector and the perpendicular bisector
intersect.
P
C
B
−−
33a. Step 1 Graph AC.
−−
−−
The ⊥ bisector of AC is ⊥ to AC at its midpoint
−−
Step 2 Find the midpoint of AC.
x_
y + y2
1 + x2 _
, 1
2
2
−−
-3 + 3 -2 + 6
midpoint of AC = _ , _ = (0, 2)
2
2
Step 3 Find the slope of the perpendicular
bisector.
y2 - y1
slope = _
x2 - x1
- (-2)
−− 6
_ = 4_
slope of AC = _ = 8
6
3
3 - (-3)
Since the slopes of ⊥ lines are opposite
3.
reciprocals, the slope of the ⊥ bisector is - _
4
Step 4 Use point-slope form to write an equation.
−−
_ and passes
The ⊥ bisector of AC has slope - 3
4
through (0, 2).
y - y 1 = m(x - x 1)
_ (x – 0)
y-2=- 3
4
_
y = - 3x + 2
4
b. There are 2 points on the ⊥ bisector that are 4 mi
−−
dist. from the midpoint of AC.
−−
c. Distance of warehouse from midpoint of AC = 4
mi.
−−
Distance of midpoint of AC from A
(
(
)
2
2
= 3 + 4 = 5 mi.
Distance of warehouse from A =
B
Y
32. Possible answer: By stating that the point must be
in the int. of the ∠, the thm. implies that it must be
in the same plane as the ∠. It is possible for a point
to be equidistant from the sides of an ∠ but to lie in
a different plane. In the diagram, ∠ABC is in plane
Z, and P is equidistant from the sides of ∠ABC, but
P does not lie in plane Z. Thus P cannot be on the
bisector of the ∠, because the bisector must lie in
the same plane as the ∠.
A
P
CHALLENGE AND EXTEND
are both
and from P to BC
38a. The dist. from P to BA
, and
and BC
2 √5. So P is equidistant from BA
therefore by the Converse of the Angle Bisector
Theorem, P is on the bisector of ∠ABC.
)
b. Possible answer: y = 3x - 6.
39. The distance of a point (x, y) from x-axis is ⎪y⎥, and
its distance from y-axis is ⎪x⎥. So locus is ⎪y⎥ = ⎪x⎥,
or the lines y = x and y = -x.
40.
1.
2.
3.
4.
5.
Statements
−−
−−
, VZ ⊥ YZ
,
VX ⊥ YX
VX = VZ
∠VXY and ∠VZY are
rt. .
−− −−
YV YV
YXV YZV
∠XYV ∠ZYV
bisects ∠XYZ.
6. YV
Reasons
1. Given
2. Def. of ⊥
3. Reflex. Prop. of 4. HL
5. CPCTC
6. Def. of ∠ bisector
42 + 5 2 ≈ 6.4 mi.
96
Holt McDougal Geometry
−−
−−
41. It is given that KN is the perpendicular bisector of JL
−−
and LN is the perpendicular bisector of
−−
KM. By the Perpendicular Bisector Theorem, JK =
KL and KL = ML. Thus JK = ML by the Trans. Prop.
−− −−
of =. By the definition of segs., JK ML. By the
Seg. Add. Post., JR + RL = JL and .MT + TK =
MK. By the definition of the perpendicular bisector,
−−
R is the midpoint of JL and T is the midpoint of
−−
−− −−
−− −−
MK. Thus JR RL and MT TK. By the definition
of cong segs., JR = RL and MT = TK. By Subst.,
JR + JR = JL and MT + MT = MK. It is given
−− −−
that JR MT. So JR = MT by definition of segs.
By Subst., JR + JR = MK. By the Trans. Prop. of =,
−− −−
JL = MK, so JL MK by the definition of segs.
−− −−
By the Reflex. Prop. of , JM JM. Therefore
JKM MLJ by SSS, and ∠JKM ∠MLJ
by CPCTC.
2.
y
x O
H x
y G
Step 1 Graph the .
Step 2 Find equations for two perpendicular
bisectors. Since two sides of lie along the axes,
use the graph to find the perpendicular bisectors of
these two sides. the perpendicular bisector of GO
is y = -4.5, and the perpendicular bisector of OH
is x = 4.
Step 3 Find the intersection of the two equations.
The lines y = -4.5 and x = 4 intersect at (4, -4.5),
the circumcenter of GOH.
SPIRAL REVIEW
3a. X is the incenter of PQR. By the Incenter
Theorem, X is euqidistant from the sides of PQR.
−−
The distance from X to PR is 19.2, so
−−
the distance from X to PQ is also 19.2.
42. C
2
4-2 =_
= _
43. slope of RS
1+4
5
-5 + 1 _
= _
=2
slope of VT
-7 - 3
5
The slopes are the same, so the lines are parallel.
-5 - 2 = _
7
= _
44. slope of RV
3
-7 + 4
5
-1 - 4 = -_
= _
slope of ST
2
3-1
The slopes are not the same, so the lines are not
parallel. The product of the slopes is not -1, so the
lines are not perpendicular.
b. m∠PRQ = 2m∠PRX
m∠PRQ = 2(12°) = 24°
m∠RQP + m∠PRQ + m∠QPR = 180°
52 + 24 + m∠RQP = 180°
m∠RQP = 104°
1 m∠RQP
m∠PQX = _
2
1 (104°) = 52°
m∠PQX = _
2
4. By the Incenter Theorem, the incenter of a is
equidistant from the sides of the . Draw the formed by the streets and draw the ∠ bisectors
to find the incenter, point M. The city should place
the monument at point M.
3
-1 - 2 = -_
= _
45. slope of RT
7
3+4
-5
2
7
_
_
=
slope of VR
=
3
-7 + 4
The product of the slopes is -1, so the lines are
perpendicular.
-9 - (-1)
46. m = _ = -8 47. y - y 1 = m(x - x 1)
2-1
y + 15 = -0.5(x - 10)
y - y 1 = m(x - x 1)
y + 15 = -0.5x + 5
y + 1 = -8(x - 1)
1 x - 10
y = -8x + 7
y = -_
2
5
0
5
_
_
48. m =
=
4
0 - (-4)
y = mx + b
5x + 5
y=_
4
M
THINK AND DISCUSS
1. Possible answer:
5-2 BISECTORS OF TRIANGLES,
PAGES 307–313
CHECK IT OUT!
2. Q; P. Possible answer: the incenter is always inside
the , so Q cannot be the incenter. Therefore
P must be the incenter, and Q must be the
circumcenter.
1a. GM = MJ = 14.5
b. GK = KH = 18.6
c. Z is circumcenter of GHJ, By the Circumcenter
Theorem, Z is equidistant from the vertices of
GHJ.
JZ = GZ = 19.9
97
Holt McDougal Geometry
3.
#IRCUMCENTER
9. F is the incenter of CDE. By the Incenter
Theorem, F is equaldistant from the sides of
−−
CDE. The distance from F to DE is 42.1, so the
−−
distance from F to CD is also 42.1.
)NCENTER
$EFINITION
4HE PT OF
CONCURRENCY OF
THE ʡBISECTORS
4HE PT OF
CONCURRENCY OF
THE ȜBISECTORS
$ISTANCE
%QUIDISTANT FROM
THE VERTICES OF
THE ̱
%QUIDISTANT FROM
THE SIDES OF
THE ̱
,OCATION )NSIDE
/UTSIDE OR /N
#AN BE INSIDE
OUTSIDE OR ON
THE ̱
)NSIDE THE ̱
10. m∠DCE = 2m∠FCD
m∠DCE = 2(17°) = 34°
m∠DCE + m∠CDE + m∠CED = 180°
34 + 54 + m∠CED = 180°
m∠CED = 92°
1
_
m∠FED = m∠CED
2
1 (92°) = 46°
m∠FED = _
2
11. The largest possible in the int. of the is its
inscribed , and the center of the inscribed is the
incenter. Draw the and its ∠ bisectors. Center the
at E, the point of concurrency of the ∠ bisectors.
EXERCISES
GUIDED PRACTICE
1. They do not intersect at a single point.
2. circumscribed about
3. N is the circumcenter of PQR. By the
Circumcenter Theorem, N is equidistant from
vertices of PQR. NR = NP = 5.64
x
4. RV = PV = 5.47
O
L
x x
Step 1 Graph .
Step 2 Find equations for the two perpendicular
bisectors. Since the two sides of the lie along
the axes, use the graph to find the perpendicular
bisectors of these two sides. The perpendicular
bisector of KO is y = 6, and the perpendicular
bisector of OL is x = 2.
Step 3 Find the intersection of the two equations.
The lines y = 6 and x = 2 intersect at (2, 6), the
circumcenter of KOL.
x 15. AY = YB = 63.9
19. m∠TSR = 2m∠JSR
m∠TSR = 2(14°) = 28°
m∠TSR + m∠SRT + m∠RTS = 180°
28 + 42 + m∠TSR = 180°
m∠TSR = 110°
1
_
m∠RTJ = m∠TSR
2
_
m∠RTJ = 1 (110°) = 55°
2
20. By the Circumcenter Theorem, the circumcenter of
the is equidistant from the vertices. Draw the formed by the cities, and draw the perpendicular
bisectors of the sides. The main office should be
located at M, the circumcenter.
x
O
14. DB = AD = 62.8
18. J is the incenter of RST. By the Incenter
Theorem, J is equaldistant from the sides of RST.
−−
The distance from J to ST is 8.37, so the distance
−−
from J to RS is also 8.37.
A
13. YC = YB = 63.9
17. Step 1 Write equations of the perpendicular bisectors
−−
−−−
of OV and OW.
−−
The perpendicular bisector of OV is y = 9.5; the
−−
perpendicular bisector of WO is x = -1.5.
Step 2 Find the circumcenter of the .
The circumcenter is at the intersection of the
perpendicular bisectors, (-1.5, 9.5).
y
8.
12. CF = FA = 59.7
16. Step 1 Write equations of the perpendicular bisectors
−−−
−−
of MO and NO.
−−−
The perpendicular bisector of MO is x = -2.5; the
−−
perpendicular bisector of NO is y = 7.
Step 2 Find the circumcenter of the .
The circumcenter is at the intersection of the
perpendicular bisectors, (-2.5, 7).
ÓäÂ
PRACTICE AND PROBLEM SOLVING
y
K
y xäÂ
6. N is the circumcenter of PQR. By the
Circumcenter Theorem, N is equidistant from
vertices of PQR.
QN = NP = 5.64
7.
*
+
5. TR = QT = 3.95
y B
Step 1 Graph the .
Step 2 Find equations for the two perpendicular
bisectors. Since the two sides of the lie along
the axes, use the graph to find the perpendicular
bisectors of these two sides. The perpendicular
bisector of AO is x = -3.5, and the perpendicular
bisector of OL is y = -5.
Step 3 Find the intersection of the two equations.
The lines x = -3.5 and y = -5 intersect at
(-3.5, -5), the circumcenter of AOB.
98
Holt McDougal Geometry
21. Possible answer: if ∠JML is a rt. ∠, then m∠MJL
+ m∠MLJ = 90° because the acute of a rt. −−
are comp. Since M is the incenter of JKL, JM
−−
and LM are ∠ bisectors of JKL. So by the def.
of ∠ bisector, m∠KJL = 2m∠MJL and m∠KLJ =
2m∠MLJ. By subst., m∠KJL + m∠KLJ = 2(m∠MJL
+ m∠MLJ) = 2(90°) = 180°. But by the Sum
Theorem, m∠K = 180° - (m∠KJL + m∠KLJ) =
180° - 180° = 0°. This would mean that JKL is
not a . Therefore ∠JML cannot be a rt. ∠.
36.
4.
5.
6.
7.
1. Given
2. Def. of ∠
bisector
3. Reflex. Prop.
of 4. SAS
5. CPCTC
6. Lin. Pair Thm.
7. supp.
→ rt. 8. Def. of rt. ∠
PQS RQS
∠PSQ ∠RSQ
∠PSQ and ∠RSQ are supp.
∠PSQ and ∠RSQ are rt. .
8. ∠PSQ = ∠RSQ = 90°
−−
9. QS ⊥ PR
−− −−
10. PS RS
−−
11. S is midpoint of PR.
S
H
22. The angle bisector; m∠BAE = m∠EAC
−− −−
23. The perpendicular bisector; AD = BD, AD ⊥ DG
−− −−
and BD ⊥ DG
9. Def. of ⊥
10. CPCTC
11. Def. of
midpoint
12. Def. of the
perpendicular
bisector
12. QS is the perpendicular
−−
bisector of PR.
24. The angle bisector; m∠ABG = m∠GBC
−−
−−
25. The angle bisector; since AE and BG are ∠
−−
bisectors, R is the incenter. CR intersects the
incenter, so it is an ∠ the bisector.
37a. The new store is at the circumcenter of ABC.
−−
The perpendicular bisector of AB is x = 4.
−− 3
−−
3 .
; the midpoint of AC is 2, _
The slope of AC is _
4
2
−−
The perpendicular bisector of AC is
3 = -_
4 (x - 2).
y-_
3
2
3 = -_
4 (4 - 2)
At the intersection, x = 4 and y - _
3
2
8
9 - 16 = -_
7.
= -_, so y = _
3
6
6
7 .
The new store is located at 4, -_
6
b. outside, since y > 0 for all int. poins. of the , but
7 <0
-_
6
c. distance from each store = distance from store C
7 = 4_
1 ≈ 4.2 mi
= 3 - -_
6
6
38. Possible answers: Similarities: Both are circles.
Both intersect the triangle in exactly 3 points.
Differences: The inscribed circle is smaller than
the circumscribed circle. Except for the points of
intersection, the inscribed circle lies inside the
triangle, while the circumscribed circle lies outside.
The center of the inscribed circle always lies inside
the triangle, while the center of the circumscribed
circle may be inside, outside, or on the triangle.
The center of the inscribed circle is the point of
concurrency of the angle bisectors, while the center
of the circumscribed circle is the point of concurrency
of the perpendicular bisectors.
( )
27. neither
28. never
29. sometimes
Reasons
−− −−
3. QS QS
P
26. neither
Statements
−−
−− −−
1. QS bisects ∠PQR, PQ RQ.
2. ∠PQS ∠RQS
30. sometimes
(
31. never
( )
32. sometimes
−−
−−
33. The slope of OA is 2; the midpoint of OA is (2, 4).
−−
The perpendicular bisector of OA is
1 (x - 2).
y - 4 = -_
2
−−
The perpendicular bisector of OB is x = 4.
1 (4 - 2) =
At the intersection, x = 4 and y - 4 = - _
2
-1, so y = 3. The circumcenter is at (4, 3).
−−
34. The perpendicular bisector of OY is y = 6.
−−
−−
The slope of OZ is = 1; the midpoint of OZ is (3, 3).
−−
The perpendicular bisector of OZ is
y - 3 = -(x - 3).
At the intersection, y = 6 and 6 - 3 = 3 = -x + 3, so
x = 0. The circumcenter is at (0, 6).
35a. ∠ Bisector Theorem
)
39a. Check students’ constructions.
b. Check students’ constructions.
TEST PREP
40. B;
PX = PY by the Incenter Theorem.
41. F;
m = 1, y + 2 = x - 5, or y = x - 7.
b. the bisector of ∠B
c. PX = PZ
99
Holt McDougal Geometry
−−
1-4 =_
1 = opposite reciprocal of
52. slope of MY = _
2
-4 - 2
2; so Y is on the perpendicular bisector
42. 14.75
KN = MN
5z - 4 = z + 11
4z = 15
z = 3.75
LN = MN = 3.75 + 11 = 14.75
−− -8 - 4
53. slope of MZ = _
= -3 ≠ opposite reciprocal
-2 - 2
of 2;
so Z is not on the ⊥ bisector
CHALLENGE AND EXTEND
43a. Possible answer:
−−
Given: M is the midpoint of QR.
Prove: PM = QM = RM
Proof: The coordinates of M are
0 + 2a _
2b + 0
_
= (a, b).
,
2
2
(
5-3 MEDIANS AND ALTITUDES OF
TRIANGLES, PAGES 314–320
CHECK IT OUT!
)
1a.
By the Distance Formula,
PM = √
(a - 0) 2 + (b - 0) 2
2
2
2. 3; 4; possible answer: the x-coordinate of the
centroid is the average of the x-coordinates of
the vertices of the , and the y-coordinate of the
centroid is the average of the y-coordinates of the
vertices of the .
−−
3. Possible answer: An equation of altitude to JK is
1 x + 3. It is true that 4 = -_
1 (-2) + 3, so
y = -_
2
2
(-2, 4) is a solution of this equation. Therefore this
altitude passes through the orthocenter.
and
- 2a) 2 + (b - 0) 2
√(a
2
(-a) 2 + b 2 = a + b2.
= √
RM =
Therefore, PM = QM = RM.
b. Possible answer: The circumcenter of a rt. is the
midpoint of the hyp.
44. Let C be the circumcenter of the . Given: AC = 28
1 AC.
cm; so by the properties of 30-60-90 , BC = _
2
So AB = AC + BC
3 AC
=_
2
3 (28) = 42 cm.
=_
2
SPIRAL REVIEW
t =_
10
45. _
26
65
65t = 260
t= 4
47.
_2 KW + ZW = KW
3
1 KW
ZW = _
3
1 KW
7=_
2 LX
b. LZ = _
3
2 (8.1)
=_
3
= 5.4
3
21 = KW
a +b ,
2
2
QM = √(a - 0) + (b - 2b)
2
a 2 + (-b) 2 = a + b2,
= √
=
KZ + ZW = KW
THINK AND DISCUSS
1. Possible answer: is isosc.
B
2.5 = _
6
46. _
x
1.75
2.5x = 10.5
x = 4.2
A
D
C
2. Possible answer: is a rt. .
K
420 = _
7
_
y
2
840 = 7y
y = 120
J
3. The ratio of the length of the longer segment to the
length of the shorter segment is 2 : 1.
48. m∠AFB + m∠BFE = 180°
55 + m∠BFE = 180°
m∠BFE = 125°
4.
49. m∠AFB + m∠BFD + m∠DFE = 180°
55 + 90 + m∠DFE = 180°
m∠DFE = 35°
So m∠BFC = m∠DFE = 35°
50. m∠BFC + m∠CFE = m∠BFD + m∠DFE
35 + m∠CFE = 90 + 35
m∠CFE = 90°
−− _
−−
51. slope of ST = 8 = -2; midpoint of ST = M(2, 4)
-4
−− 4 - 3
1 = opposite reciprocal
= -_
slope of MX = _
2
2-0
of 2; so X is on the perpendicular bisector
L
$EFINITION
,OCATION )NSIDE
/UTSIDE OR /N
#ENTROID
/RTHOCENTER
4HE PT OF
CONCURRENCY OF
THEMEDIANS
4HE PT OF
CONCURRENCY OF
THEALTITUDES
)NSIDE THE ̱
#AN BE INSIDE
OUTSIDE OR ON
THE ̱
EXERCISES
GUIDED PRACTICE
1. centroid
100
2. altitude
Holt McDougal Geometry
2 VX
3. VW = _
3
2 (204) = 136
=_
3
5.
1 VX
4. WX = _
3
1 (204) = 68
=_
3
2 RY
RW = _
3
2 RY
104 = _
3
_3 (104) = RY
2
RY = 156
9.
)
)
x K
M
10.
W
P
y
Q
x R
y x
x
Step 1 Graph the .
Step 2 Find an equation of the line containing the
−− −−
altitude from P to QR. QR is horizontal, the altitude
is vertical, so the equation is x = -5.
Step 3 Find an equation of the line containing the
−−
altitude from Q to PR.
−−
5 - 8 = -1.
Slope of PR = _
-2 + 5
Equation is y - 5 = x - 4, or y = x + 1.
Step 4 Solve the system to find the coordinates of
the orthocenter.
x = -5 and y = -5 + 1 = -4.
The coordinates of the orthocenter are (-5, -4).
Step 1 Graph the .
Step 2 Find an equation of the line containing the
−−
−−
altitude from L to KM. Since KM is horizontal, the
altitude is vertical, so the equation is x = 4.
Step 3 Find an equation of the line containing the
−−
altitude from K to LM.
-2 - 6 = -2.
Slope of LM = _
8-4
1 (x - 2).
Equation is y + 2 = _
2
Step 4 Solve the system to find the coordinates of
the orthocenter.
1 (4 - 2) = 1, so y = -1.
x = 4 and y + 2 = _
2
The coordinates of the orthocenter are (4, -1).
Step 1 Graph the .
Step 2 Find an equation of the line containing the
−−
−−
altitude from W to UV. Since UV is vertical, the
altitude is horizontal, so the equation is y = -3.
Step 3 Find an equation of the line containing the
−−−
altitude from U to VW.
−−− -3 - 6
Slope of VW = _
= -1.
5+4
Equation is y + 9 = x + 4, or y = x - 5.
Step 4 Solve the system to find the coordinates of
the orthocenter.
y = -3 and -3 = x - 5, so x = 2.
The coordinates of the orthocenter are (2, -3).
x
U
x ÞÊ ÊÓÊÊÚÊ­ÝÓ®
Ó
y x x
1 RW
6. WY = _
2
1 (104) = 52
=_
2
7. 1 Understand the Problem
Answer will be the coordinates of the centroid of
the . Important information is the location of
vertices, A(0, 2), B(7, 4), and C(5. 0).
2 Make a Plan
The centroid of the is the point of intersection of the
three medians. So write the equations for two medians
and find their point of intersection.
3 Solve
−−
Let M be the midpoint of AB and N be the midpoint
−−
of BC.
0+7 2+4
M = _ , _ = (3.5, 3)
2
2
5+7 _
0+4
_
= (6, 2)
N=
,
2
2
−−
AN is horizontal. Its equation is y = 2.
−−−
3 - 0 = -2. Its equation is
Slope of CM = _
3.5 - 5
y = -2(x - 5).
At the centroid, y = 2 = -2(x - 5), so
x = 5 + (-1) = 4. The coordinates of the centroid
are D(4, 2).
4 Look Back
−−
−−
Let L be the midpoint of AC. Equation for BL is
2
_
y - 4 = (x - 7), which intersects y = 2 at (4, 2).
3
y
8.
L
£
(
(
y
V
11.
y
x y D
E
x
C
Step 1 Graph the .
Step 2 Find an equation of the line containing the
−− −−
altitude from E to CD. CD is vertical, the altitude is
horizontal, so the equation is y = 2.
Step 3 Find an equation of the line containing the
−− −−
altitude from C to DE. DE is horizontal, altitude is
vertical, so the equation is x = -1.
Step 4 y = 2 and x = -1.
The coordinates of the orthocenter are (-1, 2).
101
Holt McDougal Geometry
PRACTICE AND PROBLEM SOLVING
20. Step 1 Find an equation of the line containing the
−−
altitude through A. BC is vertical, the altitude is
horizontal, so the equation is y = -3.
Step 2 Find an equation of the line containing the
altitude through C.
−− 5 + 3
Slope of AB is _ = 2.
8-4 1
(x - 8).
Equation is y + 8 = -_
16. Support should be attached at the centroid. Equation
2
of the median through (4, 0) is x = 4. The median
Step 3 Find the coordinates of the orthocenter.
through (0, 10) also passes through
1 (x - 8),
y = -3, so -3 + 8 = 5 = -_
2
0 + 14
4+8 _
10
7
1
_
_
_
or x = -10 + 8 = -2.
,
=- .
= (6, 7), and has slope
2
2
2
0-6
The coordinates are (-2, -3).
1x +
Equation of the second median is y = -_
3 GP
1 GP
21. GL = _
22. PL = _
2
2
2
1
_
10. At intersection, x = 4, so y = - (4) + 10 = 8.
3 (8) = 12
1 (8) = 4
2
=_
=_
2
2
The coordinates of the centroid are (4, 8).
1 HC
12. PC = _
3
1 (10.8) = 3.6
=_
3
14. JA = 3PA
= 3(2.9) = 8.7
(
2 Hp
13. HC = _
3
2 (10.8) = 7.2
=_
3
15. JP = 2PA
= 2(2.9) = 5.8
)
17. Step 1 Find an equation of the line containing the
−−
altitude through X. YZ is vertical, the altitude is
horizontal, so the equation is y = -2.
Step 2 Find an equation of the line containing the
altitude through Z.
−− 10 + 2
3.
Slope of XY is _ = _
2
6+2
2 (x - 6).
Equation is y + 6 = - _
3
Step 3 Find the coordinates of the orthocenter.
2 (x - 6),
y = -2, so -2 + 6 = 4 = - _
3
or x = -6 + 6 = 0.
The coordinates are (0, -2).
18. Step 1 Find an equation of the line containing the
−−
altitude through J. GH is horizontal, the altitude is
vertical, so the equation is x = 4.
Step 2 Find an equation of the line containing the
altitude through H.
−− -1 - 5
Slope of GJ is _
= -1.
4+2
Equation is y - 5 = x - 6.
Step 3 Find the coordinates of the orthocenter.
x = 4, so y - 5 = 4 - 6, or y = 5 - 2 = 3.
The coordinates are (4, 3).
19. Step 1 Find an equation of the line containing the
−−
altitude through T. RS is horizontal, the altitude is
vertical, so the equation is x = -2.
Step 2 Find an equation of the line containing the
−−
altitude through R. ST is vertical, the altitude is
horizontal, so the equation is y = 9.
Step 3 Find the coordinates of the orthocenter.
x = -2 and y = 9.
The coordinates are (-2, 9).
23. HL = LJ = 5
−−
−− −−
−−
24. GL is the perpendicular bisector of HJ, so GJ GH.
GJ = GH
= 2GK
= 2(6.5) = 13
25. P = GJ + GH + HJ
= 2GH + 2LJ
= 2(13) + 2(5) = 36 units
1 (HJ)(GL)
26. A = _
2
1 (10)(12) = 60 square units
=_
2
)
(
1 (8 + 2 + 5), _
1 (-1 + 7 - 3) = (5, 1)
28. G = ( _
)
3
3
1 (0 + 14 + 16), _
1 (-4 + 6 - 8) = (10, -2)
27. G = _
3
3
29. PZ = 2ZX
= 2(27) = 54
30. PX = 3ZX
= 3(27) = 81
31. Step 1 Find n.
2n + 17 = 54
2n = 54 - 17
n = 18.5
Step 2 Find QZ.
QZ = 4n - 26
= 4(18.5) - 26 = 48
1 (QZ)
32. YZ = _
2
1 (48) = 24
=_
2
33. Possible answer: the
perpendicular bisector
of base; the bisector of
vertex ∠; the median to
the base; the altitude to
the base
102
Holt McDougal Geometry
34. sometimes
TEST PREP
35. always
41. D
42. G;
I, III true since incenter,
centroid always inside
II false since obtuse
43. D
CHALLENGE AND EXTEND
36. never
44a. Possible answer: ABC is equil., and is the
−−−
perpendicular bisector of BC. Since ABC is
−− −−
equil., AB AC by definition. So AB = AC
by the definition of segs. Therefore by the
Converse of the Perpendicular Bisector Theorem,
A is on line . Similarly, B is on the perpendicular
−−
bisector of AC, and C is on the perpendicular
−−
bisector of AB.
37. always
B
38.
Statements
−−
−−
1. PS and RT are medians
−− −−
of PQR. PS RT
2. PS = RT
2 RT
2 PS = _
3. _
3
3
2 PS, RZ = _
2 RT
4. PZ = _
3
3
5. PZ = RZ
−− −−
6. PZ RZ
7. ∠SPR ∠TRP
−− −−
8. PR PR
9. PTR RSP
10. ∠QPR ∠QRP
−− −−
11. PQ RQ
12. PQR is an isosc .
Reasons
D
1. Given
2. Def. of segs.
3. Mult. Prop. of =
5. Subst.
6. Def. of segs.
7. Isosc. Thm.
8. Reflex. Prop.
of 9. SAS
10. CPCTC
11. Con. of Isosc. Theorem
12. Def. of isosc. b. DG =
) (
(3) (3)
_8 2 + _8 2
8 √2
≈ 3.8 mi
=_
3
−−
c. Perpendicular from G crosses EF at H(4, 4),
2
2
4 + _
4
distance = _
3
3
4 √2
≈ 1.9 mi
=_
3
() ()
)
A
C
b. Possible answer: By the definition of the
−− −−
perpendicular bisector, BD CD. So D is the
−−
−−
midpoint of BC by definition, and AD is a median
of ABC by the definition of median. Therefore
contains the median of ABC through A. Also
−−
by the definition of the perpendicular bisector, AD
−−
−−
⊥ BC. So AD is the altitude of ABC by the
definition. Therefore contains the altitude of
ABC through A. Again by the definition of the
−− −− −− −−
perpendicular bisector, BD CD. AB AC by the
−− −−
definition of equil. , and AD AD by the Reflex.
Prop. of . So ABD ACD by SSS. Then
−−
∠DAB ∠DAC by CPCTC, and AD is the bisector
of ∠BAC by the definition of ∠ bisector. Therefore
contains the ∠ bisector of ABC through A. The
same reason can be applied to the other two ⊥
bisectors.
4. Centroid Thm.
39. Possible answer: The centroid of a is also called
its center of gravity because the weight of the shape is evenly distributed in every direction from
this point. This means the shape will rest in a
horizontal position when supported at this point.
1 (0 + 0 + 8), _
1 (0 + 8 + 0) = 2_
2 , 2_
2
40a. G = _
3
3
3 3
(
Ű
c. Possible answer: The perpendicular bisectors of
a are concurrent at the circumcenter, and the
∠ bisectors are concurrent at the incenter. The
medians of a are concurrent at the centroid,
and the altitudes of a are concurrent at the
orthocenter. But in an equil. , the perpendicular
bisector through a given vertex also contains the
∠ bisector, the median, and the altitude through
that vertex. So the points of concurrency must all
be the same point That is, the circumcenter, the
incenter, the centroid, and the orthocenter in an
equil. are the same point.
c ; slope of ST = _
c ;
45a. slope of RS = _
b
b-a
slope of RT = 0.
−−
−−
b
b. Since ⊥ RS, slope of = -_. Since m ⊥ ST,
c
−−
b-a
a - b . Since n ⊥ RT
slope of m = -_ = _
, n is
c
c
a vertical line, and its slope is undefined.
103
Holt McDougal Geometry
c. An equation of is
b
y - 0 = - _ (x - a)
c
b
ab
_
y=- x+_
c
c
1 AE
3. HF = _
2
1 (1550) = 775
=_
2
The distance she measures between H and F
is 775 m.
An equation of m is
a - b (x - 0)
y-0=_
c
a
b x
_
y=
c
An equation of n is x = b.
(
ab - b
d. b, _
c
2
THINK AND DISCUSS
)
−−
1. The endpoints of XY are not the midpoints of
the sides of the .
e. Since the equation of line n is x = b and the
x-coordinate of P is b, P lies on n.
2.
f. Lines , m, and n are concurrent at P.
$EFINITION A SEG JOINING
THE MDPTS OF SIDES
OF A ̱
4RIANGLE
-IDSEGMENT
SPIRAL REVIEW
46. Let x, y be prices of peanuts and popcorn.
x = y + 0.75 or y = x - 0.75
5x + 3y = 21.75
5x + 3(x - 0.75) = 21.75
5x + 3x - 2.25 = 21.75
8x = 24
x=3
Price of peanuts is $3.00.
47. F; Possible answer: a rectangle with width 5 and
length 8.
.ONEXAMPLE
%XAMPLE
B
D
A
B
E
-IDSEGMENT
D
C
E
A
C
EXERCISES
GUIDED PRACTICE
49. KL = 2KP
= 2(7.0) = 14.0
48. T
0ROPERTIES ȡ TO THE
THIRD SIDE HALF THE
LENGTH OF THE THIRD SIDE
1. midpoints
2. The midpoints are S(-1, 4), T(4, 6);
−− 2
−−
2;
4 =_
slope of ST = _
; slope of PR = _
10
5
5
−− −−
since the slopes are =, ST PR.
2
2 + 5 2 = √
29 ;
ST = 50. QJ = QL = 9.1
51. m∠ JLQ = m∠LJQ = 36°
m∠JQL + m∠LJQ + m∠ JLQ = 180°
m∠JQL + 36 + 36 = 180°
m∠JQL = 108°
2
4 + 10 2 = √
116 = 2 √
29 ;
PR = 1
_
and ST = PR.
2
1 XY
3. NM = _
4. XZ = 2LM
2
= 2(5.6) = 11.2
1 (10.2) = 5.1
=_
2
CONSTRUCTION
1. Check students’ constructions.
2. Possible answer: The orthocenter of an acute is
inside the .
The orthocenter of an obtuse is outside the .
The orthocenter of a rt. is the vertex of the rt. ∠.
1 XZ
5. NZ = _
2
1 (11.2) = 5.6
=_
2
5-4 THE TRIANGLE MIDSEGMENT
THEOREM, PAGES 322–327
6. m∠LMN = m∠MNZ = 29°
CHECK IT OUT!
7. m∠YXZ = m∠MNZ = 29°
8. m∠XLM + m∠LMN = 180
m∠XLM + 29 = 180
m∠XLM = 151°
1
_
9. CD < XZ = 15 ft = 5 yd
2
The width of the 2nd floor is less than 5 yd.
1. The midpoints are M(1, 1), N(3, 4);
−− 6 _
−−− 3
; slope of RS = _
= 3;
slope of MN = _
2
4
2
−−− −−
since the slopes are =, MN RS.
2
2 + 3 2 = √
13 ;
MN = 2
2
4 + 6 = √
52 = 2 √
13 ;
1
_
and MN = RS.
2
1 KL
2a. JL = 2PN
b. PM = _
2
= 2(36) = 72
1 (97) = 48.5
=_
2
c. m∠MLK = m∠JMP = 102°
RS =
PRACTICE AND PROBLEM SOLVING
10. The midpoints are D(-4, 3), E(0, 4);
−− 2 _
−− 1
slope of DE = _
; slope of CB = _
= 1;
4
4
−− −−8
since the slopes are =, DE CB.
2
1 + 4 2 = √
17 ;
DE = 2
2 + 8 2 = √
68 = 2 √
17 ;
CB = 1 CB.
and DE = _
2
104
Holt McDougal Geometry
1 GH
12. RQ = _
2
1 (27) = 13.5
=_
2
14. m∠PQR = m∠QRJ
= 55°
11. GJ = 2PQ
= 2(19) = 38
1 GJ
13. RJ = _
2
1 (38) = 19
=_
2
15. m∠HGJ = m∠QRJ 16. m∠GPQ + m∠HGJ = 180°
= 55°
m∠GPQ + 55 = 180°
m∠GPQ = 125°
−−
17. Yes; DE is a midsegment of ABC, so its length is
1 ft, or 2 _
1 ft, which is 27 in. This is less than
half of 4 _
2
4
30 in. So the carpenter can use the 30 in. timber to
make the crossbar.
19. P = KL + LM + KM
1 GH + _
1 HJ
=7+_
2
2
1 (12) + _
1 (8)
=7+_
2
2
= 17
18. P = GH + HJ + GJ
= 12 + 2LJ + 2KL
= 12 + 2(4) + 2(7)
= 34
22. 2(n - 9) = 35
2n - 18 = 35
2n = 53
n = 26.5
23. 2(4n + 5) = 74
8n + 10 = 74
8n = 64
n=8
24. 2n - 23 = 2(9.5)
2n - 23 = 19
2n = 42
n = 21
1 GH
32. FJ = _
33. m∠DCG = m∠CBA
2
= 57°
1 CG
=_
4
1 (16.5) = 4.125
=_
4
34. m∠GHE = m∠HCD
= m∠ABC
= 57°
35. m∠FJG + m∠GHE = 180
m∠FJG + 57 = 180
m∠FJG = 123°
b. XA = CX = 3.5 mi, BC = 2BY = 8 mi
trip length = WX + XA + AB + BC + CX + XW
= 2.25 + 3.5 + 9 + 8 + 3.5 + 2.25
= 28.5 mi
26. 2(5n) = 8n + 10
10n = 8n + 10
2n = 10
n =5
−−
27. B; possible answer: in ABC, DE is a midsegment
−−
and BC is the side to it. By the Midsegment
Theorem, the length of a midsegment is half the
1 BC.
length of the side, so DE = _
2
28. ∠D ∠FZY ∠YXE ∠ZYX;
∠E ∠ZYF ∠DXZ ∠XZY;
∠F ∠XYE ∠DZX ∠ZXY
25. 6n = 2(n + 8)
6n = 2n + 16
4n = 16
n=4
0 + 2a _
0 + 2b
= (a, b)
,
(_
2
2 )
2a + 2c 2b + 0
b. N = (_, _) = (a + c, b)
2
2
38a. M =
−−
0-0 =0
c. slope of PR = _
2c - 0
−−−
b-b =0
slope of MN = _
a+c-a
−−
−−−
Slopes of MN and PR are =, so MN PR.
−−
d. PR = 2c; MN = a + c - a = c; the length of PR is
−−−
1 PR.
twice length of MN, so MN = _
2
E
Y
X
1 DC
31. EH = _
2
1 CB
=_
2
1 (22) = 11
= _
2
36. Yes; possible answer: let x be the length of each
side of an isosc. . By the Midsegment
Theorem, the length of the midsegment to each
1 x. Since these two midsegments
of those sides is _
2
are equal in length, they are .
1 XY
37a. WX = _
2
1 AB
=_
4
_
= 1 (9) = 2.25 mi
4
20. The perimeter of GHJ is twice the perimeter of
KLM.
21. 3n = 2(54)
3n = 108
n = 36
1 AB
30. CG = _
2
1 (33) = 16.5
=_
2
TEST PREP
D
Z
F
39. D;
29. Possible answer: about 18 parking spaces; the new
street is along the midsegment of the triangle plot
of land. The length of the street is half of 440 ft, or
220 ft. Estimate the quotient 220 ÷ 23 by rounding
220 to 225 and 23 to 25. Since 225 ÷ 25 = 9, city
can put about 9 parking spaces on one side of the
street.
So the total number of parking spaces is about 2(9),
or 18.
105
RT = 2PQ
4x - 27 = 2(x + 9)
4x - 27 = 2x + 18
2x = 45
x = 22.5
RT = 4(22.5) - 27
= 63 m
40. H
41. D
Holt McDougal Geometry
53. NX = 2XS
= 2(3) = 6
CHALLENGE AND EXTEND
42. Let the coordinates of the vertices be (u, v), (w, x),
(y, z). By the Midpoint Formula,
v + x = 2(3) = 6
u + w = 2(-6) = -12
x + z = 2(1) = 2
w + y = 2(2) = 4
v + z = 2(-3) = -6
u + y = 2(0) = 0
2w = -12 + 4 - 0
2x = 6 + 2 - (-6)
2w = -8
2x = 14
w = -4
x =7
So (w, x) = (-4, 7).
u + (-4) = -12
v+7=6
u = -8
v = -1
So (u, v) = (-8, -1).
-4 + y = 4
7+z=2
y=8
z = -5
So (y, z) = (8, -5).
55. NP = 2NR
= 2(4.5) = 9
CONSTRUCTION
1 AC
1. XY = _
2
2. Possible answer: Find m∠BXY and m∠BAC
and confirm that they are =. This means the two
segments are by the Converse of the Corr.
Post.
READY TO GO ON? PAGE 329
1. PQ = 2PR
= 2(4.8) = 9.6
43. The midsegment is equilateral and equiangular
44. n 2 - 3 = 2(39)
n 2 = 81
n = ±9
45.
2
n - 6n + 3 = 2(3n - 16)
n 2 - 6n + 3 = 6n - 32
n 2 - 12n + 35 = 0
(n - 7)(n - 5) = 0
n = 7 or 5
3(7) - 16 = 5 > 0
3(5) - 16 = -1 < 0
So n = 7 is the only possible
solution.
2. JM = ML = 58
32
16
8
b. length of midsegment 8
1
= (length of midsegment 4) _
2
1
1
_
_
=
=4
16
4
()
( )
c. length of midsegment
n
n
1 = 64 _
1 = 26 - n
n = AB _
2
2
()
()
SPIRAL REVIEW
2% + 3%
48. concentration = _ = 2.5%
2
3(2%) + 1(3%)
__
= 2.25%
49. concentration =
3+1
50. G(-3, -2) → G(-3, 2) = G(-3 + 0, -2 + 4)
H(0, 0) → H(0 + 0, 0 + 4) = H(0, 4)
J(4, 1) → J(4 + 0, 1 + 4) = J(4, 5)
K(1, -2) → K(1 + 0, -2 + 4) = K(1, 2)
51. G(-3, -2) → G(1, -4) = G[-3 + 4, -2 + (-2)]
H(0, 0) → H[0 + 4, 0 + (-2)] = H(4, -2)
J(4, 1) → J[4 + 4, 1 + (-2)] = J(8, -1)
K(1, -2) → K[1 + 4, -2 + (-2)] = K(5, -4)
52. G(-3, -2) → G(3, 0) = G(-3 + 6, -2 + 2)
H(0, 0) → H(0 + 6, 0 + 2) = H(6, 2)
J(4, 1) → J(4 + 6, 1 + 2) = J(10, 3)
K(1, -2) → K(1 + 6, -2 + 2) = K(7, 0)
AB = AC
5z + 16 = 8z - 5
21 = 3z
7=z
AC = 8(7) - 5 = 51
equation is y + 1 = -2(x - 3).
5. PS = PT = 83.9
XT = RX = 46.7
6. m∠GJK + m∠KJH
+ m∠JHK + m∠KHL + m∠LGJ = 180
2m∠GJK + 2(16) + 50 = 180
2m∠GJK = 98
m∠GJK = 49°
−−
The distance from K to HJ = KL = 21.
4
4
3.
−−− 4 _
4. Slope of MN = _
= 1 , so the slope of the
8
2
perpendicular bisector = -2;
46. QXY XPZ YZR ZYX;
1 (area of PQR)
area of XYZ = _
4
47a. Number of
1
2
3
4
Midsegment
Length of
Midsegment
3 MX
54. MR = _
2
3 (5.5) = 8.25
=_
2
7. The equations of the two perpendicular bisectors
are x = 4.5 and y = -2. So C = (4.5, -2).
1 BD
8. BW = _
3
1 WE
CW = _
1
_
2
= (87) = 29
1 (38) = 19
3
=_
CE = 3CW
2
= 3(19) = 57
1 (0 + 8 + 10), _
1 (4 + 0 + 8) = (6, 4)
9. G = _
3
3
−−
10. PS is horizontal, the altitude is vertical, so the
−− 4
equation is x = 4; the slope of SV = _
= 1, so the
4
slope of the altitude to it is -1; the equation of this
altitude is y - 4 = -(x - 2); at he orthocenter O,
x = 4, so y = 4 - (4 - 2) = 2, and O = (4, 2).
(
)
1 JM
11. ZV = _
2
= RM = 45
m∠RZV = m∠PVZ = 36°
PM = 2ZR
= 2(53) = 106
12. XY = 2MN
= 2(39) = 78
The distance across the pond is 78 m.
106
Holt McDougal Geometry
3.
5-5 INDIRECT PROOF AND
INEQUALITIES IN ONE TRIANGLE,
PAGES 332–339
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.
C
A
CHECK IT OUT!
1. Possible answer:
Given: RST
Prove: RST cannot have 2 rt .
Proof: Assume that RST has 2 rt . Let ∠R and
∠S be the rt. . By the def. of rt. ∠, m∠R = 90° and
m∠S = 90°. By the Sum Thm., m∠R + m∠S +
m∠T = 180°. But then 90 + 90 + m∠T = 180° by
subst., so m∠T = 0°. However, a cannot have
an ∠ with a measure of 0°. So there is no RST,
which contradicts the given information. This means
the assumption is false, and RST cannot have
2 rt. .
−−
2a. The shortest side is AC, so the smallest ∠ is ∠B.
−−
The longest side is AB, so the greatest ∠ is ∠C.
from smallest to greatest are ∠B, ∠A, ∠C.
B
4HEOREM )F BC AB THEN
MȜA MȜC
)F BC AC THEN
MȜA MȜB
)F AC BC THEN
MȜB MȜA
)F AC AB THEN
MȜB MȜC
)F AB BC THEN
MȜC MȜA
)F AB AC THEN
MȜC MȜB
4HEOREM )F MȜA MȜC
THEN BC AB
)F MȜA MȜB
THEN BC AC
)F MȜB MȜA
THEN AC BC
)F MȜB MȜC
THEN AC AB
)F MȜC MȜA
THEN AB BC
)F MȜC MȜB
THEN AB AC
4RIANGLE
)NEQUALITY
4HEOREM
AB BC A#
BC AC A"
AB AC BC
EXERCISES
GUIDED PRACTICE
1. Possible answer: To prove something indirectly,
you assume the opposite of what you are trying to
prove. Then you use logic to lead to a contradiction
of given information, a definition, a postulate, or a
previously proven theorem. You can then conclude
that the assumption was false and the original
statement is true.
3a. 8 + 13 21
21 ≯ 21
No; 8 + 3 = 21, which is not greater than the third
side length.
b. 6.2 + 7 9
6.2 + 9 7
7 + 9 6.2
13.2 > 9
15.2 > 7
16 > 6.2
Yes; the sum of each pair of the lengths is greater
than the third length.
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 .
c. When t = 4, t - 2 = 2, 4t = 16, t 2 + 1 = 17.
2 + 16 17
2 + 17 16
16 + 17 2
18 > 17
19 > 16
33 > 2
Yes; the sum of each pair of the lengths is greater
than the third length.
4. Let s be the length of the 3rd side. Apply the Inequal. Theorem.
s + 17 > 22
s + 22 > 17
17 + 22 > s
s >5
s > -5
39 > s
Combine the inequals. So 5 < s < 39. the length of
the 3rd side is > 5 in. and < 39 in.
3. Possible answer:
−−
Given: PQR is an isosc. with base PR.
Prove: PQR cannot have a base ∠ that is a rt. ∠.
Proof: Assume that PQR has a base ∠ that is a
rt. ∠. Let ∠P be the rt. ∠. By the Isosc. Theorem,
∠P ∠R, so ∠R is also a rt. ∠. By the definition of
rt. , m∠P = m∠R = 90°. By the Sum Theorem,
m∠P + m∠Q + m∠R = 180°. By Subst. m∠Q =
0°. However, a cannot have an ∠ with a measure
of 0°. So there is no PQR, which contradicts the
given information. This means the assumption is
false, and therefore PQR can not have a base ∠
that is rt.
−−
4. The shortest side is PQ, so the smallest ∠ is ∠R.
−−
The longest side is PR, so the greatest ∠ is ∠Q.
from smallest to greatest are ∠R, ∠P, ∠Q.
5. Let d be the distance from Seguin to Johnson City.
d + 22 > 50
d + 50 > 22
22 + 50 > d
d > 28
d > -28
72 > d
28 < d < 72
The distance from Seguin to Johnson City is
> 28 mi and < 72 mi.
THINK AND DISCUSS
1. No; possible answer: the student must consider 2
cases and assume that either the ∠ is acute or the
∠ is rt.
5. m∠Z = 180 - (39 + 46) = 95°
−−
The smallest ∠ is ∠X, so the shortest side is YZ.
−−
The greatest ∠ is ∠Z, so the longest side is XY.
−− −− −−
Sides from shortest to longest are YZ, XZ, XY.
2. Possible answers: 2 cm, 4 cm, 5 cm; 2 cm, 4 cm,
8 cm
107
Holt McDougal Geometry
6. 4 + 7 10
4 + 10 7
7 + 10 4
11 > 10 14 > 7 17 > 4 Yes; the sum of each pair of 2 lengths is greater
than the third length.
PRACTICE AND PROBLEM SOLVING
16. Possible answer:
−−
−−
Given: ABC is scalene. XZ and YZ are
midsegments of ABC.
Prove: ABC cannot have 2 midsegments.
Proof: Assume that ABC does have 2
−−
−−
midsegments. Let XZ and YZ be the
midsegments. By the def. of segs., XZ = YZ. By
7. 2 + 9 12
11 ≯ 12
No; 2 + 9 = 11, which is not greater than the third
side length.
1
BC and
the Mid segment Thm., XZ = __
2
8. 3.5 + 3.5 6
3.5 + 6 3.5
7>6 9.5 > 3.5 Yes; the sum of each pair of 2 lengths is greater
than the third length.
1
1
1
BA. So __
BC = __
BA by subst. But then
YZ = __
2
2
2
−− −−
BC = BA, and by the def. of segs., BC BA.
However, a scalene by def. has no sides. So
ABC is not scalene, which contradicts the given
information. This means the assumption is false,
and therefore
a scalene cannot have 2 midsegments.
9. 1.1 + 1.7 3
2.8 ≯ 3
No; 1.1 + 1.7 = 2.8, which is not greater than the
third side length.
10. When x = 5, 3x = 15, 2x - 1 = 9, x 2 = 25.
9 + 15 25
24 ≯ 25
No; when x = 5, the value of 3x is 15, the value of
2x - 1 is 9 and the value of x 2 is 25. 15 + 9 = 24,
which is not greater than the third side length.
B
X
A
11. When c = 2, 7c + 6 = 20, 10c - 7 = 13, 3c 2 = 12.
12 + 13 20
12 + 20 13
13 + 20 12
25 > 20 32 > 13 33 > 12 Yes; when c = 2, the value of 7c + 6 is 20, the
value of 10c - 7 is 13, and the value of 3c 2 is 12.
The sum of each pair of 2 lengths is greater than
the third length.
12. Let s be the length of the 3rd side. Apply the Inequal. Theorem.
s + 8 > 12
s + 12 > 8
8 + 12 > s
s>4
s > -4
20 > s
Combine the inequals. So 4 < s < 20. The length of
the 3rd side is > 4 mm and < 20 mm.
13. Let s be the length of the 3rd side. Apply the Inequal. Theorem.
s + 16 > 16
16 + 16 > s
s>0
32 > s
Combine the inequals. So 0 < s < 32. The length of
the 3rd side is > 0 ft and < 32 ft.
14. Let s be the length of the 3rd side. Apply the Inequal. Theorem.
s + 11.4 > 12
s + 12 > 11.4
11.4 + 12 > s
s > 0.6
s > -0.6
23.4 > s
Combine the inequals. So 0.6 < s < 23.4. the
length of the 3rd side is > 0.6 cm and < 23.4 cm.
15a. The longest side is opposite the greatest ∠. So the
longest side is between the refrigerator and the
stove.
b. No; 4 + 5 = 9 ≯ 9. By the Inequal. Theorem,
a cannot have these side lengths.
Y
Z
C
17. Possible answer:
Given: ∠J and ∠K are supp.
Prove: ∠J and ∠K cannot both be obtuse.
Proof: Assume that ∠J and ∠K are both obtuse.
Then m∠J > 90° and m∠K > 90° by the definition
of obtuse. Add the 2 inequals., m∠J + m∠K > 180°.
However, by the definition of supp. , m∠J + m∠K
= 180°. So m∠J + m∠K > 180° contradicts the
given information. This means the assumption is
false, and therefore a pair of supp. cannot both
be obtuse.
−−
18. The shortest side is KL, so the smallest ∠ is ∠J.
−−
The longest side is JL, so the greatest ∠ is ∠K.
from smallest to greatest are ∠J, ∠L, ∠K.
19. m∠S = 90, m∠T = 90 - m∠R = 24°
−−
The smallest ∠ is ∠T, so the shortest side is RS.
−−
The greatest ∠ is ∠S, so the longest side is RT.
−− −− −−
Sides from shortest to longest are RS, ST, RT.
20. 6 + 10 15
16 > 15 Yes; the sum of each pair of 2 lengths is greater
than the third length.
21. 14 + 18 32
32 ≯ 32
No; 14 + 18 = 32, which is not greater than the
third side length.
22. 5.8 + 5.8 11.9
11.6 ≯ 11.9
No; 5.8 + 5.8 = 11.6, which is not greater than the
third side length.
23. 41.9 + 62.5 103
104.4 > 103 Yes; the sum of each pair of 2 lengths is greater
than the third.
108
Holt McDougal Geometry
24. When z = 6, z + 8 = 14, 3z + 5 = 23, 4z - 11 = 13.
13 + 14 23
27 > 23 Yes; when z = 6, the value of z + 8 is 14, the value
of 3z + 5 is 23, and the value of 4z - 11 is 13. The
sum of each pair of 2 lengths is greater than the
third side length.
25. When m = 3, m + 11 = 14, 8m = 24, m 2 + 1 = 10.
10 + 14 24
24 ≯ 24
No; when m = 3, the value of m + 11 = 14, the
value of 8m is 24, and the valu e of m 2 + 1 is 10.
the sum of 14 and 10 is 24, which is not greater
than the third side length.
26. b + c > s
s + 4 > 19
s > 15
15 yd < s < 23 yd
27. s + 23 > 28
s>5
5 km < s < 51 km
4 + 19 > s
23 > s
23 + 28 > s
51 > s
28. s + 3.8 > 9.2
3.8 + 9.2 > s
s > 5.4
13.0 > s
5.4 cm < s < 13.0 cm
29. s + 1.89 > 3.07
s > 1.18
1.18 m < s < 4.96 m
1.89 + 3.07 > s
4.96 > s
5
1 > 3_
30. s + 2_
8
8
1
s > 1_
2
3 in.
1 in. < s < 5_
1_
2
4
5 >s
1 + 3_
2_
8
8
3>s
5_
4
1
5 > 6_
31. s + 3_
6
2
_
s > 22
3
1 ft
2 ft < s < 10_
2_
5 + 6_
1 >s
3_
6
2
1>s
10_
3
3
3
−− −− −− −− −−
32. AD, BD, AB, BC, CD; possible answer: in ABD,
m∠ABD = 50°. In BCD, m∠DBC = 74°. In
ABD, the order of the tubes from shortest to
−− −− −−
longest is AD, BD, AB. In BCD, the order of the
−− −− −−
tubes from shortest to longest is BD, BC, CD. So
AD < BD < AB, and BD < BC < CD. Since AB =
50.8 and BC = 54.1, it is also true that AB < BC.
−− −− −− −− −−
So AD < BD < AB < BC < CD.
33. a > 7.5, where a is the length of a leg. Possible
answer: By the Inequal. Thm., a + a > 15 and a
+ 15 > a. The solution of the first inequality is
a > 7.5. The second inequality simplifies to 15 > 0,
which is always a true statement.
34. Step 1 Find x.
2x + 5x - 1 = 90
7x = 91
x = 13
Step 2 Find ∠ measures and order sides.
m∠A = 90°, m∠B = 5(13) - 1 = 64°,
m∠C = 2(13) = 26°
−− −− −−
m∠C < m∠B < m∠A, so order is AB, AC, BC.
35. Step 1 Find x.
4.5x - 5 + 10x - 2 + 5x - 8 = 180
19.5x = 195
x = 10
Step 2 Find ∠ measures and order sides.
m∠D = 4.5(10) - 5 = 40°, m∠E = 10(10) - 2 = 98°,
m∠F = 5(10) - 8 = 42°
−− −− −−
m∠D < m∠F < m∠E, so order is EF, DE, DF.
36. A rt. ∠ cannot be an acute ∠. So the 1st and the 3rd
statements contradict each other.
37. An obtuse ∠ measures > 90°. So the 2nd and the
3rd statements contradict each other.
38. If 1st statement is true, JK = LK. So the 1st and the
3rd statements contradict each other.
39. 2 line segs. cannot be both ⊥ and . So the 1st and
the 3rd statements contradict each other.
40. A figure cannot be both a and a quad. So the 2nd
and the 3rd statements contradict each other.
41. 4 is not a prime number, so no multiple of 4 is
prime. So the 2nd and the 3rd statements contradict
each other.
42. m∠P > m∠PQS, so QS > PS.
43. m∠PSQ = 180 - (54 + 75) = 51°
m∠PSQ < m∠P, so PQ < QS.
44. m∠R < m∠RSQ, so QS < QR.
45. m∠RQS = 180 - (51 + 78) = 51° = m∠R
By Converse of Isosc. Theorem, QS = RS.
46. PQ < QS and QS = RS, so PQ < RS.
47. RS = QS and QS > PS, so RS > PS.
48. AE > BA, so m∠ABE > m∠BEA.
49. CE > BC, so m∠CBE > m∠CEB.
50. CD = DE, so by Isosc. Theorem,
m∠DCE = m∠DEC.
51. DE < CE, so m∠DCE < m∠CDE.
52. AE < BE, so m∠ABE < m∠EAB.
53. BE = CE, so by Isosc. Theorem,
m∠EBC = m∠ECB.
54. JK = 6 2 + 8 2 = 10; KL = 5 2 + 8 2 ≈ 9.4;
JL = ⎪-3 - 8⎥ = 11
KL < JK < JL, so order is ∠J, ∠L, ∠K.
55. JK = ⎪-10 - 2⎥ = 12; KL =
12 2 + 7 2 ≈ 13.9;
2
2
JL = 12 + 5 = 13
JK < JL < KL, so order is ∠L, ∠K, ∠J.
56. JK =
1 2 + 7 2 ≈ 7.1; KL =
2
6 2 + 4 2 ≈ 7.2;
2
JL = 7 + 3 ≈ 7.6
JK < KL < JL, so order is ∠L, ∠J, ∠K.
57. JK =
10 2 + 7 2 ≈ 12.2; KL =
2 2 + 11 2 ≈ 11.2;
2
2
JL = 12 + 4 ≈ 12.6
KL < JK < JL, so order is ∠J, ∠L, ∠K.
109
Holt McDougal Geometry
66. Possible answer:
Given: P is in the int. of XYZ.
Prove: XY + XP + PZ > YZ.
Proof: By the Inequal. Theorem, PY + PZ >
YZ and XY + XP > YP. Since PZ > 0, the second
inequal. is equivalent to XY + XP + PZ > YP + PZ.
But then YP + PZ > YZ, so XY + XP + PZ > YZ by
the Trans. Prop of Inequal.
58. Possible answer: Assume that the client committed
the burglary. A person who commits a burglary
must be present at the scene when the crime is
committed. However, a witness saw the client
in a different city at the time that the crime was
committed. This means that the assumption that the
client committed the burglary is false. Therefore the
client did not commit the burglary.
X
59a. AR + 400 > 600
AR > 200 mi
400 + 600 > AR
1000 mi > AR
200 = 0.4 h. Time to travel
Time to travel 200 mi is _
500
1000 = 2 h. So the range of time is
1000 mi is _
500
0.4 h < t < 2 h.
P
Y
Z
67a. definition of segs.
b. No; AR < 1000, so by the Inequal. Theorem,
AM must be less than 1800.
60. Step 1 Write and solve 2 inequals. for n.
n+6>8
6+8>n
n>2
14 > n
Step 2 Combine the inequals.
2 < n < 14
b. Isosc. Theorem
c. definition of d. m∠1 + m∠3
e. subst.
f. m∠S
g. trans. Prop. of Inequal.
68a. ABC
61. Step 1 Write and solve 2 inequals. for n.
2n + 5 > 7
5 + 7 > 2n
2n > 2
12 > 2n
n >1
6 >n
Step 2 Combine the inequals.
1<n<6
b. AD
c. Isosc. Theorem
d. definition of e. m∠3
f. subst.
g. in , longer side is
opp. larger ∠
h. subst.
i. AC + BC > AB
69. Possible answer: A rt. has a rt. ∠ and 2 acute .
By definition, the rt. ∠ has the greatest measure.
Since the hyp. is the side opposite the rt. ∠, the
hyp. is the longest side by Thm 5-5-2. Similarly, the
diagonal of a square forms 2 rt. , with the diagonal
being the hyp. of each. Since the diagonal is longer
than the leg lengths in both , the diagonal is
longer than the side length of the square.
62. Step 1 Write and solve 2 inequals. for n.
n+1+3>6
3+6>n+1
n>2
8 >n
Step 2 Combine the inequal.s
2<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
TEST PREP
70. A;
3+3=6>5 64. Step 1 Write and solve 2 inequals. for n. Use the
fact that n + 2 < n + 3.
n + 2 + n + 3 > 3n - 2
3n - 2 + n + 2 > n + 3
2n + 5 > 3n - 2
4n > n + 3
7>n
3n > 3
n>1
Step 2 Combine the inequals.
1<n<7
65. Step 1 Write and solve 2 inequals. for n. Use fact
that n < n + 2.
2n + 1 + n > n + 2
n + n + 2 > 2n + 1
3n + 1 > n + 2
2 > 1 always true
2n > 1
n > 0.5
71. H;
GH + HJ < GJ
contradicts the Inequal. Theorem
72. C;
−−
∠S must be the largest ∠, so RT is the longest side.
CHALLENGE AND EXTEND
()
5
73. The total number of choices is
= 10. The
3
choices that form a :
1 + 3 = 4 ≯ 5, 7, or 9
3+5=8>7 1 + 5 = 6 ≯ 7 or 9
3+5=8≯9
1+7=9≯9
3 + 7 = 10 > 9 5 + 7 = 12 > 9 3 or 30%.
is possible for 3 choices. So prob. = _
10
2
p
b. _
74a. √2 is rational
q2
2
c. 2 q
2
2
d. (2x) = 4x
2
1 p 2 and p 2 is divisible by 4
e. q = _
2
110
Holt McDougal Geometry
75.
1.
2.
3.
4.
5.
Statements
−−
PX ⊥ ; Y is any point
on other than X.
m∠1 = 90°
∠1 is a rt. ∠.
XPY is a rt. .
∠2 and ∠P are comp.
6. 90° = m∠2 + m∠P
7. 90° > m∠2
8. m∠1 > m∠2
9. PY > PX
3a.
Reasons
Statements
Reasons
1. C is the midpoint
−−
of BD;
m∠1 = m∠2
m∠3 > m∠4
−− −−
2. BC DC
3. ∠1 ∠2
−− −−
4. AC EC
1. Given
2. Def. of ⊥
3. Def. of rt. ∠
4. Def. of rt. 5. Acute of rt. are comp.
6. Def. of comp. 7. Comparison
Prop. of Inequal.
8. Subst.
9. In , longer side
is opp. larger ∠
1. Given
2. Def. of midpoint
3. Def. of 4. Con. of Isosc. Thm.
5. Hinge Thm.
5. AB > ED
b.
Statements
Reasons
1. ∠SRT ∠STR,
TU > RU
−− −−
2. ST SR
SPIRAL REVIEW
-2 - 2 = -2 77.
y - y 1 = m(x - x 1)
76. slope = _
-1 + 3
y - 0 = 2[x - (-3)]
y - y 1 = m(x - x 1)
y - 0 = 2(x + 3)
y - 2 = -2[x - (-3)]
y = 2x + 6
y - 2 = -2(x + 3)
-2x
+
y=6
y - 2 = -2x - 6
2x + y = -4
1. Given
−− −−
3. SU SU
4. m∠TSU > m∠RSU
2. Con. of Isosc. Thm.
3. Reflex. Prop. of 4. Con. of the Hinge
Thm.
THINK AND DISCUSS
1. Possible answer: kitchen tongs
2
78. QP = 5(-1) - 2 = 3, ST = -1 + 7 = 6,
SU = 3(-1) 2 + 1 = 4, so PQR TUS by SSS.
2. No; in this case, 2 sides of the 1st are to 2
sides of the 2nd , but the given ∠ measures are
not the measures of included between the sides. Thus you cannot apply the Hinge Theorem.
2
79. BC = 6 - 5(6) + 4 = 10, EF = 2(6) - 1 = 11,
m∠ABC = 14(6) + 18 = 102°, so ABC EFD
by SAS.
3.
80. Equation of altitude from S is y = 3. Slope of RS is
3 - 5 = -_
1 , so slope of altitude from T is 2;
_
2
4-0
equation is y - 1 = 2x, or y = 2x + 1. At O, y = 3
and therefore 3 = 2x + 1, so x = 1; thus O = (1, 3).
)NEQUALITIES IN 4WO 4RIANGLES
B
A
Y
C
X
Z
81. The altitudes from N and P lie along the x- and
y-axes, respectively. Therefore O = (0, 0).
(INGE 4HEOREM
)F AB Ɂ XY, AC Ɂ XZ,
AND MȜA MȜX, THEN
BC YZ.
5-6 INEQUALITIES IN TWO TRIANGLES,
PAGES 340–345
#ONVERSE OF (INGE 4HEOREM
)F AB Ɂ XY, AC Ɂ XZ,
AND BC YZ, THEN
MȜA MȜX.
CHECK IT OUT!
1a. Compare the side lengths in EFG and EHG.
EF > EH
EG = EG
FG = HG
By the Converse of the Hinge Theorem, m∠EGF >
m∠EGH.
EXERCISES
GUIDED PRACTICE
b. Compare the sides and the in ABD and CBD.
AD = CD
BD = BD
m∠CDB > m∠ADB
By the Hinge Theorem, BC > AB.
2. The ∠ of swing at full speed is greater than the ∠ of
swing at low speed.
1. Compare the sides and the in ABC and XYZ.
AB = YZ
BC = XY
m∠B < m∠Y
By the Hinge Theorem, AC < XZ.
2. Compare the side lengths in SRT and QRT.
RT = RT
RS = RQ
ST > QT
By the Converse of Hinge Theorem, m∠SRT >
m∠QRT.
3. Compare the sides and in KLM and KNM.
KM = KM
LM = NM
m∠KML > m∠KMN
By the Hinge Theorem, KL > KN.
111
Holt McDougal Geometry
2z + 7 > 0
13. 2z + 7 < 72
2z > -7
2z < 65
z > -3.5
z < 32.5
Combining, -3.5 < z < 32.5.
4. Step 1 Compare the side lengths in . By the
Converse of the Hinge Theorem,
2x + 8 < 25
2x < 17
x < 8.5.
Step 2 Since (2x + 8)° is an angle in a ,
2x + 8 > 0
2x > -8
x > -4.
Step 3 Combine the inequals.
The range of values is -4 < x < 8.5.
14. 4z - 6 < z + 11
4z - 6 > 0
3z < 17
4z > 6
3
17
z <_
z>_
2
3
3 <z<_
17 .
Combining, _
2
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.
5. Step 1 Compare the sides and the in . By the
Hinge Theorem,
5x - 6 < 9
5x < 15
x < 3.
Step 2 Since 5x - 6 is a length,
5x - 6 > 0
5x > 6
x > 1.2.
Step 3 Combine the inequals.
The range of values is 1.2 < x < 3.
16.
6. Step 1 Compare the sides and the in . By the
Hinge Theorem,
2x - 5 < x + 7
x < 12.
Step 2 Since 2x - 5 is a length,
2x - 5 > 0
2x > 5
x > 2.5.
Step 3 Combine the inequals.
The range of values is 2.5 < x < 12.
5. JQ + QP = JP,
NP + QP = NQ
6. JP > NQ
7. m∠K > m∠M
17. BC = YZ
19. m∠QPR > m∠QRP
21. m∠RSP = m∠RPS
7. The 2nd position; the lengths of the upper and lower
arm are the same in both positions, but the distance
from the shoulder to the wrist is greater in the 2nd
position. So the included ∠ measure is greater by
the Converse of the Hinge Theorem.
8.
Statements
−−
1. FH is a median of DFG;
m∠DHF > m∠GHF
−−
2. H is midpoint of DG.
−− −−
3. DH GH
−− −−
4. FH FH
5. DF > GF
Reasons
1. Given
2.
3.
4.
5.
Def. of median
Def. of midpoint
Reflex. Prop. of Hinge Thm.
PRACTICE AND PROBLEM SOLVING
9. BC = CD, CA = CA, AD > AB; by Converse of
Hinge Theorem, m∠DCA > m∠BCA.
10. GH = KL, HJ = LM, GJ < KM; by Converse of
Hinge Theorem, m∠GHJ < m∠KLM.
11. ST = UV, SU = SU, m∠UST > m∠SUV; by Hinge
Theorem, TU > SV.
Statements
−− −−− −− −−−
1. JK NM, KP MQ,
JQ > NP
−− −−
2. QP QP
3. QP = QP
4. JQ + QP > NP + QP
Reasons
1. Given
2. Reflex. Prop. of 3. Def. of segs.
4. Add. Prop. of
Inequal.
5. Segment Add.
Post.
6. Subst.
7. Con. of the Hinge
Thm.
18. m∠QRP < m∠SRP
20.m∠PRS < m∠RSP
22. m∠QPR > m∠RPS
23. m∠PSR < m∠PQR
24. Corr. sides are , and the included are ∠B and
∠E. By the Hinge Theorem, m∠B > m∠E → AC >
DF.
−− −−
−− −−
25. SR ST by definition, and SV SV. So by the
Converse of the Hinge Theorem, RV < TV →
m∠RSV < m∠TSV.
26. Corr. sides are , and the included are ∠G and
∠K. m∠G = 90° > m∠K, so by the Hinge Theorem,
HJ > LM.
−− −−
−− −−
27. YM MZ by definition, and XM XM. So by the
Converse of the Hinge Theorem, YX > ZX →
m∠YMX > m∠ZMX.
28. Possible answer: As the angle made by a door
hinge gets larger, the width of the door opening
increases. As the angle made by the hinge gets
smaller, the width of the door opening decreases.
This is like the side opposite an angle in a triangle
getting larger as the measure of the angle increases
or getting smaller as the angle decreases.
4z - 12 > 0
12. 4z - 12 < 16
4z < 28
4z > 12
z<7
z>3
Combining, 3 < z < 7.
112
Holt McDougal Geometry
29. Possible answer:
Similarities: Both the SAS Post. and the Hinge
Theorem concern the relationship between 2 .
Both involve 2 sides and the included ∠ of each .
Differences: To apply the SAS Post., you must
know that 2 sides and the included ∠ of one are
to 2 sides and the included ∠ of the 2nd . To
apply the Hinge Theorem, you must know that 2
sides of one are to 2 sides of the 2nd , but
the included are ≠ in measure. The SAS Post.
allows you to conclude that the 2 are ; then by
CPCTC, you can show that the sides opposite the are . The Hinge Theorem involves 2 that are
; in this case, the sides opposite the included are ≠ in length, and the exact relationship between
the lengths is determined by the sizes of the
included .
−− −− −− −−
30a. Newton Springs; NS HS, SJ SJ, and
m∠NSJ < m∠HSJ, so NJ < JH by the Hinge
Theorem.
b. By Inequal. Theorem,
NJ + SJ > SN
NJ + 182 > 300
NJ > 118 mi
Min. distance = SN + NJ
> 300 + 118 = 418 mi
TEST PREP
31. D;
0 < 3x - 9 < 2x + 1
9 < 3x or 3 < x, and x < 10
3 < x < 10
32. H;
−−
D lies on AB; AD = DB by the definition of median.
33. Group A is closer to the camp.
Possible answer: The 6.5-mi and 4-mi paths
together with the distance lines back to the camp
form 2 . 2 sides of 1 are to 2 sides of the
other . In the for Group A, the measure of the
included ∠ is 90° + 35° = 125°. In the for Group
B, the measure of the included ∠ is 90° + 45° =
135°. By the Hinge Theorem, the side opposite the
125° ∠ is shorter than the side opposite the 135° ∠.
So Group A is closer to the camp.
CHALLENGE AND EXTEND
−−
−−
b. Locate point Q on AC so that BQ bisects ∠PBC.
By the definition of ∠ bisector, ∠QBC ∠QBP. It
−− −−
−− −−
is given that BC EF. Since BP EF from part a,
−− −−
BC BP by the Trans. Prop. of . By the Reflex.
−− −−−
Prop. of , BQ BQ. So BQP BQC by SAS,
−− −−
and QP QC by CPCTC.
c. AQ + QP > AP by the Inequal. Theorem in
AQP. AQ + QC = AC by the Segment Add.
−− −−
Post. From part b, QP QC, so QP = QC by
the definition of segs. Thus AQ + QC > AP by
−−
subst., and so AC > AP by subst. From part a, AP
−−
DF. So by the definition of segs., AP = DF.
Therefore AC > DF by subst.
SPIRAL REVIEW
36. range: 5 - 0.5 = 4.5
mode: 2
37. range: 99 - 85 = 14
mode: none
38. range: 9 - 4 = 5
modes: 4, 5, 7
39. m∠2 = 3(5) + 21 = 36°,
m∠6 = 7(5) + 1 = 36° = m∠2; m
of the Corr. Post.
n by the Converse
40. m∠4 = 2(7) + 34 = 48°,
m∠7 = 15(7) + 27 = 132°; so m∠4 + m∠7 = 180°;
m n by the Converse of the Same-Side Int. Theorem.
41. By Similar Triangles
Theorem:
1 AB
DF = _
2
= AE = 2.5
42. BC = 2DE
= 2(2.3) = 4.6
43. m∠BFD = 180 - m∠CFD
= 180 - m∠CBA
= 180 - 95 = 85°
5-7 THE PYTHAGOREAN THEOREM,
PAGES 348–355
CHECK IT OUT!
1a. a 2 + b 2 = c 2
42 + 82 = x 2
80 = x 2
√80 = x
x = √(16)(5) = 4 √5
b.
34. Step 1 Apply Hinge Theorem.
−− −− −−
By Converse of Isosc. Theorem, VZ VY; VX
−−
VX; m∠XVZ > m∠XVY. So XZ > XY.
Step 2 Write and solve 2 inequals.
5x + 15 > 8x - 6
8x - 6 > 0
21 > 3x
8x > 6
7>x
x > 0.75
Step 3 Combine the inequals.
0.75 < x < 7
a 2 + b 2= c 2
x 2 + 12 2 = (x + 4) 2
x 2 + 144 = x 2 + 8x + 16
128 = 8x
x = 16
35a. Locate point P outside ABC so that ∠ABP −− −−
−−
∠DEF and BP EF. It is given that AB −− −−
−−
DE, so ABP DEF by SAS. Thus AP DF
by CPCTC.
113
Holt McDougal Geometry
2. Let y be the distance in ft from the foot of the ladder
to the base of the wall. Then 4y is the distance in ft
from the top of the ladder to the base of the wall.
THINK AND DISCUSS
1. The greatest number is substituted for c. The other
2 numbers are substtituted for a and b in any order.
a 2 + b 2= c 2
(4y) 2 + y 2 = 30 2
17y 2 = 900
2
900
y =_
17
900
y= _
17
900 ≈ 29 ft 1 in.
4y = 4 _
17
2. Possible answer: The sum of the areas of the
2 smaller squares equals the area of the
largest square. So
3 2 + 4 2 = 5 2, or 9 + 16 = 25.
3. Must be nonzero and whole numbers, and must
satisfy the equation a 2 + b 2 = c 2
√
√
3a.
b.
4.
2
2
2
a +b =c
2
8 + 10 = c 2
164 = c 2
c = √
164 = 2 √41
The side lengths do not form a Pythagorean triple
because 2 √
41 is not a whole number.
2
a2 + b2 = c2
24 2 + b 2 = 26 2
b 2 = 100
b = 10
The side lengths are nonzero whole numbers that
satisfy the equation a 2 + b 2 = c 2, so they form a
Pythagorean triple.
c.
2
2
2
a +b =c
2
1 + 2.4 = c 2
6.76 = c 2
c = 2.6
The side lengths do not form a Pythagorean triple
because 2.4 and 2.6 are not whole numbers.
d.
a2 + b2 = c2
16 2 + 30 2 = c 2
1156 = c 2
c = 34
The side lengths are nonzero whole numbers that
satisfy the equation a 2 + b 2 = c 2, so they form a
Pythagorean triple.
0YTHAGOREAN
2ELATIONSHIPS
0YTH 4HM )N A RṮ
THE SUM OF THE SQUARES OF
THE LEG LENGTHS EQUALS THE
SQUARE OF THE HYPOTENUSE
#ONV OF THE 0YTH
4HM )F THE SUM OF
THE SQUARES OF SIDE
LENGTHS OF A ̱
EQUALS THE SQUARE OF
THE THIRD SIDE LENGTH
THEN THE ̱ IS A RT ̱
0YTH )NEQUAL 4HM )N A ̱
WITH c AS THE LENGTH OF THE
LONGEST SIDE IF c Êa Êb THE
̱IS OBTUSE BUT IF c Êa Êb THE ̱IS ACUTE
EXERCISES
GUIDED PRACTICE
1. No; although it is true that (2.7) 2 + (3.6) 2 = (4.5) 2,
the numbers 2.7, 3,6, and 4.5 are not whole
numbers.
2. a 2 + b 2 = c 2
32 + 92 = x2
90 = x 2
√
90 = x
x = √
(9)(10)
= 3 √
10
2
4.
4a. Step 1 Determine if the measures form a .
By the Inequal. Theorem, 7, 12, and 16 can be
the side lengths of a .
Step 2 Classify the .
c2 a2 + b2
16 2 7 2 + 12 2
256 49 + 144
256 > 193
Since c 2 > a 2 + b 2, is obtuse.
2
2
2
3. a + b = c
2
2
x + 7 = 11 2
x 2 = 72
x = √
72
x = √
(36)(2)
√
=6 2
2
2
2
a +b =c
2
2
2
(x - 2) + 8 = x
x 2 - 4x + 4 + 64 = x 2
-4x + 68 = 0
68 = 4x
x = 17
5. Let the width and the height of the monitor be
w = 5x and h = 4x, respectively.
a2 + b2= c2
2
2
2
(4x) + (5x) = 19
2
41x = 361
2
361
x =_
41
361
x= _
41
61 ≈ 14.8 in.
w = 5x = 5 3_
41
√
√
361 ≈ 11.9 in.
h = 4x = 4 √_
41
b. Step 1 Determine if the measures form a .
Since 11 + 18 = 29 ≯ 34, these cannot be the side
lengths of a .
c. Step 1 Determine if the measures form a .
By the Inequal. Theorem, 3.8, 4.1, and 5.2 can
be the side lengths of a .
Step 2 Classify the .
c2 a2 + b2
5.2 2 3.8 2 + 4.1 2
27.04 14.44 + 16.81
27.04 < 31.25
Since c 2 < a 2 + b 2, is acute.
2
2
2
6. a + b = c
2
2
4 + 5 = c2
41 = c 2
c = √
41
The side lengths do not form a Pythagorean triple
is not a whole number.
because √41
114
Holt McDougal Geometry
7. a 2 + b 2 = c 2
12 2 + b 2 = 20 2
b 2 = 256
b = 16
The side lengths are nonzero whole numbers that
satisfy the equation a 2 + b 2 = c 2, so they form a
Pythagorean triple.
8.
a2 + b2 = c2
1.5 2 + b 2 = 1.7 2
b 2 = 0.64
b = 0.8
The side lengths do not form a Pythagorean triple
because they are not whole numbers.
9. Step 1 Determine if the measures form a .
By the Inequal. Theorem, 7, 10, and 12 can be
the side lengths of a .
Step 2 Classify the .
c2 a2 + b2
12 2 7 2 + 10 2
144 49 + 100
144 < 149
Since c 2 < a 2 + b 2, is acute.
10. Step 1 Determine if the measures form a .
By the Inequal. Theorem, 9, 11, and 15 can be
the side lengths of a .
Step 2 Classify the .
c2 a2 + b2
15 2 9 2 + 11 2
225 81 + 121
225 > 202
Since c 2 > a 2 + b 2, is obtuse.
11. Step 1 Determine if the measures form a .
By the Inequal. Theorem, 9, 40, and 41 can be
the side lengths of a .
Step 2 Classify the .
c2 a2 + b2
41 2 9 2 + 40 2
1681 81 + 1600
1681 = 1681
Since c 2 = a 2 + b 2, is a rt. .
12. Step 1 Determine if measures form a .
3 = 3_
1 + 1_
1 ≯ 3_
1 , these cannot be
Since 1_
2
4
4
4
the side lengths of a .
13. Step 1 Determine if the measures form a .
By the Inequal. Theorem, 5.9, 6, and 8.4 can be
the side lengths of a .
Step 2 Classify the .
c2 a2 + b2
8.4 2 5.9 2 + 6 2
70.56 34.81 + 36
70.56 < 70.81
Since c 2 < a 2 + b 2, is acute.
14. Step 1 Determine if the measures form a .
can
By the Inequal. Theorem, 11, 13, and 7 √6
be the side lengths of a .
Step 2 Classify the .
c2 a2 + b2
(7 √6 ) 2 11 2 + 13 2
294 121 + 169
294 > 290
Since c 2 > a 2 + b 2, is obtuse.
PRACTICE AND PROBLEM SOLVING
15. 6 2 + 8 2 = x 2
100 = x 2
x = 10
16. 9 2 + x 2 = 13 2
81 + x 2 = 169
x 2 = 88
x = √
88 = 2 √
22
2
2
2
17. x + 7 = (x + 1)
2
2
x + 49 = x + 2x + 1
48 = 2x
x = 24
18. (3x) 2 + (5x) 2 = 8 2
34x 2 = 64
32
x2 = _
17
32 ≈ 4 ft 1 in.
3x = 3 _
17
_
5x = 5 32 ≈ 6 ft 10 in.
17
2
2
2
19. 2.5 + b = 6.5
2
6.25 + b = 42.25
b 2 = 36
b =6
The side lengths cannot form a because 2.5 and
6.5 are not whole numbers.
20. 15 2 + 20 2 = c 2
625 = c 2
c = 25
Yes; the three side lengths are nonzero whole
numbers that satisfy a 2 + b 2 = c 2.
2
2
2
21. 2 + b = 7
2
4 + b = 49
b 2 = 45
b = √
45 = 3 √
5
is
The side lengths cannot form a because 3 √5
not a whole number.
22. 10 + 12 = 22 > 15
15 2 10 2 + 12 2
225 < 244
The side lengths form
an acute .
24. 9 + 14 = 23 > 17
17 2 9 2 + 14 2
289 > 277
The side lengths form
an obtuse .
115
23. 8 + 13 = 21 ≯ 23
The side lengths can
not form a .
1 + 2 = 3_
1 > 2_
1
25. 1_
2
2
2
2
2
1
1
_
_
2
1
+ 22
2
2
1 = 6_
1
6_
4
4
The side lengths form
a rt. .
( ) ( )
Holt McDougal Geometry
26. 0.7 + 1.1 = 1.8 > 1.7 27. 7 + 12 = 19 > 6 √
5 2
1.7 2 0.7 2 + 1.1 2
(6 √5 ) 7 2 + 12 2
2.89 > 1.7
180 < 193
The side lengths form
The side lengths form
an obtuse .
an acute .
28. Possible answer: Shape the rope into a with side
lengths of 3, 4, and 5. Because 3 2 + 4 2 = 5 2, the is a rt. with the rt. ∠ opposite the side 5.
2
2
2
29. B; (x + 3) + 4 = (x + 6x + 9) + 16. In the
solution shown, the 6x term was omitted.
30. Let a and b be the horizontal leg lengths of the leftand right-hand .
92 + b2 = 152
b2 = 144
b = 12
a + b = 25
a = 25 - 12 = 13
132 + 92 = x 2
250 = x 2
x = √
250
= 5 √
10
31. Let a and b be the horizontal leg lengths of the left- and
right-hand .
2
2
2
a + 6 = 10
2
a = 64
a=8
x = a + b = 8 + √13
2
2
2
6 +b =7
2
b = 13
b = √
13
32. Let d be the length of 33. Let h be the height of
the shared side.
the .
2
22 + d 2 = 72
3 2 + h 2 = ( √
34 )
2
2
d = 45
h = 25
x 2 + 52 = d2
x 2 + h 2 = 11 2
x 2 = 20
x 2 = 96
x = √
x = √
20 = 2 √5
96 = 4 √
6
34. Let h be the height of the .
5 2 + h 2 = 13 2
h 2 = 144
(x + 5) 2 + h 2 = 20 2
2
x + 10x + 25 + 144 = 400
x 2 + 10x - 231= 0
(x - 11)(x + 21) = 0
Since x > 0, the only possible solution is x = 11.
35. Let b be the base length of each .
18 2 + (2b) 2 = 30 2
4b 2 = 576
b 2 = 144
2
2
18 + b = x 2
468 = x 2
x = √
468 = 6 √13
2
2
2
36. 3963 + x = (3963 + 250)
2
x = 2,044,000
x = √
2,044,000 ≈ 1430 mi
37. Possible answer:
Outer figure: The length of each side is a + b, so
the outer figure has 4 sides. Each ∠ is a rt. ∠
from one of the rt. , so the outer figure has 4 rt. .
By definition, it is a square.
Inner figure: The length of each side is c, so the
inner figure has 4 sides. The 2 acute of a rt. are comp., so the measure of each ∠ in the inner
figure is 90°. Therefore the inner figure has 4 rt. .
By definition, it is a square.
38. Let b be the base of
the .
8 2 + b 2 = 17 2
b 2 = 225
b = 15
P = 8 + 15 + 17
= 40 units
1 (15)(8)
A=_
2
= 60 square units
39. Let 2b be the base of
the .
62 + b2 = 82
b 2 = 28
b = 2 √
7
2b = 4 √
7
P = 8 + 8 + 4 √
7
= 16 + 4 √
7 units
1 (4 √
7 )(6)
A=_
2
= 12 √
7 square units
40. Let h be the height of 41. Let h be the height and
c be the 3rd side length
the .
of the .
4 2 + h 2 = 12 2
h 2 = 128
32 + h2 = 52
h = √
128
h 2 = 16
= 8 √
2
h =4
P = 12 + 12 + 8
62 + 42 = c2
= 32 units
52 = c 2
1
_
√
A = (8)8 2
c = √
52
2
= 2 √
13
2 square units
= 32 √
P = 5 + (3 + 6) + 2 √13
= 14 + 2 √
13 units
1 (3 + 6)(4)
A=_
2
= 18 square units
42. Let a + b = 15 be the 2nd side length and c be the
3rd side length of the .
2
2
2
a + 12 = 15
2
a = 81
a=9
b = 15 - 9 = 6
6 2 + 12 2 = c 2
180 = c 2
c = √
180 = 6 √5
P = 15 + 15 + 6 √5
= 30 + 6 √
5 units
1 (15)(12) = 90 square units
A=_
2
43. P = 4 + 5 + 5 + 8 = 22 units
1 (a + b)h
A=_
2
1 (5 + 8)(4)
=_
2
= 26 square units
44. Possible answer: When you use Pythagorean
Theorem, you know that the is a rt. . You
substitute the known values into a 2 + b 2 = c 2
and solve for the unknown side length. When you
use the Converse of Pythagorean Theorem, you
are trying to find out whether a given is a rt. .
Usually all side lengths are known. You substitute all
the values into a 2 + b 2 = c 2 to determine whether
the resulting equation is true. If it is true, then you
know that the is a rt. .
116
Holt McDougal Geometry
45. Draw PQR with ∠R as the rt. ∠, leg lengths of a
and b, and hyp. length of x. In ABC, it is given
that a 2 + b 2 = c 2. In PQR, a 2 + b 2 = x 2 by the
Pythagorean Theorem. Since a 2 + b 2 = c 2, and
2
2
2
2
2
a + b = x , it follows by subst. that x = c .
Take the positive square root of both sides, and
x = c. So AB = PQ, BC = QR, and AC = PR.
−− −− −− −−
By the definition of segs., AB PQ, BC QR ,
−− −−
and AC PR . Then ABC PQR by SSS,
and ∠C ∠R by CPCTC. By the definition of rt.
∠, m∠R = 90°. So by the definition of , m∠C
= 90°. Therefore ∠C is a rt. ∠ by definition, and
ABC is a rt. by definition.
CHALLENGE AND EXTEND
52. Let 3 points be A(-1, 2), B(-10, 5), and C(-4, k).
2
2
2
2
2
2
AB = 9 + 3 = 90
AC = 3 + (k - 2)
2
2
2
2
BC = 6 + (k - 5)
= 9 + k - 4k + 4
2
=
k 2 - 4k + 13
= 36 + k - 10k + 25
2
= k - 10k + 61
2
2
2
If AB + BC = AC ,
90 + k 2 - 10k + 61 = k 2 - 4k + 13
138 = 6k
k = 23
If AB 2 + AC 2 = BC 2,
90 + k 2 - 4k + 13 = k 2 - 10k + 61
6k = -42
k = -7
2
2
If AC + BC = AB 2,
k 2 - 4k + 13 + k 2 - 10k + 61 = 90
2k 2 - 14k - 16 = 0
k 2 - 7k - 8 = 0
(k - 8)(k + 1) = 0
k = 8 or -1
So k = -7, -1, 8, or 23
b. JL = x 2 - x 1,
LK = y 2 - y 1
46a. (x 2, y 1)
2
c. √n + 1 ; possible answer: the length of the hyp. is
the square root of the whole number 1 greater than
the number of the , or √n + 1 .
2
2
c. JK = JL + LK
= (x 2 - x 1) 2 + (y 2 - y 1) 2
2
2
JK = (x 2 - x 1) + (y 2 - y 1)
2
2
2
47a. KR + 500 = 1300
2
KR = 1,440,000
KR = 1200 mi
KM 2 + 390 2 = 1200 2
KM 2 = 1,287,900
KM ≈ 1135 mi
SK + KM SR + RM
500 + 1135 1300 + 390
1635 < 1690
She should fly first to King City.
b.
2
2
53. By the Inequal. Theorem, a + b > c.
By the Pythagorean Theorem, c =
By subst., a + b >
a + b2.
c
a
b
54. c 2 = a 2 + b 2
TEST PREP
c=
2
a + b2
1 hc
1 ab = _
A=_
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
51a. PA 2 = 1 2 + 1 2 = 2
PA = √2
PC 2 = 1 2 + PB 2 = 4
PC = 2
PE 2 = 1 2 + PD 2 = 6
PE = √6
a + b2.
2
2
SM
SR + RM
1360 2 1300 2 + 390 2
1,849,600 > 1,842,100
So by the Pythagorean Inequals. Theorem,
m∠SRM > 90°.
49. B;
7 2 + 24 2 25 2
625 = 625
2
2
2
ab = hc
ab = h
ab
_
=h
2
a + b2
50. H;
11 2 7 2 + 9 2
121 ≯ 130
2
2
2
PB = 1 + PA = 3
PB = √3
PD 2 = 1 2 + PC 2 = 5
PD = √5
PF 2 = 1 2 + PE 2 = 7
PF = √7
2
a + b2
55a. No; possible answer: let a = 3, b = 4, and c = 5.
So a + 1 = 4, b + 1 = 5, and c + 1 = 6. 3, 4, and
5 form a Pythagorean triple, but 4, 5 and 6 do not
because 4 2 + 5 2 ≠ 6 2.
b. Yes; possible answer: if a, b and c form a
Pythagorean triple, a 2 + b 2 = c 2 is true. Multiply
both sides by 4 to get the equation 4a 2 + 4b 2 =
4c 2. This is equivalent to (2a) 2 + (2b) 2 = (2c) 2. So
by def., 2a, 2b, and 2c also form a Pythagorean
triple.
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. No; possible answer: let a = 3, b = 4, and c =
5. So a 2 = 9, b 2 = 16, and c 2 = 25. 3, 4, and 5
form a Pythagorean triple, but 9, 16, and 25 do not
because 9 2 + 16 2 ≠ 25 2.
d. No; possible answer: let a = 3, b = 4, and c = 5.
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.
117
Holt McDougal Geometry
4. Step 1 Divide the equil. into two 30°-60°-90° .
The height of the frame is the length of the longer
leg.
Step 2 Find the length x of the shorter leg.
30 = x √
3
30 = x
_
√
3
3
30 √
_
=x
3
=x
10 √3
Step 3 Find the length s of each side of the frame.
s = 20 √
3 ≈ 34.6 cm
SPIRAL REVIEW
56. (4 + x)12 - (4x + 1)6 = 0
48 + 12x - 24x - 6 = 0
42 - 12x = 0
42 = 12x
x = 3.5
2x
5
_
57.
=x
58. 4x + 3(x + 2) = -3(x + 3)
3
4x + 3x + 6 = -3x - 9
2x - 5 = 3x
10x = -15
-5 = x
3
x = -1.5 = - _
2
59. By the Midpoint Formula, the coordinates of M are
(a, b). By the Distance Formula,
2
AM = (a - 0) 2 + (b - 0) 2 = a + b 2 and
THINK AND DISCUSS
1. Possible answer: The is a rt. , so the measure
of one ∠ is 90°, and the other 2 acute are comp.
The is isosc., so its base are . So the
measure of each of the base is 45°.
2
- a) 2 + (2b - b) 2 = a + b2.
MB = (0
So by subst., AM = MB.
−− −− −− −−−
60. JK NP, JL NM, KL < MP
m∠J > 0°
m∠J < m∠N
4x - 6 > 0
4x - 6 < 68
4x > 6
4x < 74
x > 1.5
x < 18.5
So 1.5 < x < 18.5.
−− −− −− −−
61. BA BC, BD BD, m∠ABD < m∠CBD
3x + 1 < 7
3x + 1 > 0
3x < 6
3x > -1
1
x <2
x < -_
3
1
_
So - < x < 2.
3
2. In figure I, use the relationship x = 2(8). In figure II,
first use the relationship 8 = √
3 (shorter leg), and
then use the relationship x = 2(shorter leg).
3.
3PECIAL 2IGHT
4RIANGLES
ƒ ƒ ƒ ̱
s
ƒ
s Ȗе
s
s
5-8 APPLYING SPECIAL RIGHT
TRIANGLES, PAGES 356–362
CHECK IT OUT!
1a. The is an isosc. rt.
, which is a 45°-45°90° .
) √
x = (10 √2
2 = 20
c. 24 = 2x
12 = x
y = x √
3
y = 12 √
3
ƒ
s
ƒ
ƒ
s Ȗе
EXERCISES
GUIDED PRACTICE
b. The is an isosc. rt.
, which is a 45°-45°90° .
16 = x √2
16 = x
_
√2
2
16 √
_
=x
2
x = 8 √2
1. The is an isosc. rt.
, which is a
45°-45°-90° .
x = 14 √
2
3. The is an isosc. rt.
, which is a
45°-45°-90° .
9 √
2 = x √
2
x=9
2. Tessa needs a 45°-45°-90° with hyp. of length
[C + 2(8)] cm and leg length of 42 cm.
C + 2(8) = 42 √2
C = -16 + 42 √
2 ≈ 43 cm
3 = 2x
3a. 18 √
=x
9 √3
y = x √
3
y = (9 √
3 ) √
3
y = 27
ƒ ƒ ƒ ̱
4. The sign forms a right .
Using the Pyth. Thm., we get
d = √
19.5 2+ 19.5 2
d ≈ 27.6 in.
b. x = 5 √3
y = 2(5) = 10
d.
2. The is an isosc. rt.
, which is a
45°-45°-90° .
12 = x √
2
12 = x
_
√
2
√
2
12
_=x
2
2
x = 6 √
5. 6 = 2x
3=x
y = x √
3
y = 3 √
3
3 ) √
3
7. x = (7 √
x = 21
y = 2(7 √
3)
y = 14 √
3
9 = y √
3
9 =y
_
√
3
3
9 √
_
=y
6.
15 = x √
3
15 = x
_
√
3
3
15 √
_
=x
3
5 √
3=x
y = 2x
y = 10 √
3
3
3 √
3 =y
x = 2y
x = 6 √3
118
Holt McDougal Geometry
8. Step 1 Divide the equil. into two 30°-60°-90° .
ƒ
The height of the frame is
the length of the longer leg.
h
Step 2 Find the length x
of the shorter leg.
5(2.25) = 2x
5.625 = x
Step 3 Find the length h of the longer leg.
h = 5.625 √
3 ≈ 9.75 in.
IN
x
ƒ
PRACTICE AND PROBLEM SOLVING
9. The is an isosc. rt.
, which is a
45°-45°-90° .
15 = x √
2
15 = x
_
√2
15 √2
_
=x
2
10. The is an isosc. rt.
, which is a
45°-45°-90° .
2 ) √
2=8
x = (4 √
11. The is an isosc. rt.
, which is a
45°-45°-90° .
18 √
2 = x √
2
18 = x
12. The tabletop is a 45°-45°-90° .
48 = w √2
= 2w
48 √2
≈ 33.9 in.
w = 24 √2
13. x = 2(24) = 48
y = 24 √3
15.
3 = 2x
14. 10 √
5 √
3 =x
y = x √
3
√
(
= 5 3 ) √
3 = 15
3
2 = x √
2 √
3 = 3x
3
2 √
_
=x
3
4 √3
y = 2x = _
3
()
22. Let s be the leg length.
18 = s √
2
9 √
2=s
1 = 36 nails
Hyp.: 18 ÷ _
2
1 = 72 √
2) ÷ _
2 ≈ 102 nails
Legs: 2(9 √
4
The total is approximately 138 nails.
23. No; possible answer: if the ∠ measures are in ratio
1 : 2 : 3 , then the measures of the angles are
30°-60°-90°, and the is a 30°-60°-90° . Assume
the length of the shortest leg is 1. Then the length of
the hyp. is 2, and the length of the longer leg is √
3.
So the side lengths would be in the ratio 1 : √
3 : 2.
24.
y
Q
x
R
−−
Let P = (x, y). QR is the hyp.
−−
From the diagram, QR is a 45° ∠ to the axes. P is in
−−
quad II → PQ is horizontal → y = y-coordinate of Q
−−
= 6; PR is vertical → x = x-coordinate of R = -6.
So P = (-6, 6).
b. Length of the dog walk = x + 12 + x
= 9 + 12 + 9 = 30 ft
18. 28 = 2a
12 = a √2
= 2a
12 √2
14 = a
b = a √
6 √
2 =a
3 = 14 √
3
b = a = 6 √
P=a+b+c
2
= 14 + 14 √
P=a+b+c
3 + 28
+ 6 √2
+ 12
= 6 √2
= (42 + 14 √
3 ) cm
) in.
1
_
= (12 + 12 √2
A = ab
2
1 ab
A=_
_
2
= 1 (14)(14 √
3)
2
1
_
)
= (6 √
2 )(6 √2
2
3 cm
= 98 √
2
2
= 36 in.
P
16a. The ramp forms a 30°-60°-90° . Let the length of
the ramp be x.
x = 2(4.5) = 9 ft
17.
()
s √
18 = s √
20. h = _
2
3
2
√
18 2 = 2s
= 2 √
3
9 √
2=s
P
= 3s
P = 4s
= 3(4) = 12 ft
= 4(9 √
2 ) = 36 √
2m
1 sh
2
A=_
A=s
2
2
= (9 √
2)
_
)
= 1 (4)(2 √3
= 81(2) = 162 m 2
2
2
3 ft
= 4 √
s
_
√
P = 3s
3
21.
h=
2
) = 60 √3
yd
= 3(20 √3
2h = s √
3
1
_
A = sh
60 = s √
3
2
60 √
3 = 3s
_
)(30)
= 1 (20 √3
20 √
3=s
2
3 yd 2
= 300 √
19.
25.
T
y
P
x
S
−−
Let P = (x, y). PT is the hyp.
−−
From the diagram, ST is a 45° ∠ to the axes. P is in
−−
quad I → PT is horizontal → y = y-coordinate of
2=
6 2 + 6 2 √
2
T = 3; PT = ST √
(
)
= √
144 = 12;
x = (x-coordinate of T ) + 12 = -2 + 12 = 10.
So P = (10, 3).
119
Holt McDougal Geometry
26.
y
P
35. Step 1 Identify the pattern.
2 times the length of
The length of each hyp. is _
√
3
x
the previous hyp. The length of the first hyp. is 2.
Step 2 Find x.
32
2 4(2) = _
x= _
9
√
3
( )
X
W
−−
Let P = (x, y). PX is the hyp.
−−−
From the diagram, P is in quad. II → PW is vertical
→ x = x-coordinate of W = -1; PX = WX √
3=
; y = (y-coordinate of W ) + 5 √
5 √3
3 = -4 + 5 √
3.
√
)
(
So P = -1, -4 + 5 3 .
27.
y
Y
36a. Let f be the length
of the face diagonal.
2.
Then f = e √
e = 1: f = √
2 , so
e2 + f 2
d = 3e 2
= √
=e 3
= √4
+8
= √
12 = 2 √3
e = 3: f = 3 √
2 , so
e2 + f 2
d = x
=
e2 + f 2
e 2 + 2e 2
= √
1 + 2 = √3
e = 2: f = 2 √
2 , so
e2 + f 2
d = Z
b. d =
= √9
+ 18
= √
27 = 3 √3
0
−−
Let P = (x, y). PY is the hyp.
−−
From the diagram, P is in quad. IV → PZ is vertical
→ x = x-coordinate of Z = 5; PZ = YZ √
3 = 12 √
3;
= 10 - 12 √
y = (y-coordinate of Z ) - 12 √3
3.
So P = (5, 10 - 12 √
3 ).
28. Possible answer: Both types of are rt. . In each
one, there is a unique relationship among the side
lengths. For each type of , if you know 1 side
length, you can find the other 2.
29a. NB = 2NL
= 2(320) = 640 mi
≈ 453 mi
b. IN = NL √2
c. BI = BL - IL
= NL √
3 - NL
- 320 ≈ 234 mi
= 320 √3
37. Possible answer:
Given: ABC is a 30°-60°-90° with m∠A = 30°
−−
and m∠B = 60°. CD is the altitude to the hyp.
Prove: AD = 3DB
−−
Proof: It is given that CD is the altitude to the hyp.
−− −−
Thus CD ⊥ AB by the definition of altitude. So
∠ADC and ∠BDC are rt. by the definition of ⊥,
and ADC and BDC are rt. by definition. It is
given that m∠A = 30° and m∠B = 60°. Since the
acute of a rt. are comp., m∠DCA = 60° and
m∠DCB = 30° by Subtr. Prop. of =. So ADC and
BDC are both 30°-60°-90° . By the 30°-60°-90°
(DB).
Theorem, AD = √
3 (DC) and DC = √3
(DB)). This simplifies to
By subst., AD = √
3 ( √3
AD = 3DB.
TEST PREP
30. C
D ƒ
31. F;
(5, 12, 13) is a
Pythagorean triple, and
5 + 13 = 18.
32. B;
33. 32 = 2w
24 = a √
2
w = 16
a = 12 √
2 ≈ 17.0 in.
= w √
3 = 16 √3
A = w
= 16 √
3 (16)
= 443.4 in. 2
CHALLENGE AND EXTEND
34. Step 1 Identify the pattern.
times the length of
The length of each hyp. is √2
the previous hyp.
Step 2 Write and solve an equation for x.
4
4 = ( √
2) x
4 = 4x
x=1
A
ƒ
B
C
SPIRAL REVIEW
39. y = x 2 - 10x - 2
38. y = x 2 + 4x + 0
2
2
= (x + 2) + 0 - 2
= (x - 5) 2 - 2 - 5 2
2
= (x + 2) - 4
= (x - 5) 2 - 27
Axis of symmetry:
Axis of symmetry: x = 5
x = -2
40. y = x 2 + 7x + 15
= (x + 3.5) 2 + 15 - 3.5 2
= (x + 3.5) 2 + 2.75
Axis of symmetry: x = -3.5
41. m∠ADB - 180 - 70 = 110° is obtuse. So ADB is
obtuse.
42. m∠DBC = 180 - (60 + 70) = 50°. All 3
acute, so BDC is acute.
120
are
Holt McDougal Geometry
−− −− −− −−
7. PQ ST, QR TV, and m∠Q > m∠T.
By the Hinge Theorem, PR > SV.
−− −− −− −−
8. JK JM, JL JL, and KL < ML.
By the Converse of the Hinge Theorem,
m∠KJL < m∠MJL.
−− −− −− −−
9. AD BC, BD BD, and m∠ADB < m∠DBC.
By the Hinge Theorem,
AB > 0
AB < CD
4x - 13 > 0
4x - 13 < 15
4x > 13
4x < 28
x > 3.25
x<7
3.25 < x < 7
43. m∠ABC = m∠ABD + m∠DBC
= 180 - (30 + 110) + 50
= 90°
∠ABC is a rt. ∠, so ABC is a rt. .
44. ∠PSQ and ∠PQS are comp. By the Converse of the
is the bisector of ∠PQR.
∠ Bisector Theorem, QS
So m∠PQR = 2m∠PQS
= 2(90 - m∠PSQ)
= 2(90 - 65) = 50°.
45. ∠QTV and ∠VTS are supp., and ∠TQV and ∠QTV
are comp. By the Converse of the ∠ Bisector
is the bisector of ∠PQR. So
Theorem, QS
m∠VTS = 180 - m∠QTV
= 180 - (90 - m∠TQV)
= 180 - (90 - m∠PQS)
= 180 - (90 - 42) = 132°.
10. x 2 = 5 2 + 9 2
x 2 = 106
x = √106
46. By the ∠ Bisector Theorem, PS = SR and TU = TV.
Substitute in the given equation.
SR = 3TU
PS = 3TV
7.5 = 3TV
TV = 2.5
READY TO GO ON? PAGE 365
1. Possible answer:
Given: ∠A and ∠B are supplementary. ∠A is an
acute angle.
Prove: ∠B cannot be an acute angle.
Proof: Assume that ∠B is an acute angle. By the
def. of acute, m∠A < 90° and m∠B < 90°. When
the 2 inequalities are added. m∠A + m∠B < 180°.
However, by the def. of supp., m∠A + m∠B =
180°. So m∠A + m∠B < 180° contradicts the given
information, and the assumption that ∠B is an acute
∠ is false. Therefore ∠B cannot be acute.
−−
2. KM is the shortest side, so ∠L is the least ∠.
−−
KL is the longest side, so ∠M is the greatest ∠.
From smallest to greatest, the order is ∠L, ∠K, ∠M.
12. 10 + 12 = 22 > 16
The side lengths can form a .
16 2 10 2 + 12 2
256 100 + 144
256 > 244
The is obtuse.
13. Length of the walkway = √50 2 + 80 2
= √8900 ≈ 94 ft 4 in.
14. Length of the shorter leg of a 30°-60°-90° is
36 ÷ 2 = 18 in. So h = 18 √3 ≈ 31 in.
15. x = 8 √2
16.
22 = x √2
22 √2 = 2x
11 √2 = x
17. 5 √3 = x √3
5=x
y = 2x
= 2(5) = 10
STUDY GUIDE: REVIEW, PAGES 366–369
3. m∠D = 90 - 48 = 42°, m∠E = 90°
−−
∠D is the least ∠, so EF is the shortest side.
−−
∠E is the greatest ∠, so DF is the longest side.
−− −− −−
From shortest to longest, the order is EF, DE, DF.
4. No; possible answer: the sum of 8.3 and 10.5
is 18.8, which is not greater than 18.8. By the
Inequality Thm., a cannot have these side
lengths.
2
2
2
11. a + 9 = 11
2
a + 81 = 121
a 2 = 40
a = √40 = 2 √10
The side lengths do not form a
Pythagorean triple, because
2 √10 is not a whole number.
1. equidistant
2. midsegment
3. incenter
4. locus
LESSON 5-1
5. BD = 2CD = 2(3.7) = 7.4
6.
5. Yes; possible answer: when s = 4, the value of 4s
is 16, the value of s + 10 is 14, and the value of s 2
is 16. The sum of each pair of 2 lengths is greater
than the third length. So a can have sides with
these lengths.
XY = YZ
3n + 5 = 8n - 9
14 = 5n
n = 2.8
YZ = 8(2.8) - 9 = 13.4
7. HT = FT = 5.8
8. m∠MNV = m∠PNV
2z + 10 = 4z - 6
16 = 2z
z=8
m∠MNP = 2m∠MNV
= 2[2(8°) + 10°] = 52°
6. Let d be the distance from the theater to the zoo.
d + 9 > 16
9 + 16 > d
d > 16 - 9 = 7
25 > d
Range of the distances: greater than 7 km and less
than 25 km.
121
Holt McDougal Geometry
−−
9. The midpoint of AB is (1, 0);
−− -10
slope of AB = _
= -1, so the slope of the
10
perpendicular bisector is 1;
the equation of the perpendicular bisector is
y = x -1.
−−
10. The midpoint of XY is (4, 6);
−− _
8
slope of XY = = 4, so the slope of the
2
perpendicular bisector is -0.25;
the equation of the perpendicular bisector is
y - 6 = -0.25(x - 4).
11. No; to apply the Converse of the Angle Bisector
−− −−
Theorem, you need to know that AP ⊥ AB and
−− −−
CP ⊥ CB.
−− −− −− −−
−− −−
12. Yes; since AP ⊥ AB, CP ⊥ CB, and AP CP,
P is on the bisector of ∠ABC by the Converse
of the Angle Bisector Theorem.
LESSON 5-2
13. GY = HY = 42.2
14. GP = JP = 46
15. GJ = 2GX
16. PH = JP = 46
= 2(28.8) = 57.6
−−
−−−
17. distance from A to UV = distance from A to UW
= 18
18. m∠WVU + m∠VUW + m∠UWV = 180
2m∠WVA + 2(20)+ 66 = 180
2m∠WVA = 74
m∠WVA = 37°
−−−
19. MO is vertical, so the equation of the horizontal
perpendicular bisector is y = 3;
−−
NO is horizontal, so the equation of the vertical
perpendicular bisector is x = 4.
The circumcenter is at (4, 3).
−−
20. OR is vertical, so the equation of the horizontal
perpendicular bisector is y = -3.5;
−−
OS is horizontal, so the equation of the vertical
perpendicular bisector is x = -6.
The circumcenter is at (-6, -3.5).
LESSON 5-3
2 DB
21. DZ = _
22. DB = 3ZB
3
24.6 = 3ZB
2 (24.6) = 16.4
=_
ZB = 8.2
3
23. EZ = 2ZC
24. EC = 3ZC
11.6 = 2ZC
= 3(5.8) = 17.4
ZC = 5.8
−−
25. JK is vertical, so the equation of the altitude from L
is y = 0;
−−
KL is horizontal, so the equation of the altitude from
J is x = -6.
The orthocenter is at (-6, 0).
−−
26. AB is horizontal, so the equation of the altitude from
C is x = 1;
−−
AC is vertical, so the equation of the altitude from B
is y = 2.
The orthocenter is at (1, 2).
−−
27. RT is horizontal, so the equation of the altitude from
S is x = 7;
−−
5 = 1, so the equation of the altitude
RS has slope _
5
from T is y - 3 = -(x - 8).
At the orthocenter, x = 7 and y - 3 = -(7 - 8) = 1
→ y = 4, so the orthocenter is at (7, 4).
−−
28. XY is horizontal, so the equation of the altitude from
Z is x = 3;
−−
6 = -1, so the equation of the altitude
XZ has slope _
-6
from Y is y - 2 = x - 5 or y = x - 3.
At the orthocenter, x = 3 and y = x - 3 = 0, so the
orthocenter is at (3, 0).
(
)
1 (0 + 3 + 6), _
1 (4 + 8 + 0) = (3, 4)
29. G = _
3
3
LESSON 5-4
1 XY
30. BC = _
2
1 (70.2) = 35.1
=_
2
1 XZ
32. XC = _
2
= AB = 32.4
31. XZ = 2AB
= 2(32.4) = 64.8
33. m∠BCZ = m∠ABC
= 42°
34. m∠BAX = 180° - m∠ABC
= 180° - 42° = 138°
35. m∠YXZ = m∠BCZ = 42°
−−− 2
36. V = (-1, -1); W = (6, 1); slope of VW = _
;
7
−− _
4
2
_
slope of GJ =
= ; since the slopes are the
14 7
−−− −−
same, VW GJ.
VW =
GJ =
2 2 + 7 2 = √53 ;
1 GJ.
4 2 + 14 2 = 2 √53 , so VW = _
2
LESSON 5-5
−−
37. ∠A is the smallest ∠, so BC is the shortest side;
−−
∠C is the largest ∠, so AB is the longest side;
−− −− −−
From shortest to longest, the order is BC, AC, AB.
−−
38. GH is the shortest side, so ∠F is the smallest ∠;
−−
FH is the longest side, so ∠G is the largest ∠;
From smallest to largest, the order is ∠F, ∠H, ∠G.
39. x + 4.5 > 13.5
4.5 + 13.5 > x
x>9
18 > x
Range of the values: > 9 cm and < 18 cm
40. 6.2 + 8.1 14.2
14.3 > 14.2
Yes; possible answer: the sum of each pair of
2 lengths is greater than the third length.
41. z + z 3z
2z ≯ 3z
No; possible answer: when z = 5, the value of 3z
is 15. So the 3 lengths are 5, 5, and 15. the sum of
5 and 5 is 10, which is not greater than 15. By the
Inequality Thm., a cannot have these side
lengths.
122
Holt McDougal Geometry
42. Possible answer:
Given: ABC
Prove: ABC cannot have 2 obtuse .
Proof: Assume that ABC has 2 obtuse . Let ∠A
and ∠B be the obtuse . By the definition of obtuse,
m∠A > 90° and m∠B > 90°. If the 2 inequalities
are added, m∠A + m∠B > 180°. However, by the
Sum Theorem, m∠A + m∠B + m∠C = 180°.
So m∠A + m∠B = 180° - m∠C. But then 180°
- m∠C > 180° by subst., and thus m∠C < 0°. A cannot have an ∠ with a measure less than 0°. So
the assumption that ABC has 2 obtuse is false.
Therefore a cannot have 2 obtuse .
LESSON 5-8
55. 45°-45°-90° x = 26 √2
56. 45°-45°-90° 12 = x √2
12 √2 = 2x
x = 6 √2
57. 45°-45°-90° x = (16 √2 ) √2 = 32
58. 30°-60°-90° 48 = 2x
x = 24
y = 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
14 √3
x=_
3
y = 2x
14 √3
28 √3
=2 _ =_
3
3
LESSON 5-6
−− −− −− −−
43. PQ QR, QS QS, and m∠PQS < m∠RQS.
By the Hinge Theorem, PS < RS.
−− −− −− −−
44. BC DC, AC AC, and AB < AD.
By the Converse of the Hinge Theorem,
m∠BCA < m∠DCA.
45. m∠GFH < m∠EFH
5n + 7 < 22
5n < 15
n<3
-1.4 < n < 3
46.
XZ < JK
4n - 11 < 39
4n < 50
n < 12.5
2.75 < n < 12.5
m∠GFH > 0
5n + 7 > 0
5n > -7
n > -1.4
XZ > 0
4n - 11 > 0
4n > 11
n > 2.75
(
61. The diagonal forms two 45°-45°-90°
30 = s √2
30 √2 = 2s
s = 15 √2 ≈ 21 ft 3 in.
62. The altitude forms two 30°-60°-90°
legs measure 9 ft.
h = 9 √3 ≈ 15 ft 7 in.
.
. The shorter
CHAPTER TEST, PAGE 370
1. KL = JK = 9.8
2. m∠WXY = 2m∠WXZ = 2(17) = 34°
3.
LESSON 5-7
47. a 2 + b 2 = c 2
22 + 62 = x 2
40 = x 2
x = 2 √10
)
48. a 2 + b 2 = c 2
x 2 + 8 2 = 14 2
x 2 = 132
x = 2 √33
49.
2
2
2
a +b =c
2
2
x + (4.5) = (7.5) 2
x 2 = 36
x =6
The side lengths do not form a Pythagorean triple
because 4.5 and 7.5 are not whole numbers.
50.
a 2 + b 2= c 2
24 2 + 32 2 = x 2
1600 = x 2
x = 40
The side lengths form a Pythagorean triple because
they are nonzero whole numbers that satisfy
52. 11 + 14 = 25 ≯ 27
The 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 = 15.21
The side lengths can
form a rt. .
54. 2 + 3.7 = 5.7 > 4.1
4.1 2 2 2 + 3.7 2
16.81 < 17.69
The side lengths can
form an acute .
4. RS = 2MS
= 2(3.4) = 6.8
RQ = SQ = 4.9
5. m∠DEF + m∠EFD + m∠FDE = 180
2m∠GEF + 2(25) + 42 = 180
2m∠GEF = 88
m∠GEF = 44°
−−
−−
distance from G to DF = distance from G to DE
= 3.7
2
_
6. XW = XC
3
2 (261) = 174
=_
3
1
_
BW = ZW
2
1 (118) = 59
=_
2
BZ = 3BW
= 3(59) = 177
−−
7. JK is vertical, so the equation of the horizontal
altitude is y = 4;
−− -6
slope of KL is _
= -1, so the slope of the altitude
6
is 1, and its equation is y - 2 = x + 5, or y - 7 = x.
At the orthocenter, y = 4 and x = 4 - 7 = -3.
The orthocenter is at (-3, 4).
a 2 + b 2 = c 2.
51. 9 + 12 = 21 > 16
16 2 9 2 + 12 2
256 > 225
The side lengths can
form an obtuse .
AC = BC
2n + 9 = 5n - 9
18 = 3n
n =6
BC = 5(6) - 9 = 21
123
Holt McDougal Geometry
1 HJ
8. PR = _
2
= QJ = 51
GJ = 2PQ
= 2(74) = 148
m∠GRP = m∠GJH = 71°
9. Possible answer:
Given: ∠1 and ∠2 form a lin. pair.
Prove: ∠1 and ∠2 cannot both be obtuse .
Proof: Assume ∠1 and ∠2 are both obtuse . By
the definition of obtuse, m∠1 > 90° and m∠2 > 90°.
If the 2 inequalities are added, m∠1 + m∠2 > 180°.
However, by the Lin. Pair Theorem, ∠1 and ∠2 are
supp. . By the definition of supp. , m∠1 + m∠2 =
180°. So m∠1 + m∠2 > 180° contradicts the given
information. The assumption that ∠1 and ∠2 are
both obtuse is false. Therefore ∠1 and ∠2 cannot
both be obtuse.
−−
10. BH is the shortest side, so ∠E is the smallest ∠.
−−
BE is the longest side, so ∠H is the largest ∠.
From smallest to largest, the order is ∠E, ∠B, ∠H.
−−
11. ∠R is the smallest ∠, so TY is the shortest side.
−−
∠Y is the largest ∠, so RT is the longest side.
−− −− −−
From shortest to longest, the order is TY, RY, RT.
−− −− −− −−
14. DH KH, HN HN, and DN < KN.
m∠DHN > 0°
m∠DHN < m∠KHN
4x - 10 > 0
4x - 10 < 24
4x > 10
4x < 34
x > 2.5
x < 8.5
2.5 < x < 8.5
15. x 2 + 21 2 = 24 2
x 2 = 135
x = 3 √
15
The side lengths do not form a Pythagorean triple
because 3 √
15 is not a whole number.
16. 18 + 20 = 38 > 27
27 2 18 2 + 20 2
729 > 724
The side lengths can
form an obtuse .
18.
2
20 = x √
20 √
2 = 2x
x = 10 √
2
20.
8 = x √
3
8 √
3 = 3x
8 √3
x=_
3
y = 2x
3
3
8 √
16 √
=2 _ =_
3
3
12. AC + 114 > 247
114 + 247 > AC
AC > 133
361 > AC
Range of the distance: > 133 mi and < 361 mi.
−− −− −− −−
13. PS PZ, PV PV, and SV < ZV.
By the Converse of the Hinge Theorem,
m∠SPV < m∠ZPV.
17. c 2 = 62 2 + 82 2
= 10,568
c = √
10,568
≈ 102 ft 10 in.
19. 32 = 2x
x = 16
y = x √
3 = 16 √
3
( )
124
Holt McDougal Geometry