Document 1805783

Chapter 5,
continued
Mixed Review for TEKS (p. 317)
4. G; Total length of the blue segments:
1. A; Because nBGF > nBGE by the Hypotenuse-Leg
LM 1 LZ 1 MZ 1 NP 1 NZ 1 PZ
Congruence Theorem. So BE 5 BF 5 3 centimeters,
so AE 5 AB 2 EB 5 10 2 3 5 7 centimeters.
5 30 1 }2(73) 1 }2 (91) 1 30 1 }2 (80) 1 }2 (100)
(AG)2 5 (EG)2 1 (AE)2
5 232 feet
1
82 5 (EG)2 1 72
1
64 5 (EG) 1 49
AB 1 BC 1 BD 1 CD 1 AD
15 5 (EG)2
5 91 1 80 1 2(30) 1 100 1 73
}
Ï15 5 EG
5 404 feet
3.9 ø EG
So, 232 1 404 5 636 feet of railroad ties are needed to
form the garden.
The radius of the wheel is about 3.9 centimeters.
2. J; In the graph, points A, B, and C are the endpoints of
5. By the Converse of the Perpendicular Bisector Theorem,
the target is on the perpendicular bisector of the segment
along the edge of the football field. This perpendicular
bisector forms a right triangle with a leg of 12 feet and a
hypotenuse of 15 feet.
the midsegments of nGHJ.
y
H(6, 8)
B(8, 6)
152 5 a2 1 122
J(10, 4)
2
G(2, 2) C(6, 3)
225 5 a2 1 144
x
22
81 5 a2
}
Equation of the line containing AB:
1
625
m5}
5 }4
824
95a
The target is 9 feet from the edge of the football field.
6.
y 5 mx 1 b
1
5 5 }4 (4) 1 b
3x 2 7
L
45b
x 1 10
T
K
U
8x 2 18
1
1
2
y 5 }4x 1 4
} (JL) 5 TU
}
Equation of the line containing BC:
326
23
3
m5}
5}
5 }2
22
628
1
2
}(4x 1 6) 5 3x 2 7
2x 1 3 5 3x 2 7
10 5 x
y 5 mx 1 b
1
LU 5 }2(KL)
3
6 5 }2 (8) 1 b
1
26 5 b
5 }2(8x 2 18)
3
y 5 }2 x 2 6
5 4x 2 9
}
Equation of the line containing AC:
325
J
4x 1 6
V
22
m5}
5}
5 21
2
624
y 5 mx 1 b
5 5 21(4) 1 b
95b
y 5 2x 1 9
So, the line y 5 2x 1 14 does not contain one of these
midsegments.
3. C; nZSP > nZSQ and nZTQ > nZTR by the SAS
Congruence Postulate. Because corresponding parts of
} }
congruent triangles are congruent, PZ > QZ and
} }
QZ > RZ, so PZ 5 QZ and QZ 5 RZ. By the Transitive
Property of Equality, PZ 5 QZ 5 RZ, which means that
point Z is equidistant from the vertices of nPQR.
Geometry
Worked-Out Solution Key
5 4(10) 2 9
5 31
}
The length of LU is 31 units.
Lesson 5.4
Investigating Geometry Activity 5.4 (p. 318)
1. Sample answer:
Length of segment from AD 5
vertex to midpoint of
27 mm
opposite side
BF 5
18 mm
CE 5
36 mm
Length of segment from AP 5
vertex to P
18 mm
BP 5
12 mm
CP 5
24 mm
Length of segment from PD 5
P to midpoint
9 mm
PF 5
6 mm
PE 5
12 mm
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
A(4, 5)
1
Total length of red segments:
2
134
1
Chapter 5,
continued
2. The length of the segment from a vertex to P is twice
2. A perpendicular bisector is perpendicular to a side of
the length of the segment from P to the midpoint of the
opposite side.
the triangle at its midpoint and doesn’t necessarily pass
through a vertex. An altitude is a perpendicular segment
from a vertex to the opposite side, but not necessarily at
the midpoint of the opposite side. A median is a segment
from a vertex to the midpoint of the opposite side, but is
not necessarily perpendicular to that side.
2
3. The length of the segment from a vertex to P is } the
3
length of the segment from the vertex to the midpoint of
the opposite side.
5.4 Guided Practice (pp. 320–321)
2
2
1. PC 5 } SC 5 } (2100 ft) 5 1400 ft
3
3
3. FC 5 AF 5 12
2
4. BG 5 } BF
3
2
PS 5 SC 2 PC 5 2100 ft 2 1400 ft 5 700 ft
2. TC 5 BT 5 1000 ft
BC 5 BT 1 TC 5 1000 ft 1 1000 ft 5 2000 ft
2
3. PT 1 PA 5 TA
PA 5 }3 TA
2
6 5 }3 BF
5 }3 (15)
9 5 BF
5 10
6. GE 5 AE 2 AG 5 15 2 10 5 5
2
7. D; AM 5 } AQ
3
2
2
800 ft 1 }3 TA 5 TA
AM 5 }3 (AM 1 30)
1
800 ft 5 }3 TA
2
AM 5 }3 AM 1 20
2400 ft 5 TA
1
3
} AM 5 20
2
PA 5 }3 (2400 ft) 5 1600 ft
4.
2
5. AG 5 } AE
3
P (orthocenter)
AM 5 60
1
2
5 1 11 5 1 (23)
8. a. P }, } 5 P(8, 1)
2
2
For U(21, 1) and P(8, 1), UP 5 8 2 (21) 5 9 units.
2
B
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
A
D
So, the centroid Q is }3(9) 5 6 units to the right of U
}
on UP.
C
Q(21 1 6, 1) 5 Q(5, 1)
1
that ŽABD > ŽCBD because corresponding parts
s are >. Also ŽABD and ŽCBD are adjacent.
of > n
}
Therefore, BD is an angle bisector.
}
6. OQ is also a perpendicular bisector, an angle bisector,
and a median.
P(2h, 0)
2
11 1 (21) 23 1 1
b. R }, } 5 R(5, 21)
2
2
5. Once you prove that nABD > nCBD, you would state
SQ 5 5 2 1 5 4 units
SR 5 5 2 (21) 5 6 units
2
2
4 5 }3(6), so SQ 5 }3 SR.
9.
y
B(5, 6)
5.4 Exercises (pp. 322–325)
Skill Practice
1. The point of concurrency of the perpendicular bisectors
of a triangle is the circumcenter. The circumcenter is
inside the triangle when the triangle is acute, on the
triangle when the triangle is right, and outside the
triangle when the triangle is obtuse.
The point of concurrency of the angle bisectors of a
triangle is the incenter. The incenter always lies inside
the triangle. The point of concurrency of the medians
of a triangle is the centroid. The centroid is always
inside the triangle. The point of concurrency of the lines
containing the altitudes of a triangle is the orthocenter.
The orthocenter is inside the triangle when the triangle
is acute, on the triangle when the triangle is right, and
outside the triangle when the triangle is obtuse.
A(1, 2)
P(3, 2)
1
x
1
C(5, 2)
515 622
}
Midpoint of BC: F 1 }
,}
5 F(5, 2)
2
2 2
AF 5 5 2 (21) 5 6
}
2
So, the centroid P is }3(6) 5 4 units right from A on AF.
P(21 1 4, 2) 5 P(3, 2)
Geometry
Worked-Out Solution Key
135
Chapter 5,
10.
y
A(0, 4)
continued
2
16. You cannot conclude that NT 5 } NQ because the point
3
B(3, 10)
of concurrency shown is not the centroid of the triangle.
The segments shown are altitudes, not medians.
}
17. Altitude, because YW is a perpendicular segment from
}
V to XZ
P(3, 4)
18. Angle bisector, because ŽXYZ is bisected.
}
19. Median, because W is the midpoint of XZ.
20. Perpendicular bisector, angle bisector, median, and
F(3, 1)
x
1
altitude, because nXYW > nZYW by SAS.
21. Perpendicular bisector, angle bisector, median and
C(6, 2)
1
2
0 1 6 4 1 (22)
Midpoint of AC: F }
,}
5 F(3, 1)
2
2
}
BF 5 10 2 1 5 9 units.
2
So, the centroid P is }3 (9) 5 6 units down from B
}
on BF. P(3, 10 2 6) 5 P(3, 4)
y
B(6, 6)
P
1
F(6, 0)
1
C(12, 0) x
2
}
The centroid P is }3 (6) 5 4 units down from B on BF.
P(6, 6 2 4) 5 P(6, 2)
12. Sample answer: Draw obtuse scalene nABC.
y
22. Perpendicular bisector, angle bisector, median and
altitude, because n XYW > nZYW by HL, so ŽXYZ is
} }
bisected and YW > XZ at its midpoint.
23. By Theorem 3.9, mŽBDC 5 mŽBDA 5 908.
} } } }
So, ŽBDC > ŽBDA. Because BC > BA, BD > BD,
} }
then nBCD > nBAD by HL. Therefore, DC > BA,
ŽCBD > ŽABD, DC 5 BA 5 6, and
mŽABD 5 mŽCBD 5 228.
} } } }
24. Because BC > BA, BD > BD, AD 5 DC then
n ABD > nCBD by SSS. So, ŽCDB > ŽADB. By the
definition of a linear pair, mŽCDB 5 mŽADB 5 908.
mŽABD 5 mŽCBD 5 228
11. Sample answer:
A(0, 0)
} }
altitude, because ŽXYZ is bisected and YW > XZ at
its midpoint.
P
1
25. 3; }EJ 5 KJ
3
y
B(5, 6)
EJ 5 3KJ
2
1
26. 2; DK 5 }DH; KH 5 }DH
3
3
1
DK 5 2 }3 DH 5 2KH
2
3
27. }; KF 5 }FG
3
2
3
} KF 5 FG
2
1
2
A(1, 2)
P(3, 2)
1
x
1
C(5, 2)
28. Any isosceles triangle can be placed in the coordinate
B
1
1
A(0, 0)
}
D
}
C
}
x
Extend AB and CB. Draw and extend DB, the altitude
}
through B. Use the reciprocal of the slope of CB to draw
###$
an altitude through A that intersects DB. The intersection
}
P is the orthocenter. CP is the altitude through C.
}
}
13. No; no; yes; D is not at the midpoint of AC, so BD cannot
}
be a perpendicular bisector or a median. Because BD >
} }
AC, BD is an altitude.
}
} }
14. Yes; yes; yes; D is at the midpoint of AC and BD > AC,
}
so BD is a perpendicular bisector, a median, and
an altitude.
}
}
}
15. No; yes; no; BD is not perpendicular to AC, so BD cannot
be a perpendicular bisector or a median, but because D is
} }
the midpoint of AC, BD is a median.
136
Geometry
Worked-Out Solution Key
plane with its base on the x-axis, centered at the origin
and the opposite vertex on the y-axis. This is because in
an isosceles triangle, the perpendicular bisector, angle
bisector, median, and altitude from the vertex angle to
the base are all the same segment, which is perpendicular
to the base at its midpoint.
29–31. Sample answers:
29. Equilateral n ABC: All 3 altitudes intersect at P
inside n ABC.
B
P
A
C
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
Chapter 5,
continued
30. Right scalene n ABC: All 3 altitudes intersect at vertex A
36.
K
of the right angle.
h
A
J
B
N
10
17
M
9
L
1
9
1
9
Area of nJKM 5 }2 (9)(h) 5 }2 h
C
Area of nLKM 5 }2 (9)(h) 5 }2 h
31. Obtuse isosceles nABC: All 3 altitudes intersect at
vertex D outside nABC.
For nKNL, h2 1 (NM 1 9)2 5 172
For nKNM, h2 1 (NM)2
D
5 102
18(NM) 1 81 5 189
NM 5 6
}
h 5 Ï 102 2 62 5 8
B
9
Area of nJKM 5 Area of nLKM 5 }2 3 8 5 36
A
The areas of the two triangles are the same. Any two
triangles determined in this way have the same height
C
1
2
416 911
}
32. Midpoint J of GH: J }, } 5 J(5, 5)
2
2
}
So, median FJ from F(2, 5) to J(5, 5) passes
through P(4, 5).
2
FJ 5 3 and FP 5 2, so FP 5 }3 FJ.
}
Therefore, Theorem 5.8 holds for FJ.
1
and congruent bases. So the area A 5 }2 bh will always be
the same.
Problem Solving
37.
Midpoint L of FG: L 1 }
,}
5 L(3, 7)
2
2 2
214 519
A
B
} 721
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
5 22
Slope of median HL: }
326
}
So, median HL passes through P(4, 5) because
moving L(3, 7) with a rise of 22 and run of 1 you
have (3 1 1, 7 1 [22]) 5 (4, 5).
}}
}
}
HP 5 Ï(4 2 6)2 1 (5 2 1)2 5 Ï20 5 2Ï 5
}}
}
You should place the rod at B because the medians
intersect at B. A triangle will balance at the intersection
of the medians.
38.
y
y
(n, h)
(0, h)
}
HL 5 Ï(3 2 6)2 1 (7 2 1)2 5 Ï45 5 3Ï5
2
}
So, HP 5 }3 HL. Therefore, Theorem 5.8 holds for HL.
2
33.
BD 5 }3 BF
2
4x 1 5 5 }3(9x)
4x 1 5 5 6x
5 5 2x
5
2
}5x
1
35. } AD 5 DE
2
1
2
} (5x) 5 3x 2 2
1
3
34.
}GC 5 GD
1
3
}(3x 1 3) 5 2x 2 8
x 1 1 5 2x 2 8
95x
x
(2n, 0)
(n, 0)
x
(0, 0) (2n, 0)
1
39. The base is } (9) 5 4.5 inches and the height is 3 inches.
2
1
So, the area is A 5 }2 (4.5)(3) 5 6.75 in.
The special 3 inch segment is an altitude.
40. The orthocenter in a right triangle is at the vertex where
the legs intersect because the legs are also altitudes. A
right triangle is the only type of triangle with a point of
concurrency at a vertex.
5
2
} x 5 3x 2 2
1
2 5 }2 x
45x
Geometry
Worked-Out Solution Key
137
Chapter 5,
b.
y
8
(4, 8)
3
(0, 2)
(24, 2)
2
x
(0, 24)
3
y15 3x 2 4
y35 2 2 x 2 4
midpoint of side contained in y1:
0 1 4 24 1 8
,}
5 (2, 2)
1}
2
2 2
midpoint of side contained in y2:
1
Statements
y25 4 x 1 5
2
24 1 4 2 1 8
}, } 5 (0, 5)
2
2
Reasons
1. nABC is equilateral,
}
BD is an altitude of
n ABC.
1. Given
2. AB 5 BC
2. Definition of
equilateral triangle
3. B is on the
perpendicular bisector
}
of AC.
} }
4. BD > AC
}
5. BD is a perpendicular
}
bisector of AC.
3. Converse of
Perpendicular
Bisector Theorem
4. Definition of altitude
5. Perpendicular
Postulate
43. a. Answers will vary.
midpoint of side contained in y3:
24 1 0
,}
1}
2 5 (22, 21)
2
2
b. The areas of the blue, green, and red triangles are
The centroid is (0, 2).
c. Because their areas are the same, the weights of three
2 1 (24)
the same.
42. a. Given: nABC is equilateral.
such triangles cut out of some material of uniform
thickness would be the same. This helps to support the
possibility that the weight is balanced at the centroid.
B
}
BD is an altitude of nABC.
}
Prove: BD is also a
}
perpendicular bisector of AC.
44. a.
A
Statements
1. n ABC is equilateral,
}
BD is an altitude
of ABC.
} }
2. AB > BC
}
D
Reasons
1. Given
2. Definition of
equilateral triangle
3. BD > BD
3. Reflexive Property of
Segment Congruence
4. ŽADB and ŽCDB
are right angles.
4. Definition of altitude
5. n ABD and nCBD
are right triangles.
5. Definition of right
triangle
6. n ABD > nCBD
} }
7. AD > CD
6. HL
}
9. BD is a perpendicular
}
bisector of AC.
Q(8, 12)
C
}
8. AD 5 CD
y
s
7. Corr. parts of > n
are >.
8. Definition of
congruent segments
9. Definition of
perpendicular bisector
1
1
P(0, 0)
R(14, 0) x
} 12 2 0 3
Slope of PQ: }
5 }2
820
}
Because it is perpendicular to PQ, the slope of the
}
altitude from R to PQ is the negative reciprocal of the
2
}
slope of PQ, or 2}3 .
}
b. Equation of altitude from R to PQ:
y 5 mx 1 b
2
0 5 2}3 (14) 1 b
28
3
}5b
2
28
y 5 2}3 x 1 }
3
}
Equation of altitude from Q to PR:
x58
}
Equation of altitude from P to QR:
} 0 2 12
Slope of QR: }
5 22
14 2 8
} 1
Slope of altitude from P to QR: }2
1
The y-intercept is 0, so y 5 }2 x.
138
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
41.
continued
Chapter 5,
continued
c. Prove R, N, and S are collinear.
The system of equations is:
28
2
y 5 2}3x 1 }
3
Statements
x58
1
y 5 }2 x
1
Because x 5 8, y 5 }2 (8) 5 4.
The orthocenter is (8, 4).
c. To find the circumcenter, you would find the
midpoint of each side of the triangle and draw each
perpendicular bisector. Then you would write an
equation for each perpendicular bisector and find the
ordered pair that is a solution of each of the three
equations.
}
}
45. Given: LP and MQ are medians of scalene nLMN.
} }
Point R is on ###$
LP such that LP > PR. Point S is on
} }
####$
MQ such that MQ > QS.
} }
a. Prove: NS > NR
S
L
Q
1. Proven in part (b).
2. There is exactly one line
through N parallel to
}
LM, so @##$
RN and @##$
SN are
the same line.
2. Parallel Postulate
3. R, N, and S
are collinear.
3. Definition of
collinear points
Mixed Review for TAKS
46. D; Each time the number of wind chimes sold increases
by 1, the profit increases by $13. So, there is a linear
relationship between number sold and the profit, whose
rate of change is $13. This relationship can be written as
an equation in the form p 5 13w 1 b, where p is Sylvia’s
profit and w is the number of wind chimes she sells.
Substitute values from the table to find b.
p 5 13w 1 b
O
M
Reasons
} } } }
1. NS i LM, NR i LM
P
N
212 5 13(1) 1 b
225 5 b
So, the equation p 5 13w 2 25 represents Sylvia’s profit,
p, when she sells w wind chimes.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
R
Statements
} }
1. LP, MQ are medians
} }
of nLMN, LP > PR,
} }
MQ >QS
Reasons
1. Given
3
2. Definition of median
4. ŽLQM > ŽNQS,
ŽLPM > ŽRPN
4. Vertical Angles
Congruence Theorem
5. nLQM > nNQS
nLPM > nRPN
} } } }
6. NS > LM, NR > ML
5. SAS
}
51 5 4 + 12 1 3
751 5 74 + 12 1 3
751 5 (74)12 + 73
3. Definition of midpoint
According to the pattern, 748 5 (74)12 has a 1 in the ones
place. So, the digit in the ones place in 751 is the same as
the digit in the ones place in 73, which is 3.
Quiz 5.3–5.4 (p. 325)
1. 3x 2 4 5 2x 1 6
s
6. Corr. parts of > n
are >.
7. Transitive Property
of Congruence
}
divide 51 by 4 to get a remainder of 3.
51 4 4 5 12 1 }4
2. Q is the midpoint
}
of LN, P is the
}
midpoint of MN.
} } } }
3. LQ > NQ, MP > NP
} }
7. NS > NR
47. G; The pattern repeats itself every fourth power. So,
}
b. Prove: NS and NR are both parallel to LM.
x 5 10
Angle Bisector Theorem
2. 2x 1 2 5 x 1 7
x55
Concurrency of Angle Bisectors of a Triangle Theorem
3. LY 5 LX 5 15
Statements
Reasons
1. nLQM > nNQS,
nLPM > nRPN
1. Proven in part (a).
2. ŽQLM > ŽQNS,
ŽLMP > ŽRNP
}i} }i}
3. NS LM, NR LM
s
2. Corr. parts of > n
are >.
2
4. YP 5 }YN
3
2
3. Alt. Int. Angles
Converse
12 5 }3YN
18 5 YN
1
1
5. LP 5 }LZ 5 }(18) 5 6
3
3
Geometry
Worked-Out Solution Key
139
Chapter 5,
continued
Technology Activity 5.4 (pp. 326–327)
7. The orthocenter and circumcenter are sometimes outside
the triangle. Both lie outside the triangle only when it is
obtuse. The incenter and centroid never lie outside the
triangle. The incenter is always in the interior of the three
angles of the triangle so it is always inside the triangle.
The centroid is always two-thirds of the distance from
each vertex to the midpoint of the opposite side, a point
that is always inside the triangle.
1.
F
C
B
E
A
D
8.
C
G
D
Yes, points D, E, and F are collinear.
E
2.
A
B
F
F
C
B
E
Both the incenter and the centroid are contained by
the triangle formed by the midsegments. They remain
contained when the shape of the triangle is changed.
D
A
Yes, points D, E, and F remain collinear.
3.
5.5 Guided Practice (pp. 329–330)
} }
}
1. ST, RS, and RT, because the side opposite the smallest
}
}
angle is ST and the side opposite the largest angle is RT.
F
2.
C
B
378
E
A
8 ft
D
10 ft
538
6 ft
4. G is not collinear with the three collinear points
D, E, and F.
3.
5.
15 in.
F
G A
11 in.
x in.
11 in.
15 in.
x in.
E
C
B
D
11 1 15 > x
26 > x
x 1 11 > 15
x>4
The length of the third side must be greater than 4 inches
and less than 26 inches.
D, E, and F remain collinear. G is collinear with
D, E, and F only when the triangle is isosceles.
6.
C
D
E
A
B
F
Yes, D, E, and F remain collinear no matter how the
triangle’s vertices are moved.
140
Geometry
Worked-Out Solution Key
5.5 Exercises (pp. 331–334)
Skill Practice
}
1. BC is opposite ŽA.
}
AC is opposite ŽB.
}
AB is opposite ŽC.
2. By Theorem 5.11, you can tell that the longest side
is opposite the largest angle and the shortest side is
opposite the smallest angle.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
G
Lesson 5.5