Document 1806979

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
Chapter 5,
continued
3.
18. Yes, using the Triangle Inequality Theorem:
28 1 34 > 39
Blue
d
Re
62 > 39 Red
Blue
The longest side is opposite the largest angle and the
shortest side is opposite the smallest angle.
4.
ue
Bl
Bl
Red
Blue
The longest side is opposite the obtuse angle and the two
congruent legs are opposite the two congruent angles.
} } }
} } }
6. AB, BC, AC
7. XY, YZ, XZ
ŽC, ŽA, ŽB
} } }
8. RT, ST, RS
ŽZ, ŽX, ŽY
} } }
9. KL, JL, JK
ŽS, ŽR, ŽT
ŽJ, ŽK, ŽL
Choice B:
21. x 1 5 > 12
ŽM, ŽP, ŽN
x>7
11. mŽD 5 1808 2 (908 1 338) 5 578
DF, FG, DG
ŽG, ŽD, ŽF
12. C;
mŽR 5 1808 2 (658 1 568) 5 598
Because mŽR > mŽT, ST > RS. Therefore, ST > 8.
1238
3m
9m
13 > 9 17 > x
x>1
7>x
The length of the third side must be greater than 1 meter
and less than 7 meters.
12 1 18 > x
30 > x
The length of the third side must be greater than 6 feet
and less than 30 feet.
24. x 1 10 > 23
12 ft
x > 13
718
198
17 > 15 314>x
x>6
418
35 ft
5 1 8
>9
5 1 12 > x
22. x 1 3 > 4
23. x 1 12 > 18
7m
9 1 8
>5
The length of the third side must be greater than 7 inches
and less than 17 inches.
} } }
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
13ò15
14 > 8 NP, MN, MP
10 1 23 > x
33 > x
The length of the third side must be greater than 13 yards
and less than 33 yards.
37 ft
718
9 ft, 4 ft, 15 ft
9 1 4
> 15
9 ft, 5 ft, 8 ft
9 1 5
>8
} } }
15.
245 > 35 20. B;
3 yd 5 9 ft
10. mŽm 5 1808 2 (1278 1 298) 5 248
14.
155 > 125 3 ft 5 1 yd
Blue
168
35 1 120 > 125
Choice A:
ue
Red
13.
73 > 28 120 1 125 > 35
The longest side is opposite the 908 angle and the
shortest side is the leg opposite the smallest angle.
5.
67 > 34 19. Yes, using the Triangle Inequality Theorem:
160 > 120 Red
Blue
34 1 39 > 28
35 1 125 > 120
Blue
d
Re
28 1 39 > 34
11 in.
25. 2 ft 5 24 in.
618
24 inches, 40 inches
13 in.
488
14 in.
16. Yes, using the Triangle Inequality Theorem:
6 1 7
> 11
13 > 11 6 1 11 >7
7 1 11 >6
17 > 7 17. No, using the Triangle Inequality Theorem:
3 1 6
>9
18 > 6 x 1 24 > 40
24 1 40 > x
x > 16
64 > x
The length of the third side must be greater than
16 inches and less than 64 inches.
26. x 1 25 > 25
x>0
25 1 25 > x
50 > x
The length of the third side must be less than 50 meters.
9ò9
Geometry
Worked-Out Solution Key
141
Chapter 5,
continued
27. By the Exterior Angle Theorem, mŽ1 5 mŽA 1 mŽB.
34. (6x 2 11) 1 (3x 2 1) > 2x 1 3
Therefore, mŽ1 > mŽA and mŽ1 > mŽB.
9x 2 12 > 2x 1 3
28. ŽE is an exterior angle of nDEG and ŽD is a
7x > 15
nonadjacent interior angle. Therefore, mŽE > mŽD.
15
x>}
7
29. By Theorem 5.10, ŽQNP must be the largest angle
of nNPQ because it is opposite the longest side.
Because ŽQNP and ŽLNM are vertical,
mŽQNP 5 mŽLNM 5 598. Then by the Triangle Sum
Theorem, 598 1 mŽP 1 mŽQ 5 1808;
so mŽP 1 mŽQ 5 1218. Then 1218 > 2mŽQ,
mŽQq60.58 by Theorem 5.10. This contradicts the
fact that ŽQNP must be the largest angle of nNPQ.
Therefore, the diagram must be incorrect.
30. The hypotenuse of a right triangle is the side opposite the
right angle. The right angle is the largest angle of a right
triangle. Therefore, by Theorem 5.11 the hypotenuse
must always be longer than either leg.
31.
PQ 1 QR > PR
Ï58 1 2Ï13 > 5Ï 2
}
13 > x
The value of x must satisfy all three inequalities, so x
15
and less than 13.
must be greater than }
7
Because mŽWYX 5 mŽXYZ,
mŽWXY 1 mŽW 5 mŽZ 1 mŽYXZ.
Then, because mŽW > mŽYXZ, mŽZ > mŽWXY.
Ï58 1 5Ï2 > 2Ï 13
}
So, ŽWXY is the smallest angle.
14.7 > 7.2 Yes, it is possible to build a triangle with the given
side lengths.
}
PQ 5 Ï58 ø 7.6
From smallest to greatest, the angles are:
ŽWXY, ŽZ, ŽYXZ, ŽWYX > ŽXYZ, ŽW
36. In nGJH:
x13>4
}
QR 5 2Ï13 ø 7.2
}
PR 5 5Ï2 ø 7.1
From the least to greatest measure, the angles are
ŽQ, ŽP, and ŽR.
32.
ST 1 TU > SU
}
}
Ï29 1 2Ï17 > 13.9
13.6 ò13.9
No, it is not possible to build a triangle with the given
side lengths.
33. (x 1 11) 1 (2x 1 10) > 5x 2 9
3x 1 21 > 5x 2 9
30 > 2x
15 > x
(x 1 11) 1 (5x 2 9) > 2x 1 10
6x 1 2 > 2x 1 10
4x > 8
x>2
(2x 1 10) 1 (5x 2 9) > x 1 11
7x 1 1 > x 1 11
6x > 10
5
x > }3
The value of x must satisfy all three inequalities, so x
must be greater than 2 and less than 15.
142
5x 1 2 > 6x 2 11
nXYZ: ŽZ, ŽYXZ, ŽXYZ
}
}
(3x 2 1) 1 (2x 1 3) > 6x 2 11
nWXY: ŽWXY, ŽWYX, ŽW
14.3 > 7.6 PQ 1 PR > QR
}
7
x > }5
triangle, the angles are as follows.
> Ï 58
5Ï2 1 2Ï 13 }
5x > 7
35. By Theorem 5.10, from the smallest to largest in each
}
14.8 > 7.1 PR 1 QR > PQ
}
8x 2 8 > 3x 2 1
Geometry
Worked-Out Solution Key
314>x
x>1
7>x
}
The length of GH must be greater than 1 and less than 7.
In nFJH:
y14>5
415>y
y>1
9>y
}
The length of FH must be greater than 1 and less than 9.
In nFJG:
z13>5
315>z
z>2
8>z
}
The length of FG must be greater than 2 and less than 8.
P5x1y1z
P>11112
P>4
71918>P
24 > P
So, the perimeter of nHGF must be between 4 and 24.
Problem Solving
37. mŽR > mŽQPR and mŽQ > mŽQPR by Theorem 5.10.
mŽR 5 mŽQ by the Base Angles Theorem.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
(6x 2 11) 1 (2x 1 3) > 3x 2 1
Chapter 5,
continued
38. You know that AB is greater than 35 yards and less than
43. Sample answer:
50 yards. To get a closer estimate, you could sighting
point C closer to A and sighting point D farther from A
so the two angles are closer to 458.
39. a. By the Triangle Inequality Theorem, you know that
489 km 1 565 km > x km, or 1054 km > x km.
Therefore, x cannot be 1080.
5 cm, 5 cm, 14 cm
2 cm, 9 cm, 13 cm
6 cm, 6 cm, 12 cm
44. x 1 359 > 709
359 1 709 > x
x > 350
1068 > x
b. No, by the Triangle Inequality Theorem, you know
that 489 km 1 x km > 565 km, or x km > 76 km.
Therefore, x cannot be 40.
c. x 1 489 > 565
The range of possible distances you might have to walk is
greater than 350 yards and less than 1068 yards.
45.
2 mi
489 1 565 > x
x > 76
d. Because Ž2 is the smallest angle, the distance must be
less than 489 kilometers.
3
3
1
2 }4 > x
Special cases for collinear locations:
2 mi.
House
3
4
Grocery
Store
mi.
Grocery
Store
mi.
Library
The distance from your house to the grocery store is
Try the Pythagorean Theorem:
72 1 72 0 102
1
greater than or equal to 1 }4 miles and less than or equal
98 Þ 100
3
The longest side is too long to satisfy the Pythagorean
Theorem, so the angle is obtuse. So, the triangle is obtuse.
6.5cm
9 cm
6.5cm
8 cm
8 cm
1cm
8 cm
6 cm
Obtuse isosceles
Acute isosceles
42. Sample answer:
4cm
9cm
8 cm
7cm
Obtuse triangle
to 2 }4 miles.
46. The congruent sides are shorter than the base in an
isosceles triangle when the measure of the angle
opposite the base is greater than 608. The congruent sides
are longer than the base when the measure of the angle
opposite the base is less than 608.
47. Given: n ABC
B
Prove: (1) AB 1 BC > AC
9cm
10 cm
3
4
Library
7 cm
11 cm
mi
3
4
House
10 cm
3
4
}12>x
x > 1 }4
41. Sample answer:
Acute isosceles
2 mi
2 mi.
greater than the longest side length, then the side
lengths form a triangle.
9 cm
mi
x 1 }4 > 2
40. Yes. If the sum of the shortest two side lengths is
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3
4
x mi
1054 > x
The range of possible distances from Granite Peak
to Fort Peck Lake is between 76 kilometers and
1054 kilometers.
7 cm
x mi
Acute triangle
2 3
(2) AC 1 BC > AB
(3) AB 1 AC > BC
1
D
A
C
}
One side, say BC, is longer than or at least as long as
each of the other sides. Then (1) and (2) are true.
8 cm
6 cm
2
6 1 82 0 102
100 5 100 This is a right triangle.
Geometry
Worked-Out Solution Key
143
Chapter 5,
continued
Proof of (3):
1. n ABC
1. Given
Statements
}
1. AX, BY, and CZ are
medians of n ABC.
}
2. Extend AC to D so
} }
that AB > AD.
2. Ruler Postulate
2. AM 5 }3 AX,
3. AD 1 AC 5 DC
3. Segment Addition
Postulate
Reasons
Reasons
}}
1. Given
2
2. Concurrency of
Medians of a Triangle
Theorem
2
CM 5 }3 CZ,
2
BM 5 }3 BY
4. Ž1 > Ž2
4. Base Angles Theorem
5. mŽDBC > mŽ2
5. Protractor Postulate
6. mŽDBC > mŽ1
6. Substitution
7. DC > BC
7. If one angle of a triangle
is larger than another
angle, then the side
opposite the larger angle
is longer than the side
opposite the smaller
angle.
8. AD 1 AC > BC
3. AM 1 BM > AB,
AM 1 CM > AC,
BM 1 CM > BC
2
2
3
3
2
2
} BY 1 } CZ > BC
3
3
4
AB 1 AC 1 BC
A
6. AX 1 BY 1 CZ >
6. Multiplication Property
of Inequality
7. AX 1 BY 1 CZ >
7. Rewrite right side as
a sum.
1
} (AB 1 AC 1 BC) 1
2
1
} (AB 1 AC 1 BC)
4
of n ABC.
1
Statements
5. If a > b and c > d, then
a 1 c > b 1 d.
5. }3 (AX 1 BY 1 CZ) >
3
} (AB 1 AC 1 BC)
4
48. a. Given: AD is a median
B
D
C
8. If a > b and c is
positive, then
a > b 2 c.
8. AX 1 BY 1 CZ >
1
}(AB 1 AC 1 BC)
2
Reasons
}
4. Substitution
} AX 1 } CZ > AC,
9. Substitution
Prove: AD < }2 (AB 1 BC 1 AC)
2
2
4. }3 AX 1 }3 BY > AB,
8. Substitution
9. AB 1 AC > BC
3. Triangle Inequality
Theorem
1. AD is a median of
n ABC.
1. Given
2. BD 1 DC 5 BC
2. Segment Addition
Postulate
3. AB 1 BD > AD
AC 1 DC > AD
3. Triangle Inequality
Theorem
4. AB 1 AC 1 BD 1
DC > AD 1 AD
4. If a > b and c > d,
then a 1 c > b 1 d.
5. AB 1 AC 1 BC >
AD 1 AD
5. Substitution Property
6. AB 1 AC 1 BC > 2AD
6. Simplify.
2. By the Converse of the Hinge Theorem, ŽSPQ is larger
7. Division Property
of Inequality
3.
Mixed Review for TAKS
1
7. }2 (AB 1 AC 1 BC) > AD
} }
49. A;
P(star tile and then blue tile) 5 P(star tile) + P(blue tile)
3
than ŽRPQ.
Group C
1.2 mi
40
}
1
140
N
A
Group A
2 mi
M
Z
X
Geometry
Worked-Out Solution Key
135
2 mi
B
1.2 mi
Group B
144
1
5.6 Guided Practice (pp. 336–338)
}
}
1. By the Hinge Theorem, RQ is longer than SQ.
Prove: AX 1 BY 1 CZ > }2 (AB 1 BC 1 AC)
C
3
Lesson 5.6
b. Given: AX, BY, and CZ are medians of n ABC.
Y
1
5}
5 }6 + }6 5 }
36
12
150
2 mi
1.2 mi
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Statements