1 2

Chapter 4
}}
1. Ž A is an obtuse angle because mŽ A > 908.
2. Ž B is a right angle because mŽ B 5 908.
}}
}
}}
}
AC 5 Ï(23 2 0)2 1 (3 2 0)2 5 Ï18 ø 4.24
BC 5 Ï(23 2 3)2 1 (3 2 3)2 5 Ï36 5 6
3. Ž C is an acute angle because mŽ C < 908.
Because AB 5 AC, n ABC is an isosceles triangle.
4. Ž D is an obtuse angle because mŽ D > 908.
} 320
51
Slope AB 5 }
320
5. 70 1 2y 5 180
2x 5 5x 2 54
6.
2y 5 110
23x 5 254
y 5 55
x 5 18
7. 40 1 x 1 65 5 180
x 1 105 5 180
2 1 (21) 25 1 (22)
7
1
8. M }, } 5 M }, 2}
2
2
2
2
1
2
1
3
24 1 1 7 1 (25)
9. M }, } 5 M 2}, 1
2
2
2
2
25 5 x
mŽ1 1 408 1 3(25)8 5 1808
1
mŽ1 5 658
4. mŽ A 1 mŽB 1 mŽC 5 1808
2
x8 1 2x8 1 3x8 5 1808
3
}
The midpoint of PQ is 1 2}2 , 1 2.
6x 5 180
h1h k10
2h k
k
10. M }, } 5 M }, } 5 M h, }
2
2
2 2
2
1
2
1
}
} }
The product of the slopes is 1(21) 5 21, so AB > AC
and ŽBAC is a right angle. Therefore, n ABC is a right
isosceles triangle.
50 5 2x
2
7
}
1
The midpoint of PQ is 1 }2 , 2}2 2.
1
320
}
Slope AC 5 }
5 21
23 2 0
3. 408 1 3x8 5 (5x 2 10)8
x 5 75
2
1
1
2
x 5 30
mŽA 5 x 5 308
mŽB 5 2x 5 608
2
k
The midpoint of PQ is h, }2 .
11. Ž2 > Ž3 by the Vertical Angles Congruence Theorem.
12. Ž1 > Ž4 by the Corresponding Angles Postulate.
13. Ž2 > Ž6 by the Alternate Interior Angles Theorem.
14. Ž3 > Ž5 by the Alternate Exterior Angles Theorem.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
2. AB 5 Ï (3 2 0)2 1 (3 2 0)2 5 Ï 18 ø 4.24
Prerequisite Skills (p. 214)
Lesson 4.1
mŽC 5 3x 5 908
5. 2x8 1 (x 2 6)8 5 908
3x 5 96
x 5 32
Therefore, the measures of the two acute angles are
2(32)8 5 648 and (32 2 6)8 5 268.
6. 908 1 608 5 1508
Investigating Geometry Activity 4.1 (p. 216)
1. The measure of the third angle of a triangle can be found
by subtracting the sum of the other two angles from 1808.
2. Sample answer:
By the Exterior Angle Theorem, the angle between the
staircase and the extended segment is 1508.
4.1 Exercises (pp. 221–224)
Skill Practice
26°
1. C; The triangle is right because it contains a 908 angle.
45°
53°
2. E; The triangle is equilateral because all sides are the
same length.
64°
45°
37°
The two acute angles of each triangle sum to 908, so the
angles are complementary.
4.1 Guided Practice (pp. 218–220)
3. F; The triangle is equiangular because all angles have the
same measure.
4. A; The triangle is isosceles because two sides are the
same length.
5. B; The triangle is scalene because each side has a
1. Sample answer:
different length.
6. D; The triangle is obtuse because it contains an angle
with measure greater than 908.
7. Sample answer: A right triangle cannot also be obtuse
5
4
because the sum of the other two angles cannot be greater
than 908.
6
Geometry
Worked-Out Solution Key
87
Chapter 4,
continued
nXYZ is a right isosceles triangle.
8. X
13.
y A(1, 9)
45°
B(4, 8)
45°
Z
Y
9.
C(2, 5)
nLMN is an equiangular
equilateral triangle.
M
60°
1
x
1
L 60°
10. J
60° N
130°
nJKH is an obtuse scalene triangle.
K
}
}}
}
AC 5 Ï(2 2 1)2 1 (5 2 9)2 5 Ï17 ø 4.12
BC 5 Ï(4 2 2)2 1 (8 2 5)2 5 Ï13 ø 3.61
20°
y
}
}}
AB 5 Ï(4 2 1) 1 (8 2 9)2 5 Ï10 ø 3.16
2
30°
11.
}}
So, there are no equal sides.
H
1
} 829
5 2}3
Slope AB 5 }
421
C(2, 7)
} 529
Slope AC 5 }
5 24
221
} 528 3
Slope BC 5 }
5 }2
224
B(6, 3)
So, there are no right angles. Therefore, n ABC is a
scalene triangle.
1
21
x
}}
}
}}
}
}}
}
AB 5 Ï(6 2 2)2 1 (3 2 3)2 5 Ï16 5 4
AC 5 Ï(2 2 2)2 1 (7 2 3)2 5 Ï16 5 4
BC 5 Ï(2 2 6) 1 (7 2 3) 5 Ï32 ø 5.66
2
2
So, AB 5 AC.
}
}
} }
CA is vertical and BA is horizontal. So, CA> BA and
Ž CAB is a right angle. Therefore, n ABC is a right
isosceles triangle.
12.
14. 608 1 608 1 x8 5 1808
x 5 60
Because each angle measures 608, the triangle is
equiangular.
15. x8 1 3x8 1 608 5 1808
4x 5 120
x 5 30
The measures of the angles are 308, 608, and 3(30) 5 908.
So, the triangle is a right triangle.
y
B(6, 9)
16. x8 5 648 1 708
(3, 3)
A
x 5 134
2
22
The angles in the triangle measure 648, 708, and
180 2 134 5 468, so the triangle is acute.
x
C(6, 23)
}}
}
AB 5 Ï(6 2 3)2 1 (9 2 3)2 5 Ï45 ø 6.71
}}
}
}}
}
AC 5 Ï(6 2 3) 1 (23 2 3) 5 Ï45 ø 6.71
2
2
BC 5 Ï(6 2 6)2 1 (23 2 9)2 5 Ï144 ø 12
So AB 5 AC.
}
923
52
Slope BA 5 }
623
} 23 2 3
Slope CA 5 }
5 22
623
17. x8 1 458 5 (2x 2 2)8
2x 5 247
x 5 47
2(47) 2 2 5 928
18. 248 1 (2x 1 18)8 5 (3x 1 6)8
36 5 x
3(36) 1 6 5 1148
19. 908 1 x8 1 (3x 1 2)8 5 1808
4x 5 88
} 23 2 9
Slope BC 5 }
is undefined.
626
So, there are no right angles. Therefore, n ABC is an
isosceles triangle.
x 5 22
mŽ1 5 90 1 3(22) 1 2 5 1588
20. Sample answer: By the Corollary to the Triangle Sum
Theorem, the acute angles must sum to 908. So you
would solve 3x 1 2x 5 90 for x, then use substitution to
find each angle measure.
21. mŽ1 1 408 5 908
mŽ1 5 508
88
Geometry
Worked-Out Solution Key
22. mŽ2 5 908 1 408
mŽ2 5 1308
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
A(2, 3)
Chapter 4,
continued
23. mŽ3 5 mŽ1
24. mŽ4 5 mŽ2
mŽ3 5 508
b. Sample answer: The three lines are y 5 ax 1 b,
y 5 x 1 2, and y 5 4x 2 7. If a 5 0 and b 5 5, then
all three lines will intersect at only one point, (3, 5),
so no triangle is formed.
mŽ4 5 1308
25. mŽ5 1 mŽ3 5 908
mŽ5 1 508 5 908
4
1
4
41
c. y 5 } x 1 }, y 5 2} x 1 }, y 5 21
3
3
3
3
mŽ5 5 408
26. mŽ6 1 mŽ4 1 208 5 1808
y
A (5, 7)
mŽ6 1 1308 1 208 5 1808
mŽ6 5 308
2
27. Let mŽ P 5 mŽ R 5 x8. Then mŽ Q 5 2x8.
x8 1 x8 1 2x8 5 1808
C
(11, 21)
(21, 21)
4x 5 180
}}
}
AB 5 Ï(21 2 5)2 1 (21 2 7)2 5 Ï100 5 10
x 5 45
}}
}}}
So, AB 5 AC, and the triangle is isosceles.
mŽ E 5 (3x 2 30)8.
7x 5 210
x 5 30
So, mŽ G 5 308, mŽ F 5 3(30)8 5 908, and
mŽ E 5 (3(30) 2 30)8 5 608.
29. The second statement in incorrect. Being isosceles does
not guarantee three congruent sides, only two. So, if
n ABC is equilateral, then it is isosceles as well.
30. By the Exterior Angle Theorem, the measure of the
exterior angle is equal to the sum of the measures
of the two nonadjacent interior angles.
So, mŽ1 5 808 1 508 5 1308.
31. B; If a triangle has two acute exterior angles, then it has
two obtuse interior angles. This is not possible because
the sum of the angles in the triangle must be 1808.
32. x8 5 438, so x 5 43.
y8 5 1808 2 438 2 1058, so y 5 32.
33. x8 5 1188, so x 5 118.
y8 5 1808 2 628 2 228, so y 5 96.
34. x8 5 1808 2 708 2 258, so x 5 85.
y8 5 1808 2 958 2 208, so y 5 65.
35. x8 5 1808 2 908 2 648, so x 5 26.
y8 5 1808 2 908 2 268, so y 5 64.
36. x8 5 478 1 158, so x 5 62.
y8 5 1808 2 908 2 628, so y 5 28.
37. y8 5 1808 2 908 2 (358 1 188), so y 5 37.
}
BC 5 Ï(11 2 (21))2 1 (21 2 (21))2 5 Ï144 5 12
28. Let mŽ G 5 x8. Then mŽ F 5 3x8 and
x8 1 3x8 1 (3x 2 30)8 5 1808
}
AC 5 Ï(11 2 5)2 1 (21 2 7)2 5 Ï 100 5 10
So, mŽ P 5 458, mŽ R 5 458, and mŽQ 5 2(45)8 5 908.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x
24 B
Problem Solving
8 ft
40.
39°
6.5 ft
54°
87°
5 ft
Because each side is a different length, the triangle is
scalene. Also, because each angle measures less than 908,
the triangle is acute.
41. The side lengths are 2 inches because all sides must be
6
equal in an equilateral triangle and }3 5 2. The angle
measures will always be 608 in any equiangular triangle.
42. You could bend the strip again at 6 inches, so the
sides would be 6, 6, and 8 inches. Or, you could bend
the strip again at 7 inches, so the sides would be 6, 7,
and 7 inches.
43. C; An angle cannot measure 08 or 1808, but it must
measure between 08 and 1808.
44. mŽ6 1 mŽ3 5 1808
45. mŽ5 5 mŽ2 1 mŽ3
mŽ6 1 658 5 1808
mŽ5 5 508 1 658 5 1158
mŽ6 5 1158
46. mŽ1 1 mŽ2 5 1808
mŽ1 1 508 5 1808
mŽ1 5 1308
47. mŽ2 1 mŽ3 1 mŽ4 5 1808
508 1 658 1 mŽ4 5 1808
mŽ4 5 658
x8 5 1808 2 908 2 188 2 378, so x 5 35.
38. No. Sample answer: If an angle in a triangle is really
close to zero, then the sum of the remaining two
angles would be almost 1808. If these two angles were
congruent, then the measure of each would be less than
908, and the triangle would not be obtuse.
39. a. Sample answer: The three lines will always form a
triangle as long as they do not all intersect at the same
point and no two lines are parallel.
Geometry
Worked-Out Solution Key
89
Chapter 4,
continued
48. Given: n ABC is a right triangle.
51. Sample answer: Mary and Tom both reasoned correctly,
Prove: Ž A and Ž B are complementary.
C
but the initial plan is not correct. The measure of the
exterior angle should be 1008 1 508 5 1508, not 1458.
52. a. If AB 5 AC 5 x and BC 5 2x 2 4:
x 1 x 1 2x 2 4 5 32
B
4x 5 36
Statements
Reasons
1. n ABC is a right
triangle.
1. Given
2. mŽC 5 908
2. Definition of right angle
x59
If AB 5 x and BC 5 AC 5 2x 2 4:
x 1 (2x 2 4) 1 (2x 2 4) 5 32
5x 5 40
x58
3. mŽA 1 mŽB 1 mŽC 3. Triangle Sum Theorem
5 1808
The two possible values for x are 8 and 9.
b. If AB 5 AC 5 x and BC 5 2x 2 4:
4. mŽA 1 mŽB 1 908
5 1808
4. Substitution Property of
Equality
5. mŽA 1 mŽB 5 908
5. Subtraction Property of
Equality
4x 5 16
6. ŽA and ŽB are
complementary.
6. Definition of
complementary angles
If AB 5 x and BC 5 AC 5 2x 2 4:
}
}
}
}
}
}
x 1 x 1 2x 2 4 5 12
x54
x 1 (2x 2 4) 1 (2x 2 4) 5 12
5x 5 20
49. a. 2Ï 2x 1 5Ï 2x 1 2Ï 2x 5 180
x54
b. 2Ï 2x 1 5Ï 2x 1 2Ï 2x 5 180
}
9Ï 2x 5 180
There is only one possible value for x, which is 4.
}
53. Given: nABC, points D and E
Ï2x 5 20
The measures of the angles are 2(20)8 5 408,
5(20)8 5 1008, and 2(20)8 5 408.
Prove: mŽ1 1 mŽ2 1 mŽ3 5 1808
B
2
c. The triangle is obtuse because it contains an angle
greater than 908.
A
50. Given: n ABC, exterior angle ŽBCD
Prove: mŽBCD 5 mŽ CBA 1 mŽ BAC
B
A
1
C
Reasons
1. nABC, exterior angle
ŽBCD
1. Given
2. mŽ ACD 5 1808
2. Definition of a straight
angle
3. mŽ ACB 1 mŽ BCD 5
mŽACD
3. Angle Addition
Postulate
4. mŽACB 1 mŽ BCD 5
1808
4. Substitution Property
of Equality
5. mŽ BCD 5
1808 2 mŽACB
5. Subtraction Property
of Equality
6. mŽACB 1 mŽ CBA 1
mŽBAC 5 1808
6. Triangle Sum Theorem
7. mŽCBA 1 mŽ BAC 5
1808 2 mŽ ACB
7. Subtraction Property
of Equality
8. mŽBCD 5
mŽ CBA 1 mŽBAC
8. Transitive Property of
Equality
Geometry
Worked-Out Solution Key
3 4 5
C
E
Statements
1. n ABC, }
AB i }
CD
Reasons
1. Given
2. mŽ ACE 5 1808
2. Definition of straight
angle
3. mŽ3 1 mŽ4 1 mŽ5
5 mŽ ACE
3. Angle Addition Postulate
4. mŽ3 1 mŽ4 1 mŽ5
5 1808
4. Substitution Property of
Equality
5. Ž1 > Ž5
5. Corresponding Angles
Postulate
6. Ž2 > Ž4
6. Alternate Interior Angles
Theorem
7. mŽ1 5 mŽ5,
mŽ2 5 mŽ4
7. Definition of congruent
angles
8. mŽ3 1 mŽ2 1 mŽ1
5 1808
8. Substitution Property of
Equality
D
Statements
90
1
D
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
A
Chapter 4,
continued
Mixed Review for TAKS
2. Sample answer: To prove that two triangles are
congruent, you need to show that all corresponding
sides and angles are congruent.
} } } } } }
3. AB > DE, BC > EF, AC > DF,
54. C;
(23, 2), (4, 3)
322
4 2 (23)
1
m 5 } 5 }7
Ž A > ŽD, ŽB > ŽE, ŽC > ŽF
Sample answer: nCBA > nFED
} } } } } } }
y 5 mx 1 b
1
2 5 }7 (23) 1 b
ŽG > ŽQ, ŽH > ŽR, ŽJ > ŽS, ŽK > ŽT
3
2 5 2}7 1 b
17
}5b
7
Sample answer: KJHG > TSRQ
17
1
So, the line y 5 }7 x 1 }
contains the points (23, 2)
7
and (4, 3).
5. mŽY 5 mŽN 5 1248
6. mŽM 5 mŽX 5 338
7. YX 5 NM 5 8
8. YZ > NL
}
9. nLNM > nZYX
}
10. nYXZ > nNML
11. nXYZ > nZWX because all corresponding sides and
angles are congruent.
55. J;
Only points C(22, 23) and D(22, 2) satisfy the
12. The triangles cannot be proven congruent because
satisfies the condition y q 22. So, point D(22, 2)
13. nBAG > nCDF because all corresponding sides and
3
condition x < 2}2. Of these two points, only point D
3
satisfies the conditions x < 2}2 and y q 2 2.
4.2 Guided Practice (pp. 226–227)
1. Corresponding angles: ŽA > ŽC, ŽB > ŽD,
ŽH > ŽF, ŽG > ŽE
} } } } } }
Corresponding sides: AB > CD, BG > DE, GH > EF,
} }
HA > FC
}
BC À DF and only one pair of corresponding angles
are shown congruent.
angles are congruent.
14. VWXYZ > KLMNJ because all corresponding sides and
15. mŽ M 5 1808 2 708 2 908 5 208
mŽ M 5 mŽ X, so x 5 20.
16. mŽ C 5 1808 2 808 2 458 5 558
mŽ C 5 mŽ R, so 558 5 5x8, or x 5 11.
17. The student has only shown that the corresponding
angles are congruent. The student still needs to show
that all corresponding sides are congruent, which they
are not.
ŽH > ŽF
2.
}
angles are congruent.
Lesson 4.2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
4. GH > QR, HJ > RS, JK > ST, KG > TQ,
mŽH 5 mŽF
(4x 1 5)8 5 1058
18. Sample answer:
4x 1 5 5 105
y
C(7, 9)
4x 5 100
N(8, 8)
x 5 25
mŽH 5 4(25) 1 5 5 1058
3. The sides of nPTS are congruent to the corresponding
sides of nRTQ by the indicated markings.
Ž PTS > Ž RTQ by the Vertical Angles Theorem. Also,
Ž P > Ž R and Ž S > Ž Q by the Alternate Interior
Angles Theorem. Because all corresponding sides and
angles are congruent, nPTS > nRTQ.
4. mŽ DCN 5 758 by the Third Angles Theorem.
5. To show that nNDC > nNSR, you need to know that
}
}
} }
DC > SR and DN > SN.
K
}
B(7, 2)
L(3, 1)
M(8, 1)
21
x
nLMN > nABC
19. 12x 1 4y 5 40
17x 2 y 5 50
12x 1 4y 5 40
34
68x 2 4y 5 200
5 240
x53
12(3) 1 4y 5 40 l y 5 1
So, x 5 3 and y 5 1.
Skill Practice
J
1
80x
4.2 Exercises (pp. 228–232)
1.
A(2, 2)
S
L
} }
T
} }
R
}
JK > RS, KL > ST, JL > RT,
Ž J > Ž R, Ž K > Ž S, Ž L > Ž T
Geometry
Worked-Out Solution Key
91