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Chapter 5
Chapter 5 Maintaining Mathematical Proficiency (p. 229)
8.
−4 + 0 1 + 7
1. M —, — = M(−2, 4)
2
2
(
)
6x + 1 = 3
——
AQ = √[ 0 − (−4) ]2 + (7 − 1)2
—
—
4x + 1 = 3 − 2x
4x + 1 + 2x = 3 − 2x + 2x
6x + 1 − 1 = 3 − 1
—
= √(4)2 + (6)2 = √ 16 + 36 = √52 ≈ 7.2 units
(
3 + 9 6 + (−2)
2
2
6x = 2
) ( )
x = —13
——
GH = √(9 − 3)2 + (−2 − 6)2
—
——
( −12+ 8 −22+ 0 ) ( 72 −22 ) ( 72 )
3. M —, — = M —, — = M —, −1
UV =
=
———
√[ 8 − (−1) ] + [ 0 − (−2) ]
——
—
√(8 + 1)2 + (2)2 = √92 + 4
—
—
= √81 + 4 = √ 85 ≈ 9.2 units
4.
7x + 12 = 3x
7x − 7x + 12 = 3x − 7x
12 = − 4x
−3 = x
The solution is x = −3.
5.
14 − 6t = t
14 − 6t + 6t = t + 6t
14 = 7t
2=t
The solution is t = 2.
6.
5p + 10 = 8p + 1
5p + 10 − 10 = 8p + 1 − 10
5p = 8p − 9
5p − 8p = 8p − 9 − 8p
−3p = −9
The solution is x = —13.
—
= √(6)2 + (−8)2 = √ 36 + 64 = √ 100 = 10 units
2
2
6
—6 x = —6
12 4
2 2
2. M —, — = M —, — = M(6, 2)
2
9.
z − 2 = 4 + 9z
z − 2 + 2 = 4 + 9z + 2
z = 6 + 9z
z − 9z = 6 + 9z − 9z
−8z = 6
6
−8
—z = —
−8
−8
6
z=—
−8
3
z = −—
4
3
The solution is z = −—4.
10. yes; The length can be found using the Pythagorean
Theorem.
Chapter 5 Mathematical Practices (p. 230)
1. theorem; It is the Slopes of Perpendicular Lines Theorem
(Thm. 3.14) studied in Section 3.5.
2. theorem; It is the Linear Pair Perpendicular Theorem
(Thm. 3.10) studied in Section 3.4.
3. definition; This is the definition of perpendicular lines.
4. postulate; This is the Two Point Postulate (Post. 2.1) studied
in Section 2.3. In Euclidean geometry, it is assumed, not
proved, to be true.
p=3
The solution is p = 3.
7.
w + 13 = 11w − 7
w + 13 − w = 11w − 7 − w
13 = 10w − 7
13 + 7 = 10w − 7 + 7
20 = 10w
2=w
The solution is w = 2.
5.1 Explorations (p. 231)
1. a. Check students’ work.
b. Check students’ work.
c. The sum of the interior angle measures of all triangles
is 180°.
d. Check students’ work; The sum of the measures of the
interior angles of a triangle is 180°.
2. a. Check students’ work.
b. Check students’ work.
c. Check students’ work.
d. Check students’ work; The sum is equal to the measure of
the exterior angle.
e. Check students’ work; The measure of an exterior angle is
equal to the sum of the measures of the two nonadjacent
interior angles.
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Geometry
Worked-Out Solutions
141
Chapter 5
3. The sum of the measures of the interior angles of a triangle
is 180°, and the measure of an exterior angle of a triangle
is equal to the sum of the measures of the two nonadjacent
interior angles.
4. 2x° + (x − 6)° = 90°
3x − 6 = 90
3x = 96
x = 32
4. m∠ A + m∠B + m∠C = 180°
2x° = 2(32°) = 64°
m∠C = 180° − 32° = 148°
(x − 6)° = 32° − 6° = 26°
m∠A + m∠B + 148 = 180°
The angle measurements are 26° and 64°.
m∠A + m∠B = 32°
The sum of the measures of the two nonadjacent interior
angles is 32° and the third angle which is adjacent tot he
exterior angle has a measure of 180° − 32° = 148°. These
are known because of the conjectures made in Explorations 1
and 2.
5.1 Monitoring Progress (pp. 232–235)
1. Sample answer: Obtuse isosceles triangle:
5.1 Exercises (pp. 236–238)
Vocabulary and Core Concept Check
1. no; By the Corollary to the Triangle Sum Theorem (Cor.
5.1), the acute angles of a right triangle are complementary.
Because their measures have to add up to 90°, neither angle
could have a measure greater than 90°.
2. The measure of an exterior angle of a triangle is equal to the
sum of the measures of the two nonadjacent interior angles.
Monitoring Progress and Modeling with Mathematics
3. Two sides are congruent and one angle is a right angle.
Acute scalene triangle:
So, △XYZ is a right isosceles triangle.
4. All sides are congruent and therefore, all angles are congruent.
So, △LMN is an equiangular equivalent triangle.
2. AB =
BC =
AC =
——
—
—
√(0 − 3)2 + (0 − 3)2 = √9 + 9 = √18 ≈ 4.2
——
—
——
—
—
√[ 3 − (−3) ]2 + (3 − 3)2 = √(3 + 3)2 = √62 = 6
—
√[ 0 − (−3) ]2 + (0 − 3)2 = √32 + 32 = √9 + 9
—
= √18 ≈ 4.2
Because AC = AB, that indicates that △ABC is isosceles.
3−0 3
—=—
=—=1
Slope of AB
3−0 3
0
3−3
—
Slope of BC = — = — = 0
−3 − 3 −6
3 − 0 −3
— = −—
= — = −1
Slope of AC
3−0
3
— and AC
— equals −1,
Because the product of the slopes of AB
—
—
that indicates that AB ⊥ AC , therefore △ABC is a right
triangle. So, △ABC is a right isosceles triangle.
3. 40° + 3x° = (5x − 10)°
50 = 2x
x = 25
5. None of the sides are congruent and one angle is obtuse.
So, △JKH is an obtuse scalene triangle.
6. None of the sides are congruent and all angles are acute.
So, △ABC is an acute scalene triangle.
——
—
——
—
7. AB = √ (6 − 2)2 + (3 − 3)2 = √ 42 = 4
BC = √(2 − 6)2 + (7 − 3)2 = √(−4)2 + 42
—
—
= √16 + 16 = √ 32 ≈ 5.7
——
—
AC = √(2 − 2)2 + (7 − 3)2 = √42 = 4
△ABC is isosceles because AB = AC.
3−3 0
Slope of AB = — = — = 0
6−2 4
4
7−3
Slope of BC = — = — = −1
2 − 6 −4
7−3 4
Slope of AC = — = — = undefined
2−2 0
— has a slope of 0 and AC
— has an undefined slope,
Because AB
— ⊥ AC
—. There is a right angle at ∠ A, which makes △ABC
AB
a right triangle. So, △ABC is a right isosceles triangle.
m∠ 1 + 3(25°) + 40° = 180°
m∠ 1 + 75° + 40° = 180°
m∠ 1 + 115° = 180°
m∠ 1 = 65°
142
Geometry
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Chapter 5
——
—
—
8. AB = √ (6 − 3)2 + (9 − 3)2 = √ 32 + 62 = √ 9 + 36
11. m∠ 1 + 78° + 31° = 180°
—
= √45 ≈ 6.7
m∠ 1 + 109° = 180°
——
——
BC = √(6 − 6)2 + [ 9 − (−3) ]2 = √ (0)2 + (9 + 3)2
=
—
√ 122
= 12
——
——
AC = √ (6 − 3)2 +[ 3 − (−3) ]2 = √32 + (3 + 3)2
12. m∠ 1 + 30° + 40° = 180°
—
—
= √9 + 36 = √ 45 ≈ 6.7
m∠ 1 + 70° = 180°
△ABC is isosceles because AB = AC.
9−3 6
Slope of AB = — = — = 2
6−3 3
−3 − 9 −12
Slope of BC = — = — = undefined
6−6
0
−3 − 3 −6
Slope of AC = — = — = −2
6−3
3
m∠ 1 = 110°
The triangle is an obtuse triangle.
13. m∠ 1 + 38° + 90° = 180°
m∠ 1 + 128° = 180°
m∠ 1 = 52°
None of the slopes of the sides of the triangle are negative
reciprocals of each other. So, the triangle is not a right
triangle. △ABC is an isosceles triangle.
——
—
m∠ 1 = 71°
The triangle is an acute triangle.
—
The triangle is a right triangle.
14. m∠ 1 + 60° + 60° = 180°
m∠ 1 + 120° = 180°
9. AB = √ (4 − 1)2 + (8 − 9)2 = √ 32 + (−1)2 = √ 9 + 1
m∠ 1 = 60°
—
= √10 ≈ 3.2
The triangle is an equiangular triangle.
——
BC = √(2 −
4)2
—
+ (5 −
8)2
——
=√
= √13 ≈ 3.6
——
(−2)2
+
(−3)2
—
= √4 + 9
—
AC = √ (2 − 1)2 + (5 − 9)2 = √ 12 + (−4)2
—
—
= √1 + 16 = √ 17 ≈ 4.1
15. m∠ 2 = 75° + 64° = 139°
16. x° + 45° = (2x − 2)°
x = 2x − 47
−x = −47
△ABC is a scalene triangle.
8 − 9 −1
Slope of AB = — = —
4−1
3
5 − 8 −3 3
Slope of BC = — = — = —
2 − 4 −2 2
5 − 9 −4
Slope of AC = — = — = −4
2−1
1
x = 47
The exterior angle has a measure of 92°.
17. 24° + (2x + 18)° = (3x + 6)°
42 + 2x = 3x + 6
None of the slopes of the sides of the triangle are negative
reciprocals of each other. So, the triangle is not a right
triangle. △ABC is a scalene triangle.
———
—
= √4 + 36 = √ 40 ≈ 6.3
———
——
BC = √(3 − 0)2 + [ −2 − (−3) ]2 = √ (3)2 + (−2 + 3)2
—
—
= √9 + 1 = √ 10 ≈ 3.2
———
——
AC = √ [ 3 − (−2) ]2 + (−2 − 3)2 = √ (3 + 2)2 + (−5)2
—
2x = 3x −6
−x = −36
—
10. AB = √ [ 0 − (−2) ]2 + (−3 − 3)2 = √ 22 + (−6)2
—
⋅
(2x − 2)° = 2 47 − 2 = 94 − 2 = 92
—
= √25 + 25 = √ 50 ≈ 7.1
△ABC is a scalene triangle.
−6
−3 − 3
Slope of AB = — = — = −3
0 − (−2)
2
−2 − (−3) −2 + 3 1
Slope of BC = — = — = —
3−0
3
3
−5
−5
−2 − 3
Slope of AC = — = — = — = −1
3 − (−2) 3 + 2
5
Because AB has a slope of −3 and BC has a slope of —13 ,
—⊥ —
AB
BC. There is a right angle at ∠ B, which makes △ABC a
right triangle. So, △ABC is a right scalene triangle.
⋅
x = 36
(3x + 6)° = 3 36 + 6 = 114
The exterior angle has a measure of 114°.
18. (x + 8)° + 4x° = (7x − 16)°
5x + 8 = 7x − 16
−2x + 8 = −16
−2x = −24
x = 12
⋅
(7x − 16)° = 7 12 − 16 = 84 − 16 = 68
The exterior angle has a measure of 68°.
19. 3x° + 2x° = 90°
5x = 90
x = 18
⋅
⋅
3x° = 3 18 = 54
2x° = 2 18 = 36
The two acute angles measure 36° and 54°.
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Geometry
Worked-Out Solutions
143
Chapter 5
27. The sum of the measures of the interior angles of the triangle
20. (3x + 2)° + x° = 90°
is 180°, not 360°. 115° + 39° + m∠ 1 = 180°
4x + 2 = 90
154° + m∠ 1 = 180°
4x = 88
m∠ 1 = 26°
x = 22
⋅
(3x + 2)° = 3 22 + 2 = 68
The two acute angles measure 22° and 68°.
of the measures of the two nonadjacent interior angles.
21. (11x − 2)° + (6x + 7)° = 90°
m∠ 1 = 80° + 50° = 130°
17x + 5 = 90
29. m∠ 1 = 90° − 40° = 50°
17x = 85
⋅
x=5
(11x − 2)° = 11 5 − 2 = 55 − 2 = 53
⋅
(6x + 7)° = 6 5 + 7 = 30 + 7 = 37
The two acute angles measure 37° and 53°.
22. (19x − 1)° + (13x − 5)° = 90°
30. m∠ 2 = 180° − 50° = 130°
31. m∠ 3 = m∠ 1 = 50°
32. m∠ 4 = m∠ 2 = 130°
33. m∠ 5 = 90° − 50° = 40°
34. m∠ 6 = 180° − 40° = 140°
32x − 6 = 90
35. m∠ 7 = 90°
32x = 96
⋅
⋅
28. The measure of the exterior angle should be equal to the sum
36. m∠ 8 = 180° − 40° = 140°
x=3
(19x − 1)° = 19 3 − 1 = 57 − 1 = 56
(13x − 5)° = 13 3 − 5 = 39 − 5 = 34
The two acute angles measure 34° and 56°.
23. x° + 5x° = 90°
6x = 90
37. None of the sides and none of the angles of the triangle are
equal, therefore it is a scalene triangle. When you measure
the three angles of the triangle, you find that they are all
acute. Therefore the triangle is acute. So, the triangle is an
acute scalene triangle.
38. The following sets of angle measures could form a triangle:
x = 15
B. 96° + 74° + 10° = 180°
⋅
5x° = 5 15 = 75
The two acute angles measure 15° and 75°.
24. x° + 8x° = 90°
D. 101° + 41° + 38° = 180°
E. 90° + 45° + 45° = 180°
F. 84° + 62° + 34° = 180°
9x = 90
39. 2
x = 10
⋅ 6 + x = 20
12 + x = 20
⋅
8x° = 8 10 = 80
The two acute angles measure 10° and 80°.
25. x° + [ 3(x + 8) ]° = 90°
x + 3x + 24 = 90
4x = 66
x = 16.5
x=8
2x + 6 = 20
2x = 14
x=7
You could make another bend 6 inches from the first bend
and leave the last side 8 inches long, or you could make
another bend 7 inches from the first bend and then the last
side will also be 7 inches long.
40. Sample answer:
⋅
3(x + 8)° = 3(16.5 + 8) = 3 24.5 = 73.5
1
GO TEAM!
The two acute angles measure 16.5° and 73.5°.
2
3
26. x° + [2(x − 12)]° = 90°
x + 2x − 24 = 90
3x = 114
x = 38
⋅
[2(x − 12)]° = 2(38 − 12) = 2 26 = 52
The two acute angles measure 38° and 52°.
144
Geometry
Worked-Out Solutions
When a triangular pennant is on a stick, the two angles
on the top edge of the pennant (∠ 1 and ∠ 2) should have
measures such that their sum is equal to the measure of the
angle formed by the stick and the bottom edge of the
pennant (∠ 3).
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Chapter 5
41. Given △ABC is a right triangle.
43. It is possible to draw an obtuse isosceles triangle.
A
Sample answer:
Prove ∠ A and ∠ B are
complementary.
C
B
STATEMENTS
REASONS
1. △ABC is a right triangle.
1. Given
2. ∠ C is a right angle.
2. Given (marked
in diagram)
3. m∠ C = 90°
3. Definition of a
right angle
4. m∠ A + m∠ B + m∠ C = 180°
4. Triangle Sum
Theorem
(Thm. 5.1)
5. m∠ A + m∠ B + 90° = 180°
44. It is possible to draw a right isosceles triangle. It will have
angle measurements of 45°, 45°, and 90°.
A right equilateral triangle is not possible, because the
hypotenuse must be longer than either leg in a right triangle.
5. Substitution
Property of
Equality
6. m∠ A + m∠ B = 90°
45. a. AB + AB + BC = Perimeter
x + x + 2x − 4 = 32
6. Subtraction
Property of
Equality
7. ∠ A and ∠ B are
complementary.
An obtuse equilateral triangle is not possible, because when
two sides form an obtuse angle the third side that connects
them must be longer than the other two.
4x − 4 = 32
4x = 36
x=9
7. Definition of
complementary
angles
AB + AB + BC = Perimeter
x + 2x − 4 + 2x − 4 = 32
5x − 8 = 32
42. Given △ABC, exterior ∠ BCD
5x = 40
B
Prove m∠ A + m∠ B = m∠ BCD
x=8
A
C
STATEMENTS
REASONS
1. △ABC, exterior ∠ BCD
1. Given
2. m∠ A + m∠ B + m∠ BCA
= 180°
2. Triangle Sum
Theorem
(Thm. 5.1)
3. ∠ BCA and ∠ BCD form a
linear pair.
3. Definition of
linear pair
4. m∠ BCA + m∠ BCD = 180°
4. Linear Pair
Postulate
(Post 2.8)
5. m∠ A + m∠ B + m∠ BCA
= m∠ BCA + m∠ BCD
5. Transitive
Property of
Equality
6. m∠ A + m∠ B = m∠ BCD
6. Subtraction
Property of
Equality
D
b. AB + AB + BC = Perimeter
x + x + 2x − 4 = 12
4x − 4 = 12
4x = 16
x=4
AB + AB + BC = Perimeter
x + 2x − 4 + 2x − 4 = 12
5x − 8 = 12
5x = 20
x=4
There is one value for x, x = 4.
46. a. The triangle appears to have three congruent sides and
three congruent angles. So, the triangle is equiangular,
acute, equilateral, and isosceles.
b. The triangle appears to have two congruent sides, no right
angles, and no obtuse angle. So, the triangle is acute and
isosceles.
c. The triangle appears to have one obtuse angle and no
congruent sides. So, the triangle is obtuse and scalene.
d. The triangle appears to have a right angle and no
congruent sides. So, the triangle is right and scalene.
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Geometry
Worked-Out Solutions
145
Chapter 5
47. Exterior angle = Sum of the two nonadjacent interior angles
?
A. 100° = 62° + 38°
?
B. 81° = 57° + 24°
100° = 100° ✓
?
F. 149° = 101° + 48°
81° = 81° ✓
53. Given ⃖⃗
AB ⃖⃗
CD
Prove m∠ 1 + m∠ 2 + m∠ 3 = 180°
B
2
149° = 149° ✓
1
A
D
3
48. no; According to the Exterior Angle Theorem (Thm. 5.2),
the measure of an exterior angle in a triangle is always equal
to the sum of the measures of the two nonadjacent interior
angles.
4
5
C
E
STATEMENTS
REASONS
1. ⃖⃗
AB ⃖⃗
CD
1. Given (marked
in diagram)
x = 43. Use the Exterior Angle theorem (Thm. 5.2) to find y.
2. ∠ ACD and ∠ 5 form
a linear pair.
2. Definition of linear
pair
75° = y° + 43°
3. m∠ ACD + ∠ 5 = 180°
3. Linear Pair Postulate
(Post. 2.8)
4. m∠ 3 + m∠ 4 = m∠ ACD
4. Angle Addition
Postulate (Post. 1.4)
Use the Exterior Angle Theorem (Thm. 5.2) to find the value
of y.
5. m∠ 3 + m∠ 4 + m∠ 5
= 180°
5. Substitution Property
of Equality
118° = y° + 22°
6. m∠ 1 ≅ m∠ 5
6. Corresponding Angle
Theorem (Thm. 3.1)
7. m∠ 2 ≅ m∠ 4
7. Alternate Interior
Angles Theorem
(Thm 3.2)
8. m∠ 1 = m∠ 5,
m∠ 2 = m∠ 4
8. Definition of
congruent angles
9. m∠ 3 + m∠ 2 + m∠ 1
= 180°
9. Substitution Property
of Equality
49. By the Alternate Interior Angles Theorem (Thm. 3.2),
32 = y
So, x = 43 and y = 32.
50. By the Corresponding Angle Theorem (Thm. 3.1), x = 118.
96 = y
So, x = 118 and y = 96.
51. The sum of the measures of the two acute angles of a right
triangle is 90°.
y° + 25° = 90°
y = 65
Use the Exterior Angle Theorem (Thm. 5.2) to find the value
of x.
x° = 65° + 20° = 85°
So, x = 85 and y = 65.
52. The sum of the measures of the two acute angles of a right
Maintaining Mathematical Proficiency
54. (5x − 27)° = (3x + 1)°
2x − 27 = 1
2x = 28
triangle is 90°.
x° + 64° = 90°
x = 26
By the Alternate Interior Angles Theorem (Thm. 3.2), the
unmarked angle of the triangle containing y° measures 26°.
Because both triangles contain a right angle and an angle
measuring 26°, y° must be 64°. So, x = 26 and y = 64.
x = 14
3(14) + 1 = 42 + 1 = 43
So, m∠ KHL = 43°.
55. m∠ ABC = m∠ GHK
(6x + 2)° = (5x − 27)° + (3x + 1)°
6x + 2 = 8x − 26
−2x = −28
⋅
x = 14
6 14 + 2 = 86
So, m∠ ABC = 86°.
56. 5y − 8 = 3y
2y = 8
y=4
⋅
3y = 3 4 = 12
So, GH = 12.
146
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Chapter 5
57.
3z + 6 = 8z − 9
−5z + 6 = −9
−5z = −15
z=3
⋅
3z + 6 = 3 3 + 6 = 9 + 6 = 15
So, BC = 15.
—
—
— —
— —
3. From the diagram, PS ≅ RQ, PT ≅ RT , and ST ≅ QT . Also,
by the Vertical Angles Congruence Theorem (Thm. 2.6),
— RQ
—, and
∠ PTS ≅ ∠ RTQ. From the diagram, PS
∠ P ≅ ∠ R and ∠ S ≅ ∠ Q by the Alternate Interior Angles
Theorem (Thm. 3.2). Because all corresponding parts are
congruent △PTS ≅ △RTQ.
4. ∠ NSR ≅ ∠ NDC and ∠ CND ≅ ∠ RNS, so by the Third
5.2 Explorations (p. 239)
Angles Theorem (Thm. 5.4), ∠ SRN ≅ ∠ DCN. So,
∠ DCN = 75°.
1. translation, reflection, rotation; A rigid motion maps
each part of a figure to a corresponding part of its image.
Because rigid motions preserve length and angle measure,
corresponding parts of congruent figures are congruent. In
congruent triangles, this means that the corresponding sides
and corresponding angles are congruent, which is sufficient
to say that the triangles are congruent.
5. The additional information that is needed to conclude that
— ≅ RS
— or DN
— ≅ SN
—.
△NDC ≅ △NSR is that CD
5.2 Exercises (pp. 243–244)
Vocabulary and Core Concept Check
1. To show that two triangles are congruent, you need to show
2. a. Sample answer: Reflect △ABC in the x-axis and translate
3 units right.
b. Sample answer: Rotate △ABC 180° about the origin.
c. Sample answer: Rotate △ABC 270° counterclockwise
about the origin and translate 3 units down.
d. Sample answer: Reflect △ABC in the line y = x.
3. Look at the orientation of the original triangle and decide
which rigid motion or composition of rigid motions will
result in the same orientation as the second triangle. Then, if
necessary, use a translation to move the first triangle so that
it coincides with the second.
4. Sample answer: Reflect △ABC in the y-axis and translate
3 units right and 2 units down.
4
F
−2
y
C
A
E
−2
2. “Is △JLK ≅ △STR?” is different. Because corresponding
angles and sides are not congruent, △JLK is not congruent
to △STR. for the other three questions, corresponding angles
are congruent and corresponding sides are congruent. So, the
triangles are congruent.
Monitoring Progress and Modeling with Mathematics
— —— —— —
3. corresponding sides: AB ≅ DE , BC ≅ EF , AC ≅ DF
corresponding angles: ∠ A ≅ ∠ D, ∠ B ≅ ∠ E, ∠ C ≅ ∠ F
Sample answer: △CBA ≅ △FED
— —— —— —
4. corresponding sides: GH ≅ QR , HJ ≅ RS , JK ≅ ST ,
—
GK ≅ QT
corresponding angles: ∠ G ≅ ∠ Q, ∠ H ≅ ∠ R, ∠ J ≅ ∠ S,
∠K ≅ ∠T
Sample answer: GHJK ≅ QRST
B
D
that all corresponding parts are congruent. If two triangles
have the same side lengths and angle measures, then they
must be the same size and shape.
4 x
5. ∠ N ≅ ∠ Y, so m∠ Y = 124°.
6. ∠ X ≅ ∠ M, so m∠ M = 33°.
5.2 Monitoring Progress (pp. 241–242)
— —— —— —
1. corresponding sides: AB ≅ CD , BG ≅ DE , GH ≅ EF ,
—
—
AH ≅ CF
corresponding angles: ∠ A ≅ ∠ C, ∠ B ≅ ∠ D, ∠ G ≅ ∠ E,
∠H ≅ ∠F
2. 4x + 5 = 105
4x = 100
x = 25
7. ∠ Z ≅ ∠ L and m∠ L = 180° − 124° − 33° = 23°. So,
m∠ Z = 23°.
— —
8. MN ≅ XY , so XY = 8.
9. 135 = 10x + 65
70 = 10x
7=x
4y − 4 = 28
4y = 32
y=8
So, x = 7 and y = 8.
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Geometry
Worked-Out Solutions
147
Chapter 5
10. m∠ N = 180° − (142° + 24°) = 180° − 166° = 14°
16. Prove △ABG ≅ △DCF
B
m∠ N = m∠ U
14° = (2x − 50)°
A
64 = 2x
x = 32
STATEMENTS
AB ≅ —
DC , —
AF ≅ —
DG ,
1. —
—
—
BE ≅ CE ≅ —
EF ≅ —
EG ,
∠ B ≅ ∠ C, ∠ A ≅ ∠ D
NP = US
2x − y = 13
2(32) − y = 13
64 − y = 13
y = 51
So, x = 32 and y = 51.
— —— —— —— —— —
11. VZ ≅ KJ , ZY ≅ JN , YX ≅ NM , XW ≅ ML , VW ≅ KL
∠ V ≅ ∠ K, ∠ Z ≅ ∠ J, ∠ Y ≅ ∠ N, ∠ X ≅ ∠ M, ∠ W ≅ ∠ L
Because all corresponding parts of the polygons are
congruent, VZYXW ≅ KJNML.
— — — —
12. From the diagram WX ≅ YZ and XY ≅ ZW . By the Reflexive
— ≅ WY
—. Also
Property of Congruence (Thm. 2.1), WY
from the diagram, ∠ X ≅ ∠ Z. Then, from the markings
— and XW
— ZY
—. You can conclude
in the diagram, —
XY ZW
that ∠ XYW ≅ ∠ ZWY and ∠ XWY ≅ ∠ ZYW by the
Alternate Interior Angles Theorem (Thm. 3.2). Because all
corresponding parts are congruent, △WXY ≅ △YZW.
B
D
STATEMENTS
— DC
—, E is the
1. AB
— and BD
—.
midpoint of AC
3. AF + FG = AG,
DG + FG = DF,
BE + EG = BG,
CE + EF = CF
3. Segment Addition
Postulate (Post. 1.2)
4. AF = DG,
BE = CE = EF = EG
4. Definition of congruent
segments
5. DG + FG = AG,
BE + EG = CF
5. Substitution Property of
Equality
6. DF = AG,
BG = CF
6. Transitive Property of
Equality
7. —
DF ≅ —
AG ,
—
—
BG ≅ CF
7. Definition of congruent
segments
8. △ABG ≅ △DCF
8. All corresponding parts
are congruent.
148
18. In order to conclude that triangles are congruent, the sides
REASONS
1. Given
3. ∠ BAE ≅ ∠ DCE,
∠ ABE ≅ ∠ CDE
3. Alternate Interior
Angles Theorem
(Thm. 3.2)
Geometry
Worked-Out Solutions
m∠ S = m∠ Y
C
2. Vertical Angles
Congruence Theorem
(Thm. 2.6)
5. △AEB ≅ △CED
∠S ≅ ∠Y
must also be congruent; △MNP is not congruent to △RSP
because the corresponding sides are not congruent.
2. ∠ AEB ≅ ∠ CED
4. —
AE ≅ —
EC , —
BE ≅ —
DE
1. Given (marked in
diagram)
m∠ S = 90° − 42° = 48°
E
Prove △AEB ≅ △CED
REASONS
corresponding parts are matched up correctly.
(Thm. 5.4), ∠ C ≅ ∠ 1. By the Triangle Sum Theorem
(Thm. 5.1), m∠ 1 = 180° − 45° − 80° = 55°.
A
D
17. The congruence statement should be used to ensure that
14. ∠ B ≅ ∠ Q and ∠ A ≅ ∠ S, so by the Third Angles Theorem
— and BD
—.
E is the midpoint of AC
G
2. Third Angles Theorem
(Thm. 5.4)
13. ∠ L ≅ ∠ Z and ∠ N ≅ ∠ Y, so by the Third Angles Theorem
(Thm. 5.4), ∠ 1 ≅ ∠ M. By the Triangle Sum Theorem
(Thm. 5.1), m∠ 1 = 180° − 90° − 70° = 20°.
F
2. ∠ BGA ≅ ∠ CFD
−y = −51
— —— —
15. Given AB DC , AB ≅ DC ,
C
E
4. Definition of midpoint
5. All corresponding
parts are congruent.
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Chapter 5
19. Given ∠ A ≅ ∠ D and ∠ B ≅ ∠ E
23. m∠ L + m∠ M + m∠ N = 180°
Prove ∠ C ≅ ∠ F
40° + 90° + m∠ N = 180°
130° + m∠ N = 180°
E
B
m∠ N = 50°
m∠ N = m∠ R
A
C
D
F
STATEMENTS
REASONS
1. ∠ A ≅ ∠ D, ∠ B ≅ ∠ E
1. Given
2. m∠ A = m∠ D,
m∠ B = m∠ E
2. Definition of
congruent angles
3. m∠ A + m∠ B + m∠ C
= 180°, m∠ D + m∠ E
+ m∠ F = 180°
3. Triangle Sum
Theorem (Thm. 5.1)
4. m∠ D + m∠ E + m∠ F
= m∠ A + m∠ B + m∠ C
4. Transitive Property
of Equality
5. m∠ D + m∠ E + m∠ F
= m∠ D + m∠ E + m∠ C
5. Substitution
Property of Equality
6. m∠ F = m∠ C
6. Subtraction
Property of Equality
7. ∠ F = ∠ C
7. Definition of
congruent angles
8. ∠ C = ∠ F
8. Symmetric Property
m∠ L = m∠ P
50° = (2x + 4y)°
40° = (17x − y)°
So, a system of equations is
2x + 4y = 50 .
17x − y = 40
Solve the second equation for y to get y = 17x − 40.
Substitute this for y in the first equation and solve for x.
2x + 4(17x − 40) = 50
2x + 68x − 160 = 50
70x − 160 = 50
70x = 210
x=3
Substitute 3 for x in the second equation and solve for y.
⋅
y = 17 3 − 40
y = 51 − 40 = 11
So, x = 3 and y = 11.
24. m∠ S + m∠ T + m∠ U = 180°
130° + 28° + (4x + y)° = 180°
158 + 4x + y = 180
4x + y = 22
20. Sample answer:
m∠ Y = m∠ T
(8x − 6y)° = 28°
So, a system of equations is 4x + y = 22 .
8x − 6y = 28
The Transitive Property of Triangle Congruence (Thm. 5.3)
guarantees that all the triangles are congruent.
— —— —— —
21. corresponding sides: JK ≅ XY , KL ≅ YZ , JL ≅ XZ
corresponding angles: ∠ J ≅ ∠ X, ∠ K ≅ ∠ Y, ∠ L ≅ ∠ Z
Solve the first equation for y to get y = 22 − 4x. Substitute
this for y in the second equation and solve for x.
8x − 6(22 − 4x) = 28
8x − 132 + 24x = 28
32x − 132 = 28
22. a. They are congruent because corresponding parts of
32x = 160
congruent figures are congruent.
b. They are congruent because they are both supplementary
to congruent angles.
c. ∠ GEB is also a right angle, and all right angles are
congruent.
— —
d. yes; From parts (a)–(c), you know that BE ≅ DE ,
∠ ABE ≅ ∠ CDE, ∠ GBE ≅ ∠ GDE, and
— ≅ GE
— by the Reflexive
∠ GEB ≅ ∠ GED. Also, GE
Property of Congruence (Thm. 2.1), ∠ BGE ≅ ∠ DGE
— ≅ DG
—
by the Third Angles Theorem (Thm. 5.4), and BG
from the diagram markings. So, △BEG ≅ △DEG because
all corresponding parts are congruent.
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x=5
Substitute 5 for x in the first equation and solve for y.
4(5) + y = 22
20 + y = 22
y=2
So, x = 5 and y = 2.
25. A rigid motion maps each part of a figure to a corresponding
part of its image. Because rigid motions preserve length and
angle measure, corresponding parts of congruent figures are
congruent, which means that the corresponding sides and
corresponding angles are congruent.
Geometry
Worked-Out Solutions
149
Chapter 5
Maintaining Mathematical Proficiency
26. ∠ Z ≅ ∠ W
— —
27. ∠ N ≅ ∠ T, RS ≅ PQ
— —
28. ∠ J ≅ ∠ M, JK ≅ KM (K is the midpoint), and
m∠ LKM = 90°.
— —— —
29. ∠ D ≅ ∠ H, DF GH , DE ≅ HI , ∠ DFE ≅ ∠ HGI
5.3 Explorations (p. 245)
1. a. Check students’ work.
b. Check students’ work.
2.
STATEMENTS
REASONS
1. ∠ B and ∠ D are right
angles,
—
BA ≅ —
DA ≅ —
DC ≅ —
BC ,
R is the midpoint of —
BA ,
U is the midpoint of —
DA ,
T is the midpoint of —
DC ,
and S is the midpoint of
—
BC .
1. Given
2. ∠ B ≅ ∠ D
2. Right Angles
Congruence
Theorem (Thm. 2.3)
3. BA = DA = DC = BC
3. Definition of
congruent segments
4. BA = BR + RA,
DA = DU + UA,
DC = DT + TC,
BC = BS + SC
4. Segment Addition
Postulate (Post. 1.2)
5. BR + RA = DU + UA =
DT + TC = BS + SC
5. Transitive Property
of Equality
c. BC ≈ 1.95 units, m∠ B ≈ 98.79°, m∠ C ≈ 41.21°
d. Check students’ work. If two sides and the included angle
of a triangle are congruent to two sides and the included
angle of another triangle, then the triangles are congruent.
2. Two triangles can be proved congruent if two pairs of
corresponding sides and corresponding included angles
are congruent.
3. Start with two triangles so that two sides and the included
angle of one triangle are congruent to two sides and the
included angle of another triangle. Then show that one
triangle can be translated until it coincides with the other
triangle by a composition of rigid motions.
5.3 Monitoring Progress (pp. 247–248)
— ≅ BC
— ≅ CD
— ≅ AD
—, ∠ A, ∠ B, ∠ C,
Given ABCD is a square, AB
∠ D are right angles. R, S, T, and U are midpoints of the
— ⊥ SU
— and SV
— ≅ VU
—.
sides ABCD. RT
S
B
R
A
C
T
V
U
D
— ≅ VU
—, RT
— ⊥ SU
—
1. SV
VR ≅ —
VR
2. —
7. BR = RA, DU = UA,
DT = TC, BS = SC
7. Definition of
congruent segments
8. BR + BR = DU + DU =
DT + DT = BS + BS
8. Substitution
Property of Equality
⋅
⋅
9. 2 BR = 2 DU
= 2 DT = 2 BS
⋅
1. Given
2. Reflexive Property of
Congruence (Thm. 2.1)
3. Definition of perpendicular
lines
4. ∠ SVR ≅ ∠ UVR
4. Right Angles Congruence
Theorem (Thm. 2.3)
5. SAS Congruence Theorem
(Thm. 5.5)
10. BR = DU = DT = BS
11. —
BR ≅ —
DU ≅ —
DT ≅ —
BS
Geometry
Worked-Out Solutions
10. Division Property
of Equality
11. Definition
of congruent
segments
12. SAS Congruence
Theorem
(Thm. 5.5)
— —
3. Given DA ≅ DG and
D
∠ ADR ≅ ∠ GDR
Prove △DRA ≅ △DRG
A
STATEMENTS
— ≅ DG
—,
1. DA
∠ ADR ≅ ∠ GDR
2. —
DR ≅ —
DR
3. △DRA ≅ △DRG
150
9. Distributive
Property
⋅
REASONS
3. ∠ SVR and ∠ UVR
are right angles.
5. △SVR ≅ △UVR
6. Definition of
midpoint
12. △BSR ≅ △DUT
1. Prove △SVR ≅ △UVR
STATEMENTS
BR ≅ —
RA , —
DU ≅ —
UA ,
6. —
—
—
—
DT ≅ TC , BS ≅ —
SC
R
G
REASONS
1. Given
2. Reflexive Property of
Congruence (Thm. 2.1)
3. SAS Congruence Theorem
(Thm. 5.5)
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Chapter 5
— —— —
16. Given AB ≅ CD , AB CD
5.3 Exercises (pp. 249–250)
A
Prove △ABC ≅ △CDA
Vocabulary and Core Concept Check
D
1
1. An included angle is an angle formed by two adjacent
2
consecutive sides of a triangle.
B
2. If two sides and the included angle of one triangle are
congruent to two sides and the included angle of a second
triangle, then the two triangles are congruent.
Monitoring Progress and Modeling with Mathematics
— —
3. ∠ JKL is the included angle between JK and KL .
—
—
—
—
—
—
—
—
—
—
4. ∠ PKL is the included angle between PK and LK .
STATEMENTS
REASONS
— ≅ CD
—, AB
— CD
—
1. AB
1. Given
2. —
AC ≅ —
AC
2. Reflexive Property of
Congruence (Thm. 2.1)
3. ∠ 1 ≅ ∠ 2
3. Alternate Interior Angle
Congruence Theorem
(Thm. 3.2)
4. △ABC ≅ △CDA
4. SAS Congruence Theorem
(Thm. 5.5)
5. ∠ KLP is the included angle between LP and LK .
6. ∠ LJK is the included angle between JL and JK .
7. ∠ KLJ is the included angle between KL and JL .
8. ∠ KPL is the included angle between KP and PL .
C
D
17. Given C is the midpoint
— and BD
—.
of AE
9. no; The congruent angles are not the included angle.
Prove △ABC ≅ △EDC
A
E
C
10. yes; Two pairs of sides and the included angles are
B
congruent.
11. no; One of the congruent angles is not the included angle.
12. no; The congruent angles are not the included angle.
13. yes; Two pairs of sides and the included angles are
congruent.
14. no; ∠ NKM and ∠ KML are congruent by the Alternate
Interior Angles Theorem (Thm. 3.2) but they are not the
included angles.
—
— ≅ TP
—
SP
15. Given PQ bisects ∠ SPT,
S
T
STATEMENTS
REASONS
1.
1. Given
— bisects ∠ SPT.
PQ
1. C is the midpoint of
— and BD
—.
AE
1. Given
2. ∠ ACB ≅ ∠ ECD
2. Vertical Angles Congruence
Theorem (Thm. 2.6)
3. —
AC ≅ —
EC ,
—
BC ≅ —
DC
3. Definition of midpoint
4. SAS Congruence Theorem
(Thm. 5.5)
— —— —
18. Given PT ≅ RT , QT ≅ ST
Q
—
SP ≅ —
TP ,
REASONS
4. △ABC ≅ △EDC
P
Prove △SPQ ≅ △TPQ
STATEMENTS
P
Prove △PQT ≅ △RST
Q
T
S
R
STATEMENTS
REASONS
2. Reflexive Property of
Congruence (Thm. 2.1)
— ≅ RT
—,
1. PT
—
—
QT ≅ ST
1. Given
3. ∠ SPQ ≅ ∠ TPQ
3. Definition of angle bisector
2. ∠ PTQ ≅ ∠ RTS
4. △SPQ ≅ △TPQ
4. SAS Congruence Theorem
(Thm. 5.5)
2. Vertical Angles Congruence
Theorem (Thm. 2.6)
3. △PQT ≅ △RST
3. SAS Congruence Theorem
(Thm. 5.5)
2. —
PQ ≅ —
PQ
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Geometry
Worked-Out Solutions
151
Chapter 5
— —
19. △SRT ≅ △URT; RT ≅ RT by the Reflexive Property of
Congruence (Thm. 2.1). Also, because all points on a circle
— ≅ RU
—. It is given
are the same distance from the center, RS
that ∠ SRT ≅ ∠ URT. So, △SRT and △URT are congruent
by the SAS Congruence Theorem (Thm. 5.5).
20. △BAD ≅ △DCB; Because the sides of the square are
— ≅ DC
— and AD
— ≅ CB
—. Also, because the
congruent, BA
angles of the square are congruent, ∠ A ≅ ∠ C. So, △BAD
and △DCB are congruent by the SAS Congruence Theorem
(Thm. 5.5).
21. △STU ≅ △UVR; Because the sides of the pentagon are
— ≅ UV
— and TU
— ≅ VR
—. Also, because the angles
congruent, ST
of the pentagon are congruent ∠ T ≅ ∠ V. So, △STU and
△UVR are congruent by the SAS Congruence Theorem
(Thm. 5.5).
26. In order to prove △ABC ≅ △DBC, you will need to know
that ∠ ACB ≅ ∠ DCB.
27. Given △ABC, △BCD,
and △CDE are
isosceles triangles
and ∠ B ≅ ∠ D.
B
A
D
C
E
Prove △ABC ≅ △CDE
Because △ABC, △BCD, and △CDE are isosceles triangles,
— ≅ BC
—, BC
— ≅ CD
—, and CD
— ≅ DE
—.
you know that AB
So, by the Transitive Property of Congruence (Thm. 2.1),
— ≅ CD
— and BC
— ≅ DE
—. It is given that ∠ B ≅ ∠ D,
AB
So △ABC ≅ △CDE by the SAS Congruence
Theorem (Thm. 5.5).
28. SSS, ASA, AAS, and SAS all correspond to congruence
theorems.
22. △NMK ≅ △NLK; Because all points on a circle are the
— ≅ NL
— and KM
— ≅ KL
—.
same distance from the center, NM
— ⊥ MN
— and KL
— ⊥ NL
—, ∠ M and ∠ L are right
Because MK
angles by definition of perpendicular lines, which means
that ∠ M ≅ ∠ L by the Right Angles Congruence Theorem
(Thm. 2.3). So, △NMK and △NLK are congruent by the
SAS Congruence Theorem (Thm. 5.3.)
—
Counterexample for SSA
—
23. Construct side DE so that it is congruent to AC . Construct
∠ D, with vertex D and side ⃗
DE so that it is congruent
— so that it is congruent to AB
—. Draw
to ∠ A. Construct DF
△DFE. By the SAS Congruence Theorem (Thm. 5.5),
△ABC ≅ △DFE.
F
C
D
A
B
E
—
—
24. Construct side DE so that it is congruent to AC . Construct
∠ D, with vertex D and side ⃗
DE so that it is congruent
— so that it is congruent to AB
—. Draw
to ∠ A. Construct DF
△DFE. By the SAS Congruence Theorem (Thm. 5.5),
△ABC ≅ △DFE.
E
C
Counterexample for AAA
29. Prove △ABC ≅ △DEC
STATEMENTS
REASONS
— ≅ DC
—,
1. AC
—
—
BC ≅ EC
1. Given (marked in diagram)
2. ∠ ACB ≅ ∠ DCE
2. Vertical Angles Congruence
Theorem (Thm. 2.6)
3. △ABC ≅ △DEC
3. SAS Congruence Theorem
(Thm. 5.5)
AC = CD
4y = 2x + 12
y=
A
B
D
BC = CE
4y − 6 = 2x + 6
1
—2 x
3y + 1 = 4x
(
)
3 —12x + 3 + 1 = 4x
+3
1.5x + 9 + 1 = 4x
1.5x + 10 = 4x
F
10 = 2.5x
25. △XYZ and △WYZ are congruent so either the expressions for
— and WZ
— or the expressions for XY
— and WY
— should be set
XZ
equal to each other because they are corresponding sides.
y = —12 4 + 3 = 2 + 3 = 5
5x − 5 = 3x + 9
So, x = 4 and y = 5.
2x − 5 = 9
2x = 14
x=7
152
Geometry
Worked-Out Solutions
x=4
⋅
—
—
30. no; When you construct AB and AC, you have to construct
them at an angle that is congruent to ∠ A. Otherwise, when
you construct an angle congruent to ∠ C, you might not get a
—.
third segment that is congruent to BC
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Chapter 5
31. Given A reflection in line K
maps point A to A′,
a reflection in line m
maps A′ to A″, and
m∠ MPK = x°.
m
A″
M
A′
Prove The angle of rotation
is 2x°.
P
STATEMENTS
1. Given
2. Line k is the perpendicular
bisector of —
AA′ , and line m is the
perpendicular bisector of —
A′A″ .
2. Definition of
reflection
4. ∠ AKP ≅ ∠ A′KP,
∠ A′MP ≅ ∠ A″MP
5. —
KP ≅ —
KP
6. △AKP ≅ △A′KP,
△A′MP ≅ △A″MP
7. —
AP ≅ —
A′P , —
A′P ≅ —
A″P ,
∠ APK ≅ ∠ A′PK,
∠ A′PM ≅ ∠ A″PM
8. —
AP ≅ —
A″P
9. m∠ APK = m∠ A′PK,
m∠ A′PM = m∠ A″PM
12. m∠ APA″ = 2(m∠ A′PK +
m∠ A′PM)
12. Distributive
Property
13. m∠ APA″ = 2(m∠ MPK)
13. Substitution
Property of
Equality
14. m∠ APA″ = 2(x°) = 2x°
14. Substitution
Property of
Equality
15. Point A is rotated about point P,
and the angle of rotation is 2x°.
15. Definition of
rotation
A
Maintaining Mathematical Proficiency
32. Two sides are equivalent and one angle is a right angle.
So, the triangle is a right isosceles triangle.
33. Two sides are equivalent and one angle is obtuse. So, the
triangle is an obtuse isosceles triangle.
3. Definition of
perpendicular
bisector
4. Right Angles
Congruence
Theorem
(Thm. 2.3)
5. Reflexive
Property of
Congruence
(Thm. 2.1)
6. SAS Congruence
Theorem
(Thm. 5.5)
7. Corresponding
parts of congruent
triangles are
congruent.
8. Transitive
Property of
Congruence
(Thm. 2.1)
9. Definition of
congruent angles
10. m∠ MPK = m∠ A′PK +
m∠ A′PM, m∠ APA″ =
m∠ APK + m∠ A′PK +
m∠ A′PM + m∠ A″PM
10. Angle Addition
Postulate
(Post. 1.4)
11. m∠ APA″ = m∠ A′PK +
m∠ A′PK + m∠ A′PM +
m∠ A′PM
11. Substitution
Property of
Equality
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All rights reserved.
REASONS
REASONS
1. A reflection in line k maps
point A to A′, a reflection
in line m maps A′ to A″, and
m∠ MPK = x°.
3. —
AK ≅ —
KA′ , ∠ AKP and
∠ A′KP are right angles,
—
A′M ≅ —
MA″ , and ∠ A′MP
and ∠ A″MP are right angles.
k
K
STATEMENTS
34. All sides are congruent, and therefore all angles are
congruent. So, the triangle is a equiangular equilateral
triangle.
35. None of the sides are congruent and one angle is obtuse.
So, the triangle is an obtuse scalene triangle.
5.4 Explorations (p. 251)
1. a. Check students’ work.
b. Check students’ work.
c. Because all points on a circle are the same distance from
— ≅ AC
—.
the center, AB
d. ∠ B ≅ ∠ C
e. Check students’ work.
If two angles of a triangle are congruent, then the angles
opposite them are congruent.
f. If two angles of a triangle are congruent, then the sides
opposite them are congruent. The converse is true.
2. In an isosceles triangle, two sides are congruent and the
angles opposite them are congruent.
3. Draw the angle bisector of the included angle between the
congruent sides to divide the given isosceles triangle into
two triangles. Use the SAS Congruence Theorem (Thm. 5.5)
to show that these two triangles are congruent. Then, use
properties of congruent triangles to show that the two angles
opposite the shared sides are congruent.
For the converse, draw the angle bisector of the angle that is
not congruent to the other two. This divides the given triangle
into two triangles that have two pairs of corresponding
congruent angles. The third pair of angles are congruent by
the Third Angles Theorem (Thm. 5.4). Also, the angle bisector
is congruent to itself by the Reflexive Property of Congruence
(Thm. 2.1). So, the triangles are congruent, and the sides
opposite the congruent angles in the original triangle
are congruent.
Geometry
Worked-Out Solutions
153
Chapter 5
5.4 Monitoring Progress (pp. 253–255)
—
7. △ABC is an equiangular triangle and, therefore, an
equilateral triangle, So, x = 12.
1. If HG ≅ HK, then ∠ HKG ≅ ∠ G.
— —
2. If ∠ KHJ ≅ ∠ KJH, then KJ ≅ KH .
3. The triangle is equiangular; therefore, by Corollary to the
Converse of the Base Angles Theorem (Cor. 5.3) the length
— is 5 units.
of ST
4. The triangle on the right is an equilateral triangle. Each
angle has a measure of 60°, therefore, x = 60. The triangle
on the left is an isosceles triangle. Both base angles are
90° − 60° = 30°. So, y = 180 − 2 30 = 120.
⋅
— —
5. From Example 4, you know that PT ≅ QT and ∠ 1 ≅ ∠ 2.
— ≅ QR
— and ∠ QPS ≅ ∠ PQR. By definition
It is stated that PS
of congruent segments and angles, m∠ 1 = m∠ 2 and
m∠ QPS ≅ m∠ PQR. Also, by the Angle Addition Postulate
(Post 1.4), m∠ 1 + m∠ TPS + m∠ QPS and
m∠ 2 + m∠ TQR + m∠ PQR. By substituting m∠ 1 for
m∠ 2 and m∠ QPS for m∠ PQR, you get m∠ 1 + m∠ TQR =
m∠ QPS. Then, by the Transitive Property of Equality,
m∠ 1 + m∠ TPS = m∠ 1 + m∠ TQR. So, by the
Subtraction Property of Equality, m∠ TPS = m∠ TQR.
Because ∠ TPS = ∠ TQR by definition, you can conclude
that △PTS ≅ △QTR by the SAS Congruence Theorem
(Thm. 5.5).
5.4 Exercises (pp. 256–258)
Vocabulary and Core Concept Check
1. The vertex angle is the angle formed by the congruent sides,
8. △MLN is an equiangular triangle and, therefore, an
equilateral triangle, So, x = 16.
9. △RST is an equilateral triangle and, therefore, an
equiangular triangle, So, x = 60.
10. △DEF is an equilateral triangle and, therefore, each angle is
3x° = 60°, or x = 20.
11. The pennant is an isosceles triangle. So, x = 79 and
⋅
y = 180 − 2 79 = 180 − 158 = 22.
12. Sample answer:
7 cm
Reuse
Reduce
Recycle
7 cm
13. The triangle in the center is equilateral; therefore, each angle
measures 60°. So, x = 60. Because the top and bottom lines
are parallel, the alternate interior angles have a measure of
60°. By the Base Angles Theorem (Thm. 5.6), y = 60.
14. The vertex angle of the isosceles triangle measures
180° − 40° = 140°. To find the base angles:
2x° + 140° = 180°
or legs, of an isosceles triangle.
2x = 40
x = 20
2. The base angles of an isosceles triangle are opposite the
congruent sides, and they are congruent by the Base Angles
Theorem (Thm. 5.6).
Monitoring Progress and Modeling with Mathematics
— —
3. If AE ≅ DE , then ∠ D ≅ ∠ A by the Base Angles Theorem
(Thm. 5.6).
— —
4. If AB ≅ EB , then ∠ AEB ≅ ∠ A by the Base Angles Theorem
(Thm. 5.6).
— —
5. If ∠ D ≅ ∠ CED, then EC ≅ DC , by the Converse of the
Base Angles Theorem (Thm. 5.7).
— —
6. If ∠ EBC ≅ ∠ ECB, then EC ≅ EB , by the Converse of the
Base Angles Theorem (Thm. 5.7).
154
Geometry
Worked-Out Solutions
7 cm
To find y:
y° + x° = 90°
y + 20 = 90
y = 70
So, x = 20 and y = 70.
15. The triangle on the left is equiangular and, therefore, equilateral.
8y = 40
y=5
The triangle on the right is isosceles and the vertex angle
measures 180° − 60° = 120°. So, by the Base Angles
180° − 120° 60°
Theorem (Thm. 5.6), —— = — = 30°. So, x = 30
2
2
and y = 5.
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Chapter 5
16. By the Converse of Base Angles Theorem (Thm. 5.7):
3x − 5 = y + 12
3x − 5 − 12 = y
y = 3x − 17
The triangle on the right is equiangular and, therefore,
equilateral.
3x − 5 = 5y − 4
3x − 5 = 5(3x − 17) − 4
3x − 5 = 15x − 85 − 4
3x − 5 = 15x − 89
−12x = − 84
x=7
y = 3x − 17
⋅
y = 3 7 − 17
y = 21 − 17 = 4
So, x = 7 and y = 4.
17. Draw a segment with length 3 inches. Draw an arc with
center at one endpoint and radius 3 inches. Draw an arc with
center at the other endpoint and radius 3 inches. Connect
the intersection of the arcs with two segments to form an
equilateral triangle.
20. a. Because △ABD and △CBD are congruent and equilateral,
— ≅ CB
—. So △ABC is isosceles.
you know that AB
b. Because △ABC is isosceles, ∠ BAE ≅ ∠ BCE by the Base
Angels Theorem (Thm. 5.6).
c. By the Reflexive Property of Congruence (Thm. 2.1),
—. Because △ABD and △CBD are congruent and
BE ≅ BE
equilateral, and also equiangular by the Corollary to the
Base Angels Theorem (Cor. 5.2), you can conclude that
— ≅ CB
— as explained in part (a).
∠ ABE ≅ ∠ CBE. Also, AB
So, by the SAS Congruence Theorem (Thm 5.5),
△ABE and △CBE.
—
d. m∠ ABE + m∠ CBE = m∠ ABC
Angle Addition Postulate (Post. 1.4)
m∠ ABE = 60°, m∠ CBE = 60°
Definition of equiangular triangle
m∠ ABC = 60° + 60° = 120°
Substitution Property of Equality
m∠ ABC + m∠ BAE + m∠ BCE = 180°
Triangle Sum Theorem (Thm. 5.1)
120° + m∠ BAE + m∠ BCE = 180°
Substitution Property of Equality
m∠ BAE + m∠ BCE = 60°
Subtraction Property of Equality
m∠ BAE = m∠ BCE
Corresponding parts of congruent triangles are
congruent.
m∠ BAE + m∠ BCE = 60°
3 in.
18. Draw a segment with length 1.25 inches. Draw an arc with
center at one endpoint and radius 1.25 inches. Draw an arc
with center at the other endpoint and radius 1.25 inches.
Connect the intersection of the arcs with two segments to
form an equilateral triangle.
Substitution Property of Equality
2m∠ BAE = 60°
Simplify.
m∠ BAE = 30°
Division Property of Equality
21. a. Each edge is made out of the same number of sides of the
original equilateral triangle.
b. The areas of the first four triangles in the pattern are
1.25 in.
19. When two angles of a triangle are congruent, the sides
opposite the angles are congruent; Because ∠ A ≅ ∠ C,
— ≅ BC
—. So BC = 5.
AB
1 square unit, 4 square units, 9 square units, and
16 square units.
c. Triangle 1 has an area of 12 = 1, Triangle 2 has an area of
22 = 4, Triangle 3 has an area of 32 = 9, and so on. So, by
deductive reasoning, you can predict that Triangle n has
an area of n2.
Seventh triangle: 72 = 49 square units.
—
22. A, C; If the base of isosceles △XYZ is YZ , then the legs are
— and XZ
— and XY
— ≅ XZ
— and ∠ Y ≅ ∠ Z.
XY
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Geometry
Worked-Out Solutions
155
Chapter 5
29. no; The two sides that are congruent can form an obtuse
23. x + 4 = 4x + 1
angle or a right angle.
4 = 3x + 1
3 = 3x
30. no; The sum of the angles of a triangle is always 180°. So, if
all three angles are congruent, then they will always be
180°
— = 60°.
3
1=x
Perimeter = 7 + 1 + 4 + 4 + 1 = 17 inches
24. 2x − 3 = x + 5
31.
x=8
21 − x = x + 5
3t = 5t − 12 or 3t = t + 20 or 5t − 12 = t + 20
−2t = −12
2t = 20
4t = 32
t=6
t = 10
t=8
The values of t could be 6, 8, and 10.
16 = 2x
8=x
32.
21 − x = 2x − 3
x°
24 = 3x
8=x
⋅
Perimeter = (2 8 − 3) + (8 + 5) + (21 − 8)
= 13 + 13 + 13 = 39 inches
25. By the Reflexive Property of Congruence (Thm. 2.1), the
yellow triangle and the yellow-orange triangle share a
congruent side. Because the triangles are all isosceles, by
the Transitive Property of Congruence (Thm. 2.1), the
yellow-orange triangle and the orange triangle share a side
that is congruent to the one shared by the yellow triangle and
the yellow-orange triangle. This reasoning can be continued
around the wheel, so the legs of the isosceles triangles are
all congruent. Because you are given that the vertex angles
are all congruent, you can conclude that the yellow triangle
is congruent to the purple triangle by the SAS Congruence
Theorem (Thm. 5.5).
26. 180° − 30° = 150°
150
= 75
—
2
The measures of the base angles are each 75°.
27.
yellowgreen
yellow
green
orange
redorange
blue
red
bluepurple
purple
redpurple
The three sides of the triangle are congruent. So, the triangle
is an equiangular equilateral triangle.
28. Every fourth color is a triad: Yellow-green, blue-purple, red-
orange; green, purple, orange; and blue-green, red-purple,
yellow-orange.
156
Geometry
Worked-Out Solutions
The vertex angle is (180 − x)°
and the base angles are
180° − (180 − x)°
x °
—— = — .
2
2
()
33. If the base angles are x°, then the vertex angle is (180 − 2x)°,
or [2(90 − x)]°. Because 2(90 − x) is divisible by 2, the
vertex angle is even when the angles are whole numbers.
34. a. ∠ XVY, ∠ UXV; ∠ WUX ≅ ∠ XVY because they are both
vertex angles of congruent isosceles triangles. Also,
m∠ UXV + m∠ VXY = m∠ UXY by the Angle Addition
Postulate (Post. 1.4), and m∠ UXY = m∠ WUX + m∠ UWX
by the Exterior Angle Theorem (Thm. 5.2). So, by the
Transitive Property of Equality, m∠ UXV + m∠ VXY
= m∠ WUX + m∠ UWX. Also, m∠ UWX = m∠ VXY
because they are base angles of congruent isosceles
triangles. By substituting m∠ UWX for m∠ VXY, you get
m∠ UXV + m∠ UWX = m∠ WUX + m∠ UWX. By the
Subtraction Property of Equality, m∠ UXV = m∠ WUX,
so ∠ UXV ≅ ∠ WUX.
b. Because the triangles are congruent isosceles triangles
yelloworange
bluegreen
x°
The base angles are
(180 − x)° and the vertex
angle is 180° − 2(180 − x)°
= (2x − 180)°.
— ≅ VY
— ≅ UW
— ≅ VX
—, then
and from part (a), UX
—≅—
—. So, the distance between points U
WX
XY ≅ UV
and V is 8 meters.
35. a. 2.1 mi; by the Exterior Angle Theorem (Thm. 5.2),
m∠ L = 70° − 35° = 35°. Because m∠ SRL = 35°
= m∠ RLS, by definition of congruent angles,
∠ SRL ≅ ∠ RLS. So, by the Converse of the Base Angles
— ≅ SL
—. So, SL = RS = 2.1 miles.
Theorem (Thm. 5.7), RS
b. Find the point on the shore line that has an angle of 45°
from the boat. Then, measure the distance that the boat
travels until the angle is 90°. That distance is the same as
the distance between the boat and the shore line because
the triangle formed is an isosceles right triangle.
36. no; The sum of the angle measures of a very large spherical
triangle will be greater than 180°, but for smaller spherical
triangles, the sum will be closer to 180°.
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Chapter 5
37. Given △ABC is equilateral.
41. Given △ABC is equilateral.
A
D
B
STATEMENTS
REASONS
1. △ABC is
equilateral.
1. Given
2. —
AB ≅ —
AC ,
— ≅ BC
—,
AB
—
—
AC ≅ BC
A
∠ CAD ≅ ∠ ABE ≅ ∠ BCF
Prove △ABC is equiangular.
C
Prove △DEF is equilateral.
STATEMENTS
2. Definition of equilateral
triangle
3. ∠ B ≅ ∠ C,
∠ A ≅ ∠ C,
∠A ≅ ∠B
3. Base Angles Theorem
(Thm. 5.6)
4. △ABC is
equiangular.
4. Definition of equiangular
triangle
— —— —
38. a. By the markings, AE ≅ DE , AB ≅ DC , and
∠ BAE ≅ ∠ CDE. So, △ABE ≅ △DCE by SAS
Congruence Theorem (Thm. 5.5).
b. △AED and △BEC are isosceles triangles.
c. ∠ EDA, ∠ BCA, and ∠ CBD are all congruent to ∠ EAD.
A
39. Given △ABC is equiangular.
Prove △ABC is equilateral.
B
C
B
E
F
REASONS
1. △ABC is equilateral.
∠ CAD ≅ ∠ ABE
≅ ∠ BCF
1. Given
2. △ABC is equiangular.
2. Corollary to the Base
Angles Theorem (Cor. 5.2)
3. ∠ ABC ≅ ∠ BCA
≅ ∠ BAC
3. Definition of equiangular
triangle
4. m∠ CAD = m∠ ABE =
m∠ BCF, m∠ ABC =
m∠ BCA = m∠ BAC
4. Definition of congruent
angles
5. m∠ ABC = m∠ ABE +
m∠ EBC, m∠ BCA =
m∠ BCF + m∠ ACF,
m∠ BAC = m∠ CAD +
m∠ BAD
5. Angle Addition Postulate
(Post. 1.4)
6. m∠ ABE + m∠ EBC =
m∠ BCF + m∠ ACF =
m∠ CAD + m∠ BAD
6. Substitution Property of
Equality
7. m∠ ABE + m∠ EBC =
m∠ ABE + m∠ ACF =
m∠ ABE + m∠ BAD
7. Substitution Property of
Equality
STATEMENTS
REASONS
1. △ABC is
equiangular.
1. Given
8. m∠ EBC = m∠ ACF =
m∠ BAD
8. Subtraction Property of
Equality
2. ∠ B ≅ ∠ C,
∠ A ≅ ∠ C,
∠A ≅ ∠B
2. Definition of equiangular
triangle
9. ∠ EBC ≅ ∠ ACF
≅ ∠ BAD
9. Definition of congruent
angles
3. —
AB ≅ —
AC ,
—
AB ≅ —
BC ,
—
AC ≅ —
BC
3. Converse of the Base Angles
Theorem (Thm. 5.7)
4. △ABC is
equilateral.
4. Definition of equilateral
triangle
40. no; The distance between point T(0, 6) and a point on
y = x is equal to the distance between U(6, 0) and y = x
(using the same point). So any point V on y = x will be
the same distance from T and U. Therefore, TV = VU
and △TVU is an isosceles triangle. Unless V(3, 3) is the
third point, in which case, T, V, and U are collinear and
perpendicular to y = x.
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C
10. ∠ FEB ≅ ∠ DFC
≅ ∠ EDA
10. Third Angles Theorem
(Thm. 5.4)
11. ∠ FEB and ∠ FED
are supplementary,
∠ DFC and ∠ EFD are
supplementary, and
∠ EDA and ∠ FDE are
supplementary.
11. Linear Pair Postulate
(Post. 2.8)
12. ∠ FED ≅ ∠ EFD
≅ ∠ FDE
12. Congruent Supplements
Theorem (Thm. 2.4)
13. △DEF is equiangular.
13. Definition of equiangular
triangle
14. △DEF is equilateral.
14. Corollary to the Converse
of the Base Angles
Theorem (Cor. 5.3)
Geometry
Worked-Out Solutions
157
Chapter 5
Maintaining Mathematical Proficiency
—
—
42. Reflexive Property of Congruence (Theorem 2.1): SE ≅ SE
43. Symmetric Property of Congruence (Theorem 2.1):
— —
— ≅ JK
—.
If JK ≅ RS
, then RS
44. Transitive Property of Congruence (Theorem 2.1):
— ≅ PQ
—, and PQ
— ≅ UV
—, then EF
— ≅ UV
—.
If EF
5.1–5.4 What Did You Learn? (p. 259)
5. corresponding angles: ∠ Q ≅ ∠ W, ∠ R ≅ ∠ X, ∠ S ≅ ∠ Y,
∠T ≅ ∠Z
— ≅ WX
—, RS
— ≅ XY
—, ST
— ≅ YZ
—,
corresponding sides: QR
—
—
QT ≅ WZ
Sample answer: RSTQ ≅ XYZW
6. no; The congruent angles are not the included angle.
7. yes; △GHF ≅ △KHJ by the SAS Congruence Theorem
(Thm. 5.5).
1. You are given a diagram of the triangle made from the
segments that connect each person to the other two, along
with the length of each segment; the people are all standing
on the same stage (plane), so the points are coplanar; You are
asked to classify the triangle by its sides and by measuring
its angles.
STATEMENTS
1. —
GH ≅ —
HK ,
—
FH ≅ —
HJ
3. Sample answer: a large triangle made up of 9 small triangles,
a hexagon, a parallelogram
2. Vertical Angles Congruence
Theorem (Thm. 2.6)
3. △GHF ≅ △KHJ
3. SAS Congruence Theorem
(Thm. 5.5)
8. yes; △LMP ≅ △NMP by the SAS Congruence Theorem
(Thm. 5.5).
STATEMENTS
1. —
LM ≅ —
NM ,
∠ LMP ≅ ∠ NMP
2. —
MP ≅ —
MP
5.1–5.4 Quiz (p. 260)
1. x° = 30° + 80° = 110°; The exterior angle measures 110°.
3. △LMP ≅ △NMP
2.(5x + 2)° + 6x° = 90°
11x = 88
x=8
y° = 180° − (5x + 2)°
⋅
= 180° − (5 8 + 2)°
= 180° − 42° = 138°
The exterior angle measures 138°.
3. 29° + (12x + 26)° = (15x + 34)°
55 + 12x = 15x + 34
21 = 3x
1. Given
2. ∠ GHF ≅ ∠ KHJ
2. There is a pair of congruent triangles, so all pairs of
corresponding sides and angles are congruent. By the
Triangle Sum Theorem (Thm. 5.1), the three angles in
△LMN have measures that add up to 180°. You are given two
measures, so you can find the third using this theorem. The
measure of ∠ P is equal to the measure of its corresponding
angle, ∠ L. The measure of ∠ R is equal to the measure of
its corresponding angle, ∠ N. Once you write this system of
equations, you can solve for the values of the variables.
REASONS
REASONS
1. Given
2. Reflexive Property of
Congruence (Thm. 2.1)
3. SAS Congruence Theorem
(Thm. 5.5)
— —
9. If VW ≅ WX , then ∠ VXW ≅ ∠ XVW by the Base Angles
Theorem (Thm. 5.6).
— —
10. If XZ ≅ XY , then ∠ XYZ ≅ ∠ XZY by the Base Angles
Theorem (Thm. 5.6).
— —
11. If ∠ ZVX ≅ ∠ ZXV, then XZ ≅ VZ , by the Converse of the
Base Angles Theorem (Thm. 5.7).
— —
12. If ∠ XYZ ≅ ∠ ZXY, then XZ ≅ YZ , by the Converse of the
Base Angles Theorem (Thm. 5.7).
7=x
⋅
(15x + 34)° = (15 7 + 34)° = (105 + 34)° = 139°
The exterior angle measures 139°.
4. corresponding angles: ∠ C ≅ ∠ F, ∠ A ≅ ∠ D, ∠ B ≅ ∠ E
—≅—
— ≅ DE
—, CB
— ≅ FE
—
corresponding sides: CA
FD, AB
Sample answer: △CAB ≅ △FDE
158
Geometry
Worked-Out Solutions
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Chapter 5
5.5 Explorations (p. 261)
DE = QR
13.
5y − 7 = 38
1. a. Check students’ work.
5y = 45
b. Check students’ work.
—
c. AB = 2 because AB has one endpoint at the origin and one
y=9
endpoint on a circle with a radius of 2 units.
—
AC = 3 because AC has one endpoint at the origin and
one endpoint on a circle with a radius of 3 units. BC = 4
because it was created that way.
m∠ S = m∠ F
m∠ F = 180° − (123° + 29°) = 28°
2x + 2 = 28
2x = 26
d. m∠ A = 104.43°, m∠ B = 46.61°, m∠ C = 28.96°
x = 13
e. Check students’ work; If two triangles have three pairs
So, x = 13 and y = 9.
of congruent sides, then they will have three pairs of
congruent angles.
14. 5x − 1 = 24
2. When the corresponding sides of two triangles are congruent,
5x = 25
the corresponding angles are also congruent.
x=5
3. Use rigid transformations to map triangles.
6y = 120
y = 20
5.5 Monitoring Progress (pp. 263–265)
— —— —
So, x = 5 and y = 20.
15.
1. yes; From the diagram markings, DF ≅ HJ , FG ≅ JK , and
— ≅ HK
—. So, △DFG ≅ △HJK by the SSS Congruence
DG
Theorem (Thm. 5.8).
x = 4[ (90 − x) − 5 ]
x = 360 − 4x − 20
—
—
2. no; AB corresponds with CD , but they are not the same
5x = 340
measure. In order for two triangles to be congruent, all pairs
of corresponding sides must be congruent.
x = 68
90° − 68° = 22°
— —— —
3. yes; From the diagram markings, QP ≅ RS , PT ≅ ST , and
The acute angles are 68° and 22°.
16. a. Triangle 1 has a right angle. So, it is a right triangle.
— ≅ RT
—. So, △QPT ≅ △RST by the SSS Congruence
QT
Theorem (Thm. 5.8).
It appears that triangle 2 has an obtuse angle. So, it is
an obtuse triangle. It appears that triangle 3 has three
nonequivalent acute angles. So, it is an acute triangle.
4. not stable; This square is not stable because there are many
Triangle 4 has three congruent angles. So, it is an
equiangular triangle.
5. stable; The diagonal support in this figure forms triangles
b. Triangle 4 has three congruent sides. So, it is an
equilateral triangle. It appears that triangle 5 has three
noncongruent sides. So, it is a scalene triangle.
Triangle 6 has two congruent sides. So, it is an isosceles
triangle.
c. yes;
A
E
7
B
D
F
STATEMENTS
REASONS
1.
1. Given
— —
—
AB ≅ —
DE , BC ≅ EF ,
∠B ≅ ∠E
2. △ABC ≅ △DEF
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with fixed side lengths. By the SSS Congruence Theorem
(Thm. 5.8), these triangles cannot change shape, so the figure
is stable.
6. not stable; The diagonal support in this figure forms a
triangle and a quadrilateral. The triangle would be stable,
but the quadrilateral is not because there are many possible
quadrilaterals with the given side lengths.
7. Redraw △ABC and △DCB.
8
C
possible quadrilaterals with the given side lengths.
C
B
B
A
C
D
2. SAS Congruence
Theorem (Thm. 5.5)
Geometry
Worked-Out Solutions
159
Chapter 5
— —
8. Given AC ≅ DB , ∠ ABC and ∠ DCB are right angles.
REASONS
— ≅ DB
—, ∠ ABC
1. AC
and ∠ DCB are
right angles.
1. Given
CB ≅ —
BC
2. —
— —
by the Reflexive Property of Congruence (Thm. 2.1). So, it
should say △JKL ≅ △LMJ or △JKL ≅ △JML by the SSS
Congruence Theorem (Thm. 5.8).
Prove △ABC ≅ △DCB
STATEMENTS
— — — —
10. no; You are given that JK ≅ KL ≅ LM ≅ MJ . Also, JL ≅ JL
11. yes; The diagonal supports in this figure form triangles
with fixed side lengths. By the SSS Congruence Theorem
(Thm. 5.8), these triangles cannot change shape, so the figure
is stable.
2. Reflexive Property of
Congruence (Thm. 2.1)
12. no; The support in this figure forms two quadrilaterals, which
3. △ABC and △DCB
are right triangles.
3. Definition of a right triangle
4. △ABC ≅ △DCB
4. HL Congruence Theorem
(Thm. 5.9)
are not stable because there are many possible quadrilaterals
with the given side lengths.
— —
— ⊥ AD
—, CD
— ⊥ AD
—
AB
13. Given AC ≅ DB ,
A
B
Prove △BAD ≅ △CDA
D
5.5 Exercises (pp. 266–268)
D
C
Vocabulary and Core Concept Check
1. The side opposite the right angle is called the hypotenuse of
the right triangle.
A
2. Three of the triangles are confirmed as right triangles. The
second triangle from the left is the only triangle with legs
that are not the legs of a right triangle.
STATEMENTS
REASONS
— ≅ DB
—,
1. AC
1. Given
2.
Monitoring Progress and Modeling with Mathematics
— —— —— —
3. yes; AB ≅ DB , BC ≅ BE , AC ≅ DE
—
4. no; You cannot tell for sure from the diagram whether PS and
— are congruent.
RS
— —— —
5. yes; ∠ B and ∠ E are right angles, AB ≅ FE , AC ≅ FD
6. no; the hypotenuses are not marked as congruent.
— —— —
— —
7. no; You are given that RS ≅ PQ , ST ≅ QT, and RT ≅ PT .
So, it should say △RST ≅ △PQT by the SSS Congruence
Theorem (Thm. 5.8).
— —
— —
8. yes; You are given that AB ≅ CD and AD ≅ CB . Also
— ≅ BD
— by the Reflexive Property of Congruence
BD
(Thm. 2.1). So, △ABD ≅ △CDB by the SSS Congruence
Theorem (Thm. 5.8).
— — — —
—
—
DF ≅ DF by the Reflexive Property of Congruence
—
AB ⊥ —
AD ,
—
CD ⊥ —
AD
—
AD ≅ —
AD
2. Reflexive Property of
Congruence (Thm. 2.1)
3. ∠ BAD and ∠ CDA
are right angles.
3. Definition of perpendicular
lines
4. △BAD and △CDA
are right triangles.
4. Definition of a right triangle
5. △BAD ≅ △CDA
5. HL Congruence Theorem
(Thm. 5.9)
—
14. Given G is the midpoint of EH ,
— ≅ GI
—,
FG
E
F
∠ E and ∠ H are right angles.
Prove △EFG ≅ △HIG
G
H
I
9. yes; You are given that EF ≅ GF and DE ≅ DG . Also
(Thm. 2.1). So, △DEF ≅ △DGF by the SSS Congruence
Theorem (Thm. 5.8).
160
Geometry
Worked-Out Solutions
G
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Chapter 5
STATEMENTS
REASONS
1. G is the midpoint
— ≅ GI
—,
of —
EH , FG
∠ E and ∠ H are
right angles.
1. Given
2. —
EG ≅ —
HG
—
—
to the length of QR . Use this length to draw an arc. Draw
17. Construct a side that is congruent to QS . Open your compass
an arc with radius RS. Complete the triangle. By the SSS
Congruence Theorem (Thm. 5.8), the two triangles are
congruent.
R
2. Definition of midpoint
3. △EFG and △HIG
are right triangles.
3. Definition of a right triangle
4. △EFG ≅ △HIG
4. HL Congruence Theorem
(Thm. 5.9)
Q
— —— —
15. Given LM ≅ JK , MJ ≅ KL
—
—
to the length of QR . Use this length to draw an arc. Draw
18. Construct a side that is congruent to QS . Open your compass
K
L
J
M
Prove △LMJ ≅ △JKL
REASONS
— ≅ JK
—, MJ
— ≅ KL
—
1. LM
1. Given
3. △LMJ ≅ △JKL
— —
16. Given WX ≅ VZ ,
Prove △VWX ≅ △WVZ
V
2.
—
WV ≅ —
WV
X
Y
Z
1. Given
△TUV ≅ △ZYX by the SSS Congruence Theorem
(Thm. 5.8).
1
20. When you substitute —4 for x, KL ≠ LM.
6x = 4x + 4
2x = 4
2. Reflexive Property of
Congruence (Thm. 2.1)
3. Definition of congruent
segments
4. WZ = WY + YZ,
VX = VY + YX
4. Segment Addition
Postulate (Post. 1.2)
5. VX = WY + YZ
5. Substitution Property
of Equality
6. VX = WZ
6. Transitive Property of
Equality
8. △VWX ≅ △WVZ
19. The order of the points in the congruence statement should
reflect the corresponding sides and angles.
REASONS
3. WY = VY, YZ = YX
7. —
VX ≅ —
WZ
S
3. SSS Congruence Theorem
(Thm. 5.8)
W
— ≅ VZ
—, WY
—≅—
1. WX
VY ,
—
YZ ≅ —
YX
Q
2. Reflexive Property of
Congruence (Thm. 2.1)
— ≅ VY
—,
WY
—
—
YZ ≅ YX
STATEMENTS
an arc with radius RS. Complete the triangle. By the SSS
Congruence Theorem (Thm. 5.8), the two triangles are
congruent.
R
STATEMENTS
2. —
JL ≅ —
JL
S
x=2
2x + 1 = 3x − 1
1=x−1
2=x
21. no; The sides of a triangle do not have to be congruent to
each other, but each side of one triangle must be congruent to
the corresponding side of the other triangle.
7. Definition of congruent
segments
8. SSS Congruence Theorem
(Thm. 5.8)
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Worked-Out Solutions
161
Chapter 5
— — — —— —
22. Given HF ≅ FS ≅ ST ≅ TH ; FT ≅SH ;
———
——
26. AB = √ [ 3 − (−2) ]2 + (−3 − 1)2 = √ (3 + 2)2 + (−4)2
S
∠ H, ∠ F, ∠ S, and ∠ T
are right angles.
—
—
—
= √ 52 + 16 = √25 + 16 = √41 ≈ 6.4
——
Prove △HFS ≅ △FST ≅ △STH
T
F
——
BC = √ (7 − 3)2 + [ 5 − (−3) ]2 = √(4)2 + (5 + 3)2
—
—
—
= √ 16 + 82 = √16 + 64 = √80 ≈ 8.9
——
——
AC = √ [ 7 − (−2) ]2 + (5 − 1)2 = √(7 + 2)2 + (4)2
—
—
—
= √ 92 + 42 = √81 + 16 = √97 ≈ 9.8
——
H
——
STATEMENTS
REASONS
— ≅ FS
— ≅ ST
— ≅ TH
—;
1. HF
DE = √ (8 − 3)2 + (2 − 6)2 = √ (5)2 + (−4)2
—
—
= √ 25 + 16 = √41 ≈ 6.4
1. Given
EF = √ (10 − 8)2 + (11 − 2)2 = √ 22 + 92
—
FT ≅ —
SH ; ∠ H, ∠ F, ∠ S,
——
—
—
= √ 4 + 81 = √85 ≈ 9.2
and ∠ T are right angles.
——
2. Reflexive Property of
Congruence (Thm. 2.1)
3. △HFS, △FST, and
△STH are right triangles.
3. Definition of a
right triangle
4. △HFS ≅ △FST ≅
△STH
4. HL Congruence
Theorem (Thm. 5.9)
—
—
= √ 49 + 25 = √74 ≈ 8.6
The triangles are not congruent.
——
—
—
= √ 36 + 25 = √61 ≈ 7.8
——
— ≅ ML
—.
JL
b. SAS Congruence Theorem (Thm. 5.5); By definition of
— ≅ MK
—. Also, LK
— ≅ LK
—, by the Reflexive
midpoint, JK
Property of Congruence (Thm. 2.1), and ∠ JKL ≅ ∠ MKL
by the Right Angles Congruence Theorem (Thm. 2.3).
24. a. To use the SSS Congruence Theorem (Thm. 5.8) to prove
— ≅ CD
—.
△ABC ≅ △CDE, you need to know AB
—
—
= √ 9 + 25 = √34 ≈ 5.8
△ABC ≅ △CDE, you need to know that ∠ B and ∠ D are
right angles.
—
—
———
+ (6 +
=
DE = √(5 −
5)2
√62 + 82
+ (1 −
=√
—
(−6)2
——
—
= √36 = 6
—
EF = √(13 − 5)2 + (1 − 1)2 = √ 82 = 8
——
——
DF = √(13 − 5)2 + (1 − 7)2 = √(8)2 + (−6)2
—
—
—
—
= √ 9 + (−5)2 = √9 + 25 = √34 ≈ 5.8
———
——
DF = √ (9 − 0)2 + [ −1 − (−1) ]2 = √92 + (−1 + 1)2
—
AB = DE, BC = EF, and AC = DE, so △ABC ≅ △DEF.
———
——
28. AB = √ [ −5 − (−5) ]2 + (2 − 7)2 = √ (−5 + 5)2 + (−5)2
—
—
——
—
AC = √ [ 0 − (−5) ]2 + (2 − 7)2 = √52 + (−5)2
—
—
= √ 25 + 25 = √50 ≈ 7.1
—
——
—
——
—
——
—
—
EF = √(4 − 0)2 + (1 − 1)2 = √42 = √16 = 4
—
7)2
——
DE = √(0 − 0)2 + (1 − 6)2 = √(−5)2 = √25 = 5
—
= √36 + 64 = √ 100 = 10
——
—
EF = √ (9 − 6)2 + [ −1 − (−6) ]2 = √32 + (−1 + 6)2
——
√[ 4 − (−2) ]2 + [ 6 − (−2) ]2
—
—
—
———
BC = √[ 0 − (−5) ]2 + (2 − 2)2 = √ 52 = √25 = 5
—
BC = √(4 − 4)2 + [ 6 − (−2) ]2 = √(6 + 2)2 = √82 = 8
= √(4 +
——
= √ 36 + (−5)2 = √36 + 25 = √61 ≈ 7.8
= √ 25 = 5
= √(4 + 2)2 + (−2 + 2)2 = √ 62 = 6
2)2
———
DE = √ (6 − 0)2 + [ −6 − (−1) ]2 + √62 + (−6 + 1)2
—
———
25. AB = √ [ 4 − (−2) ]2 + (−2 − (−2))2
2)2
—
—
AC = √(9 − 0)2 + (0 − 0)2 = √92 = √9 = 9
= √ 81 = 9
b. To use the HL Congruence Theorem (Thm. 5.9) to prove
——
—
BC = √ (9 − 6)2 + (0 − 5)2 = √ 32 + (−5)2
——
——
—
27. AB = √ (6 − 0)2 + (5 − 0)2 = √ 62 + 52
23. a. You need to know that the hypotenuses are congruent:
——
—
DF = √ (10 − 3)2 + (11 − 6)2 = √ (7)2 + (5)2
2. —
SH ≅ —
SH
AC =
—
—
= √64 + 36 = √ 100 = 10
AB = DE, BC = EF, and AC = DF, so △ABC ≅ △DEF.
DF = √ (4 − 0)2 + (1 − 6)2 = √ 42 + (−5)2
—
—
= √ 16 + 25 = √41 ≈ 6.4
The triangles are not congruent.
29. yes; You could use the string to measure the length of each
side on one triangle and compare it to the length of the
corresponding side of the other triangle to determine whether
SSS Congruence Theorem (Thm. 5.8) applies.
30. SAS Congruence Theorem (Thm. 5.5), HL Congruence
Theorem (Thm. 5.9), SSS Congruence Theorem (Thm. 5.8)
162
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Chapter 5
— —
31. both; JL ≅ JL by the Reflexive Property of Congruence
(Thm. 2.1), and the other two pairs of sides are marked as
congruent. So, the SSS Congruence Theorem (Thm. 5.8) can
be used. Also, because ∠ M and ∠ K are right angles, they
are both right triangles, and the legs and hypotenuses are
congruent. So, the HL Congruence Theorem (Thm. 5.9) can
be used.
32. yes; They would have to be formed from circles that were at
the same angle with each other. So, all corresponding parts
would be congruent.
33. Sample answer:
L
36. 5x = 3x + 10
5x − 2 = 4x + 3
2x = 10
x=5
x=5
⋅
⋅
BD = 4 ⋅ 5 + 3 = 23
CD = 3 ⋅ 5 + 10 = 25
— ——
AB = 5 5 = 25
AC = 5 5 − 2 = 25 − 2 = 23
—, and BC
— ≅ CB
— (Reflexive
For x = 5, AB ≅ CD , AC ≅ BD
Property of Congruence, Thm. 2.1). Then, by the SSS
Congruence Theorem (Thm. 5.8), △ABC ≅ △DCB.
5x = 4x + 3
S
x=3
⋅
⋅
BD = 4 ⋅ 3 + 3 = 15
CD = 3 ⋅ 3 + 10 = 19
— —
AB = 5 3 = 15
AC = 5 3 − 2 = 15 − 2 = 13
M
N
T
U
34. Sample answer:
L
For x = 3, AC ≅ CD , so the triangles are not congruent.
S
5x − 2 = 3x + 10
2x = 12
x=6
M
N
T
U
— —
35. a. BD ≅ BD by the Reflexive Property of Congruence
— ≅ CB
— and that ∠ ADB
(Thm. 2.1). It is given that AB
and ∠ CDB are right angles. So, △ABC and △CBD are
right triangles and are congruent by the HL Congruence
Theorem (Thm. 5.9).
— — — —— —
⋅
⋅
BD = 4 ⋅ 6 + 3 = 27
CD = 3 ⋅ 6 + 10 = 28
— —
AB = 5 6 = 30
AC = 5 6 − 2 = 30 − 2 = 28
For x = 6, AB ≅ BD , so the triangles are not congruent.
x = 5 is the only solution that yields △ABC ≅ △DCB.
b. yes; Because AB ≅ CB ≅ CE ≅ FE , BD ≅ EG ,
and they are all right triangles, it can be shown that
△ABD ≅ △CBD ≅ △CEG ≅ △FEG by the HL
Congruence Theorem (Thm. 5.9).
Maintaining Mathematical Proficiency
—
—
— —
—
—
— —
37. DF corresponds to AC . So, AC ≅ DF .
38. BC corresponds to EF . So, EF ≅ BC .
39. ∠ E corresponds to ∠ B. So, ∠ B ≅ ∠ E.
40. ∠ C corresponds to ∠ F. So, ∠ F ≅ ∠ C.
5.6 Explorations (p. 269)
1. a. Check students’ work.
b. Check students’ work.
—
—
—
—
— of
AB ≅ AB . Also, BC of △ABC is congruent to BD
c. The side that △ABC and △ABD share is AB . So,
△ABD. The nonincluded angle that these two triangles
share is ∠ BAC.
d. no; The third pair of sides are not congruent.
e. no; These two triangles provide a counterexample for
SSA. They have two pairs of congruent sides and a pair
of nonincluded congruent angles, but the triangles are not
congruent.
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Geometry
Worked-Out Solutions
163
Chapter 5
2.
Possible congruence
theorem
Valid or
not valid?
SSS
Valid
SSA
Not valid
SAS
Valid
AAS
Valid
ASA
Valid
AAA
Not valid
— —— —
— ≅ DC
—
AC
2. Given AB ⊥ AD , DE ⊥ AD ,
C
−5
B′
A
−4
D
C
B
STATEMENTS
REASONS
— ⊥ AD
—, DE
— ⊥ AD
—,
1. AB
—
—
AC ≅ DC
1. Given
2. ∠ BAC and ∠ EDC are
right angles.
2. Definition of
perpendicular lines
3. ∠ BAC ≅ ∠ EDC
3. Right Angles
Congruence Theorem
(Thm. 2.3)
4. ∠ ACB ≅ ∠ DCE
4. Vertical Angles
Congruence Theorem
(Thm. 2.6)
5. △ABC ≅ △DEC
5. ASA Congruence
Theorem (Thm. 5.10)
y
B
C′
A
Prove △ABC ≅ △DEC
Sample answer: A counterexample for SSA is given in
Exploration 1.
A counterexample for AAA is shown here.
6
5
4
E
5 x
A′
— —
3. Given ∠ S ≅ ∠ U, RS ≅ VU
R
U
Prove △RST ≅ △VUT
In this example, each pair of corresponding angles is
congruent, but the corresponding sides are not congruent.
T
S
3. In order to determine that two triangles are congruent, one of
the following must be true.
V
STATEMENTS
REASONS
1. ∠ S ≅ ∠ U,
— ≅ VU
—
RS
1. Given
Two pairs of corresponding angles and the pair of included
sides are congruent (ASA).
2. ∠ RTS ≅ ∠ VTU
2. Vertical Angles Congruence
Theorem (Thm. 2.6)
Two pairs of corresponding angles and one pair of
nonincluded sides are congruent (AAS).
3. △RST ≅ △VUT
3. AAS Congruence Theorem
(Thm. 5.11)
All three pairs of corresponding sides are congruent (SSS).
Two pairs of corresponding sides and the pair of included
angles are congruent (SAS).
The hypotenuses and one pair of corresponding legs of two
right triangles are congruent (HL).
4. yes; Sample answer:
In the diagram, △ABD ≅ △ACD by the
HL Congruence Theorem (Thm. 5.9), the
SSS Congruence Theorem (Thm. 5.8),
and the SAS Congruence Theorem
(Thm. 5.5).
A
5.6 Exercises (pp. 274–276)
Vocabulary and Core Concept Check
1. Both theorems are used to prove that two triangles are
B
D
C
5.6 Monitoring Progress (pp. 271–273)
congruent, and both require two pairs of corresponding
angles to be congruent. In order to use the AAS Congruence
Theorem (Thm. 5.11), one pair of corresponding
nonincluded sides must also be congruent. In order to
use the ASA Congruent Theorem (Thm. 5.10), the pair of
corresponding included sides must be congruent.
1. yes; By the Alternate Interior Angles Theorem (Thm. 3.2),
∠ 1 ≅ ∠ 3 and ∠ 4 ≅ ∠ 2, and by the Reflexive Property of
— ≅ WY
—. So, △WXY ≅ △YZW by
Congruence (Thm. 2.1), WY
the ASA Congruence Theorem (Thm. 5.10).
164
Geometry
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2. You need to know that one pair of corresponding sides are
congruent.
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Chapter 5
Modeling Progress and Modeling with Mathematics
—
—
—
—
—
—
—
NL ⊥ NQ , NL ⊥ MP , QM PL
17. Given M is the midpoint of NL ,
3. yes; △ABC ≅ △QRS by the AAS Congruence Theorem
Prove △NQM ≅ △MPL
(Thm. 5.11).
4. yes; △ABC ≅ △DBC by the AAS Congruence Theorem
Q
P
N
M
(Thm. 5.11).
5. no; Two sides and a nonincluded angle are not sufficient to
determine congruence.
6. yes; △RSV ≅ △UTV by the ASA Congruence Theorem
(Thm. 5.10).
— —
7. Given GH ≅ MN , ∠ G ≅ ∠ M, ∠ F ≅ ∠ L
— —
8. Given FG ≅ LM , ∠ G ≅ ∠ M, ∠ F ≅ ∠ L
9. yes; △ABC ≅ △DEF by the ASA Congruence Theorem
(Thm. 5.10).
10. no; The congruence statements follow the pattern SSA,
L
STATEMENTS
REASONS
—,
1. M is the midpoint of NL
—
—
—
—
NL ⊥ NQ , NL ⊥ MP ,
— PL
—
QM
1. Given
2. ∠ QNM and ∠ PML are
right angles.
2. Definition of
perpendicular lines
3. ∠ QNM ≅ ∠ PML
3. Right Angles
Congruence Theorem
(Thm. 2.3)
4. ∠ QMN ≅ ∠ PLM
4. Corresponding Angles
Theorem (Thm. 3.1)
which is not sufficient to conclude that the triangles are
congruent.
—
—
11. no; AC and DE are not corresponding sides.
12. yes; △ABC ≅ △DEF by the AAS Congruence Theorem
(Thm. 5.11).
—
13. Construct a side that is congruent to DF . Construct an angle
that is congruent to ∠ D and a second angle that is congruent
to ∠ F. Complete the triangle. By the ASA Congruence
Theorem (Thm. 5.10), △DEF ≅ △ACB.
5. —
NM ≅ —
ML
5. Definition of midpoint
6. △NQM ≅ △MPL
— —
18. Given AJ ≅ KC ,
B
∠ BJK ≅ ∠ BKJ,
∠A ≅ ∠C
Prove △ABK ≅ △CBJ
—
14. Construct a side that is congruent to JK . Construct an angle
that is congruent to ∠ J and construct a second angle that
is congruent to ∠ K. Complete the triangle. By the ASA
Congruence Theorem (Thm. 5.10), △JKL ≅ △BAC.
15. In the congruence statement, the vertices should be
in corresponding order; △JKL ≅ △FGH by the ASA
Congruence Theorem (Thm. 5.10).
16. The diagram shows ∠ Q ≅ ∠ V, ∠ R ≅ ∠ W, and
— ≅ VW
—. Because QR
— and VW
— are included sides,
QR
△QRS ≅ △VWX by the ASA Congruence Theorem
(Thm. 5.10).
STATEMENTS
—≅—
1. AJ
KC ,
∠ BJK ≅ ∠ BKJ,
∠A ≅ ∠C
A
J
K
C
REASONS
1. Given
2. AJ = KC
2. Definition of congruent
segments
3. JC = JK + KC,
AK = AJ + JK
3. Segment Addition Postulate
(Post. 1.2)
4. AK = KC + JK
4. Substitution Property of
Equality
5. AK = JK + KC
5. Commutative Property of
Addition
6. AK = JC
6. Transitive Property of Equality
7. —
AK ≅ —
JC
8. △ABK ≅ △CBJ
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6. ASA Congruence
Theorem (Thm. 5.10)
7. Definition of congruent
segments
8. ASA Congruence Theorem
(Thm. 5.10)
Geometry
Worked-Out Solutions
165
Chapter 5
—
—
19. Given VW ≅ UW ,
Z
∠X ≅ ∠Z
V
Prove △XWV ≅ △ZWU
U
W
STATEMENTS
REASONS
—,
1. —
VW ≅ UW
∠X ≅ ∠Z
1. Given
2. ∠ W ≅ ∠ W
3. △XWV ≅ △ZWU
X
Y
25. By finding the values of x and y in each triangle and solving
for the angle measurements, △ABC ≅ △DBC by the ASA
Congruence Theorem (Thm. 5.10) or the AAS Congruence
Theorem (Thm. 5.11). Both triangles have a common side,
— ≅ BC
—.
BC
△ABC:
(2x − 8)° + (5x + 10)° + (8x − 32)° = 180°
15x − 30 = 180
15x = 210
2. Reflexive Property of
Congruence (Thm. 2.2)
⋅
⋅
m∠ BCA = 5 ⋅ 14 + 10 = 80°
m∠ CAB = 2 14 − 8 = 28 − 8 = 20°
m∠ ABC = 8 14 − 32 = 80°
3. ASA Congruence Theorem
(Thm. 5.11)
20. Given ∠ NKM ≅ ∠ LMK,
L
△DBC:
N
(y − 6)° + (3y + 2)° + (4y − 24)° = 180°
8y − 28 = 180
∠L ≅ ∠N
Prove △NMK ≅ △LKM
STATEMENTS
1. ∠ NKM ≅ ∠ LMK,
∠L ≅ ∠N
2. —
KM ≅ —
MK
3. △NMK ≅ △LKM
x = 14
8y = 208
K
y = 26
M
⋅
⋅
REASONS
m∠ DBC = 4 26 − 24 = 80°
1. Given
m∠ BCD = 3 26 + 2 = 80°
2. Reflexive Property of
Congruence (Thm. 2.1)
3. AAS Congruence Theorem
(Thm. 5.11)
21. You are given two right triangles, so the triangles have
congruent right angles by the Right Angles Congruence
Theorem (Thm. 2.3). Because another pair of angles and a
pair of corresponding nonincluded sides (the hypotenuses)
are congruent, the triangles are congruent by the AAS
Congruence Theorem (Thm. 5.11).
m∠ CDB = 26 − 6 = 20°
— ≅ BC
—, ∠ BCA ≅ ∠ BCD
∠ BAC ≅ ∠ CDB, BC
△ABC ≅ △DBC by the ASA Congruence Theorem
(Thm 5.10).
— ≅ BC
—, ∠ BCA ≅ ∠ BCD
∠ BAC ≅ ∠ CDB, BC
△ABC ≅ △DBC by the AAS Congruence Theorem
(Thm 5.11).
— ≅ BC
—, ∠ ABC ≅ ∠ DBC
∠ BAC ≅ ∠ BDC, BC
△ABC ≅ △DBC by the AAS Congruence Theorem
(Thm 5.11).
— —
26. C; ∠ TSV ≅ ∠ VUW (given), ST ≅ UV (given), and
∠ STV ≅ ∠ UVW because they are right angles.
22. You are given two right triangles, so the triangles have
congruent right angles by the Right Angles Congruence
Theorem (Thm. 2.3). Because both pairs of legs are
congruent, and the congruent right angles are the included
angles, the triangles are congruent by the SAS Congruence
Theorem (Thm. 5.5).
23. You are given two right triangles, so the triangles have
congruent right angles by the Right Angles Congruence
Theorem (Thm. 2.3). There is also another pair of congruent
corresponding angles and a pair of congruent corresponding
sides. If the pair of congruent sides is the included side,
then the triangles are congruent by the ASA Congruence
Theorem (Thm. 5.10). If the pair of congruent sides is a
nonincluded pair, then the triangles are congruent by the
AAS Congruence Theorem (Thm. 5.11).
24. D; To prove △JKL ≅ △MNL using ASA Congruence
Theorem (Thm. 5.10), you know that ∠ L ≅ ∠ L and
— ≅ LJ
—. So, the additional information needed is
LM
∠ LKJ ≅ ∠ LNM.
166
Geometry
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Chapter 5
27. Given ∠ B ≅ ∠ C
b. Because △ABD ≅ △CBD and corresponding parts of
A
— ≅ AC
—
Prove AB
congruent triangles are congruent, you can conclude
— —
that AD ≅ CD
, which means that △ACD is isosceles by
definition.
c. no; For instance, because △ACD is isosceles, the girl sees
her toes at the bottom of the mirror. This remains true as
she moves backwards because △ACD remains isosceles.
30. △PTS ≅ △RTQ, △PTQ ≅ △RTS, △SQR ≅ △QSP,
B
STATEMENTS
D
C
REASONS
1. Draw —
AD , the angle
bisector of ∠ ABC.
1. Construction of angle
bisector
2. ∠ CAD ≅ ∠ BAD
2. Definition of angle bisector
3. ∠ B ≅ ∠ C
3. Given
4. —
AD ≅ —
AD
4. Reflexive Property of
Congruence (Thm. 2.1)
5. △ABD ≅ △ACD
5. AAS Congruence Theorem
(Thm. 5.11)
6. —
AB ≅ —
AC
6. Corresponding parts of
congruent triangles are
congruent.
28. yes; If you are using the AAS Congruence Theorem
(Thm. 5.11), you have enough information to conclude
that two pairs of angles are congruent and one pair of
non-included sides is congruent. By the Third Angles
Theorem (Thm. 5.4), the other pair of angles is congruent.
Now, choose the third pair of angles, the pair of congruent
sides, and the other pair of angles so that the pair of
congruent sides are included sides. So, you can use the ASA
Congruence Theorem (Thm. 5.10) to show that the triangles
are congruent.
29. a. Given ∠ CDB ≅ ∠ ADB, DB ⊥ AC
Prove △ABD ≅ △CBD
△SRP ≅ △QPR; Because ∠ PTS ≅ ∠ RTQ by the Vertical
Angles Congruence Theorem (Thm. 2.6), △PTS ≅ △RTQ
by the SAS Congruence Theorem (Thm. 5.5); Because
∠ PTQ ≅ ∠ RTS by the Vertical Angles Congruence
Theorem (Thm. 2.6), △PTQ ≅ △RTS by the SAS
— intersects
Congruence Theorem (Thm. 5.5); Transversal SQ
—
—
parallel sides PS and QR to create congruent alternate
— also
interior angles ∠ SQR and ∠ QSP, and transversal SQ
—
—
intersects parallel sides PQ and SR to create congruent
— ≅ SQ
—
alternate interior angles ∠ QSR and ∠ SQP. Also, SQ
by the Reflexive Property of Congruence (Thm. 2.1). So,
△SQR ≅ △QSP by the ASA Congruence Theorem (Thm.
— intersects parallel sides PS
— and QR
— to
5.10); Transversal PR
create congruent alternate interior angles ∠ SPR and ∠ QRP,
— also intersects parallel sides PQ
— and SR
— to
and transversal PR
create congruent alternate interior angles ∠ QPR and ∠ SRP.
— ≅ PR
— by the Reflexive Property of Congruence
Also, PR
(Thm. 2.1). So, △SRP ≅ △QPR by the ASA Congruence
Theorem (Thm. 5.10).
31. Two triangles with congruent
corresponding angles can be
similar triangles without being
congruent. This is based on
dilating a triangle. The angles
are equal and corresponding sides
are proportional.
B
C
A
E
D
F
32. yes; Both triangles will always have three vertices. So, each
STATEMENTS
REASONS
1. ∠ CDB ≅ ∠ ADB,
DB ⊥ AC
1. Given
2. ∠ ABD ≅ ∠ CBD
are right angles.
2. Definition of perpendicular
lines
3. ∠ ABD ≅ ∠ CBD
3. Right Angles Congruence
Theorem (Thm. 2.3)
4. —
BD ≅ —
BD
5. △ABD ≅ △CBD
vertex of one triangle can be mapped to one vertex of the other
triangle so that each vertex is only used once, and adjacent
vertices of one triangle must be adjacent in the other.
4. Reflexive Property of
Congruence (Thm. 2.1)
5. ASA Congruence Theorem
(Thm. 5.10)
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Geometry
Worked-Out Solutions
167
Chapter 5
33. a. The combinations that will make △TUV ≅ △XYZ by the
ASA Congruence Theorem (Thm. 5.10) are:
— ≅ XY
—, ∠ U ≅ ∠ Y
∠ T ≅ ∠ X, TU
— ≅ YZ
—, ∠ V ≅ ∠ Z
∠ U ≅ ∠ Y, UV
—
—
∠ T ≅ ∠ X, TV ≅ XZ , ∠ V ≅ ∠ Z
The combinations that will make △TUV ≅ △XYZ by the
AAS Congruence Theorem (Thm. 5.11) are:
5.7 Explorations (p. 277)
—). Because
measure as the distance across the river (AB
△ABC ≅ △DEC by the ASA Congruence Theorem
(Thm. 5.10), the corresponding parts of the two triangles
are also congruent.
b. Given ∠ A and ∠ D
— ≅ YZ
—, ∠ U ≅ ∠ Y, ∠ T ≅ ∠ X
UV
Prove AB = DE
— ≅ XZ
—, ∠ T ≅ ∠ X, ∠ U ≅ ∠ Y
TV
—
—, ∠ U ≅ ∠ Y, ∠ V ≅ ∠ Z
TU ≅ XY
— ≅ YZ
—, ∠ V ≅ ∠ Z, ∠ T ≅ ∠ X
UV
—
—
TV ≅ XZ , ∠ V ≅ ∠ Z, ∠ U ≅ ∠ Y
A
C
D
E
The combination that will make △TUV ≅ △XYZ by the
SSS Congruence Theorem (Thm. 5.8) is:
— ≅ XY
—, UV
— ≅ YZ
—, TV
— ≅ XZ
—
TU
STATEMENTS
1.
The combinations that will make △TUV ≅ △XYZ by the
SAS Congruence Theorem (Thm. 5.5) are:
— ≅ CD
—, ∠ A and
AC
REASONS
1. Given
∠ D are right angles.
— ≅ XY
—, ∠ T ≅ ∠ X, TV
— ≅ XZ
—
TU
2. ∠ A ≅ ∠ D
2. Right Angles Congruence
Theorem (Thm. 2.3)
— ≅ YZ
—, ∠ V ≅ ∠ Z, TV
— ≅ XZ
—
UV
3. ∠ BCA ≅ ∠ ECD
3. Vertical Angles Congruence
Theorem (Thm. 2.6)
( 6C3 = 20 ). There are 13 combinations that provide
4. △BCA ≅ △ECD
enough information. So, the probability of choosing at
random enough information to prove that the triangles are
13
congruent is —
, or 65%.
20
5. —
AB ≅ —
DE
4. ASA Congruence Theorem
(Thm. 5.10)
— ≅ YZ
—, ∠ U ≅ ∠ Y, TU
— ≅ XY
—
UV
b. There are 20 ways to choose 3 items from 6 items
Maintaining Mathematical Proficiency
1+5 0+4
6 4
34. Midpoint = —, — = —, — = (3, 2)
2
2
2 2
36.
B
are right angles.
— ≅ CD
—
AC
— ≅ XY
—, ∠ T ≅ ∠ X, ∠ V ≅ ∠ Z
TU
35.
—
1. a. The surveyor can measure DE, which will have the same
(
) ( )
−2 + 4 3 + (−1)
,
Midpoint = (
) = ( 22, 22 ) = (1, 1)
2
2
−5 + 2 −7 + (−4)
,
Midpoint = (
) = ( −32, −112 )
2
2
— —
— —
— —
—
—
37. Draw a segment. Label a point D
on the segment. Draw an arc with
center A, and label the intersection
F
points B and C. Using the same
radius, draw an arc with center D.
D
Label the point of intersection of
E
the arc and the segment as E. Draw
an arc with radius BC with center E.
⃗. So, ∠ FDE ≅ ∠ A.
Label the intersection F. Draw DF
6. AB = DE
5. Corresponding parts of
congruent triangles are
congruent.
6. Definition of congruent
segments
c. By creating a triangle on land that is congruent to a
triangle that crosses the river, you can find the distance
across the river by measuring the distance of the
corresponding congruent segment on land.
2. a. The officer’s height stays the same, he is standing
perpendicular to the ground the whole time, and he
tipped his hat the same angle in both directions. So,
△DEF ≅ △DEG by the ASA Congruence Theorem
(Thm. 5.10). Because corresponding parts of the two
— ≅ EF
—. By the definition
triangles are also congruent, EG
of congruent segments, EG equals EF, which is the width
of the river.
38. Draw a segment. Label a point D
on the segment. Draw an arc with
F
center B, and label the intersection
points A and C. Using the same
radius, draw an arc with center D.
D
E
Label the point of intersection of
the arc and the segment as E. Draw an arc with radius AC
with center E. Label the intersection F. Draw ⃗
DF. So,
∠ FDE ≅ ∠ B.
168
Geometry
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Chapter 5
— — — —
—
—
that PT ≅ UQ to prove that the triangles are congruent by
b. Given ∠ DEG is a right angle.
3. You know that TU ≅ QP and UP ≅ PU . You need to show
∠ DEF is a right angle.
∠ EDG ≅ ∠ EDF
Prove EG = EF
F
G
D
REASONS
1. ∠ EDG ≅ ∠ EDF,
∠ DEG and ∠ DEF
are right angles.
1. Given
2. ∠ DEG ≅ ∠ DEF
2. Right Angles Congruence
Theorem (Thm. 2.3)
4. △FDE ≅ △GDE
5. —
EG ≅ —
EF
6. EG = EF
—
5.7 Exercises (pp. 281–282)
STATEMENTS
3.
—
4. Segments that can be assumed congruent are AC and AB.
E
—
DE ≅ —
DE
the SSS Congruence Theorem (Thm. 5.8). △QSP ≅ △TRU
by the HL Congruence Theorem (Thm. 5.9). So
△USQ ≅ △PRT by the SAS Congruence Theorem
— ≅ UQ
—.
(Thm. 5.5) and PT
Vocabulary and Core Concept Check
1. Corresponding parts of congruent triangles are congruent.
2. Sample answer: Finding the width of a body of water, a
3. Reflexive Property of
Congruence (Thm. 2.1)
4. ASA Congruence Theorem
(Thm. 5.10)
5. Corresponding parts of
congruent triangles are
congruent.
6. Definition of congruent
segments
c. By standing perpendicular to the ground and using the
tip of your hat to gaze at two different points in such a
way that the direction of your gaze makes the same angle
with your body both times, you can create two congruent
triangles, which ensures that you are the same distance
from both points.
3. By creating a triangle that is congruent to a triangle with an
unknown side length or angle measure, you can measure
the created triangle and use it to find the unknown measure
indirectly.
4. You do not actually measure the side length or angle measure
you are trying to find. You measure the side length or angle
measure of a triangle that is congruent to the one you are
trying to find.
5.7 Monitoring Progress (pp. 278–280)
1. All three pairs of sides are congruent. So, by the SSS
Congruence Theorem (Thm. 5.8), △ABD ≅ △CBD. Because
corresponding parts of congruent triangles are congruent,
∠ A ≅ ∠ C.
2. no; As long as the rest of the steps are followed correctly,
△LKM will still be congruent to △PNM. So, the
corresponding parts will still be congruent, and you will still
—.
be able to find NP by measuring LK
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distance across a valley or canyon, or any inaccessible
distance can require indirect measurement.
Monitoring Progress and Modeling with Mathematics
3. All three pairs of sides are congruent. So, by the SSS
Congruence Theorem (Thm. 5.8), △ABC ≅ △DBC. Because
corresponding parts of congruent triangles are congruent,
∠ A ≅ ∠ D.
4. Two pairs of sides and the pair of included angles are
congruent. So, by the SAS Congruence Theorem (Thm. 5.5),
△QPR ≅ △TPS. Because corresponding parts of congruent
triangles are congruent, ∠ Q ≅ ∠ T.
5. The hypotenuses and one pair of legs of two right triangles
are congruent. So, by the HL Congruence Theorem
(Thm. 5.9), △JMK ≅ △LMK. Because corresponding parts
— ≅ LM
—.
of congruent triangles are congruent, JM
6. ∠ BAD ≅ ∠ ABC by the Alternate Interior Angles Theorem
— ≅ AD
— by the
(Thm. 3.3). From the diagram, ∠ B ≅ ∠ C. AD
Reflexive Property of Congruence (Thm. 2.1). So, by the
AAS Congruence Theorem (Thm. 5.11), △ACD ≅ △DBA.
Because corresponding parts of congruent triangles are
— ≅ DB
—.
congruent, AC
7. From the diagram, ∠ JHN ≅ ∠ KGL, ∠ N ≅ ∠ L, and
— ≅ KL
—. So, by the AAS Congruence Theorem (Thm. 5.11),
JN
△JNH ≅ △KLG. Because corresponding parts of congruent
— ≅ HJ
—.
triangles are congruent, GK
8. From the diagram, ∠ Q ≅ ∠ W ≅ ∠ RVT ≅ ∠ T and
— ≅ RT
—. So, by the AAS Congruence Theorem
VW
(Thm. 5.11), △QVW ≅ △VRT. Because corresponding
— ≅ VT
—.
parts of congruent triangles are congruent, QW
9. Use the AAS Congruence Theorem (Thm 5.11) to prove that
△FHG ≅ △GKF. Then, state that ∠ FGK ≅ ∠ GFH. Use
the Congruent Complements Theorem (Thm. 2.5) to prove
that ∠ 1 ≅ ∠ 2.
10. Use the AAS Congruence Theorem (Thm 5.11) to prove
— because
that △ABE ≅ △DCE. Then, state that —
BE ≅ CE
corresponding parts of congruent triangles are congruent.
Use the Base Angles Theorem (Thm. 5.6) to prove that
∠ 1 ≅ ∠ 2.
Geometry
Worked-Out Solutions
169
Chapter 5
11. Use the ASA Congruence Theorem (Thm 5.10) to prove
— because
that △STR ≅ △QTP. Then, state that —
PT ≅ RT
corresponding parts of congruent triangles are congruent.
Use the SAS Congruence Theorem (Thm. 5.5) to prove that
△STP ≅ △QTR. So, ∠ 1 ≅ ∠ 2.
— —— —
14. Given PA ≅ PB , QA ≅ QB
Prove ∠ QPA and ∠ QPB
are right angles.
P
A
B
Q
12. Use the ASA Congruence Theorem (Thm 5.10) to prove
— ≅ CE
— and
that △ABE ≅ △CBE. Then, state that AE
∠ BCE ≅ ∠ BAE because corresponding parts of congruent
triangles are congruent. Use the Congruent Complements
Theorem (Thm. 2.5) to state that ∠ ECD ≅ ∠ EAF. So, you
can use the SAS Congruence Theorem (Thm. 5.5) to prove
that △ECD ≅ △EAF, so ∠ 1 ≅ ∠ 2.
— —
— —
13. Given AP ≅ BP and AQ ≅ BQ
Prove ∠ AMP and ∠ BMP
are right angles.
STATEMENTS
REASONS
— ≅ PB
—, QA
— ≅ QB
—
1. PA
1. Given
PQ ≅ —
PQ
2. —
3. △APQ ≅ △BPQ
3. SSS Congruence Theorem
(Thm. 5.8)
4. ∠ QPA ≅ ∠ QPB
4. Corresponding parts of
congruent triangles are
congruent.
5. ∠ QPA and ∠ QPB
form a linear pair.
5. Definition of a linear pair
P
A
M
B
Q
STATEMENTS
— ≅ BP
—, AQ
— ≅ BQ
—
1. AP
2.
—
PQ ≅ —
PQ
1. Given
2. Reflexive Property of
Congruence (Thm. 2.1)
3. △APQ ≅ △BPQ
3. SSS Congruence Theorem
(Thm. 5.8)
4. ∠ APQ ≅ ∠ BPQ
4. Corresponding parts of
congruent triangles are
congruent.
5. —
PM ≅ —
PM
6. SAS Congruence
Theorem (Thm. 5.5)
7. ∠ AMP ≅ ∠ BMP
7. Corresponding parts of
congruent triangles are
congruent.
8. ∠ AMP and ∠ BMP
form a linear pair.
8. Definition of a linear pair
10. ∠ AMP and ∠ BMP
are right angles.
6. —
PQ ⊥ —
AB
7. ∠ QPA and ∠ QPB
are right angles.
6. Linear Pair Perpendicular
Theorem (Thm. 3.10)
7. Definition of perpendicular
lines
5. Reflexive Property of
Congruence (Thm. 2.1)
6. △APM ≅ △BPM
9. —
MP ⊥ —
AB
170
REASONS
2. Reflexive Property of
Congruence (Thm. 2.1)
9. Linear Pair Perpendicular
Theorem (Thm. 3.10)
10. Definition of
perpendicular lines
Geometry
Worked-Out Solutions
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Chapter 5
— — — —
— ≅ LM
— ≅ KM
— ≅ NM
—
JM
—
—
Prove FL ≅ HN
15. Given FG ≅ GJ ≅ HG ≅ GK ,
F
G
J
H
K
M
L
STATEMENTS
— ≅ GJ
— ≅ HG
—≅
1. FG
—
—
—
GK , JM ≅ LM ≅
REASONS
1. Given
—≅ NM
—
KM
16. Given ∠ PRU ≅ ∠ QVS,
∠ PUR ≅ ∠ QSV
≅ ∠ RUX ≅ ∠ VSY,
— ≅ VU
—
RS
Q
P
T
S
R
U
W
Prove △PUX ≅ △QSY
X
N
STATEMENTS
V
Y
REASONS
1. ∠ PRU ≅ ∠ QVS,
∠ PUR ≅ ∠ QSV ≅
∠ RUX ≅ ∠ VSY,
— ≅ VU
—
RS
1. Given
2. ∠ FGJ ≅ ∠ HGK,
∠ JML ≅ ∠ KMN
2. Vertical Angles
Congruence Theorem
(Thm. 2.6)
2. RS = VU
2. Definition of
congruent segments
3. △FGJ ≅ △HGK,
△JML ≅ △KMN
3. SAS Congruence
Theorem (Thm. 5.5)
3. RU = RS + SU,
VS = VU + SU
3. Segment Addition
Postulate (Post. 1.2)
4. ∠ F ≅ ∠ H,
∠L ≅ ∠N
4. Corresponding parts of
congruent triangles are
congruent.
4. VS = RS + SU
4. Substitution Property
of Equality
5. FG = GJ = HG
= GK
5. RU = VS
5. Definition of congruent
segments
5. Transitive Property of
Equality
6. HJ = HG + GJ,
FK = FG + GK
6. Segment Addition
Postulate (Post. 1.2)
7. FK = HG + GJ
7. Substitution Property of
Equality
8. FK = HJ
8. Transitive Property of
Equality
9. —
FK ≅ —
HJ
10. △HJN ≅ △FKL
11. —
FL ≅ —
HN
9. Definition of congruent
segments
10. AAS Congruence
Theorem (Thm. 5.11)
11. Corresponding parts of
congruent triangles are
congruent.
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6. —
RU ≅ —
VS
6. Definition of
congruent segments
7. △PUR ≅ △QSV
8. ∠P ≅ ∠ Q, —
PU ≅ —
QS
9. m∠ PUR = m∠ QSV =
m∠ RUX = m∠ VSY
7. ASA Congruence
Theorem (Thm. 5.10)
8. Corresponding parts
of congruent triangles
are congruent.
9. Definition of
congruent angles
10. m∠ PUX = m∠ PUR +
m∠ RUX, m∠ QSY =
m∠ QSV + m∠ VSY
10. Angle Addition
Postulate (Post. 1.4)
11. m∠ QSY = m∠ PUR +
m∠ RUX
11. Substitution Property
of Equality
12. m∠ PUX = m∠ QSY
12. Transitive Property of
Equality
13. ∠ PUX ≅ ∠ QSY
13. Definition of
congruent angles
14. △PUX ≅ △QSY
14. ASA Congruence
Theorem (Thm. 5.10)
Geometry
Worked-Out Solutions
171
Chapter 5
— —
— —
17. Because AC ⊥ BC and ED ⊥ BD , ∠ ACB and ∠ EDB are
—,
congruent right angles. Because B is the midpoint of CD
— ≅ BD
—. The vertical angles ∠ ABC and ∠ EBD are
BC
congruent. So, △ABC ≅ △EBD by the ASA Congruence
Theorem (Thm. 5.10). Then, because corresponding parts
— ≅ ED
—. So, you can
of congruent triangles are congruent, AC
—.
find the distance AC across the canyon by measuring ED
18. a. Because the base of the red triangle is twice the base of
20. From the Internet, the distance form Miami to Bermuda is
about 1035 miles. The distance from Bermuda to San Juan
is about 956 miles. The distance form San Juan to Miami is
about 1034 miles. The perimeter of the triangle created by
these locations is about 1034 + 956 + 1034 = 3025 miles.
To find the area of this triangle, which is close to an isosceles
triangle, let the distance between Miami and Bermuda be the
same as the distance between Miami and San Juan. Let h be
the height of the triangle.
the purple triangle, the red triangle has an area twice the
area of the purple triangle.
B
b. Because the base of the orange triangle is twice the
base of the purple triangle and the height of the orange
triangle is twice the height of the purple triangle, the
area of the orange triangle is four times the area of the
purple triangle.
— —
19. Given AD BC
1035 miles
478 miles
h
M
J
478 miles
1035 miles
—.
E is the midpoint of AC
S
Prove △AEB ≅ △CED
A
B
h2 = 10352 − 4782
h2 = 1,071,225 − 228,484
h2 = 842,741
E
—
h = √842,741 ≈ 918
The approximate height of the triangle is 918 miles.
D
C
STATEMENTS
REASONS
— BC
—, E is the
1. AD
—.
midpoint of AC
1. Given
2. —
AE ≅ —
CE
2. Definition of midpoint
3. ∠ AED ≅ ∠ BEC,
∠ AEB ≅ ∠ CED
3. Vertical Angles Congruence
Theorem (Thm. 2.6)
4. ∠ DAE ≅ ∠ BCE
4. Alternate Interior Angles
Theorem (Thm. 3.2)
5. △DAE ≅ △BCE
5. ASA Congruence Theorem
(Thm. 5.10)
6. —
DE ≅ —
BE
7. △AEB ≅ △CED
6. Corresponding parts of
congruent triangles are
congruent.
7. SAS Congruence Theorem
(Thm. 5.5)
⋅ ⋅
Area ≈ —12 956 918 = 438,804 mi2
The area of the triangle is about 439,000 square miles.
Sample answer: Three cities with approximate distances
as the three cities in the Bermuda Triangle are Erie, PA, to
Orlando, FL (970 miles); Erie, PA, to Oklahoma City, OK
(1056 miles); and Oklahoma City, OK, to Orlando, FL
(1064 miles).
Erie, PA
1056 miles
Oklahoma City, OK
970 miles
1064 miles
Orlando, FL
21. yes; You can show that WXYZ is a rectangle. This means
that the opposite sides are congruent. Because △WZY and
△YXW share a hypotenuse, the two triangles have congruent
hypotenuses and corresponding legs, which allows you to
use the HL Congruence Theorem (Thm 5.9) to prove that the
triangles are congruent.
22. a. If two triangles have the same perimeter, then they are
congruent is false. The converse is true: If two triangles
are congruent, then their perimeters are the same.
b. If two triangles are congruent, then their areas are the
same is true.
172
Geometry
Worked-Out Solutions
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Chapter 5
—
—
—
— (given). So, △ABC ≅ △GHJ by the SAS
and AB ≅ GH
23. AC ≅ GJ (given), ∠ A ≅ ∠ G because they are right angles,
d. Reflect the triangle in the x-axis, so that △ABC is still an
isosceles right triangle. So, C(3, −3).
Congruence Theorem (Thm 5.5).
— ≅ QN
— (given), and ∠ A ≅ ∠ N
∠ C ≅ ∠ Q (given), CA
because they are right angles. So, △ABC ≅ △NPQ by the
ASA Congruence Theorem (Thm. 5.10).
2
1
−2−1
Maintaining Mathematical Proficiency
——
4
3
—
——
—
—
———
—
CD = √[ (−1) − 4 ]2 + [ −2 − (−2) ]2 = √ (−1 − 4)2
—
—
= √(−5)2 = √ 25 = 5
———
AD = √[ (−1) − (−1) ]2 + (−2 − 1)2
——
—
—
= √(−1 + 1)2 + (−3)2 = √ (−3)2 = √9 = 3
Perimeter = AB + BC + CD + AD
= 5 + 3 + 5 + 3 = 16 units
———
25. JK = √ [ (−2) − (−5) ]2 + (1 − 3)2
—
=√
(3)2
—
—
+ 4 = √ 9 + 4 = √ 13 ≈ 3.6
——
——
KL = √[ 3 − (−2) ]2 + (4 − 1)2 = √(3 + 2)2 + 32
—
—
= √25 + 9 = √34 ≈ 5.8
——
JL = √[ 3 − (−5) ]2 + (4 − 3)2
——
—
—
= √(3 + 5)2 + 12 = √ 82 + 1 = √ 65 = 8.1
Perimeter = JK + KL + JL
5.8 Explorations (p. 283)
e. If C lies on the line x = 3, then the coordinates are
C(3, y). Because △ABC is an isosceles triangle,
y
y
—=—
—=—
, and mBC
.
mAC
3
−3
△ABC is a right triangle, so it must have a right angle.
— and BC
— are the congruent legs of △ABC, ∠ A
Because AC
and ∠ B are the congruent base angles by the Base Angles
Theorem (Thm. 5.6). The vertex angle, ∠ C, must be the
— ⊥ BC
— by definition of
right angle, which means that AC
perpendicular lines. By the Slopes of Perpendicular Lines
Theorem (Thm. 3.14),
y y
— — = −1 and y = ±3.
3 −3
So, the coordinates of C must be (3, 3) or (3, −3).
3. You can position the figure in a coordinate plane and then
use deductive reasoning to show that what you are trying to
prove must be true based on the coordinates of the figure.
——
——
—
——
— 2
AC = √(3 − 0)2 + (3√3 − 0)2 = √ 32 + ( 3√3 )
⋅
—
⋅
—
—
= √9 + 9 3 = √9 + 27 = √36 = 6
——
— 2
——
— 2
BC = √(6 − 3)2 + ( 0 − 3√3 ) = √32 + ( 3√3 )
—
= √9 + 9 3 = √9 + 27 = √36 = 6
—
△ABC with vertices A(0, 0), B(6, 0), C ( 3, 3√ 3 ) has side
lengths AB = 6, AC = 6, and BC = 6. By the definition of
congruent sides, △ABC is an equilateral triangle.
5.8 Monitoring Progress (pp. 284–286)
1. Another way of placing the rectangle in Example 1(a) that
is convenient for finding side lengths would be to place the
width (k) on the x-axis.
2. a. Check students’ work.
b. Check students’ work.
y
——
—
——
——
BC = √ (6 − 3)2 + (0 − 3)2 = √ 32 + (−3)2
—
—
—
= √ 9 + 9 = √18 = 3√ 2
(k, h)
(0, h)
h
AC = √ (0 − 3)2 + (0 − 3)2 = √ (−3)2 + (−3)2
—
—
—
= √ 9 + 9 = √18 = 3√ 2
3−0 3
—=—
=—=1
m AC
3−0 3
3
3−0
—=—
= — = −1
m BC
3 − 6 −3
—
— = 1 and mBC
— = −1,
Because AC = BC = 3√ 2 and mAC
AC ⊥ BC and △ABC is a right isosceles triangle.
Copyright © Big Ideas Learning, LLC
All rights reserved.
—
—
4. AB = √ (6 − 0)2 + (0 − 0)2 = √ 62 = √ 36 = 6
—
Check students’ work.
Check students’ work.
Check students’ work.
—
Using the
Distance Formula, AC = √ 9 + y2 and
— —
—
2
AB = √9 + y . AC ≅ BC , so △ABC is an isosceles
triangle.
c. AB = ∣ 6 − 0 ∣ = ∣ 6 ∣ = 6
C
—
≈ 3.6 + 5.8 + 8.1 = 17.5 units
1. a.
b.
c.
d.
90°
⋅
——
= √(−2 + 5)2 + (−2)2
B
1 2 3 4 5 6 x
−3
−4
= √52 = √ 25 = 5
BC = √(4 − 4)2 + (−2 − 1)2 = √(−3)2 = √9 = 3
A
−2
—
24. AB = √ [ 4 − (−1) ]2 + (1 − 1)2 = √ (4 + 1)2
—
y
x
(0, 0)
k
(k, 0)
Geometry
Worked-Out Solutions
173
Chapter 5
2. A square with vertices (0, 0), (m, 0), and (0, m) will have a
fourth vertex at (m, m).
5. Given Coordinates of △NPO are N(h, h), P(0, 2h), and
O(0, 0). Coordinates of △NMO are N(h, h), M(2h, 0),
and O(0, 0).
y
(0, m)
Prove △NPO ≅ △NMO
(m, m)
P(0, 2h)
y
N(h, h)
O(0, 0)
(0, 0)
(m, 0)
x
3. Use the Distance Formula to find GO and GH in order to
show they are congruent. State that ∠ OGJ ≅ ∠ HGJ by the
— ≅ GJ
— by the Reflexive
definition of angle bisector and GJ
Property of Congruence (Thm. 2.1). Then use the SAS
Congruence Theorem (Thm. 5.5) to show that
△GJO ≅ △GJH.
—
——
——
NP = √ (0 − h)2 + (2h − h)2 = √ (−h)2 + h2
—
—
= √ 2h2 = h√2
PO = √ (0 − 0)2 + (0 − 2h)2 = √ 02 + (−2h)2
—
= √ 4h2 = 2h
——
—
——
——
NM = √ (2h − h)2 + (0 − h)2 = √ h2 + (−h)2
—
—
= √ 2h2 = h√2
MO = √ (0 − 2h)2 + (0 − 0)2 = √ (−2h)2 + 02
—
= √ 4h2 = 2h
y
4.
——
M(2h, 0) x
— ≅ MN
— and PO
— ≅ MO
—, and by the Reflexive Property
So, PN
— ≅ ON
—. So, you can apply
of Congruence (Thm. 2.1), ON
the SSS Congruence Theorem (Thm. 5.8) to conclude that
△NPO ≅ △NMO.
H(m, n)
x
O(0, 0)
5.8 Exercises (pp. 287–288)
J(m, 0)
Vocabulary and Core Concept Check
Side lengths of △OHJ:
——
1. In a coordinate proof, you have to assign coordinates to
—
OH = √(m − 0)2 + (n − 0)2 = √ m2 + n2
——
—
OJ = √(m − 0)2 + (0 − 0)2 = √ m2 = m
——
—
—
HJ = √(m − m)2 + (0 − n)2 = √ (−n)2 = √n2 = n
Midpoints of each side:
m+0 n+0
m n
—= —
, — = —, —
Midpoint of OH
2
2
2 2
(
) ( )
— = m + 0, 0 + 0 = m, 0
Midpoint of OJ
( 2 2 ) (2 )
— = m + m, n + 0 = 2m, n = m, n
Midpoint of HJ
( 2 2 ) ( 2 2) ( 2)
— —
—
— —
— —
Slopes of each side:
n
n−0
—=—
=—
Slope of OH
m−0 m
0
0−0
—=—
=—=0
Slope of OJ
m−0 m
−n
0−n
Slope of —
HJ = — = — = undefined
m−m
0
— ⊥ HJ
— at angle J. Therefore, △OHJ is a right triangle.
OJ
—
vertices and write expressions for side length and the slope
of segments in order to show how sides are related; As
with other types of proofs, you still have to use deductive
reasoning and justify every conclusion with theorems,
proofs, and properties of mathematics.
2. When the triangle is positioned as shown, you are using
zeros in your expressions, so the side lengths are often the
same as one of the coordinates.
Monitoring Progress and Modeling with Mathematics
3. Sample answer: Place the legs on the x- and y-axes.
y
3
2
C(0, 2)
1
B(3, 0)
A(0, 0)
1
2
3
x
It is easy to find the lengths of horizontal and vertical
segments and distances from the origin.
174
Geometry
Worked-Out Solutions
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Chapter 5
4. Sample answer: Place one side on the x-axis and one side on
9.
the y-axis.
y
3
y
C(0, 7)
7
6
5
4
3
Z(3, 3)
W(0, 3)
8
2
2
1
1
Y(3, 0)
1
X(0, 0)
2
3
x
——
—
CD = √(0 − 9)2 + (7 − 0)2 = √(−9)2 + 72
—
—
= √81 + 49 = √ 130 ≈ 11.4
The length of the hypotenuse is about 11.4 units.
It is easy to find the lengths of horizontal and vertical
segments and distances from the origin.
5. Sample answer: Place the legs on the x- and y-axes.
D(9, 0)
1 2 3 4 5 6 7 8 9 x
O(0, 0)
10.
y
y
A(0, 50)
50
R(0, p)
40
30
20
T(p, 0) x
S(0, 0)
10
B(−30, 0)
−30
It is easy to find the lengths of horizontal and vertical
segments and distances from the origin.
C(30, 0)
−20
−10
10
20
30
x
Use the Pythagorean Theorem = + b2, where the
hypotenuse of the right triangle is one of the legs of the
isosceles triangle.
c2
6. Sample answer: Place the side with length 2m on the x-axis.
y
a2
AC2 = 502 + 302
AC2 = 2500 + 900
AC2 = 3400
—
AC = √ 3400 ≈ 58.3
L(n, p)
The length of one of the legs of the isosceles triangle is about
58.3 units.
J(0, 0)
K(2m, 0) x
It is easy to find the lengths of horizontal segments and
distances from the origin.
———
—
7. Find the lengths of OP , PM , MN , and NO to show that
— ≅ PM
— and MN
— ≅ NO
—.
OP
8. Find the coordinates of G using the Midpoint Formula. Use
these coordinates and the Distance Formula to show that
— ≅ JG
—. Show that HG
— ≅ FG
— by the definition of
OG
midpoint, and ∠ HGL ≅ ∠ FGO by the Vertical Angles
Congruence Theorem (Thm. 2.6). Then use the SAS
Congruence Theorem (Thm. 5.5) to conclude that
△GHJ ≅ △GFO.
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11.
6
y
5 L(0, 4)
4
3
M(5, 4)
2
1
O(0, 0)
N(5, 0)
1 2 3 4 5 6 x
——
—
NL = √(5 − 0)2 + (0 − 4)2 = √52 + (−4)2
—
—
= √25 + 16 = √ 41 ≈ 6.4
The length of the diagonal is about 6.4 units.
Geometry
Worked-Out Solutions
175
Chapter 5
14.
y
12.
Y(n, n)
Z(0, n)
X(n, 0)
x
F(m, 0) x
——
—
XZ = √ (n − 0)2 + (0 − n)2 = √n2 + (−n)2
—
—
Side lengths of △DEF:
—
= √n2 + n2 = √2n2 = n√ 2
——
—
——
—
——
—
—
DE = √(m − 0)2 + (n − n)2 = √m2 + 0 = √ m2 = m
—
The length of the diagonal is n√ 2 units.
—
EF = √(m − m)2 + (0 − n)2 = √0 + (−n)2 = √n2 = n
DF = √(m − 0)2 + (0 − n)2 = √m2 + n2
y
B(h, h)
Slopes of each side of △DEF:
0
n−n
—=—
=—=0
Slope of DE
m−0 m
−n
0−n
—=—
= — = undefined
Slope of EF
m−m
0
n
0−n
—=—
= −—
Slope of DF
m
m−0
C(2h, 0) x
A(0, 0)
Side lengths of △ABC:
——
—
—
—
AB = √(h − 0)2 + (h − 0)2 = √ h2 + h2 = √ 2h2 = h√2
——
—
——
—
—
—
BC = √(2h − h)2 + (0 − h)2 = √ h2 + h2 = √ 2h2 = h√2
AC = √ (2h − 0)2 + (0 − 0)2 = √ 4h2 = 2h
Slopes of each side of △ABC:
h−0 h
—=—
=—=1
Slope of AB
h−0 h
h
2h − h
—=—
= — = −1
Slope of BC
0−h
−h
0
0−0
—= —
=—=0
Slope of AC
2h − 0 2h
Midpoints of each side of △ABC:
0+h 0+h
h h
—= —
, — = —, —
Midpoint of AB
2
2
2 2
h + 2h h + 0
3h h
—
Midpoint of BC = —, — = —, —
2
2
2 2
2h + 0 0 + 0
2h
—
Midpoint of AC = —, — = —, 0 = (h, 0)
2
2
2
(
(
(
) ( )
) ( )
) ( )
— ⊥ BC
— by the Slopes of
— mBC
— = −1, AB
Because mAB
Perpendicular Lines Theorem (Thm. 3.14). So, ∠ ABC is
— ≅ BC
— because AB = BC. So, △ABC is a
a right angle. AB
right isosceles triangle.
⋅
176
E(m, n)
D(0, n)
O(0, 0)
13.
y
Geometry
Worked-Out Solutions
Midpoints of each side of △DEF:
0+m n+n
m 2n
m
—= —
, — = —, — = —, n
Midpoint of DE
2
2
2 2
2
m+m n+0
2m n
n
—
Midpoint of EF = —, — = —, — = m, —
2
2
2 2
2
0+m n+0
m n
—= —
, — = —, —
Midpoint of DF
2
2
2 2
(
(
(
) ( ) ( )
) ( ) ( )
) ( )
— is horizontal. Because
— = 0, DE
Because mDE
—
— ⊥ BC
— by
— = undefined, EF is vertical. So, DE
mEF
Postulate 3.5, and ∠ DEF is a right angle by the definition
of perpendicular lines. Also, none of the sides have the same
length. So, △DEF is a right scalene triangle.
15. The coordinates of the unlabeled vertex is N(h, k).
——
—
ON = √(h − 0)2 + (k − 0)2 = √h2 + k2
——
—
—
MN = √(2h − h)2 + (0 − k)2 = √h2 + (−k)2 = √h2 + k2
16. The coordinates of the vertices are O(0, 0), U(k , 0), R(k, k),
S(k, 2k), and T(2k, 2k).
——
——
OT = √ (2k − 0)2 + (2k − 0)2 = √(2k)2 + (2k)2
—
—
—
= √ 4k2 + 4k2 = √ 8k2 = 2k√ 2
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Chapter 5
——
—
17. Find the lengths of DE , EC , and DC of △DEC.
——
—
DE = √(2h − h)2 + (2k − 2k)2 = √ (h)2 = h
——
—
EC = √(2h − h)2 + (2k − k)2 = √ h2 + k2
——
19. The triangle formed by your position (Y), your cousins
position (C), and the campsite (O) has the coordinates
Y(500, 1200), C(1000, 0), and O(0, 0).
y
—
Y(500, 1200)
DC = √(h − h)2 + (2k − k)2 = √ k2 = k
—, OC
—, and BC
— of △BOC.
Find the lengths of BO
——
—
——
—
——
—
1000
BO = √(h − 0)2 + (0 − 0)2 = √ (h)2 = h
OC = √(h − 0)2 + (k − 0)2 = √ h2 + k2
500
BC = √(h − h)2 + (k − 0)2 = √k2 = k
Because DE = BO, EC = OC, and DC = BC, by the
— ≅ BO
—, EC
— ≅ OC
—,
definition of congruent segments DE
—
—
DC ≅ BC . By the SSS Congruence Theorem (Thm. 5.8),
△DEC ≅ △BOC.
—
18. Because H is the midpoint of DA , the coordinates of
500
1000
x
———
OY = √(500 − 0)2 + (1200 − 0)2
——
point H are
= √ 5002 + 12002
——
0 − 2h 2k + 0
H —, — = H(−h, k).
2
2
(
)
= √ 250,0000 + 1,440,000
—, the coordinates of
Because G is the midpoint of EA
point G are
0 + 2h 2k + 0
G —, — = (h, k).
2
2
(
)
———
—
+ (0 −
The distance between your position and the campsite is
1300 meters.
———
——
(−h)]2
—
= √ 1,690,000 = 1300
YC = √(1000 − 500)2 + (0 − 1200)2
——
DG = √(−2h − h)2 + (0 − k)2 = √ (−3h)2 + (−k)2
—
= √9h2 + k2
EH = √[2h −
C(1000, 0)
O(0, 0)
k)2
=√
——
(3h)2
+
(−k)2
= √9h2 + k2
Because DG = EH, by the definition of congruent segments,
— ≅ EH
—.
DG
——
= √(5002 + (−1200)2
——
= √250,000 + 1,440,000
—
= √1,690,000 = 1300
The distance between your position and your cousin’s
position is 1300 meters.
——
OH = √(1000 − 0)2 + (0 − 0)2
—
= √(1000)2
= 1000
The distance between the campsite and your cousin’s
position is 1000 meters.
— ≅ YC
—, the triangle formed by your position,
Because OY
your cousin’s position, and the campsite is an isosceles
triangle.
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Geometry
Worked-Out Solutions
177
Chapter 5
20.
23. A
y
P(0, 2)
S(−2, 1)
−4
6
−2
(−h, k)
4 x
2
y
(2h, k)
Q(3, −4)
2
−4
(−h, 0)
R(1, −5)
2
(2h, 0) x
−2
Side lengths:
——
—
PQ = √(3 − 0)2 + (−4 − 2)2 = √ 32 + (−6)2
—
—
—
= √9 + 36 = √ 45 = 3√5
———
——
QR = √(1 − 3)2 + [ −5 − (−4) ]2 = √ (−2)2 + (−5 + 4)2
—
—
—
= √ 4 + (−1)2 = √4 + 1 = √5
———
——
RS = √(−2 − 1)2 + [ 1 − (−5) ]2 = √ (−3)2 + (1 + 5)2
—
—
—
= √9 + 36 = √45 = 3√5
——
24. Sample answer: It would be easy to prove the Base Angles
Theorem (Thm. 5.6) with a coordinate proof. First, position
the given isosceles triangle, △ABC, on the coordinate plane
so that the base is on the x-axis, and one vertex is at
the origin.
x=k
y
——
PS = √(−2 − 0)2 + (1 − 2)2 = √ (−2)2 + (−1)2
—
—
= √4 + 1 = √5
A(k, m)
Slopes of the sides:
−4 − 2 −6
—=—
= — = −2
Slope of PQ
3−0
3
−5 − (−4) −5 + 4 −1 1
—=—
=—=—=—
Slope of QR
1−3
−2
−2 2
6
1 − (−5) 1 + 5
—=—
= — = — = −2
Slope of RS
−2 − 1
−3
−3
−1 1
1−2
—=—
=—=—
Slope of PS
−2 − 0 −2 2
— ≅ SR
— and SP
— ≅ RQ
—, which shows that opposite
So, PQ
— m SP
— = −1,
sides are congruent. Also, mPQ
—
—
—
—
— m RQ
— = −1.
m PQ m RQ = − 1, m SR m SP = −1, m SR
⋅
—
⋅
——
⋅
⋅
—, PQ
— ⊥ RQ , SR ⊥ SP
—, and SR
— ⊥ RQ
— by the
So, PQ ⊥ SP
Slopes of Perpendicular Lines Theorem (Thm. 3.14). So, by
definition of perpendicular lines, ∠ PSR, ∠ SRQ, ∠ RQP, and
∠ QPS are right angles. So, the quadrilateral is a rectangle.
The second friend is correct.
B(0, 0)
D(k, 0)
C(2k, 0)
x
—
This is an
isosceles triangle because BA = √ k2 + m2 and
—
2
CA = √k + m2 . Draw the line x = k that intersects △ABC
in point A(m, k) and the x-axis in the point (k, 0). Call this
— and DC
— are congruent because BD = k and
point D, BD
—
—
DC = k. AD ≅ AD by the Reflexive Property of Congruence
— is vertical and BC
— is horizontal,
(Thm. 2.1). Because AD
AD ⊥ BC is the Slopes of Perpendicular Lines Theorem
(Thm. 3.14). So, ∠ BDA and ∠ CDA are congruent right
angles. By the SAS Congruence Theorem (Thm. 5.5),
△ABD ≅ △ACD. Because corresponding parts of congruent
triangles are congruent, ∠ B ≅ ∠ C.
25. Sample answer: Reflect the triangle in the y-axis and
21. The endpoints of a segment with the origin as the midpoint
are (x, y) and (−x, −y) because
x + (−x) y + (−y)
0 0
M —, — = M —, — = M(0, 0).
2
2
2 2
(
) ( )
22. B
(5d, −5d) → (−5d, −5d) → (0, 0)
(0, −5d) → (0, −5d) → (5d, 0)
(5d, 0) → (−5d, 0) → (0, 5d)
—
y
—
26. Diagonal WU is horizontal, and diagonal TV is vertical. So,
(0, v)
by the Slopes of Perpendicular Lines Theorem (Thm. 3.14),
— ⊥ TV
—; Change the coordinates to T(0, m), U(m, 0),
WU
V(0, −m), and W(−m, 0). These coordinates can be used for
any square, and the diagonals are still horizontal and vertical.
So, the diagonals are perpendicular for any square.
(w, 0) x
(−w, 0)
translate 5d units right and 5d units up.
(0, −v)
w v
−w + 0 0 + (−v)
Midpoint = —, — = −—, −—
2 2
2
2
(
178
) (
Geometry
Worked-Out Solutions
)
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Chapter 5
27. a.
2. The theorems on page 275 correspond to other triangle
y
congruence theorems given. HA corresponds to AAS, LL
corresponds to SAS, and AL corresponds to ASA or AAS.
A(0, 2m)
3. Two sources that could be used to help solve Exercise 20 on
M(n, m)
B(0, 0)
C(2n, 0)
page 282, are a map website and a website that calculates
distances.
x
—, the coordinates of M are
Because M is the midpoint of AC
—
M(n, m).—
Using the Distance Formula,
AM = √n2 + m2 ,
—
2
2
2
2
BM = √n + m , and CM = √ n + m . So, the midpoint
of the hypotenuse of a right triangle is the same distance
from each vertex of the triangle.
b.
y
Chapter 5 Review (pp. 290–294)
1. The triangle is an acute isosceles triangle because it has two
equal sides and it appears that all angles are acute.
2. Because 46° + 86° = 132°, the measure of the exterior angle
is 132°.
3. (9x + 9)° = 45° + 5x°
4x + 9 = 45
4x = 36
R(0, m)
x=9
The exterior angle:
S(−m, 0) O(0, 0)
⋅
(9x + 9)° = 9 9 + 9
T(m, 0) x
= 81 + 9
When any two congruent right isosceles triangles are
positioned with the vertex opposite the hypotenuse on the
origin and their legs on the axes as shown in the diagram,
a triangle is formed and the hypotenuses of the original
triangles make
up two sides
of the new triangle.
—
—
SR = m√2 and TR = m√ 2 so these two sides are the
same length. So, by definition, △SRT is isosceles.
Maintaining Mathematical Proficiency
28. m∠ XYW = m∠ WYZ
= 90
The measure of the exterior angle is 90°.
4. 8x° + 7x° + 90° = 180°
15x + 90 = 180
15x = 90
x=6
⋅
⋅
8x = 8 6 = 48
7x = 7 6 = 42
(3x − 7)° = (2x + 1)°
The measure of each acute angle is 42° and 48°.
x−7=1
5. (7x° + 6)° + (6x − 7)° + 90° = 180°
x=8
13x + 89 = 180
29. m∠ XYZ = (3x − 7)° + (2x + 1)°
13x = 91
m∠ XYZ = 5x − 6
⋅
⋅
x=7
From Exercise 28, you know that x = 8.
7x + 6 = 7 7 + 6 = 49 + 6 = 55
m∠ XYZ = 5 8 − 6 = 34°
6x − 7 = 6 7 − 7 = 35
⋅
5.5 –5.8 What Did You Learn? (p. 289)
—
1. Given square ABCD with diagonal BD , as shown, prove
△BAD ≅ △DCB; In this problem, the square could
—
represent the baseball “diamond,” and then diagonal BD
would represent the distance from home plate to second
base. So, you could use this problem to prove the equivalent
of △HFS ≅ △STH. Then you could just redraw square
— this time, so that △CBA is the
ABCD with diagonal AC
equivalent of △FST. It could easily be shown that the third
triangle is congruent to the first two.
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The measure of each acute angle is 35° and 55°.
6. corresponding angles: ∠ G ≅ ∠ L, ∠ H ≅ ∠ M, ∠ J ≅ ∠ N,
∠K ≅ ∠P
— ≅ LM
—, HJ
— ≅ MN
—, JK
— ≅ NP
—,
corresponding sides: GH
—
—
GK ≅ LP
Sample answer: Another congruence statement
is KJHG ≅ PNML.
Geometry
Worked-Out Solutions
179
Chapter 5
17. yes;
7. m∠ S = m∠ T = 74°
90° + 74° + m∠ V = 180°
164° + m∠ V = 180°
m∠ V = 16°
8. no; There is enough information to prove two pairs of
congruent sides and one pair of congruent angles, but the
angle is not the included angle.
9. yes;
STATEMENTS
REASONS
1. ∠ E ≅ ∠ H,
∠ F ≅ ∠ J,
—
FG ≅ —
JK
1. Given
2. △EFG ≅ △HJK
2. AAS Congruence
Theorem (Thm. 5.11)
18. no; There is only enough information to conclude that one
STATEMENTS
1. —
WX ≅ —
YZ , WZ YX
2. —
XZ ≅ —
XZ
REASONS
1. Given
2. Reflexive Property of
Congruence (Thm. 2.1)
3. ∠ WXZ ≅ ∠ YZX
3. Alternate Interior Angles
Theorem (Thm. 3.2)
4. △WXZ ≅ △YZX
4. SAS Congruence Theorem
(Thm. 5.5)
— —
pair of angles and one pair of sides are congruent.
19. yes;
STATEMENTS
REASONS
1. ∠ PLN ≅ ∠ MLN,
∠ PNL ≅ ∠ MNL
1. Given
2. —
LN ≅ —
LN
2. Reflexive Property of
Congruence (Thm. 2.1)
3. △LPN ≅ △LMN
3. ASA Congruence
Theorem (Thm. 5.10)
10. If QP ≅ QR , then ∠ QRP ≅ ∠ P.
— —
11. If ∠ TRV ≅ ∠ TVR, then TV ≅ TR .
— —
12. If RQ ≅ RS , then ∠ RSQ ≅ ∠ RQS.
— —
13. If ∠ SRV ≅ ∠ SVR, then SV ≅ SR .
14. 8x° = 180° − 60°
pair of angles and one pair of sides are congruent.
21. By the SAS Congruence Theorem (Thm. 5.5),
△HJK ≅ △LMN. Because corresponding parts of
congruent triangles are congruent, ∠ K ≅ ∠ N.
— —
8x = 120
22. First, state that QV ≅ QV . then, use the SSS Congruence
Theorem (Thm. 5.8) to prove that △QSV ≅ △QTV. Because
corresponding parts of congruent triangles are congruent,
∠ QSV ≅ ∠ QTV. ∠ QSV ≅ ∠ 1 and ∠ QTV ≅ ∠ 2 by the
Vertical Angles Congruence Theorem (Thm. 2.6). So, by the
Transitive Property of Congruence (Thm. 2.2), ∠ 1 ≅ ∠ 2.
x = 15
5y + 1 = 26
5y = 25
y=5
So, x = 15 and y = 5.
——
—
23. OP = √ (h − 0)2 + (k − 0)2 = √ h 2 + k2
15. no; There is only enough information to conclude that two
pairs of sides are congruent.
———
—
QP = √(h − h)2 + [ (k + j) − k ]2 = √j 2 = j
——
——
——
—
——
—
QO = √(h − 0)2 + (k + j)2 = √h2 + (k + j)2
QR = √(h − 0)2 + (k − 0)2 = √h 2 + k2
16. yes;
STATEMENTS
1. —
WX ≅ —
YZ , ∠ XWZ
and ∠ ZYX are right
angles.
XZ ≅ —
XZ
2. —
180
20. no; There is only enough information to conclude that one
REASONS
1. Given
OR = √(0 − 0)2 + (0 − j)2 = √j 2 = j
— ≅ QR
— and OR
— ≅ QP
—. Also, by the Reflexive Property
So, OP
— ≅ QO
—. So, you can apply
of Congruence (Thm. 2.1), QO
the SSS Congruence Theorem (Thm. 5.8) to conclude that
△OPQ ≅ △QRO.
2. Reflexive Property of
Congruence (Thm. 2.1)
3. △WXZ and △YZX
are right triangles.
3. Definition of a right
triangle
4. △WXZ ≅ △YZX
4. HL Congruence Theorem
(Thm. 5.9)
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Chapter 5
24. Place the base along the x-axis and the vertex of the legs on
the y-axis.
— —
3. Given QR ≅ RS , ∠ P ≅ ∠ T
y
Q
S
N
B(0, k)
P
C(p, 0) x
A(−p, 0)
25.
R
Prove △SRP ≅ △QRT
y
(0, k)
(2k, k)
STATEMENTS
REASONS
— ≅ RS
—, ∠ P ≅ ∠ T
1. QR
1. Given
2. ∠ R ≅ ∠ R
2. Reflexive Property of
Congruence (Thm. 2.2)
3. △SRP ≅ △QRT
3. AAS Congruence
Theorem (Thm. 5.11)
T
4. (4x − 2)° + (3x + 8)° = 90°
7x + 6 = 90
(0, 0)
(2k, 0)
7x = 84
x
⋅
⋅
The fourth vertex of the rectangle has the vertex (2k, k).
4x − 2 = 4 12 − 2 = 48 − 2 = 46
3x + 8 = 3 12 + 8 = 36 + 8 = 44
Chapter 5 Test (p. 295)
— — — —
1. Given CA ≅ CB ≅ CD ≅ CE
A
Prove △ABC ≅ △EDC
B
STATEMENTS
REASONS
— ≅ CB
— ≅ CD
— ≅ CE
—
1. CA
1. Given
3. △ABC ≅ △EDC
if a triangle is equilateral, then it is also equiangular.
E
6. no; The Third Angles Theorem (Thm. 5.4) can be used to
prove that two triangles are equiangular, but AAA is not
sufficient to prove that the triangles are congruent. You need
to know that at least one pair of corresponding sides are
congruent.
2. Vertical Angles
Congruence Theorem
(Thm. 5.5)
7. First, use the HL Congruence Theorem (Thm. 5.9) to prove
that △ACD ≅ △BED. Because corresponding parts of
— ≅ BD
—. Then, use the
congruent triangles are congruent, AD
Base Angles Theorem (Thm. 5.6) to prove that ∠ 1 ≅ ∠ 2.
3. SAS Congruence
Theorem (Thm. 5.5)
8. Use the SSS Congruence Theorem (Thm. 5.8) to prove
——
—
—
MJ KL
2. Given JK ML ,
J
K
Prove △MJK ≅ △KLM
M
STATEMENTS
REASONS
— ML
—,
1. JK
—
—
MJ KL
1. Given
MK ≅ —
KM
2. —
The measures of the acute angles are 44° and 46°.
5. no; By the Corollary to the Base Angles Theorem (Cor. 5.2),
C
D
2. ∠ ACB ≅ ∠ ECD
x = 12
L
that △SVX ≅ △SZX. Use the Vertical Angles Congruence
Theorem (Thm. 2.6) and the SAS Congruence Theorem
(Thm. 5.5) to prove that △VXW ≅ △ZXY. Because
corresponding parts of congruent triangles are congruent,
∠ SZX ≅ ∠ SVX and ∠ W ≅ ∠ Y. Use the Segment Addition
— ≅ ZW
—. Then, use
Postulate (Post. 1.2) to show that VY
the ASA Congruence Theorem (Thm. 5.10) to prove that
△VYT ≅ △ZWR. Because corresponding parts of congruent
triangles are congruent, ∠ 1 ≅ ∠ 2.
9. yes; HL Congruence Theorem (Thm. 5.9), ASA Congruence
2. Reflexive Property of
Congruence (Thm. 2.1)
3. ∠ JKM ≅ ∠ LMK,
∠ JMK ≅ ∠ LKM
3. Alternate Interior Angles
Theorem (Thm. 3.2)
4. △MJK ≅ △KLM
4. ASA Congruence
Theorem (Thm. 5.10)
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Theorem (Thm. 5.10), AAS Congruence Theorem
(Thm. 5.11), SAS Congruence Theorem (Thm. 5.5)
Geometry
Worked-Out Solutions
181
Chapter 5
—
—
10. Use the Distance Formula to find the lengths of PQ and ST .
——
—
— —
5. a. In order to prove △ABC ≅ △DEF, show that AB ≅ DE .
——
—
—
——
—
—
PQ = √(21 − 3)2 + (30 − 30)2 = √ 182 = 18
AB = √(2 − 2)2 + (8 − 5)2 = √32 = √9 = 3
ST = √(21 − 3)2 + (0 − 0)2 = √ 182 = 18
— and DE
— have the same measure and
The segments AB
are therefore congruent. Also, from the markings in
— ≅ EF
—. So, by the SAS
the diagram, ∠ B ≅ ∠ E and BC
Congruence Theorem (Thm. 5.5), △ABC ≅ △DEF.
——
—
— ≅ ST
—. Also, the horizontal segments PQ
— and ST
— each
So, PQ
have a slope of 0, which implies that they are parallel. So,
— intersects PQ
— and ST
— to form congruent alternate interior
PS
angles, ∠ P and ∠ S. By the Vertical Angles Congruence
Theorem (Thm. 2.6), ∠ PRQ ≅ ∠ SRT. So, by the AAS
Congruence Theorem (Thm. 5.11), △PQR ≅ △STR.
11. a. The triangle shown is an isosceles triangle because it has
two congruent sides.
b. By the Triangle Sum Theorem (Thm. 5.1):
m∠ 1 + m∠ 2 + m∠ 3 = 180°
m∠ 1 + m∠ 2 + 40° = 180°
m∠ 1 + m∠ 2 = 140°
By the Base Angles Theorem (Thm. 5.6) ∠ 1 ≅ ∠ 2:
m∠ 2 + m∠ 2 = 140°
2m∠ 2 = 140°
m∠ 2 = 70°
m∠ 1 = 70°
If m∠ 3 = 40°, then m∠ 1 = 70° and m∠ 2 = 70°.
Chapter 5 Standards Assessment (pp. 296–297)
1. no; the Exterior Angle Theorem (Thm. 5.2) follows from
the Triangle Sum Theorem (Thm. 5.1). Also, the Triangle
Sum Theorem (Thm. 5.1) is used to prove the Exterior Angle
Theorem (Thm. 5.2), so you cannot use the Exterior Angle
Theorem (Thm. 5.2) to prove the Triangle Sum Theorem
(Thm. 5.1).
2. By step 1, a line through point P intersects line m in point
— ≅ QB
— ≅ PC
— ≅ PD
— and AB
— ≅ CD
—
Q. By steps 2 and 3, QA
because congruent segments were drawn with the same
— and CD
—
compass setting. So, in step 4, you see that if AB
were drawn, then △AQB and △CPD would be congruent
by the SSS Congruence Theorem (Thm. 5.8). Because
corresponding parts of congruent triangles are congruent,
⃖⃗ m by the
∠ CBD ≅ ∠ AQB, which means that PD
Corresponding Angles Converse (Thm. 3.5).
3. a. Sample answer: Reflect △JKL in the x-axis and then
translate 4 units right.
— —— —— —
b. yes; corresponding sides: JK ≅ XY , KL ≅ YZ , JL ≅ XZ ;
corresponding angles: ∠ J ≅ ∠ X, ∠ K ≅ ∠ Y, ∠ L ≅ ∠ Z
4. C;
DE = √(8 − 5)2 + (2 − 2)2 = √32 = √9 = 3
b. Rotate △ABC 90° counterclockwise about the origin
followed by a translation 13 units right.
6. yes; The coordinate rule for dilations is to multiply each
coordinate of each point by the scale factor, which is 2 in this
case. So, when you do this to the coordinates of point W, you
get (2 0, 2 0) which is still (0, 0).
⋅ ⋅
7. A, B, D
Figure A has 180° rotational symmetry. Figure B has
72° rotational symmetry. Figure D has 90° and 180°
rotational symmetry.
8. To prove quadrilateral ABCD is a rectangle, opposite
sides must be equal and right angles must be formed at
the vertices.
Side lengths:
——
—
AB = √ (4 − 2)2 + (7 − 5)2 = √ 22 + 22
—
—
—
= √ 4 + 4 = √8 = 2√ 2
——
—
BC = √ (4 − 7)2 + (7 − 4)2 = √ (−3)2 + 32
—
—
—
= √ 9 + 9 = √18 = 3√ 2
——
—
CD = √ (7 − 5)2 + (4 − 2)2 = √ 22 + 22
—
—
—
= √ 4 + 4 = √8 = 2√ 2
——
——
AD = √ (5 − 2)2 + (2 − 5)2 = √ (3)2 + (−3)2
—
—
—
= √ 9 + 9 = √18 = 3√ 2
Slopes:
7−5 2
—=—
=—=1
Slope of AB
4−2 2
3
7−4
—=—
= — = −1
Slope of BC
4 − 7 −3
4−2 2
—=—
=—=1
Slope of CD
7−5 2
2 − 5 −3
—=—
Slope of AD
= — = −1
5−2
3
— and BC
— have the same measure and the same
Sides AD
— and DC
—. So, by the Slopes of Parallel Lines
slope, as do AB
— BC
— and AB
— DC
—. Because the
Theorem (Thm. 3.13), AD
—
—
—
—, BC
— ⊥ AB
—,
product of their slopes is −1, AD ⊥ AB , AD ⊥ DC
—
—
and BC ⊥ DC . So, ABCD is a rectangle.
1
5
5−0
Slope = — = — = −—
2
−2 − 8 −10
2
x = — (10) + (−2) = 4 − 2 = 2
5
2
y = — (−5) + 5 = −2 + 5 = 3
5
⋅
⋅
So, Q(2, 3).
182
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Chapter 5
9. Prove △ABC is an equilateral triangle.
STATEMENTS
REASONS
1. AB = AC, BA = BC
1. By construction
2. AC = BC
2. Transitive Property of
Congruence (Thm. 2.1)
3. —
AB ≅ —
AC , —
BA ≅ —
BC ,
—
AC ≅ —
BC
4. △ABC is an
equilateral triangle.
3. Definition of congruent
segments
4. Definition of equilateral
triangle
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Geometry
Worked-Out Solutions
183