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CK-12-Geometry-Second-Edition-Answer-Key b v1 uj5 s1

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CK-12 Geometry - Second
Edition, Answer Key
CK-12 Foundation
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CK-12 Foundation
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Printed: February 25, 2015
iii
Contents
www.ck12.org
Contents
1
2
3
4
5
iv
Basics of Geometry, Answer Key
1.1
Geometry - Second Edition, Points, Lines, and Planes, Review Answers
1.2
Geometry - Second Edition, Segments and Distance, Review Answers .
1.3
Geometry - Second Edition, Angles and Measurement, Review Answers
1.4
Geometry - Second Edition, Midpoints and Bisectors, Review Answers .
1.5
Geometry - Second Edition, Angle Pairs, Review Answers . . . . . . .
1.6
Geometry - Second Edition, Classifying Polygons, Review Answers . .
1.7
Geometry - Second Edition, Chapter Review Answers . . . . . . . . . .
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1
2
5
7
10
13
14
17
Reasoning and Proof, Answer Key
2.1
Geometry - Second Edition, Inductive Reasoning, Review Answers . . . . . . . . . . .
2.2
Geometry - Second Edition, Conditional Statements, Review Answers . . . . . . . . .
2.3
Geometry - Second Edition, Deductive Reasoning, Review Answers . . . . . . . . . .
2.4
Geometry - Second Edition, Algebraic and Congruence Properties, Review Answers . .
2.5
Geometry - Second Edition, Proofs about Angle Pairs and Segments, Review Answers .
2.6
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Parallel and Perpendicular Lines, Answer Key
3.1
Geometry - Second Edition, Lines and Angles, Review Answers . . . . . . . . . . . .
3.2
Geometry - Second Edition, Properties of Parallel Lines, Review Answers . . . . . . .
3.3
Geometry - Second Edition, Proving Lines Parallel, Review Answers . . . . . . . . . .
3.4
Geometry - Second Edition, Properties of Perpendicular Lines, Review Answers . . . .
3.5
Geometry - Second Edition, Parallel and Perpendicular Lines in the Coordinate Plane,
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6
Geometry - Second Edition, The Distance Formula, Review Answers . . . . . . . . . .
3.7
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Review
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34
35
37
40
42
43
47
49
Triangles and Congruence, Answer key
4.1
Geometry - Second Edition, Triangle Sums, Review Answers . . . . . . . . . . . . . . . . . .
4.2
Geometry - Second Edition, Congruent Figures, Review Answers . . . . . . . . . . . . . . . .
4.3
Geometry - Second Edition, Triangle Congruence using SSS and SAS, Review Answers . . . .
4.4
Geometry - Second Edition, Triangle Congruence using ASA, AAS, and HL, Review Answers
4.5
Geometry - Second Edition, Isosceles and Equilateral Triangles, Review Answers . . . . . . .
4.6
Chapter 4 Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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50
51
53
55
58
60
63
Relationships with Triangles, Answer Key
5.1
Geometry - Second Edition, Midsegments of a Triangle, Review Answers . . . . .
5.2
Geometry - Second Edition, Perpendicular Bisectors in Triangles, Review Answers
5.3
Geometry - Second Edition, Angle Bisectors in Triangles, Review Answers . . . .
5.4
Geometry - Second Edition, Medians and Altitudes in Triangles, Review Answers .
5.5
Geometry - Second Edition, Inequalities in Triangles, Review Answers . . . . . . .
5.6
Geometry - Second Edition, Extension: Indirect Proof, Review Answers . . . . . .
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5.7
6
7
8
9
Contents
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Polygons and Quadrilaterals, Answer Key
6.1
Geometry - Second Edition, Angles in Polygons, Review Answers . . . . . . . . . . . .
6.2
Geometry - Second Edition, Properties of Parallelograms, Review Answers . . . . . . .
6.3
Geometry - Second Edition, Proving Quadrilaterals are Parallelograms, Review Answers
6.4
Geometry - Second Edition, Rectangles, Rhombuses and Squares, Review Answers . . .
6.5
Geometry - Second Edition, Trapezoids and Kites, Review Answers . . . . . . . . . . .
6.6
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Similarity, Answer Key
7.1
Geometry - Second Edition, Ratios and Proportions, Review Answers . . .
7.2
Geometry - Second Edition, Similar Polygons, Review Answers . . . . . .
7.3
Geometry - Second Edition, Similarity by AA, Review Answers . . . . . .
7.4
Geometry - Second Edition, Similarity by SSS and SAS, Review Answers .
7.5
Geometry - Second Edition, Proportionality Relationships, Review Answers
7.6
Geometry - Second Edition, Similarity Transformations, Review Answers .
7.7
Geometry - Second Edition, Extension: Self-Similarity, Review Answers . .
7.8
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . .
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92
93
94
95
96
97
99
101
103
Right Triangle Trigonometry, Answer Key
8.1
Geometry - Second Edition, The Pythagorean Theorem, Review Answers . . . . . .
8.2
Geometry - Second Edition, Converse of the Pythagorean Theorem, Review Answers
8.3
Geometry - Second Edition, Using Similar Right Triangles, Review Answers . . . . .
8.4
Geometry - Second Edition, Special Right Triangles, Review Answers . . . . . . . .
8.5
Geometry - Second Edition, Tangent, Sine and Cosine, Review Answers . . . . . . .
8.6
Geometry - Second Edition, Inverse Trigonometric Ratios, Review Answers . . . . .
8.7
Geometry - Second Edition, Extension: Laws of Sines and Cosines, Review Answers
8.8
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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104
105
106
109
111
112
113
114
115
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Circles, Answer Key
9.1
Geometry - Second Edition, Parts of Circles and Tangent Lines, Review Answers . . . . . . . .
9.2
Geometry - Second Edition, Properties of Arcs, Review Answers . . . . . . . . . . . . . . . . .
9.3
Geometry - Second Edition, Properties of Chords, Review Answers . . . . . . . . . . . . . . . .
9.4
Geometry - Second Edition, Inscribed Angles, Review Answers . . . . . . . . . . . . . . . . . .
9.5
Geometry - Second Edition, Angles of Chords, Secants, and Tangents, Review Answers . . . . .
9.6
Geometry - Second Edition, Segments of Chords, Secants, and Tangents, Review Answers . . .
9.7
Geometry - Second Edition, Extension: Writing and Graphing the Equations of Circles, Review
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.8
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 Perimeter and Area, Answer Key
10.1
Geometry - Second Edition, Triangles and Parallelograms, Review Answers . . . . . .
10.2
Geometry - Second Edition, Trapezoids, Rhombi, and Kites, Review Answers . . . . .
10.3
Geometry - Second Edition, Areas of Similar Polygons, Review Answers . . . . . . . .
10.4
Geometry - Second Edition, Circumference and Arc Length, Review Answers . . . . .
10.5
Geometry - Second Edition, Areas of Circles and Sectors, Review Answers . . . . . .
10.6
Geometry - Second Edition, Area and Perimeter of Regular Polygons, Review Answers
10.7
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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116
117
119
120
122
124
127
128
129
130
131
132
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135
136
137
139
11 Surface Area and Volume, Answer Key
140
11.1
Geometry - Second Edition, Exploring Solids, Review Answers . . . . . . . . . . . . . . . . . . 141
v
Contents
11.2
11.3
11.4
11.5
11.6
11.7
11.8
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Geometry - Second Edition, Surface Area of Prisms and Cylinders, Review Answers
Geometry - Second Edition, Surface Area of Pyramids and Cones, Review Answers .
Geometry - Second Edition, Volume of Prisms and Cylinders, Review Answers . . .
Geometry - Second Edition, Volume of Pyramids and Cones, Review Answers . . . .
Geometry - Second Edition, Surface Area and Volume of Spheres, Review Answers .
Geometry - Second Edition, Exploring Similar Solids, Review Answers . . . . . . .
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 Rigid Transformations, Answer Key
12.1
Geometry - Second Edition, Exploring Symmetry, Review Answers . . . . . . .
12.2
Geometry - Second Edition, Translations and Vectors , Review Answers . . . .
12.3
Geometry - Second Edition, Reflections, Review Answers . . . . . . . . . . . .
12.4
Geometry - Second Edition, Rotations, Review Answers . . . . . . . . . . . . .
12.5
Geometry - Second Edition, Composition of Transformations, Review Answers
12.6
Geometry - Second Edition, Extension: Tessellations, Review Answers . . . . .
12.7
Chapter Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
vi
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Chapter 1. Basics of Geometry, Answer Key
C HAPTER
1
Basics of Geometry,
Answer Key
Chapter Outline
1.1
G EOMETRY - S ECOND E DITION , P OINTS , L INES , AND P LANES , R EVIEW A N SWERS
1.2
G EOMETRY - S ECOND E DITION , S EGMENTS AND D ISTANCE , R EVIEW A NSWERS
1.3
G EOMETRY - S ECOND E DITION , A NGLES AND M EASUREMENT, R EVIEW A N SWERS
1.4
G EOMETRY - S ECOND E DITION , M IDPOINTS AND B ISECTORS , R EVIEW A N SWERS
1.5
G EOMETRY - S ECOND E DITION , A NGLE PAIRS , R EVIEW A NSWERS
1.6
G EOMETRY - S ECOND E DITION , C LASSIFYING P OLYGONS , R EVIEW A NSWERS
1.7
G EOMETRY - S ECOND E DITION , C HAPTER R EVIEW A NSWERS
1
1.1. Geometry - Second Edition, Points, Lines, and Planes, Review Answers
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1.1 Geometry - Second Edition, Points, Lines,
and Planes, Review Answers
For 1-5, answers will vary. One possible answer for each is included.
1.
2.
3.
4.
←→ ←→
←
→
←→
6. W X, YW , line m, XY and WY .
2
5.
www.ck12.org
Chapter 1. Basics of Geometry, Answer Key
7. Plane V or plane RST .
8. In addition to the pictures to the right, three planes may not intersect at all and can be parallel.
9. A circle.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
−→
←
→
PQ intersects RS at point Q.
−
→
AC and AB are coplanar and point D is not.
−→
−→
Points E and H are coplanar, but their rays, EF and GH are non-coplanar.
−
→ −
→ −
→
−
→
IJ , IK, IL, and IM with common endpoint I and J, K, L and M are non-collinear.
Always
Sometimes
Sometimes
Sometimes
Never
Always
Sometimes
Never
Always
Sometimes
#18: By definition, a point does not take up any space, it is only a location. #21: The ray is never read “BA,”
the endpoint is always stated first.
To make #15 true, they must be three non-collinear points. For #16, the two rays must lie on the same line,
which it does not state. For #20, four points could be coplanar, but you only need three points to make a plane,
so the fourth point could be in another plane. For #23, theorems can also be proven true by definitions and
previously proven theorems.
The walls, ceiling and floor are all planes. When two of them intersect the intersection is a line (i.e. the ceiling
and a wall). When two walls and either the ceiling or the floor intersect the intersection is a point.
The spokes on a wheel are segments. They intersect at a point.
Cities on a map are points and the distance between them can be found by measuring the segment connecting
the points.
29-33.
3
1.1. Geometry - Second Edition, Points, Lines, and Planes, Review Answers
4
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Chapter 1. Basics of Geometry, Answer Key
1.2 Geometry - Second Edition, Segments and
Distance, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
1.625 in
2.875 in
3.7 cm
8.2 cm
2.75 in
4.9 cm
4.625 in
8.7 cm
9.
10. O would be halfway between L and T , so that LO = OT = 8 cm
11.
a.
b. TA + AQ = T Q
c. T Q = 15 in
12.
a.
b. HM + MA = HA
c. AM = 11 cm
13. BC = 8 cm, BD = 25 cm, and CD = 17 cm
14. FE = 8 in, HG = 13 in, and FG = 17 in
15.
a.
b.
c.
d.
16.
17.
18.
19.
20.
21.
RS = 4
QS = 14
TS = 8
TV = 12
x = 3, HJ = 21, JK = 12, HK = 33
x = 11, HJ = 52, JK = 79, HK = 131
x = 1, HJ = 2 31 , JK = 5 32 , HK = 8
x = 17, HJ = 27, JK = 153, KH = 180
x = 16, HJ = 7, JK = 15, KH = 22
One possible answer.
5
1.2. Geometry - Second Edition, Segments and Distance, Review Answers
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|7 − (−6)|= 13
|−3 − 2|= 5
|0 − (−9)|= 9
|−4 − 1|= 5
Answers vary, but hopefully most students found their heights to be between 7 and 8 heads.
Answers should include some reference to the idea that multiplying and dividing by ten (according to the
prefixes) is much easier than keeping track of 12 inches in a ft, 3 ft in a yard, 5280 ft in a mile, etc.
28. Answers vary, but students should recognize that the pedometer is more likely to yield a false reading because
a person’s stride length varies. One possible way to minimize this error would be to average a person’s stride
length over a relatively long distance-i.e. count the number of steps taken in 100 m.
29. Answers vary. The cubit was the first recorded unit of measure and it was integral to the building of the
Egyptian pyramids.
30. Students should comment on the “ideal” proportions found in the human face and how these correspond to
our perception of beauty.
22.
23.
24.
25.
26.
27.
6
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Chapter 1. Basics of Geometry, Answer Key
1.3 Geometry - Second Edition, Angles and
Measurement, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
False, two angles could be 5◦ and 30◦ .
False, it is a straight angle.
True
True
False, you use a compass.
False, B is the vertex.
True
True
True
False, it is equal to the sum of the smaller angles within it.
Acute
12. Obtuse
13. Obtuse
14. Acute
15. Obtuse
7
1.3. Geometry - Second Edition, Angles and Measurement, Review Answers
16. Acute
17 & 18: Drawings should look exactly like 12 and 16, but with the appropriate arc marks.
19.
20.
21.
22.
23.
40◦
122◦
18◦
87◦
AE = CD, ED = CB, m6 EDC = 90◦ , m6 EAC = m6 ABC
24.
25. An interior point would be (2, 0).
26. An interior point would be (2, 0).
8
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28.
29.
30.
31.
32.
33.
34.
35.
36.
Chapter 1. Basics of Geometry, Answer Key
27.
m6 QOP = 100◦
m6 QOT = 130◦
m6 ROQ = 30◦
m6 SOP = 70◦
(x + 7)◦ + (2x + 19)◦ = 56◦
(3x + 26)◦ = 56◦
3x◦ = 30◦
x = 10◦
(4x − 23)◦ + (4x − 23)◦ = 130◦
(8x − 46)◦ = 130◦
8x◦ = 176◦
x = 22◦
(5x − 13)◦ + 90◦ = (16x − 55)◦
(5x + 77)◦ = (16x − 55)◦
22◦ = 11x◦
x = 2◦
(x − 9)◦ + (5x + 1)◦ = (9x − 80)◦
(6x − 8)◦ = (9x − 80)◦
72◦ = 3x◦
x = 24◦
Students should comment about the necessity to have a number of degrees in a line that is divisible by 30, 45,
60 and 90 degrees because these degree measures are prevalent in the study of geometrical figures. Basically,
setting the measure of a straight line equal to 180 degrees allows us to have more whole number degree
measures in common geometrical figures.
9
1.4. Geometry - Second Edition, Midpoints and Bisectors, Review Answers
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1.4 Geometry - Second Edition, Midpoints and
Bisectors, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
12 in
5 in
5 in
13 in
90◦
10 in
24 in
90◦
8 triangles
PS
QT ,V S
90◦
45◦
bisector
bisector
PU is a segment bisector of QT
45◦
x = 9, y = 14◦
x = 14◦
x = 20◦
d = 13◦
x = 12
a = 22◦ , x = 12◦
55◦ each
26. 37.5◦ each
10
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Chapter 1. Basics of Geometry, Answer Key
27. 3.5 cm each
28. 2 in each
29. You created a right, or 90◦ angle.
30.
31.
32.
33.
34.
(3, -5)
(1.5, -6)
(5, 5)
(-4.5, 2)
(7, 10)
11
1.4. Geometry - Second Edition, Midpoints and Bisectors, Review Answers
www.ck12.org
35. (6, 9)
36. This is incorrect. She should have written AB = CD or AB ∼
= CD.
37. This formula will give the same answer.
x1 + x2 y1 + y2
,
= (mx , my )
2
2
y1 + y2
x1 + x2
= mx and
= my
2
2
amp; x1 + x2 = 2mx and y1 + y2 = 2my
amp; x1 = 2mx − x2
For#34,
and y1 = 2my − y2
x1 = 2(3) − (−1) = 7
y1 = 2(6) − 2 = 10
38.
39. A square or a rectangle.
40. Midpoint could be used to determine where you might want to make a stop halfway through a trip (if using a
map the longitude and latitude could be used in the formula for midpoint). We often want to find the middle
of something-the middle of a wall to hang a picture, the middle of a room to divide it in half, etc.
12
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Chapter 1. Basics of Geometry, Answer Key
1.5 Geometry - Second Edition, Angle Pairs,
Review Answers
1.
a.
b.
c.
d.
45◦
8◦
71◦
(90 − z)◦
a.
b.
c.
d.
135◦
62◦
148◦
(180 − x)◦
2.
3.
4.
5.
6.
7.
6
6
6
6
JNI and 6 MNL (or 6 INM and 6 JNL)
INM and 6 MNL (or 6 INK and 6 KNL )
INJand 6 JNK
INM and 6 MNL (or 6 INK and 6 KNL)
a.
b.
c.
d.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
117◦
90◦
63◦
117◦
Always
Sometimes
Never
Always
Always
Never
Sometimes
Always
x = 7◦
x = 34◦
y = 13◦
x = 17◦
x = 15◦
y = 9◦
y = 8◦
x = 10.5◦
x = 4◦
y = 3◦
x = 67◦ , y = 40◦
x = 38◦ , y = 25◦
x = 15◦ , x = −4◦
x = 11◦ , √
x = −2◦
√
x = 1 + 102, x = 1 − 102
x = 11◦ , y = 7◦
13
1.6. Geometry - Second Edition, Classifying Polygons, Review Answers
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1.6 Geometry - Second Edition, Classifying
Polygons, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
10.
11.
12.
13.
14.
15.
16.
17.
18.
Acute scalene triangle
Equilateral and equiangular triangle
Right isosceles triangle.
Obtuse scalene triangle
Acute isosceles triangle
Obtuse isosceles triangle
No, there would be more than 180◦ in the triangle, which is impossible.
No, same reason as #7.
9.
All the angles in an equilateral triangle must be equal. So, an equilateral triangle is also an equiangular
triangle.
Concave pentagon
Convex octagon
Convex 17-gon
Convex decagon
Concave quadrilateral
Concave hexagon
A is not a polygon because the two sides do not meet at a vertex; B is not a polygon because one side is curved;
C is not a polygon because it is not closed.
2 diagonals
19. 5 diagonals
14
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Chapter 1. Basics of Geometry, Answer Key
20. A dodecagon has twelve sides, so you can draw nine diagonals from one vertex.
21. The pattern is below
TABLE 1.1:
Number of sides
3
4
5
6
7
8
9
10
11
12
Diagonals from one vertex
0
1
2
3
4
5
6
7
8
9
This shows us that the number diagonals from one vertex increase by one each time. So, for an n−gon, there are
(n − 3) diagonals from one vertex.
22. Octagon has 20 total diagonals Nonagon has 27 total diagonals Decagon has 35 total diagonals Undecagon
has 44 total diagonals Dodecagon has 54 total diagonals The pattern is 0, 2, 5, 9, 14, 20, 27, 35, 44, 54. To
find the next term you would add one more than was added previously. For example, the next term you would
add 11. The equation is n(n−3)
2 .
23. Sometimes
24. Always
25. Always
26. Never
27. Always
28. Sometimes, a square is ALWAYS a quadrilateral.
29. Sometimes, you can draw AT MOST n − 3 diagonals from one vertex.
30. Sometimes, a 5-point star is ALWAYS a decagon.
For questions 31-34 answers will vary.
31.
15
1.6. Geometry - Second Edition, Classifying Polygons, Review Answers
32.
33.
34. a rhombus or diamond
35. This triangle is to scale.
36. Use #9 to help you. It is the same construction, but do not draw the third side.
16
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Chapter 1. Basics of Geometry, Answer Key
1.7 Geometry - Second Edition, Chapter Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
E
B
L
A
H
M
F
O
J
G
I
K
D
C
N
17
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C HAPTER
2
Reasoning and Proof,
Answer Key
Chapter Outline
2.1
G EOMETRY - S ECOND E DITION , I NDUCTIVE R EASONING , R EVIEW A NSWERS
2.2
G EOMETRY - S ECOND E DITION , C ONDITIONAL S TATEMENTS , R EVIEW A N SWERS
18
2.3
G EOMETRY - S ECOND E DITION , D EDUCTIVE R EASONING , R EVIEW A NSWERS
2.4
G EOMETRY - S ECOND E DITION , A LGEBRAIC AND C ONGRUENCE P ROPERTIES ,
R EVIEW A NSWERS
2.5
G EOMETRY - S ECOND E DITION , P ROOFS ABOUT A NGLE PAIRS AND S EG MENTS , R EVIEW A NSWERS
2.6
C HAPTER R EVIEW A NSWERS
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Chapter 2. Reasoning and Proof, Answer Key
2.1 Geometry - Second Edition, Inductive Reasoning, Review Answers
1. 9, 21
2. 20, 110
3.
a.
b. there are two more points in each star than its figure number.
c. n + 2
4.
a. 10;
b. 48
c. 2n
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
20, 23; 107; 3n + 2
−19, −24; −164; −5n + 11
64, 128; 34, 359, 738, 368; 2n
12, 1; −307; −11n + 78
−12, 0; −93; odd terms: −3n + 12, even terms: −4n
6 7 35 n
7 , 8 ; 36 ; n+1
2n
12 14 70
23 , 27 ; 139 ; 4n−1
−13, 15; 71; (−1)n−1 (2n + 1)
21, −25; −137; (−1)n (4n − 3)
1 −1 −1 (−1)n
12 , 14 ; 70 ; 2n
8, 11; 73; odd terms 2n + 3, even terms −2n + 14
36, 49; 1225; n2
38, 51; the amount that is added is increasing by two with each term.
48, 63; the amount that is added is increasing by two with each term.
216, 343; the term number cubed, n3 .
8, 13; add the previous two terms together to get the current term.
There is a good chance that Tommy will see a deer, but it is not definite. He is reasoning correctly, but there
are other factors that might affect the outcome.
19
2.1. Geometry - Second Edition, Inductive Reasoning, Review Answers
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22. Maddie has experimented multiple times and recognized a pattern in her results. This is a good example of
inductive reasoning.
23. Juan does not use inductive reasoning correctly. It is important that conclusions are based on multiple
observations which establish a pattern of results. He only has one trial.
24. Answers vary-correct answers should include multiple experiments or trials which indicate a clear pattern for
outcomes.
25. Answers vary.
26. n(n+3)
2
27. (n+1)(n+2)
2
28. n(n+1)(n+2)
2
29. Students should notice that the points are collinear. Thus, they could find the rule by finding the equation of
the line using any two of the three points. The equation is y = 5x − 2.
30. The sequences in problems 5, 6 and 8 are of the same type. They can be modeled by linear equations because
they have a constant “slope” or rate of change. In other words, the same value is added or subtracted each time
to get the next term.
20
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Chapter 2. Reasoning and Proof, Answer Key
2.2 Geometry - Second Edition, Conditional
Statements, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
Hypothesis: 5 divides evenly into x. Conclusion: x ends in 0 or 5.
Hypothesis: A triangle has three congruent sides. Conclusion: It is an equilateral triangle.
Hypothesis: Three points lie in the same plane. Conclusion: The three points are coplanar.
Hypothesis: x = 3. Conclusion: x2 = 9.
Hypothesis: You take yoga. Conclusion: You are relaxed.
Hypothesis: You are a baseball player. Conclusion: You wear a hat.
Converse: If x ends in 0 or 5, then 5 divides evenly into x. True. Inverse: If 5 does not divide evenly into x,
then x does not end in 0 or 5. True. Contrapositive: If x does not end in 0 or 5, then 5 does not divide evenly
into it. True
Converse: If you are relaxed, then you take yoga. False. You could have gone to a spa. Inverse: If you do not
take yoga, then you are not relaxed. False. You can be relaxed without having had taking yoga. You could
have gone to a spa. Contrapositive: If you are not relaxed, then you did not take yoga. True
Converse: If you wear a hat, then you are a baseball player. False. You could be a cowboy or anyone else who
wears a hat. Inverse: If you are not a baseball player, then you do not wear a hat. False. Again, you could be
a cowboy. Contrapositive: If you do not wear a hat, then you are not a baseball player. True
If a triangle is equilateral, then it has three congruent sides. True. A triangle has three congruent sides if and
only if it is equilateral.
If three points are coplanar, then they lie in the same plane. True. Three points lie in the same plane if and
only if they are coplanar.
If x2 = 9, then x = 3. False. x could also be -3.
If B is the midpoint of AC, then AB = 5 and BC = 5. This is a true statement.
If AB 6= 5 and BC 6= 5, then B is not the midpoint of AC. This is true.
If B is noncollinear with A and C.
If AB 6= 5 and BC 6= 5, then B is not the midpoint of AC. It is the same as #14.
the original statement
p→q
∼ p →∼ q
∼∼ p →∼∼ q
p→q
18. the contrapositive
p→q
∼ p →∼ q
∼ q →∼ p
19. the contrapositive
p→q
q→ p
∼ q →∼ p
21
2.2. Geometry - Second Edition, Conditional Statements, Review Answers
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20. the original statement
p→q
∼ q →∼ p
∼∼ p →∼∼ q
p→q
21. If a U.S. citizen can vote, then he or she is 18 or more years old. If a U.S. citizen is 18 or more years old, then
he or she can vote.
22. If a whole number is prime, then it has exactly two distinct factors. If a whole number has exactly two distinct
factors, then it is prime.
23. If points are collinear, then there is a line that contains the points. If there is a line that contains the points,
then the points are collinear.
24. If 2x = 18, then x = 9. If x = 9, then 2x = 18.
25.
a.
b.
c.
d.
Yes.
No, x could equal -4.
No, again x could equal -4.
Yes.
a.
b.
c.
d.
Yes.
Yes.
Yes.
Yes.
a.
b.
c.
d.
Yes.
Yes.
Yes.
Yes.
a.
b.
c.
d.
Yes.
No, 6 ABC could be any value between 0 and 90 degrees.
No, again 6 ABC could be any value between 0 and 90 degrees.
Yes.
26.
27.
28.
29. Answers vary.
30. Answers vary.
22
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Chapter 2. Reasoning and Proof, Answer Key
2.3 Geometry - Second Edition, Deductive
Reasoning, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
I am a smart person. Law of Detachment
No conclusion
If a shape is a circle, then we don’t need to study it. Law of Syllogism.
You don’t text while driving. Law of Contrapositive.
It is sunny outside. Law of Detachment.
You are not wearing sunglasses. Law of Contrapositive.
My mom did not ask me to clean my room. Law of Contrapositive.
If I go to the park, I will give my dog a bath. Law of Syllogism.
This is a sound argument, but it doesn’t make sense because we know that circles exist. p → q
q→r
r→s
s→t
∴ p→t
p→q
p
∴q
p→q
∼q
∴∼ p
If I need a new watch battery, then I go to the mall. If I go to the mall, then I will shop. If I shop, then I will
buy shoes. Conclusion: If I need a new watch battery, then I will buy shoes.
If Anna’s teacher gives notes, then Anna writes them down. If Anna writes down the notes, then she can do
the homework. If Anna can do the homework, then she will get an A on the test. If Anna gets an A on the test,
her parents will take her out for ice cream. Conclusion: If Anna’s teacher gives notes, then Anna’s parents
will buy her ice cream.
Inductive; a pattern of weather was observed.
Deductive; Beth used a fact to determine what her sister would eat.
Deductive; Jeff used a fact about Nolan Ryan.
Either reasoning. Inductive; surfers observed patterns of weather and waves to determine when the best time
to surf is. Deductive; surfers could take the given statement as a fact and use that to determine when the best
time to surf is.
Inductive; observed a pattern.
Both-Inductive: Amani noticed a pattern of behavior. Deductive: Amani ruled out possible explanations until
there was only one remaining.
Deductive: The detectives narrowed their field of suspects by eliminating those who couldn’t have committed
the crime.
See the following table:
TABLE 2.1:
p
T
F
∼p
F
T
p∧ ∼ p
F
F
22. See the following table:
23
2.3. Geometry - Second Edition, Deductive Reasoning, Review Answers
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TABLE 2.2:
p
T
T
F
F
∼p
F
F
T
T
q
T
F
T
F
∼q
F
T
F
T
∼ p∨ ∼ q
F
T
T
T
q∨ ∼ q
T
T
T
T
p ∧ (p∨ ∼ q)
T
T
F
F
23. See the following table:
TABLE 2.3:
p
T
T
F
F
∼q
F
T
F
T
q
T
F
T
F
24. See the following table:
TABLE 2.4:
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
∼r
F
T
F
T
F
T
F
T
p∧q
T
T
F
F
F
F
F
F
(p ∧ q)∨ ∼ r
T
T
F
T
F
T
F
T
∼ q∨r
T
F
T
T
T
F
T
T
p ∨ (∼ q ∨ r)
T
T
T
T
T
F
T
T
25. See the following table:
TABLE 2.5:
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
26. See the following table:
24
r
T
F
T
F
T
F
T
F
∼q
F
F
T
T
F
F
T
T
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Chapter 2. Reasoning and Proof, Answer Key
TABLE 2.6:
p
T
T
T
T
F
F
F
F
27.
28.
29.
30.
31.
32.
33.
34.
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
∼r
F
T
F
T
F
T
F
T
q∨ ∼ r
T
T
F
T
T
T
F
T
p ∧ (q∨ ∼ r)
T
T
F
T
F
F
F
F
There are two more T ’s in #24. We can conclude that parenthesis placement matters.
p ∨ q ∨ r is always true except the one case when p, q, and r are all false.
True; Law of Syllogism
Not valid
True; Law of Contrapositive
Not valid
True; Law of Detachment
Not valid
25
2.4. Geometry - Second Edition, Algebraic and Congruence Properties, Review Answers
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2.4 Geometry - Second Edition, Algebraic and
Congruence Properties, Review Answers
1. 3x + 11 = −16
3x = −27
Subtraction PoE
x = −9
Division PoE
2. 7x − 3 = 3x − 35
4x − 3 = −35
Subtraction PoE
4x = −32
Addition PoE
x = −8
Division PoE
2
3. 3 g + 1 = 19
2
Subtraction PoE
3 g = 18
g = 27
Multiplication PoE
4. 21 MN = 5
MN = 10
Multiplication PoE
◦
6
5. 5m ABC = 540
m6 ABC = 108◦
Division PoE
6. 10b − 2(b + 3) = 5b
10b − 2b − 6 = 5b
Distributive Property
8b − 6 = 5b
Combine like terms
− 6 = −3b
Subtraction PoE
2=b
Division PoE
b=2
Symmetric PoE
7. 14 y + 56 = 31
3y + 10 = 4
Multiplication PoE (multiplied everything by 12)
3y = −6
Subtraction PoE
y = −2
Division PoE
8. 14 AB + 31 AB = 12 + 21 AB
3AB + 4AB = 144 + 6AB
Multiplication PoE (multiplied everything by 12)
7AB = 144 + 6AB
Combine like terms
AB = 144
Subtraction PoE
9. 3 = x
10. 12x − 32
11. x = 12
12. y + z = x + y
13. CD = 5
14. z + 4 = y − 7
15. Yes, they are collinear. 16 + 7 = 23
16. No, they are not collinear, 9 + 9 6= 16. I cannot be the midpoint.
17. 6 NOP must be an obtuse angle because it is supplementary with 56◦ , meaning that m6 NOP is 180◦ − 56◦ =
124◦ . 90◦ < 124◦ < 180◦ , so by definition 6 NOP is an obtuse angle.
18. 6 ABC ∼
= 6 DEF
6 GHI ∼
= 6 JKL; ∼
= 6 s have = measures; m6 ABC + m6 GHI = m6 DEF + m6 GHI; Substitution
19. M is the midpoint of AN, N is the midpoint MB; AM = MN, MN = NB; Transitive
20. 6 BFE or 6 BFG
←
→ ←
→
21. EF⊥BF
22. Yes, EG = FH because EF = GH and EF + FG = EG and FG + GH = FH by the Segment Addition
Postulate. FG = FG by the Reflexive Property and with substitution EF + FG = EG and FG + EF = FH.
26
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Chapter 2. Reasoning and Proof, Answer Key
Therefore, EG = FH by the Transitive Property.
23. Not necessarily, G could slide along EH.
24. See the following table:
TABLE 2.7:
Statement
1. 6 EBF ∼
= 6 HCG
6 ABE ∼
6
= DCH
2. m6 EBF = m6 HCG
m 6 ABE = m6 DCH
3. m6 ABF = m6 EBF + m6 ABE
m 6 DCG = m6 HCG + m6 DCH
4. m6 ABF = m6 EBF + m6 ABE
m 6 DCG = m6 EBF + m6 ABE
5. m6 ABF = m6 DCG
6. 6 ABF ∼
= 6 DCG
Reason
Given
∼
= angles have = measures
Angle Addition Postulate
Substitution PoE
Transitive PoE
∼
= angles have = measures
25. See the following table:
TABLE 2.8:
Statement
1. AB = CD
2. BC = BC
3. AB + BC = CD + BC
4. AC = BD
26.
27.
28.
29.
30.
31.
32.
Reason
Given
Reflexive PoE
Addition PoE
Segment Addition Postulate
No
No
Yes
Yes
No
No
See the following table:
TABLE 2.9:
Statement
1. 6 DAB is a right angle
2. m6 DAB = 90◦
3. AC bisects 6 DAB
4. m6 DAC = m6 BAC
5. m6 DAB = m6 DAC + m6 BAC
6. m6 DAB = m6 BAC + m6 BAC
7. m6 DAB = 2m6 BAC
8. 90◦ = 2m6 BAC
9. 45◦ = m6 BAC
Reason
Given
Definition of a right angle
Given
Definition of an angle bisector
Angle Addition Postulate
Substitution PoE
Combine like terms
Substitution PoE
Division PoE
27
2.4. Geometry - Second Edition, Algebraic and Congruence Properties, Review Answers
33.
TABLE 2.10:
Statement
1. 6 1 and 6 2 form a linear pair m6 1 = m6 2
2. 6 1 and 6 2 are supplementary
3. m6 1 + m6 2 = 180◦
4. m6 1 + m6 1 = 180◦
5. 2m6 1 = 180◦
6. m6 1 = 90◦
7. 6 1 is a right angle
28
Reason
Given
Linear Pair Postulate
Definition of Supplementary
Substitution
Simplify
Division PoE
Definition of a right angle
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Chapter 2. Reasoning and Proof, Answer Key
2.5 Geometry - Second Edition, Proofs about
Angle Pairs and Segments, Review Answers
1. See the following table:
TABLE 2.11:
Statement
1. AC⊥BD, 6 1 ∼
=6 4
2. m6 1 = m6 4
3. 6 ACB and 6 ACD are right angles
4. m6 ACB = 90◦
m 6 ACD = 90◦
5. m6 1 + m6 2 = m6 ACB
m 6 3 + m6 4 = m6 ACD
6. m6 1 + m6 2 = 90◦
m 6 3 + m6 4 = 90◦
7. m6 1 + m6 2 = m6 3 + m6 4
8. m6 1 + m6 2 = m6 3 + m6 1
9. m6 2 = m6 3
10. 6 2 ∼
=6 3
Reason
Given
∼
= angles have = measures
⊥ lines create right angles
Definition of right angles
Angle Addition Postulate
Substitution
Substitution
Substitution
Subtraction PoE
∼
= angles have = measures
2. See the following table:
TABLE 2.12:
Statement
1. 6 MLN ∼
= 6 OLP
2. m6 MLN = m6 OLP
3. m6 MLO = m6 MLN + m6 NLO
m 6 NLP = m6 NLO + m6 OLP
4. m6 NLP = m6 NLO + m6 MLN
5. m6 NLP = m6 MLO
6. 6 NLP ∼
= 6 MLO
Reason
Given
∼
= angles have = measures
Angle Addition Postulate
Substitution
Substitution
∼
= angles have = measures
3. See the following table:
TABLE 2.13:
Statement
1. AE⊥EC, BE⊥ED
2. 6 BED is a right angle
6 AEC is a right angle
3. m6 BED = 90◦
m 6 AEC = 90◦
4. m6 BED = m6 2 + m6 3
m 6 AEC = m6 1 + m6 2
Reason
Given
⊥ lines create right angles
Definition of a right angle
Angle Addition Postulate
29
2.5. Geometry - Second Edition, Proofs about Angle Pairs and Segments, Review Answers
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TABLE 2.13: (continued)
Statement
5. 90◦ = m6 2 + m6 3
90^◦ = m6 1 + m6 2
6. m6 2 + m6 3 = m6 1 + m6 2
7. m6 3 = m6 1
8. 6 3 ∼
=6 1
Reason
Substitution
Substitution
Subtraction PoE
∼
= angles have = measures
4. See the following table:
TABLE 2.14:
Statement
1. 6 L is supplementary to 6 M
6 P is supplementary to 6 O
6 L∼
=6 O
2. m6 L = m6 O
3. m6 L + m6 M = 180◦
m 6 P + m6 O = 180◦
4. m6 L + m6 M = m6 P + m6 O
5. m6 L + m6 M = m6 P + m6 L
6. m6 M = m6 P
7. 6 M ∼
=6 P
Reason
Given
∼
= angles have = measures
Definition of supplementary angles
Substitution
Substitution
Subtraction PoE
∼
= angles have = measures
5. See the following table:
TABLE 2.15:
Statement
1. 6 1 ∼
=6 4
2. m6 1 = m6 4
3. 6 1 and 6 2 are a linear pair
6 3 and 6 4 are a linear pair
4. 6 1 and 6 2 are supplementary
6 3 and 6 4 are supplementary
5. m6 1 + m6 2 = 180◦
m 6 3 + m6 4 = 180◦
6. m6 1 + m6 2 = m6 3 + m6 4
7. m6 1 + m6 2 = m6 3 + m6 1
8. m6 2 = m6 3
9. 6 2 ∼
=6 3
Reason
Given
∼
= angles have = measures
Given (by looking at the picture) could also be Definition of a Linear Pair
Linear Pair Postulate
Definition of supplementary angles
Substitution
Substitution
Subtraction PoE
∼
= angles have = measures
6. See the following table:
TABLE 2.16:
Statement
1. 6 C and 6 F are right angles
2. m6 C = 90◦ , m6 F = 90◦
30
Reason
Given
Definition of a right angle
www.ck12.org
Chapter 2. Reasoning and Proof, Answer Key
TABLE 2.16: (continued)
Statement
3. 90◦ + 90◦ = 180◦
4. m6 C + m6 F = 180◦
Reason
Addition of real numbers
Substitution
7. See the following table:
TABLE 2.17:
Statement
1. l⊥m
2. 6 1 and 6 2 are right angles
3. 6 1 ∼
=6 2
Reason
Given
⊥ lines create right angles.
Right Angles Theorem
8. See the following table:
TABLE 2.18:
Statement
1. m6 1 = 90◦
2. 6 1 and 6 2 are a linear pair
3. 6 1 and 6 2 are supplementary
4. m6 1 + m6 2 = 180◦
5. 90◦ + m6 2 = 180◦
6. m6 2 = 90◦
Reason
Given
Definition of a linear pair
Linear Pair Postulate
Definition of supplementary angles
Substitution
Subtraction PoE
9. See the following table:
TABLE 2.19:
Statement
1. l⊥m
2. 6 1 and 6 2 make a right angle
3. m6 1 + m6 2 = 90◦
4. 6 1 and 6 2 are complementary
Reason
Given
⊥ lines create right angles
Definition of a right angle
Definition of complementary angles
10. See the following table:
TABLE 2.20:
Statement
1. l⊥m, 6 2 ∼
=6 6
2. m6 2 = m6 6
3. 6 5 ∼
=6 2
6
4. m 5 = m6 2
5. m6 5 = m6 6
Reason
Given
∼
= angles have = measures
Vertical Angles Theorem
∼
= angles have = measures
Transitive
31
2.5. Geometry - Second Edition, Proofs about Angle Pairs and Segments, Review Answers
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
32
6
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AHM, 6 PHE and 6 GHE, 6 AHC
AM ∼
= MG,CP ∼
= PE, AH ∼
= HE, MH ∼
= HP, GH ∼
= HC
6 AMH, 6 HMG and 6 CPH, 6 HPE
6 AHC
6 MAH, 6 HAC and 6 MGH, 6 HGE
GC
AE, GC
6 AHM, 6 MHG
6 AGH ∼
= 6 HGE
Given; ∼
= angles have = measures; m6 ACE = m6 ACH + m6 ECH; m6 ACE = m6 ACH + m6 ACH; Combine
like terms; 21 m6 ACE = m6 ACH; AC is the angle bisector of 6 ACH; Definition of an angle bisector
90◦
26◦
154◦
26◦
64◦
25◦
75◦
105◦
90◦
50◦
40◦
25◦
130◦
155◦
130◦
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Chapter 2. Reasoning and Proof, Answer Key
2.6 Chapter Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
D
F
H
B
I
C
G
A
J
E
33
www.ck12.org
C HAPTER
3
Parallel and Perpendicular
Lines, Answer Key
Chapter Outline
3.1
G EOMETRY - S ECOND E DITION , L INES AND A NGLES , R EVIEW A NSWERS
3.2
G EOMETRY - S ECOND E DITION , P ROPERTIES OF PARALLEL L INES , R EVIEW
A NSWERS
3.3
G EOMETRY - S ECOND E DITION , P ROVING L INES PARALLEL , R EVIEW A N SWERS
34
3.4
G EOMETRY - S ECOND E DITION , P ROPERTIES OF P ERPENDICULAR L INES , R E VIEW A NSWERS
3.5
G EOMETRY - S ECOND E DITION , PARALLEL AND P ERPENDICULAR L INES IN
THE C OORDINATE P LANE , R EVIEW A NSWERS
3.6
G EOMETRY - S ECOND E DITION , T HE D ISTANCE F ORMULA , R EVIEW A NSWERS
3.7
C HAPTER R EVIEW A NSWERS
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Chapter 3. Parallel and Perpendicular Lines, Answer Key
3.1 Geometry - Second Edition, Lines and Angles, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
AB and EZ, XY and BW , among others
AB || VW , among others
BC ⊥ BW , among others
one, AV
one, CD
6 6
6 3
6 2
6 1
6 8
6 8
6 5
m6 3 = 55◦ (vertical angles), m6 1 = 125◦ (linear pair), m6 4 = 125◦ (linear pair)
m6 8 = 123◦ (vertical angles), m6 6 = 57◦ (linear pair), m6 7 = 57◦ (linear pair)
No, we do not know anything about line m.
No, even though they look parallel, we cannot assume it.
17.
19.
20.
21.
22.
18.
Fold the paper so that the lines match up and the crease passes through the point you drew.
Same as number 19.
One way to do this is to use the edges of the ruler as guide lines. The sides of the ruler are parallel.
Use the ruler to draw a line. Turn the ruler perpendicular to the first line (make sure it is perpendicular by
matching up a marking on the ruler to the original line. Use the ruler edge to draw the perpendicular line.
35
3.1. Geometry - Second Edition, Lines and Angles, Review Answers
www.ck12.org
23. Parallel lines are evident in the veins of the leaves of ferns and the markings on some animals and insects.
Parallel planes are illustrated by the surface of a body of water and the bottom.
24. Trees are usually perpendicular to the ground. Each leaf of a fern is perpendicular to the stem.
25. Some branches of trees are skew.
26. Any two equations in the form y = b, where b is a constant.
27. Any two equations in the form x = b, where b is a constant.
28. These two lines are parallel to each other.
←
→
←
→
29. slope of AB equals slope of CD = − 56 ; these lines are parallel
←
→
←
→
30. slope of AB = − 53 , slope of CD = 53 ; these lines are perpendicular
31. It appears that the slopes of parallel lines are the same and the slopes of perpendicular lines are opposite
reciprocals.
32. y = 2x − 11
33. y = − 53 x + 2
34. y = − 32 x + 6
35. y = 4x − 5
36
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Chapter 3. Parallel and Perpendicular Lines, Answer Key
3.2 Geometry - Second Edition, Properties of
Parallel Lines, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
Supplementary
Congruent
Congruent
Supplementary
Congruent
Supplementary
Supplementary
Same Side Interior
Alternate Interior
None
Same Side Interior
Vertical Angles
Corresponding Angles
Alternate Exterior
None
6 1, 6 3, 6 6, 6 9, 6 11, 6 14, and 6 16
x = 70◦ , y = 90◦
x = 15◦ , y = 40◦
x = 9◦ , y = 26◦
x = 21◦ , y = 17◦
x = 25◦
y = 18◦
x = 20◦
x = 31◦
y = 12◦
See the following table:
TABLE 3.1:
Statement
1. l || m
2. 6 1 ∼
=6 5
3. m6 1 = m6 5
4. 6 1 and 6 3 are supplementary
5. m6 1 + m6 3 = 180◦
6. m6 3 + m6 5 = 180◦
7. 6 3 and 6 5 are supplementary
Reason
Given
Corresponding Angles Postulate
∼
= angles have = measures
Linear Pair Postulate
Definition of Supplementary Angles
Substitution PoE
Definition of Supplementary Angles
27. See the following table:
TABLE 3.2:
Statement
1. l || m
Reason
Given
37
3.2. Geometry - Second Edition, Properties of Parallel Lines, Review Answers
TABLE 3.2: (continued)
Statement
2. 6 1 ∼
=6 5
6
3. 5 ∼
=6 8
4. 6 1 ∼
=6 8
Reason
Corresponding Angles Postulate
Vertical Angles Theorem
Transitive PoC
28. See the following table:
TABLE 3.3:
Statement
1. l || m
2. 6 4 and 6 6 are supplementary
3. m6 4 + m6 6 = 180◦
4. 6 2 ∼
= 6 6, 6 4 ∼
=6 8
6
6
5. m 2 = m 6, m6 4 = m6 8
6. m6 2 + m6 8 = 180◦
7. 6 2 and 6 8 are supplementary
Reason
Given
Same Side Interior Angles Theorem
Definition of Supplementary Angles
Corresponding Angles Postulate
∼
= angles have = measures
Substitution PoE
Definition of Supplementary Angles
29. See the following table:
TABLE 3.4:
Statement
1. l || m, s || t
2. 6 4 ∼
= 6 12
6
3. 12 ∼
= 6 10
∼
4. 6 4 = 6 10
Reason
Given
Corresponding Angles Postulate
Corresponding Angles Postulate
Transitive PoC
30. See the following table:
TABLE 3.5:
Statement
1. l || m, s || t
2. 6 2 ∼
= 6 13
3. 6 13 ∼
= 6 15
∼
6
4. 2 = 6 15
Reason
Given
Alternate Exterior Angles Theorem
Corresponding Angles Postulate
Transitive PoC
31. See the following table:
TABLE 3.6:
Statement
1. l || m, s || t
2. 6 6 ∼
=6 9
6
3. 4 ∼
=6 7
4. 6 6 and 6 7 are supplementary
38
Reason
Given
Alternate Interior Angles Theorem
Vertical Angles Theorem
Same Side Interior Angles
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Chapter 3. Parallel and Perpendicular Lines, Answer Key
TABLE 3.6: (continued)
Statement
5. 6 9 and 6 4 are supplementary
32.
33.
34.
35.
Reason
Same Angle Supplements Theorem
m6 1 = 102◦ , m6 2 = 78◦ , m6 3 = 102◦ , m6 4 = 78◦ , m6 5 = 22◦ , m6 6 = 78◦ , m6 7 = 102◦
x = 15◦ , y = 21◦
x = 37◦ , y = 28◦
The Same Side Interior Angles Theorem says that two angles are supplementary, not congruent.
39
3.3. Geometry - Second Edition, Proving Lines Parallel, Review Answers
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3.3 Geometry - Second Edition, Proving Lines
Parallel, Review Answers
1. Start by copying the same angle as in Investigation 3-1, but place the copy where the alternate interior angle
would be.
2. This question could be considered a “trick question,” because you are still copying two congruent angles, not
two supplementary ones, like asked. Indicate the consecutive interior angles with arc marks, but copy the
adjacent angle to the one that was copied in # 14.
3. Given, 6 1 ∼
= 6 3, Given, 6 2 ∼
= 6 3, Corresponding Angles Theorem, Transitive Property
∼
4. Given, 6 1 = 6 3, Given, 6 2 ∼
= 6 3, l || m
5. Give, Converse of the Alternate Interior Angles Theorem, Given, Converse of the Alternate Interior Angles
Theorem, Parallel Lines Property
6. See the following table:
TABLE 3.7:
Statement
1. m ⊥ l, n ⊥ l
2. m6 l = 90◦ , m6 2 = 90◦
3. m6 1 = m6 2
4. m || n
Reason
Given
Definition of Perpendicular Lines
Transitive Property
Converse of Corresponding Angles Theorem
7. See the following table:
TABLE 3.8:
Statement
1. 6 1 ∼
=6 3
2. m || n
40
Reason
Given
Converse of Alternate Interior Angles Theorem
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Chapter 3. Parallel and Perpendicular Lines, Answer Key
TABLE 3.8: (continued)
Statement
3. m6 3 + m6 4 = 180◦
4. m6 1 + m6 4 = 180◦
5. 6 1 and 6 4 are supplementary
Reason
Linear Pair Postulate
Substitution
Definition of Supplementary Angles
8. See the following table:
TABLE 3.9:
Statement
1. 6 2 ∼
=6 4
2. m || n
3. 6 1 ∼
=6 3
Reason
Given
Converse of Corresponding Angles Theorem
Alternate Interior Angles Theorem
9. See the following table:
TABLE 3.10:
Statement
1. 6 2 ∼
=6 3
2. m || n
3. 6 1 ∼
=6 4
Reason
Given
Converse of Corresponding Angles Theorem
Alternate Exterior Angles Theorem
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
none
yes, AK || LJ by Converse of Consecutive Interior Angles Theorem
yes, LG || KD by Converse of Corresponding Angles Theorem
none
none
yes, AD || GJ by Converse of Alternate Interior Angles Theorem
58◦
73◦
107◦
58◦
49◦
107◦
49◦
x = 30◦
x = 15◦
x = 12◦
x = 26◦
x = 5◦
Construction, the first and last lines are parallel. You might conjecture that two lines perpendicular to the same
line are parallel to each other.
29. You could prove this using any of the converse theorems learned in this section because all four angles
formed where the transversal intersects the two parallel lines are right angles. Thus, Alternate Interior Angles,
Alternate Exterior Angles and Corresponding Angles are all congruent and the Same Side Interior Angles are
supplementary.
30. These two angles should be supplementary if the lines are parallel.
41
3.4. Geometry - Second Edition, Properties of Perpendicular Lines, Review Answers
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3.4 Geometry - Second Edition, Properties of
Perpendicular Lines, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
90◦
90◦
45◦
16◦
72◦
84◦
41◦
24◦
78◦
90◦
126◦
54◦
180◦
⊥
not ⊥
not ⊥
⊥
90◦
34◦
56◦
90◦
56◦
134◦
134◦
34◦
See the following table:
TABLE 3.11:
Statement
1. l ⊥ m, l ⊥ n
2. 6 1 and 6 2 are right angles
3. m6 1 = 90◦ , m6 2 = 90◦
4. m6 1 = m6 2
5. 6 1 ∼
=6 2
6. m || n
27.
28.
29.
30.
31.
32.
42
x = 12◦
x = 9◦
x = 13.5◦
x = 8◦
x = 4◦
x = 30◦
Reason
Given
Definition of perpendicular lines
Definition of right angles
Transitive PoE
∼
= angles have = measures
Converse of the Corresponding Angles Postulate
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Chapter 3. Parallel and Perpendicular Lines, Answer Key
3.5 Geometry - Second Edition, Parallel and
Perpendicular Lines in the Coordinate
Plane, Review Answers
1.
2.
3.
4.
5.
6.
7.
1
3
-1
2
7
-2
4
undefined
Perpendicular
8. Parallel
9. Perpendicular
43
3.5. Geometry - Second Edition, Parallel and Perpendicular Lines in the Coordinate Plane, Review Answers
www.ck12.org
10. Neither
11. Perpendicular
12. Parallel
44
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Chapter 3. Parallel and Perpendicular Lines, Answer Key
13. Neither
14. Parallel
15. y = −5x − 7
45
3.5. Geometry - Second Edition, Parallel and Perpendicular Lines in the Coordinate Plane, Review Answers
www.ck12.org
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
46
y = 32 x − 5
y = 14 x + 2
y = − 32 x + 1
y = 2x + 1
y = x − 10
y = −x − 4
y = − 13 x − 4
y = − 25 x + 7
x = −1
y=8
y = −3x + 13
Perpendicular y = 23 x + 2
y = − 32 x − 4
Parallel y = − 15 x + 7
y = − 15 x − 3
Perpendicular y = x
y = −x
Neither y = −2x + 2
y = 2x − 3
⊥: y = − 43 x − 1
||: y = 43 x + 5 41
⊥: y = 3x − 3
||: y = − 13 x + 7
⊥: y = 7
||: x = −3
⊥: y = x − 4
||: y = −x + 8
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Chapter 3. Parallel and Perpendicular Lines, Answer Key
3.6 Geometry - Second Edition, The Distance
Formula, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
17.09 units
19.20 units
5 units
17.80 units
22.20 units
14.21 units
6.40 units
9.22 units
6.32 units
6.71 units
12 units
7 units
4.12 units
18.03 units
2.83 units
7.81 units
4 units
9 units
5.66 units
9.49 units
4.12 units
4.47 units
y = 21 x − 3
y = −3x + 5
y = − 23 x − 4
y = 52 x + 8
(9, -4)
(8, -1)
y = − 53 x − 6, (0, −6)
47
3.6. Geometry - Second Edition, The Distance Formula, Review Answers
www.ck12.org
30. y = 2x + 1
31. There are 12 possible answers: (-27, 9), (23, 9), (-2, -16), (-2, 34), (-17, -11), (-17, 29), (13, -11), (13, 29),
(-22, 24), (-22, -6), (18, 24), and (18, -6)
32. 1. Graph the two lines. 2. Determine the slope of a perpendicular line to the two lines. 3. Use the slope
from #2 to count from one line to the next to find a point on each line that is also on a perpendicular line. 4.
Determine coordinates of the points from #3. 5. Plug the points from #4 into the distance formula and solve.
48
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Chapter 3. Parallel and Perpendicular Lines, Answer Key
3.7 Chapter Review Answers
m6 1 = 90◦
m6 2 = 118◦
m6 3 = 90◦
m6 4 = 98◦
m6 5 = 28◦
m6 6 = 118◦
m6 7 = 128◦
m6 8 = 52◦
m6 9 = 62◦
49
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C HAPTER
4
Triangles and Congruence,
Answer key
Chapter Outline
50
4.1
G EOMETRY - S ECOND E DITION , T RIANGLE S UMS , R EVIEW A NSWERS
4.2
G EOMETRY - S ECOND E DITION , C ONGRUENT F IGURES , R EVIEW A NSWERS
4.3
G EOMETRY - S ECOND E DITION , T RIANGLE C ONGRUENCE USING SSS AND
SAS, R EVIEW A NSWERS
4.4
G EOMETRY - S ECOND E DITION , T RIANGLE C ONGRUENCE USING ASA, AAS,
AND HL, R EVIEW A NSWERS
4.5
G EOMETRY - S ECOND E DITION , I SOSCELES AND E QUILATERAL T RIANGLES ,
R EVIEW A NSWERS
4.6
C HAPTER 4 R EVIEW A NSWERS
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Chapter 4. Triangles and Congruence, Answer key
4.1 Geometry - Second Edition, Triangle Sums,
Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
43◦
121◦
41◦
86◦
61◦
51◦
13◦
60◦
70◦
118◦
68◦
116◦
161◦
141◦
135◦
a = 68◦ , b = 68◦ , c = 25◦ , d = 155◦ , e = 43.5◦ , f = 111.5◦
See the following table:
TABLE 4.1:
Statement
1. Triangle with interior and exterior angles.
2. m6 1 + m6 2 + m6 3 = 180◦
3. 6 3 and 6 4 are a linear pair, 6 2 and 6 5 are a linear
pair, and 6 1 and 6 6 are a linear pair
4. 6 3 and 6 4 are supplementary, 6 2 and 6 5 are
supplementary, and 6 1 and 6 6 are supplementary
5. m6 1 + m6 6 = 180◦ , m6 2 + m6 5 = 180◦
m 6 3 + m6 4 = 180◦
6. m6 1 + m6 6 + m6 2 + m6 5 + m6 3 + m6 4 = 540◦
7. m6 4 + m6 5 + m6 6 = 360◦
Reason
Given
Triangle Sum Theorem
Definition of a linear pair
Linear Pair Postulate
Definition of supplementary angles
Combine the 3 equations from #5.
Subtraction PoE
18. See the following table:
TABLE 4.2:
Statement
1. 4ABC with right angle B
2. m6 B = 90◦
3. m6 A + m6 B + m6 C = 180◦
4. m6 A + 90◦ + m6 C = 180◦
5. m6 A + m6 C = 90◦
6. 6 A and 6 C are complementary
Reason
Given
Definition of a right angle
Triangle Sum Theorem
Substitution
Subtraction PoE
Definition of complementary angles
51
4.1. Geometry - Second Edition, Triangle Sums, Review Answers
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
52
x = 14◦
x = 9◦
x = 22◦
x = 17◦
x = 12◦
x = 30◦
x = 25◦
x = 7◦
x = ±8◦
x = 17◦
x = 11◦
x = 7◦
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Chapter 4. Triangles and Congruence, Answer key
4.2 Geometry - Second Edition, Congruent Figures, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
R∼
= 6 U, 6 A ∼
= 6 G, 6 T ∼
= 6 H, RA ∼
= UG, AT ∼
= GH, RT ∼
= UH
∼
∼
∼
∼
∼
∼
6
6
6
6
6
B = T, I = O, G = P, BI = T O, IG = OP, BG = T P
Third Angle Theorem
90◦ , they are congruent supplements
Reflexive, FG ∼
= FG
Angle Bisector
4FGI ∼
= 4FGH
6 A∼
= 6 E and 6 B ∼
= 6 D by Alternate Interior Angles Theorem
Vertical Angles Theorem
No, we need to know if the other two sets of sides are congruent.
AC ∼
= CE and BC ∼
= CD
∼
4ABC = 4EDC
Yes, 4FGH ∼
= 4KLM
No
Yes, 4ABE ∼
= 4DCE
No
4BCD ∼
= 4ZY X
CPCTC
m6 A = m6 X = 86◦ , m6 B = m6 Z = 52◦ , m6 C = m6 Y = 42◦
m6 A = m6 C = m6 Y = m6 Z = 35◦ , m6 B = m6 X = 110◦
m6 A = m6 C = 28◦ , m6 ABE = m6 DBC = 90◦ , m6 D = m6 E = 62◦
m6 B = m6 D = 153◦ , m6 BAC = m6 ACD = 15◦ , m6 BCA = m6 CAD = 12◦
See the following table:
6
6
TABLE 4.3:
Statement
1. 6 A ∼
= 6 D, 6 B ∼
=6 E
6
6
2. m A = m D, m6 B = m6 E
3. m6 A + m6 B + m6 C = 180◦
m 6 D + m6 E + m6 F = 180◦
4. m6 A + m6 B + m6 C = m6 D + m6 E + m6 F
5. m6 A + m6 B + m6 C = m6 A + m6 B + m6 F
6. m6 C = m6 F
7. 6 C ∼
=6 F
24.
25.
26.
27.
28.
Reason
Given
∼
= angles have = measures
Triangle Sum Theorem
Substitution PoE
Substitution PoE
Subtraction PoE
∼
= angles have = measures
Transitive PoC
Reflexive PoC
Symmetric PoC
Reflexive PoC
4ABC is either isosceles or equiangular because the congruence statement tells us that 6 A ∼
= 6 B.
53
4.2. Geometry - Second Edition, Congruent Figures, Review Answers
29.
∼
∼
30. 4SMR ∼
4SMT
4T
MA
4AMR
and 4SRA ∼
=
=
=
= 4RAT ∼
= 4AT S ∼
= 4T SA
54
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Chapter 4. Triangles and Congruence, Answer key
4.3 Geometry - Second Edition, Triangle Congruence using SSS and SAS, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Yes, 4DEF ∼
= 4IGH
No, HJ and ED are not congruent because they have different tic marks
No, the angles marked are not in the same place in the triangles.
Yes, 4ABC ∼
= 4RSQ
No, this is SSA, which is not a congruence postulate
No, one triangle is SSS and the other is SAS.
Yes, 4ABC ∼
= 4FED
Yes, 4ABC ∼
= 4Y XZ
∼
AB = EF
AB ∼
= HI
6 C∼
=6 G
6 C∼
=6 K
AB ∼
= JL
AB ∼
= ON
See the following table:
TABLE 4.4:
Statement
1. AB ∼
= DC, BE ∼
= CE
∼
6
6
2. AEB = DEC
3. 4ABE ∼
= 4ACE
Reason
Given
Vertical Angles Theorem
SAS
16. See the following table:
TABLE 4.5:
Statement
1. AB ∼
= DC, AC ∼
= DB
2. BC ∼
= BC
3. 4ABC ∼
= 4DCB
Reason
Given
Reflexive PoC
SSS
17. See the following table:
TABLE 4.6:
Statement
1. B is a midpoint of DC, AB⊥DC
2. DB ∼
= BC
3. 6 ABD and 6 ABC are right angles
4. 6 ABD ∼
= 6 ABC
∼
5. AB = AB
6. 4ABD ∼
= 4ABC
Reason
Given
Definition of a midpoint
⊥ lines create 4 right angles
All right angles are ∼
=
Reflexive PoC
SAS
55
4.3. Geometry - Second Edition, Triangle Congruence using SSS and SAS, Review Answers
18. See the following table:
TABLE 4.7:
Statement
1. AB is an angle bisector of 6 DAC,
AD ∼
= AC
2. 6 DAB ∼
= 6 BAC
∼
3. AB = AB
4. 4ABD ∼
= 4ABC
Reason
Given
Definition of an Angle Bisector
Reflexive PoC
SAS
19. See the following table:
TABLE 4.8:
Statement
1. B is the midpoint of DC, AD ∼
= AC
∼
2. DB = BC
3. AB ∼
= AB
4. 4ABD ∼
= 4ABC
Reason
Given
Definition of a Midpoint
Reflexive PoC
SSS
20. See the following table:
TABLE 4.9:
Statement
1. B is the midpoint of DE and AC, 6 ABE is a right
angle
2. DB ∼
= BE, AB ∼
= BC
3. m6 ABE = 90◦
4. m6 ABE = m6 DBC
5. 4ABE ∼
= 4CBD
Reason
Given
Definition of a Midpoint
Definition of a Right Angle
Vertical Angle Theorem
SAS
21. See the following table:
TABLE 4.10:
Statement
1. DB is the angle bisector of 6 ADC,
AD ∼
= DC
6
2. ADB ∼
= 6 BDC
3. DB ∼
= DB
4. 4ABD ∼
= 4CBD
22.
23.
24.
25.
56
Yes
Yes
No
Yes
Reason
Given
Definition of an Angle Bisector
Reflexive PoC
SAS
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Chapter 4. Triangles and Congruence, Answer key
26. Check the measures of the three sides in your triangle with your ruler to make sure that they are 5cm, 3cm and
2cm. If you are having trouble, follow the directions in investigation 4-2 using these lengths.
27. Match up your construction with the original to see if they are the same.
28. Your triangle should look like this.
29 and 30. These are the two triangles you should create in these two problems.
57
4.4. Geometry - Second Edition, Triangle Congruence using ASA, AAS, and HL, Review Answers www.ck12.org
4.4 Geometry - Second Edition, Triangle Congruence using ASA, AAS, and HL, Review
Answers
1. Yes, AAS, 4ABC ∼ FDE
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
=
Yes, ASA, 4ABC ∼
= 4IHG
No
No
Yes, SAS, 4ABC ∼
= 4KLJ
No
Yes, SAS, 4RQP
Yes, HL, 4ABC ∼
= 4QPR
Yes, SAS, 4ABE ∼
= 4DBC
No
No
Yes, ASA, 4KLM ∼
= 4MNO
Yes, SSS, 4W ZY ∼
= 4Y XW
Yes, AAS, 4W XY ∼
= 4QPO
6 DBC ∼
6
= DBA because they are both right angles.
6 CDB ∼
= 6 ADB
∼
DB = DB
See the following table:
TABLE 4.11:
Statement
1. DB⊥AC,
DB is the angle bisector of 6 CDA
2. 6 DBC and 6 ADB are right angles
3. 6 DBC ∼
= 6 ADB
∼
4. 6 CDB = 6 ADB
5. DB ∼
= DB
6. 4CDB ∼
= 4ADB
Reason
Given
Definition of perpendicular
All right angles are ∼
=
Definition of an angle bisector
Reflexive PoC
ASA
19. CPCTC
20. 6 L ∼
= 6 O and 6 P ∼
= 6 N by the Alternate Interior Angles Theorem
∼
6
6
21. LMP = NMO by the Vertical Angles Theorem
22. See the following table:
TABLE 4.12:
Statement
1. LP || NO, LP ∼
= NO
∼
∼
6
6
6
2. L = O, P = 6 N
3. 4LMP ∼
= 6 OMN
23. CPCTC
24. See the following table:
58
Reason
Given
Alternate Interior Angles Theorem
ASA
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Chapter 4. Triangles and Congruence, Answer key
TABLE 4.13:
Statement
1. LP || NO, LP ∼
= NO
∼
∼
6
6
6
2. L = O, P = 6 N
3. 4LMP ∼
= 6 OMN
4. LM ∼
MO
=
5. M is the midpoint of PN.
25.
26.
27.
28.
29.
30.
31.
Reason
Given
Alternate Interior Angles
ASA
CPCTC
Definition of a midpoint
A∼
=6 N
6 C∼
=6 M
PM ∼
= MN
LM ∼
= MO or LP ∼
= NO
UT ∼
= FG
SU ∼
= FH
See the following table:
6
TABLE 4.14:
Statement
1. SV ⊥WU, T is the midpoint of SV and WU
2. 6 STW and 6 UTV are right angles
3. 6 STW ∼
= 6 UTV
∼
4. ST = TV ,W T ∼
= TU
5. 4STW ∼
= 4UTV
6. W S ∼
= UV
Reason
Given
Definition of perpendicular
All right angles are ∼
=
Definition of a midpoint
SAS
CPCTC
32. See the following table:
TABLE 4.15:
Statement
1. 6 K ∼
= 6 T, EI is the angle bisector of 6 KET
2. 6 KEI ∼
= 6 T EI
3. EI ∼
EI
=
4. 4KEI ∼
= 4T EI
5. 6 KIE ∼
= 6 T IE
6. EI is the angle bisector of 6 KIT
Reason
Given
Definition of an angle bisector
Reflexive PoC
AAS
CPCTC
Definition of an angle bisector
59
4.5. Geometry - Second Edition, Isosceles and Equilateral Triangles, Review Answers
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4.5 Geometry - Second Edition, Isosceles and
Equilateral Triangles, Review Answers
All of the constructions are drawn to scale with the appropriate arc marks.
1.
2.
3.
4.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
60
5.
x = 10, y = 7
x = 14
x = 13◦
x = 16◦
x = 7◦
x=1
y=3
y = 11◦ , x = 4◦
x = 25◦ , y = 19◦
x = 3, y = 8
Yes, 4ABC is isosceles. 4ABD is congruent to 4CBD by ASA. Therefore segments AB and BC are congruent
by CPCTC. Or, 6 A is congruent to 6 C by third angles theorem and thus the triangle is isosceles by the converse
of the Base Angles Theorem.
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a.
b.
c.
d.
18.
19.
20.
21.
22.
23.
24.
Chapter 4. Triangles and Congruence, Answer key
90◦
30◦
60◦
2
Always
Sometimes
Sometimes
Never
Always
a = 46◦ , b = 88◦ , c = 46◦ , d = 134◦ , e = 46◦ , f = 67◦ , g = 67◦
See the following table:
TABLE 4.16:
Statement
1. Isosceles 4CIS, with base angles 6 C and 6 SIO is
the angle bisector of 6 CIS
2. 6 C ∼
=6 S
6
3. CIO ∼
= 6 SIO
4. IO ∼
= IO
5. 4CIO ∼
= 4SIO
6. CO ∼
= OS
7. 6 IOC ∼
= 6 IOS
6
8. IOC and 6 IOS are supplementary
9. m6 IOC = m6 IOS = 90◦
10. IO is the perpendicular bisector of CS
Reason
Given
Base Angles Theorem
Definition of an Angle Bisector
Reflexive PoC
ASA
CPCTC
CPCTC
Linear Pair Postulate
Congruent Supplements Theorem
Definition of a ⊥ bisector (Steps 6 and 9)
25. See the following table:
TABLE 4.17:
Statement
1. Equilateral 4RST with RT ∼
= ST ∼
= RS
2. 6 R ∼
=6 S
3. 6 S ∼
=6 T
4. 6 R ∼
=6 T
5. 4RST is equilangular
Reason
Given
Base Angles Theorem
Base Angles Theorem
Transitive PoC
Definition of an Equiangular 4
26. See the following table:
TABLE 4.18:
Statement
1. Isosceles 4ICS with 6 C and 6 S, IO is the perpendicular bisector of CS
2. 6 C ∼
=6 S
3. CO ∼
= OS
4. m6 IOC = m6 IOS = 90◦
5. 4CIO ∼
= 4SIO
Reason
Given
Base Angle Theorem
Definition of a ⊥ bisector
Definition of a ⊥ bisector
ASA
61
4.5. Geometry - Second Edition, Isosceles and Equilateral Triangles, Review Answers
www.ck12.org
TABLE 4.18: (continued)
Statement
6. 6 CIO ∼
= 6 SIO
7. IO is the angle bisector of 6 CIS
Reason
CPCTC
Definition of an Angle Bisector
27. See the following table:
TABLE 4.19:
Statement
1. Isosceles 4ABC with base angles 6 B and 6 C,
Isosceles 4XY Z with base angles 6 Y and 6 Z, 6 C ∼
=
6 Z, BC ∼
= YZ
2. 6 B ∼
= 6 C, 6 Y ∼
=6 Z
6
3. 6 B ∼
Y
=
4. 4ABC ∼
= 4XY Z
Reason
Given
Base Angles Theorem
Transitive PoC
ASA
28. Bisect a 60◦ angle as shown.
29. Construct a 60◦ angle, then extend one side. The adjacent angle is 120◦ .
30. In investigations 3-2 and 3-3 you learned how to construct perpendiculars (i.e. 90◦ angles). You could make a
90◦ angle and copy your 30◦ onto it to make 120◦ . See investigation 1-2 for a review of copying an angle.
31. Method 1: Construct a 90◦ angle and bisect it. Method 2: Construct a 30◦ angle, bisect the 30◦ angle and copy
the resulting 15◦ angle onto the original 30◦ to make a total of 45◦ .
62
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Chapter 4. Triangles and Congruence, Answer key
4.6 Chapter 4 Review Answers
For 1-5, answers will vary.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
One leg and the hypotenuse from each are congruent, 4ABC ∼
= 4Y XZ
∼
Two angles and the side between them, 4ABC = EDC
Two angles and a side that is NOT between them, 4ABC ∼
= 4SRT
All three sides are congruent, 4ABC ∼
= 4CDA
Two sides and the angle between them, 4ABF ∼
= 4ECD
Linear Pair Postulate
Base Angles Theorem
Exterior Angles Theorem
Property of Equilateral Triangles
Triangle Sum Theorem
Equilateral Triangle Theorem
Property of an Isosceles Right Triangle
63
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C HAPTER
5
Relationships with
Triangles, Answer Key
Chapter Outline
5.1
G EOMETRY - S ECOND E DITION , M IDSEGMENTS OF A T RIANGLE , R EVIEW A N SWERS
5.2
G EOMETRY - S ECOND E DITION , P ERPENDICULAR B ISECTORS IN T RIANGLES ,
R EVIEW A NSWERS
5.3
G EOMETRY - S ECOND E DITION , A NGLE B ISECTORS IN T RIANGLES , R EVIEW
A NSWERS
5.4
G EOMETRY - S ECOND E DITION , M EDIANS AND A LTITUDES IN T RIANGLES ,
R EVIEW A NSWERS
5.5
G EOMETRY - S ECOND E DITION , I NEQUALITIES IN T RIANGLES , R EVIEW A N SWERS
5.6
G EOMETRY - S ECOND E DITION , E XTENSION : I NDIRECT P ROOF, R EVIEW A N SWERS
5.7
64
C HAPTER R EVIEW A NSWERS
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Chapter 5. Relationships with Triangles, Answer Key
5.1 Geometry - Second Edition, Midsegments
of a Triangle, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
RS = TU = 6
TU = 8
x = 5, TU = 10
x=4
No, we cannot say that the triangles are congruent. We do not know any angle measures.
y = 18
x = 12
x = 5.5
x=6
x = 14, y = 24
x = 6, z = 26
x = 5, y = 3
x = 1, z = 11
a. 53
b. 106
c. The perimeter of the larger triangle is double the perimeter of the midsegment triangle.
15.
16.
17.
18.
19.
(7, 1), (3, 6), (1, 3)
(3, 6), (2, 2), (-5, -3)
(2, 2), (1, -2), (-1, 1)
(5, 0), (5, -4), (2, 0)
GH = 13 , HI = 2, GI = − 12
20.
21.
22.
23.
24.
25.
(3, 4), (15,
√ -2), (-3, -8)
GH = 90 ≈ 9.49, Yes, GH is half of this side
(0, 3), (0, -5) and (-4, -1)
(-1, 4), (3, 4) and (5, -2)
a. M(0, 3), N(−1, −2), O(−4, 0);
b. slope of MN and AC = 5, slope of NO and AB = − 23 , and slope of MO and BC = 34 ;
65
5.1. Geometry - Second Edition, Midsegments of a Triangle, Review Answers
c. MN =
www.ck12.org
√
√
√
√
26 and AC = 2 26; NO = 13 and AC = 2 13; OM = 5 and BC = 10.
26.
a. M(1, 3), N(5, 2), O(2, 1);
b. slope of MN and AC = − 41 , slope of NO and AB = 12 , and slope of MO and BC = −2;
√
√
√
√
√
√
c. MN = 17 and AC = 2 17; NO = 10 and AC = 2 10; OM = 5 and BC = 2 5.
3 y1 +y3
2 y1 +y2
, M x1 +x
27. L x1 +x
2 , 2
2 , 2
28. slope of LM =
y1 +y3
y1 +y2
2 − 2
x1 +x3
x1 +x2
2 − 2
=
y3 −y2
x3 −x2
= slope of AT
y1 + y3 y1 + y2 2
x1 + x3 x1 + x2 2
29. length of LM =
+
−
−
2
2
2
2
s
2 2
x3 − x2
y3 − y2
=
+
2
2
r
1
1
(x3 − x2 )2 + (y3 − y2 )2
=
4
q4
1
2
= 2 (x3 − x2 ) + (y3 − y2 )2 = 12 AT
30. We have just proven algebraically that the midsegment (or segment which connects midpoints of sides in a
triangle) is parallel to and half the length of the third side.
s
66
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Chapter 5. Relationships with Triangles, Answer Key
5.2 Geometry - Second Edition, Perpendicular
Bisectors in Triangles, Review Answers
1.
2.
3.
4. Yes, but for #2, the circumcenter is not within the triangle.
5. For acute triangles, the circumcenter is inside the triangle. For right triangles, the circumcenter is on the
hypotenuse. For obtuse triangles, the circumcenter is outside the triangle.
6. By the definition of a perpendicular bisector, all three sides are bisected and therefore each half is congruent
and all six triangles are right triangles. Then, by the definition of a circumcenter, the distance from it to each
vertex is congruent (the hypotenuses of each triangle). Therefore, all 6 triangles are congruent by HL.
7.
8.
9.
10.
11.
12.
13.
x = 16
x=8
x=5
x = 12
x = 31◦
x = 34
a. AE = EB, AD = DB
67
5.2. Geometry - Second Edition, Perpendicular Bisectors in Triangles, Review Answers
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b. No, AC 6= CB
c. Yes, AD = DB
14.
15.
16.
17.
18.
19.
20.
No, not enough information
No, we don’t know if T is the midpoint of XY .
m = 12
(4, 2)
y=
√ −2x + 10
2 5
C is going to be on the perpendicular
In the picture, it is √
above AB, but it also could be
√ bisector of AB. √
below AB on y = −2x + 10. AB = 2 5, so AC is also 2 5. So, C will be 2 5 units above or below AB on
y = −2x + 10.
21-25. drawing
68
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Chapter 5. Relationships with Triangles, Answer Key
26.
27.
28. The perpendicular bisector of one side in a triangle is the set of all points equidistant from the endpoints of
that side. When we find the perpendicular bisector of a second side, we find all the points equidistant from
the endpoints of the second side (one of which is an endpoint of the first side as well). This means that the
intersection of these two lines is equidistant from all three vertices of the triangle. The segments connecting
this point (the circumcenter) to each vertex would be the radius of the circumscribed circle.
29. Fill in the blanks: There is exactly one circle which contains any three points.
30. See the following table:
TABLE 5.1:
Statement
←
→
1. CD is the perpendicular bisector of AB
2. D is the midpoint of AB
3. AD ∼
= DB
4. 6 CDA and 6 CDB are right angles
5. 6 CDA ∼
= 6 CDB
∼
6. CD = CD
Reason
Given
Definition of a perpendicular bisector
Definition of a midpoint
Definition of a perpendicular bisector
Definition of right angles
Reflexive PoC
69
5.2. Geometry - Second Edition, Perpendicular Bisectors in Triangles, Review Answers
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TABLE 5.1: (continued)
Statement
7. 4CDA ∼
= 4CDB
∼
8. AC = CB
Reason
SAS
CPCTC
31. See the following table:
TABLE 5.2:
Statement
1. 4ABC is a right isosceles triangle and BD is the ⊥
bisector of AC
2. D is the midpoint of AC
3. AD ∼
= DC
4. AB ∼
= BC
5. BD ∼
= BD
6. 4ABD and 4CBD are congruent.
Reason
Given
Definition of a perpendicular bisector
Definition of a midpoint
Definition of Isosceles Triangle
Reflexive Property of Congruence
SSS
32. Since 6 ABC is a right angle and 6 ABD ∼
= 6 CBD (CPCTC), each must be 45◦ . Also, since 6 ABC is a right
angle and 6 A ∼
= 6 C, by Base Angles Theorem, 6 A and 6 C = 45◦ . Therefore, by the converse of the Base
Angles Theorem, 4ABD and 4CBD are isosceles.
70
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Chapter 5. Relationships with Triangles, Answer Key
5.3 Geometry - Second Edition, Angle Bisectors in Triangles, Review Answers
1-3. Construct the incenter using investigation 5-2.
4. Yes, by definition, angle bisectors are on the interior of the angle. So, the incenter will be on the interior of all
three angles, or inside the triangle.
5. They will be the same point.
6. x = 6
7. x = 3
8. x = 8
9. x = 7
10. x = 9
11. x = 9
12. No, the line segment must be perpendicular to the sides of the angle also.
13. No, it doesn’t matter if the bisector is perpendicular to the interior ray.
14. Yes, the angles are marked congruent.
15. A is the incenter because it is on the angle bisectors. B is the circumcenter because it is equidistant to the
vertices.
16. A is the circumcenter because it is equidistant to the vertices. B is the incenter because it is equidistant to the
sides.
17. See the following table:
TABLE 5.3:
Statement
1. AD ∼
= DC
−
→
−
→
2. BA⊥AD and BC⊥DC
3.
4.
5.
6.
7.
8.
6
6
DAB and 6 DCB are right angles
DAB ∼
= 6 DCB
BD ∼
= BD
4ABD ∼
= 4CBD
6 ABD ∼
= 6 DBC
−→
BD bisects 6 ABC
18.
19.
20.
21.
Reason
Given
The shortest distance from a point to a line is perpendicular.
Definition of perpendicular lines
All right angles are congruent
Reflexive PoC
HL
CPCTC
Definition of an angle bisector
Incenter
Circumcenter
Circumcenter
Incenter
22-25. In an equilateral triangle the circumcenter and the incenter are the same point.
71
5.3. Geometry - Second Edition, Angle Bisectors in Triangles, Review Answers
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26. See diagram for 29-31.
−
→
−
→
27. slope of BA is -2 and slope of BC is 21 . The rays are perpendicular because their slopes are opposite reciprocals.
√
√
√
√
28. AB = 20 = 2 5 and BC = 20 = 2 5. They are congruent.
29-31.
−→
32. BD is the angle bisector of 6 ABC. Since AD⊥AB and CD⊥CB, 4DAB and 4DCB are right triangles. Since
we have shown that AB ∼
= BC and we know BD ∼
= BD by the reflexive property, 4DAB ∼
= 4DCB by HL. Thus,
−→
∼
6 ABD = 6 CBD by CPCTC. Now we can conclude that BD is the angle bisector of 6 ABC by definition of an
angle bisector.
72
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Chapter 5. Relationships with Triangles, Answer Key
5.4 Geometry - Second Edition, Medians and
Altitudes in Triangles, Review Answers
1-3. Use Investigation 5-3 to find the centroid.
4. The centroid will always be inside of a triangle because medians are always on the interior of a triangle.
5-7. Use Investigation 5-4 and 3-2 to find the orthocenter. For #6, the orthocenter will be outside of the triangle.
8. If a triangle is equilateral, then the incenter, circumcenter, orthocenter and centroid will all be the same point.
This is because all of the sides are equal and all the angles are equal.
9. You only have to construct two lines for each point of concurrency. That is because any two lines intersect at
one point. The fact that a third line intersects at this point does not change the location of the point.
10. y = 21 x + 2
11. y = −3x − 3
12. y = −x + 4
13. y = 31 x − 5
14. GE = 10
BE = 15
15. GF = 8
CF = 24
16. AG = 20
GD = 10
17. GC = 2x
CF = 3x
18. x = 2, AD = 27
19. See the following table:
TABLE 5.4:
Statement
1. 4ABC ∼
= 4DEF, AP and DO are altitudes
∼
2. AB = DE
3. 6 P and 6 O are right angles
4. 6 P ∼
=6 O
6
5. ABC ∼
= 6 DEF
6. 6 ABC and 6 ABP are a linear pair
6 DEF and 6 DEO are a linear pair
7. 6 ABC and 6 ABP are supplementary
6 DEF and 6 DEO are supplementary
8. 6 ABP ∼
= 6 DEO
9. 4APB ∼
= 4DOE
∼
10. AP = DO
Reason
Given
CPCTC
Definition of an altitude
All right angles are congruent
CPCTC
Definition of a linear pair
Linear Pair Postulate
Congruent Supplements Theorem
AAS
CPCTC
20. See the following table:
73
5.4. Geometry - Second Edition, Medians and Altitudes in Triangles, Review Answers
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TABLE 5.5:
Statement
1. Isosceles 4ABC with legs AB and AC
BD⊥DC and CE⊥BE
2. 6 DBC ∼
= 6 ECB
3. 6 BEC and 6 CEB are right angles
4. 6 BEC ∼
= 6 CEB
∼
5. BC = BC
6. 4BEC ∼
= 4CDB
7. BD ∼
= CE
21.
22.
23.
24.
25.
26.
27.
28.
Base Angles Theorem
Definition of perpendicular lines
All right angles are congruent
Reflexive PoC
AAS
CPCTC
M(2, 5)
y = 2x + 1
N(1, −3)
y = −4x + 1
intersection (0, 1)
Centroid
(1, -1)
(1, 3)
29. Midpoint of one side is
x1 +x2 +x3 y1 +y2 +y3
.
,
3
3
30. (1, -5)
74
Reason
Given
x1 +x2 y1 +y2
,
2 , 2
using the third vertex, the centroid is
x +x2
x3 +2( 1 2
3
)
y +y2
,
y3 +2( 1 2
3
)
=
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Chapter 5. Relationships with Triangles, Answer Key
5.5 Geometry - Second Edition, Inequalities in
Triangles, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
AB, BC, AC
BC, AB, AC
AC, BC, AB
6 B, 6 A, 6 C
6 B, 6 C, 6 A
6 C, 6 B, 6 A
No, 6 + 6 < 13
No, 1 + 2 = 3
Yes
Yes
No, 23 + 56 < 85
Yes
1 < 3rd side <17
11 < 3rd side <19
12 < 3rd side <52
Both legs must be longer than 12
0 < x < 10.3
m6 1 > m6 2 because 7 > 6
IJ, IG, GJ, GH, JH
m6 1 < m6 2, m6 3 > m6 4
a=b
a>b
a<b
d<a<e<c<b
a=b<d<e<c
x < 18
x>3
m6 C < m6 B < m6 A because AB < AC
SAS theorem doesn’t apply here since the angle is not between the pair of congruent sides.
Since the median AB bisects the side CT ,CB ∼
= BT . By the reflexive property, AB ∼
= AB. If CA > AT , then we
can use the SSS Inequality Theorem to conclude that m6 ABT < m6 ABC. Since m6 ABT and m6 ABC are also
a linear pair and must be supplementary, the smaller angle must be acute. Hence, 6 ABT is acute.
75
5.6. Geometry - Second Edition, Extension: Indirect Proof, Review Answers
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5.6 Geometry - Second Edition, Extension:
Indirect Proof, Review Answers
Answers will vary. Here are some hints.
Assume n is odd, therefore n = 2a + 1.
Use the definition of an equilateral triangle to lead you towards a contradiction.
Remember the square root of a number can be negative or positive.
Use the definition of an isosceles triangle to lead you towards a contradiction.
If x + y is even, then x + y = 2n, where n is any integer.
Use the Triangle Sum Theorem to lead you towards a contradiction.
With the assumption of the opposite of AB + BC = AC, these three lengths could make a triangle, thus making
A, B, and C non-collinear.
8. If we assume that we have an even number of nickels, then the value of the coin collection must be a multiple
of ten and we have a contradiction.
9. Assume that the last answer on the quiz is false. This implies that the fourth answer is true. If the fourth
answer is true, then the one before it (the third answer) is false. However, this contradicts the fact that the third
answer is true.
10. None. To prove this by contradiction, select each statement as the “true” statement and you will see that at
least one of the other statements will also be true. If Charlie is right, then Rebecca is also right. If Larry is
right, then Rebecca is right. If Rebecca is right, then Larry is right.
1.
2.
3.
4.
5.
6.
7.
76
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Chapter 5. Relationships with Triangles, Answer Key
5.7 Chapter Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
BE
AE
AH
CE
AG
The point of concurrency is the circumcenter and use Investigation 5-1 to help you. The circle should pass
through all the vertices of the triangle (inscribed triangle).
The point of concurrency is the incenter and use Investigation 5-2 to help you. The circle should touch all the
sides of the triangle (inscribed circle).
The point of concurrency is the centroid and it is two-thirds of the median’s length from the vertex (among
other true ratios). It is also the balancing point of a triangle.
The point of concurrency is called the orthocenter. The circumcenter and the orthocenter can lie outside a
triangle when the triangle is obtuse.
x − 7 < third side < 3x + 5
77
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C HAPTER
6
Polygons and
Quadrilaterals, Answer Key
Chapter Outline
78
6.1
G EOMETRY - S ECOND E DITION , A NGLES IN P OLYGONS , R EVIEW A NSWERS
6.2
G EOMETRY - S ECOND E DITION , P ROPERTIES OF PARALLELOGRAMS , R EVIEW
A NSWERS
6.3
G EOMETRY - S ECOND E DITION , P ROVING Q UADRILATERALS ARE PARALLEL OGRAMS , R EVIEW A NSWERS
6.4
G EOMETRY - S ECOND E DITION , R ECTANGLES , R HOMBUSES AND S QUARES ,
R EVIEW A NSWERS
6.5
G EOMETRY - S ECOND E DITION , T RAPEZOIDS AND K ITES , R EVIEW A NSWERS
6.6
C HAPTER R EVIEW A NSWERS
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Chapter 6. Polygons and Quadrilaterals, Answer Key
6.1 Geometry - Second Edition, Angles in Polygons, Review Answers
1. See the following table:
TABLE 6.1:
# of sides
3
4
5
6
7
8
9
10
11
12
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
# of 4s from one
vertex
1
2
3
4
5
6
7
8
9
10
4s × 180◦ (sum)
180◦
360◦
540◦
720◦
900◦
1080◦
1260◦
1440◦
1620◦
1800◦
Each angle in a
regular n−gon
60◦
90◦
108◦
120◦
128.57◦
135◦
140◦
144◦
147.27◦
150◦
Sum of the exterior
angles
360◦
360◦
360◦
360◦
360◦
360◦
360◦
360◦
360◦
360◦
2340◦
3780◦
26
20
157.5◦
165◦
30◦
10◦
360◦
18
30
17
24
10
11
x = 60◦
x = 90◦ , y = 20◦
x = 35◦
y = 115◦
x = 105◦
x = 51◦ , y = 108◦
x = 70◦ , y = 70◦ , z = 90◦
x = 72.5◦ , y = 107.5◦
x = 90◦ , y = 64◦
x = 52◦ , y = 128◦ , z = 123◦
larger angles are 135◦
smallest angle is 36◦
x = 117.5◦
79
6.1. Geometry - Second Edition, Angles in Polygons, Review Answers
◦
◦
30. 180◦ − (n−2)180
= 360
n
n
◦
◦
360◦
180 n−180 n+360
=
n ◦
n
360◦
360
=
n
n
31. a = 120◦ , b = 60◦ , c = 48◦ , d = 60◦ , e = 48◦ , f = 84◦ , g = 120◦ , h = 108◦ , j = 96◦
80
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Chapter 6. Polygons and Quadrilaterals, Answer Key
6.2 Geometry - Second Edition, Properties of
Parallelograms, Review Answers
m6 A = 108◦ , m6 C = 108◦ , m6 D = 72◦
m6 P = 37◦ , m6 Q = 143◦ , m6 D = 37◦
all angles are 90◦
m6 E = m6 G = (180 − x)◦ , m6 H = x◦
a = b = 53◦
c=6
d = 10, e = 14
f = 5, g = 3
h = 25◦ , j = 11◦ , k = 8◦
m = 25◦ , n = 19◦
p = 8, q = 3
r = 1, s = 2
t = 3, u = 4
96◦
85◦
43◦
42◦
12
2
64◦
42◦
(2, 1), Find the midpoint of one of the diagonals since the midpoints are the same for both
slope of EF = slope of GH = 41 ; slope of EH = slope of FG = − 52 ; Slopes of opposite sides are the same,
therefore opposite
√ sides are parallel.
√
24. EF = HG = 17; FG = EH = 29; lengths of opposite sides are the same (congruent).
25. A quadrilateral in the coordinate plane can be show to be a parallelogram by showing any one of the three
properties of parallelograms shown in questions 22-24.
26. See the following table:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
TABLE 6.2:
Statement
1. ABCD is a parallelogram with diagonal BD
2. AB || DC, AD || BC
3. 6 ABD ∼
= 6 BDC, 6 ADB ∼
= 6 DBC
∼
4. DB = DB
5. 4ABD ∼
= 4CDB
6
6. 6 A ∼
C
=
Reason
Given
Definition of a parallelogram
Alternate Interior Angles Theorem
Reflexive PoC
ASA
CPCTC
27. See the following table:
81
6.2. Geometry - Second Edition, Properties of Parallelograms, Review Answers
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TABLE 6.3:
Statement
1. ABCD is a parallelogram with diagonals BD and AC
2. AB || DC, AD || BC
3. 6 ABD ∼
= 6 BDC, 6 CAB ∼
= 6 ACD
DC
4. AB ∼
=
5. 4DEC ∼
= 4BEA
6. AE ∼
= EC, DE ∼
= EB
Reason
Given
Definition of a parallelogram
Alternate Interior Angles Theorem
Opposite Sides Theorem
ASA
CPCTC
28. See the following table:
TABLE 6.4:
Statements
1. ABCD is a parallelogram
2. m6 1 = m6 3 and m6 2 = m6 4
3. m6 1 + m6 2 + m6 3 + m6 4 = 360◦
4. m6 1 + m6 2 + m6 1 + m6 2 = 360◦
5. 2(m6 1 + m6 2) = 360◦
6. m6 1 + m6 2 = 180◦
29.
30.
31.
32.
82
w = 135◦
x = 16
y = 105◦
z = 60◦
Reasons
Given
Opposite angles congruent in parallelogram
Sum of angles in quadrilateral is 360◦
Substitution
Simplification
Division POE
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Chapter 6. Polygons and Quadrilaterals, Answer Key
6.3 Geometry - Second Edition, Proving
Quadrilaterals are Parallelograms, Review
Answers
1. No
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
Yes
Yes
Yes
No
No
Yes
No
Yes
Yes
No
No
x=5
x = 8◦ , y = 10◦
x = 4, y = 3
Yes
Yes
No
See the following table:
TABLE 6.5:
Statement
1. 6 A ∼
= 6 C, 6 D ∼
=6 B
2. m6 A = m6 C, m6 D = m6 B
3. m6 A + m6 B + m6 C + m6 D = 360◦
4. m6 A + m6 A + m6 B + m6 B = 360◦
5. 2m6 A + 2m6 B = 360◦
2m 6 A + 2m6 D = 360◦
6. m6 A + m6 B = 180◦
m 6 A + m6 D = 180◦
7. 6 A and 6 B are supplementary
6 A and 6 D are supplementary
8. AD || BC, AB || DC
9. ABCD is a parallelogram
Reason
Given
∼
= angles have = measures
Definition of a quadrilateral
Substitution PoE
Combine Like Terms
Division PoE
Definition of Supplementary Angles
Consecutive Interior Angles Converse
Definition of a Parallelogram
20. See the following table:
TABLE 6.6:
Statement
1. AE ∼
= EC, DE ∼
= EB
6
2. AED ∼
= 6 BEC
6 DEC ∼
= 6 AEB
Reason
Given
Vertical Angles Theorem
83
6.3. Geometry - Second Edition, Proving Quadrilaterals are Parallelograms, Review Answers
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TABLE 6.6: (continued)
Statement
3. 4AED ∼
= 4CEB
∼
4AEB = 4CED
4. AB ∼
= DC, AD ∼
= BC
5. ABCD is a parallelogram
Reason
SAS
CPCTC
Opposite Sides Converse
21. See the following table:
TABLE 6.7:
Statement
1. 6 ADB ∼
= 6 CBD, AD ∼
= BC
2. AD || BC
3. ABCD is a parallelogram
22.
23.
24.
25.
26.
see graph
-2√
3 5
see graph
The triangle is formed by the midsegments of the triangle formed when the parallelograms overlap. Four
congruent triangles are formed within this center triangle, which is also congruent to the three outer triangles.
27. see graph
84
Reason
Given
Alternate Interior Angles Converse
Theorem 5-10
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28.
29.
30.
31.
Chapter 6. Polygons and Quadrilaterals, Answer Key
parallelogram
slope of W X = slope of Y Z = 3; slope of XY = slope of ZW = − 12 opposite sides parallel
midpoint of diagonal YW is (1.5, 3.5); midpoint of diagonal XZ is (1.5, 3.5); midpoints bisect each other
Each side of the parallelogram is parallel to the diagonal. For example, XY || DU || ZW , so opposite sides
are parallel. They are also half the length of the diagonal so opposite sides are congruent. Either proves that
W XY Z is a parallelogram.
85
6.4. Geometry - Second Edition, Rectangles, Rhombuses and Squares, Review Answers
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6.4 Geometry - Second Edition, Rectangles,
Rhombuses and Squares, Review Answers
1.
a.
b.
c.
d.
e.
13
26
24
10
90◦
a.
b.
c.
d.
e.
12
21.4
11
54◦
90◦
a.
b.
c.
d.
90◦
90◦
45◦
45◦
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
86
Rectangle, the diagonals bisect each other and are congruent.
Rhombus, all sides are congruent and the diagonals are perpendicular.
None
Parallelogram, the diagonals bisect each other.
Square, the diagonals bisect each other, are congruent, and perpendicular.
Rectangle, all angles are right angles.
None
Square, all the angles and sides are congruent.
Parallelogram, one set of sides are parallel and congruent.
Sometimes, with the figure is a square.
Always
Sometimes, when it is a square.
Always
Sometimes, when it is a square.
Never
Square
Rhombus
Rectangle
Parallelogram
Answers will vary. One possibility: Another way to determine if a quadrilateral is a square would be to find
the length of all the sides using the distance formula. All sides must be equal. Then, find the slopes of each
side. If the adjacent sides have perpendicular slopes, then the angles are all 90◦ and thus congruent.
◦ , z = 37◦
x = 10, w = 53◦ , y = 37√
◦
◦
x = 45 , y = 90 , z = 2 2
x = y = 13, w = z = 25◦
See the following table:
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Chapter 6. Polygons and Quadrilaterals, Answer Key
TABLE 6.8:
Statements
1. ABCD is a rectangle
2. BW ∼
= WC, AY ∼
= Y D, BX ∼
= XA, CZ ∼
= ZD
3. BD = AC
4. XY is a midsegment in 4ABD
ZY is a midsegment in 4ACD
XW is a midsegment in 4ABC
W Z is a midsegment in 4BCD
5. XY = 21 BD = W Z and XW = 12 AC = Y Z
6. 12 BD = 12 AC
7. XY = W Z = Y Z = XW
8. W XY Z is a rhombus
Reasons
1. Given
2. Definition of a midpoint
3. Diagonals are congruent in a rectangle
4. Definition of a midsegment in a triangle
5. Midsegment in a triangle is half the length of the
parallel side.
6. Division POE
7. Substitution
8. Definition of a rhombus
28. Answers may vary. The quadrilateral inscribed in the rhombus will always be a rectangle because the diagonals
of a rhombus are perpendicular and the opposite sides of the inscribed quadrilateral will be parallel to the
diagonals and thus perpendicular to one another.
29. Answers may vary. First, the square is a rhombus, the inscribed quadrilateral will be a rectangle (see problem
28). Second, the diagonals of the square are congruent so the sides of the inscribed quadrilateral will be
congruent (see problem 27). Since the sides of the inscribed quadrilateral are perpendicular and congruent the
parallelogram is a square.
30.
Start by drawing a segment 2 inches long. Construct the perpendicular bisector of this segment. Mark off
points on the perpendicular bisector .75 inches from the point of intersection. Connect these points to the
endpoint of your original segment.
31.
There are an infinite number of rectangles with diagonals of length 3 inches. The picture to the left shows
three possible rectangles. Start by drawing a segment 3 inches long. Construct the perpendicular bisector of
the segment to find the midpoint. Anchor your compass at the midpoint of the segment and construct a circle
which contains the endpoints of your segment (radius 1.5 inches). Now you can draw a second diameter to
87
6.4. Geometry - Second Edition, Rectangles, Rhombuses and Squares, Review Answers
your circle and connect the endpoints to form a rectangle with diagonal length 3 inches.
88
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Chapter 6. Polygons and Quadrilaterals, Answer Key
6.5 Geometry - Second Edition, Trapezoids
and Kites, Review Answers
1.
a.
b.
c.
d.
55◦
125◦
90◦
110◦
a.
b.
c.
d.
e.
50◦
50◦
90◦
25◦
115◦
2.
3. No, if the parallel sides were congruent, then it would be a parallelogram. By the definition of a trapezoid, it
can never be a parallelogram (exactly one pair of parallel sides).
4. Yes, the diagonals do not have to bisect each other.
5. Construct two perpendicular lines to make the diagonals. One diagonal is bisected, so measure an equal length
on either side of the point of intersection on one diagonal. Mark this as two vertices. The other two vertices
are on the other diagonal. Place them anywhere on this diagonal and connect the four points to create the kite.
Answers will vary.
6. 33
7. 28
8. 8
9. 11
10. 37
11. 5
12. x = 4
√
13. x = 5, y = 73
14. x = 11, y = 17
15. y = 5◦
16. y = 45◦
17. x = 12◦ , y = 8◦
18. parallelogram
19. square
20. kite
21. trapezoid
22. None
23. isosceles trapezoid
24. rectangle
89
6.5. Geometry - Second Edition, Trapezoids and Kites, Review Answers
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25. rhombus
26. See the following table:
TABLE 6.9:
Statement
1. KE ∼
= T E and KI ∼
= TI
∼
2. EI = EI
3. 4EKI ∼
= 4ET I
4. 6 KES ∼
= 6 T ES and 6 KIS ∼
= 6 T IS
5. EI is the angle bisector of 6 KET and 6 KIT
Reason
Given
Reflexive PoC
SSS
CPCTC
Definition of an angle bisector
27. See the following table:
TABLE 6.10:
Statement
1. KE ∼
= T E and KI ∼
= TI
2. 4KET and 4KIT are isosceles triangles
3. EI is the angle bisector of 6 KET and 6 KIT
4. EI is the perpendicular bisector of KT
5. KT ⊥ EI
Reason
Given
Definition of isosceles triangles
Theorem 6-22
Isosceles Triangle Theorem
28. See the following table:
TABLE 6.11:
Statement
1. T RAP is an isosceles trapezoid with T R || AP
2. T P ∼
= RA
∼
3. AP = AP
4. 6 T PA ∼
= 6 RAP
5. 4T PA ∼
= 4RAP
∼
6. TA = RP
Reason
Given
Definition of isosceles trapezoid
Reflexive PoC
Base angles congruent in isosceles trapezoid
SAS
CPCTC
29. The sides of the parallelogram inscribed inside a kite will be parallel to the diagonals because they are
triangle midsegments. Since the diagonals in a kite are perpendicular, the sides of the parallelogram will
be perpendicular as well. The diagonals in a kite are not congruent so only opposite sides of the parallelogram
will be congruent and thus preventing the parallelogram from being a square.
30. Since the diagonals are congruent and the sides of the inscribed parallelogram are half the length of the
diagonals they are parallel to (because they are triangle midsegments), they are all congruent. This makes the
inscribed parallelogram a rhombus.
90
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Chapter 6. Polygons and Quadrilaterals, Answer Key
6.6 Chapter Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
Never
Always
Always
Sometimes
Sometimes
Never
Always
Sometimes
TABLE 6.12:
Opposite
sides ||
Diagonals ⊥
Opposite
sides ∼
=
Opposite angles ∼
=
Diagonals ∼
=
No
No
No
No,
base
6 s∼
=
Non-vertex
6 s
Yes
All 6 s ∼
=
Yes
All 6 s ∼
=
No
Yes
Trapezoid
Isosceles
Trapezoid
Kite
One set
One set
Diagonals
bisect each
other
No
No
No
No
Yes
No
Non-parallel
sides
No
Parallelogram
Rectangle
Rhombus
Square
Both sets
Both sets
Both sets
Both sets
Yes
Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
All sides ∼
=
All sides ∼
=
No
No
Yes
No
Yes
a = 64◦ , b = 118◦ , c = 82◦ , d = 99◦ , e = 106◦ , f = 88◦ , g = 150◦ , h = 56◦ , j = 74◦ , k = 136◦
91
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C HAPTER
7
Similarity, Answer Key
Chapter Outline
7.1
G EOMETRY - S ECOND E DITION , R ATIOS AND P ROPORTIONS , R EVIEW A N SWERS
7.2
G EOMETRY - S ECOND E DITION , S IMILAR P OLYGONS , R EVIEW A NSWERS
7.3
G EOMETRY - S ECOND E DITION , S IMILARITY BY AA, R EVIEW A NSWERS
7.4
G EOMETRY - S ECOND E DITION , S IMILARITY BY SSS AND SAS, R EVIEW A N SWERS
7.5
G EOMETRY - S ECOND E DITION , P ROPORTIONALITY R ELATIONSHIPS , R EVIEW
A NSWERS
7.6
G EOMETRY - S ECOND E DITION , S IMILARITY T RANSFORMATIONS , R EVIEW A N SWERS
7.7
G EOMETRY - S ECOND E DITION , E XTENSION : S ELF -S IMILARITY, R EVIEW A N SWERS
7.8
92
C HAPTER R EVIEW A NSWERS
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Chapter 7. Similarity, Answer Key
7.1 Geometry - Second Edition, Ratios and
Proportions, Review Answers
1.
a.
b.
c.
d.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
4:3
5:8
6:19
6:8:5
2:1
1:3
2:1
1:1
5:4:3
x = 18◦ , angles are 54◦ , 54◦ , 72◦
x = 3; 9, 12, 15
x = 4; 12, 20
x = 16; 64, 112
X = 4; 20, 36
x = 4; 12, 44
a+b
c+d
b = d
d(a + b) = b(c + d)
ad + bd = bc + bd
ad = bc
a−b
c−d
b = d
d(a − b) = b(c − d)
ad − bd = bc − bd
ad = bc
x = 12
x = −5
y = 16
x = 12, −12
y = −21
z = 3.75
x = 13.9 gallons
The president makes $800,000, vice president makes $600,000 and the financial officer makes $400,000.
1 32 cups water
60 marshmallows; 6 cups miniatures
False
True
True
False
28
18
7
24
93
7.2. Geometry - Second Edition, Similar Polygons, Review Answers
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7.2 Geometry - Second Edition, Similar Polygons, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
94
True
False
False
False
True
True
False
True
IG
BG
BI
6 B∼
= AT
= HT
= 6 H, 6 I ∼
= 6 A, 6 G ∼
= 6 T, HA
3
5
5 or 3
HT = 35
IG = 27
57, 95, 35 or 53
m6 E = 113◦ , m6 Q = 112◦
3
2
3 or 2
12
21
6
18
No, 32
26 6= 12
Yes, 4ABC ∼ 4NML
Yes, ABCD ∼ STUV
Yes, 4EFG ∼ 4LMN
x = 12, y = 15
31
x = 20, y = 7
≈ 14.6
a ≈ 7.4, b = 9.6
X = 6, y = 10.5
121
1:3
30u2 , 270u2 , 1 : 9, this is the ratio of the lengths squared or
1 2
3 .
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Chapter 7. Similarity, Answer Key
7.3 Geometry - Second Edition, Similarity by
AA, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
4T RI
T R, T I, AM
12
6
6, 12
4ABE ∼ 4CDE because 6 BAE ∼
= 6 DCE and 6 ABE ∼
= 6 CDE by the Alternate Interior Angles Theorem.
AE
BE
Answers will vary. One possibility: CE = DE
One possibility: 4AED and 4BEC
AC = 22.4
Only two angles are needed because of the 3rd Angle Theorem.
Congruent triangles have the same shape AND size. Similar triangles only have the same shape. Also,
congruent triangles are always similar, but similar triangles are not always congruent.
Yes, right angles are congruent and solving for the missing angle in each triangle, we find that the other two
angles are congruent as well.
FE = 43 k
k = 16
right, right, similar
Yes, 4DEG ∼ 4FDG ∼ 4FED
Yes, 4HLI ∼ 4HKJ
No only vertical angles are congruent
Yes, they are ⊥ to the same line.
Yes, the two right angles are congruent and 6 OEC and 6 NEA are vertical angles.
x = 48 f t.
Yes, we can use the Pythagorean Theorem to find EA. EA = 93.3 f t.
70 ft
29 ft 2 in
24 ft
Answers will vary. Check your answer by considering whether or not it is reasonable.
27.
28. m6 1 + m6 2 = 90◦ , therefore m6 GDF = m6 2 and m6 EDF = 6 1. This shows that the three angles in each
triangle are congruent to the three corresponding angles in each of the other triangles. Thus, they are all
similar.
29. DF
30. GD
31. FE
95
7.4. Geometry - Second Edition, Similarity by SSS and SAS, Review Answers
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7.4 Geometry - Second Edition, Similarity by
SSS and SAS, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
96
Yes, SSS. The side lengths are proportional.
No. One is much larger than the other.
There are 2.2 cm in an inch, so that is the scale factor.
There is no need. With the A and A parts of ASA we have triangles with two congruent angles. The triangles
are similar by AA.
4DFE
DF, EF, DF
DH = 7.5
4DBE
SAS
27
AB, BE, AC
7
8
Yes, 21
= 24
. This proportion will be valid as long as AC || DE.
Yes, 4ABC ∼ 4DFE, SAS
No, the angle is not between the given sides
Yes, 4ABC ∼ 4DFE, SSS
Yes, 4ABE ∼ 4DBC, SAS
15
No, 10
20 6= 25
24
No, 32
6= 16
20
x=3
x = 6, y = 3.5
The building is 10 ft tall.
The child’s shadow is 105 inches long.
The side lengths are 15, 36, 39
The radio tower
√ is 55 ft.
√
AB = BC = 11.25, AC = 3, DE = EF = 5, DF = 2
AB
BC
AC
3
DE = EF = DF = 2
Yes, 4ABC ∼ 4DEF by SSS similarity.
slope of CA = slope of LO = undefined (vertical); slope of AR = slope of OT = 0 (horizontal).
90◦ , vertical and horizontal lines are perpendicular.
T O = 6, OL = 8,CA = 4 and AR = 3; LO : CA = OT : AR = 2 : 1
Yes, by SAS similarity.
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Chapter 7. Similarity, Answer Key
7.5 Geometry - Second Edition, Proportionality Relationships, Review Answers
4ECF ∼ 4BCD
DF
CD
FE
DF, DB
14.4
21.6
16.8
45
The parallel sides are in the same ratio as the sides of the similar triangles, not the segments of the sides.
yes
no
yes
no
x=9
y = 10
y = 16
z=4
x=8
x = 2.5
a = 4.8, b = 9.6
a = 4.5, b = 4, c = 10
a = 1.8, b = 37
x = 5, y = 7
3
2 b or 1.5b
16
5 a or 3.2a
Casey mistakenly used the length of the angle bisector in the proportion rather than the other side length. The
correct proportion is a5 = 75 , thus a = 25
7.
28. The path will intersect the third side 2.25 m from the 3 m side and 3.75 m from the 5 m side.
29. a = 42m and b = 56m
30. Blanks are in red.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
TABLE 7.1:
Statement
−
→
1. AC is the angle bisector of 6 BADX, A, B are collinear
−
→ ←
→
and AC || XD
2. 6 BAC ∼
= 6 CAD
∼
6
6
3. X = BAC
4. 6 CAD ∼
= 6 ADX
∼
6
6
5. X = ADX
6. 4XAD is isosceles
7. AX ∼
= AD
8. AX = AD
BA
BC
9. AX
= CD
Reason
Given
Definition of an angle bisector
Corresponding Angles Postulate
Alternate Interior Angles Theorem
Transitive PoC
Base Angles Converse
Definition of an Isosceles Triangle
Congruent segments are also equal
Theorem 7-7
97
7.5. Geometry - Second Edition, Proportionality Relationships, Review Answers
TABLE 7.1: (continued)
Statement
BC
BA
= CD
10. AD
98
Reason
Substitution PoE
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Chapter 7. Similarity, Answer Key
7.6 Geometry - Second Edition, Similarity
Transformations, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
(2, 6)
(-8, 12)
(4.5, -6.5)
k = 23
k=9
k = 21
20, 26, 34
2 23 , 3, 5
k = 52
14
k = 11
original: 20, dilation: 80, ratio: 4:1
If k = 1, then the dilation is congruent to the original figure.
A0 (6, 12), B0 (−9, 21),C0 (−3, −6)
A0 (9, 6), B0 (−3, −12),C0 (0, −7.5)
15.
16. k = 2
17. A00 (4, 8), B00 (48, 16),C00 (40, 40)
18. k = 2
19.
√
a. √5
b. √
5
c. 3 √5
d. 2 √5
e. 4 5
20.
√
a. 5 √
5
b. 10 √5
c. 20 5
99
7.6. Geometry - Second Edition, Similarity Transformations, Review Answers
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21.
a. OA : OA0 = 1 : 2
AB : A0 B0 = 1 : 2
b. OA : OA00 = 1 : 4
AB : A00 B00 = 1 : 4
23.
24.
25.
26.
27.
22.
x=3
y = 2x + 1
(3, 7)
This point is the center of the dilation.
The scale factor is 3.
28.
0 O0
0 G0
0 D0
29. DDO
= 3, OOG
= 3 and GGD
= 3.
30. 3
31. To dilate the original figure by a scale factor of 4 make one additional tick mark with your compass.
32. To dilate the original figure by a scale factor of 12 construct the perpendicular bisectors of CG,CO and CD to
find the midpoints of the segments which will be your G0 , O0 and D0 respectively.
100
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Chapter 7. Similarity, Answer Key
7.7 Geometry - Second Edition, Extension:
Self-Similarity, Review Answers
1.
2. See the following table:
TABLE 7.2:
Stage 0
Stage 1
Stage 2
Stage 3
Stage 4
Stage 5
Number of Segments
Length of each Segment
1
2
4
8
16
32
1
Total Length of the Segments
1
1
3
1
9
1
27
1
81
1
243
2
3
4
9
8
27
16
81
32
243
3. There will be 2n segments.
4. The length of each segment will be
1
3n
units.
5.
1
6. Number of edges: 192 Edge length: 27
Perimeter:
192
27
7.
8. See the following table:
TABLE 7.3:
Color
No Color
Stage 0
0
1
Stage 1
1
8
Stage 2
9
64
Stage 3
73
512
101
7.7. Geometry - Second Edition, Extension: Self-Similarity, Review Answers
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9. Answers will vary. Many different flowers (roses) and vegetables (broccoli, cauliflower, and artichokes) are
examples of fractals in nature.
10. Answers will vary.
102
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Chapter 7. Similarity, Answer Key
7.8 Chapter Review Answers
1.
a. x = 12
b. x = 14.5
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
x = 10◦ ; 50◦ , 60◦ , 70◦
3.75 gallons
yes
no
yes, AA
yes, SSS
no
no
A0 (10.5, 3), B0 (6, 13.5),C0 (−1.5, 6)
x = 19
3
x=1
z=6
a = 5, b = 7.5
103
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C HAPTER
8
Right Triangle
Trigonometry, Answer Key
Chapter Outline
8.1
G EOMETRY - S ECOND E DITION , T HE P YTHAGOREAN T HEOREM , R EVIEW A N SWERS
8.2
G EOMETRY - S ECOND E DITION , C ONVERSE OF THE P YTHAGOREAN T HEO REM , R EVIEW A NSWERS
8.3
G EOMETRY - S ECOND E DITION , U SING S IMILAR R IGHT T RIANGLES , R EVIEW
A NSWERS
8.4
G EOMETRY - S ECOND E DITION , S PECIAL R IGHT T RIANGLES , R EVIEW A N SWERS
8.5
G EOMETRY - S ECOND E DITION , TANGENT, S INE AND C OSINE , R EVIEW A N SWERS
104
8.6
G EOMETRY - S ECOND E DITION , I NVERSE T RIGONOMETRIC R ATIOS , R EVIEW
A NSWERS
8.7
G EOMETRY - S ECOND E DITION , E XTENSION : L AWS OF S INES AND C OSINES ,
R EVIEW A NSWERS
8.8
C HAPTER R EVIEW A NSWERS
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Chapter 8. Right Triangle Trigonometry, Answer Key
8.1 Geometry - Second Edition, The
Pythagorean Theorem, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
√
505
√
9√ 5
799
12
10 √
10 14
26√
3p 41
2
2
√x + y
9 2
yes
no
no
yes
yes
no √
20 √39
14√ 429
17 287
√4
4√ 5
493
√
5 10
36.6 × 20.6
33.6
√ × 25.2
3 2
4 √s
16 3
a2 + 2ab + b2
c2 + 4 21 ab = c2 + 2ab
a2 + 2ab + b2 = c2 + 2ab, which simplifies to a2 + b2 = c2
1
1 2
2
2 (a +
b)(a 1+ b) = 2 (a1 + 2ab + b )
1
2 2 ab + 2 c = ab + 2 c
1 2
1
2
2
2
2
2
2
2
2 (a + 2ab + b ) = ab + 2 c ⇒ a + 2ab + b = 2ab + c , which simplifies to a + b = c
105
8.2. Geometry - Second Edition, Converse of the Pythagorean Theorem, Review Answers
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8.2 Geometry - Second Edition, Converse of
the Pythagorean Theorem, Review Answers
1.
a. c = 15
b. 12 < c < 5
c. 15 < c < 21
2.
a. a = 7
b. 7 < a < 24
c. 1 < c < 7
3. It is a right triangle because 8, 15, 17 is a Pythagorean triple. The “x” indicates that this set is a multiple of 8,
15, 17.
4. right
5. no
6. right
7. acute
8. right
9. obtuse
10. right
11. acute
12. acute
13. right
14. obtuse
15. obtuse
16. acute
17. obtuse
106
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Chapter 8. Right Triangle Trigonometry, Answer Key
18. One way is to use the distance formula to find the distances of all three sides and then use the converse of the
Pythagorean Theorem. The second way would be to find the slope of all three sides and determine if two sides
are perpendicular.
19. c = 13
√
20. d = 194
21. The sides of 4ABC are a multiple of 3, 4, 5 which is a right triangle. 6 A is opposite the largest side, which is
the hypotenuse, making it 90◦ .
22. See the following table:
TABLE 8.1:
Statement
1. In 4ABC, a2 + b2 < c2 , and c is the longest side. In
4LMN, 6 N is a right angle.
2. a2 + b2 = h2
3. c2 > h2
4. c > h
5. 6 C is the largest angle in 4ABC.
6. m6 N = 90◦
7. m6 C > m6 N
8. m6 C > 90◦
9. 6 C is an obtuse angle.
10. 4ABC is an obtuse triangle.
23.
24.
25.
26.
Reason
Given
Pythagorean Theorem
Transitive PoE
Take the square root of both sides
The largest angle is opposite the longest side.
Definition of a right angle
SSS Inequality Theorem
Transitive PoE
Definition of an obtuse angle.
Definition of an obtuse triangle.
right
obtuse
acute
(1, 5), (-2, -3)
27 and 28. answers vary, you can check your answer by plotting the points on graph paper and measuring with a
protractor or using the distance formula to verify the appropriate inequality.
29 and 30. While your diagram may be different because your angle at A may be different, the construction should
look something like this:
107
8.2. Geometry - Second Edition, Converse of the Pythagorean Theorem, Review Answers
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31. The sum of the angles in a triangle must be 180◦ , if 6 C is 90◦ , then both 6 A and 6 B are acute.
32. You could construct a line perpendicular to AB through 6 B (you will need to extend the segment beyond B to
do the construction). Next, select any point on this perpendicular segment and call it C. By connecting A and
C you will make 4ABC.
108
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Chapter 8. Right Triangle Trigonometry, Answer Key
8.3 Geometry - Second Edition, Using Similar
Right Triangles, Review Answers
4KML ∼
√4JML ∼ 4JKL
KM = 6√ 3
JK = 6 √7
KL√= 3 21
16 √2
15√ 7
2 √
35
14 √6
20√ 10
2 102√
x = 12√ 5
y = 5 √5
z=9 2
x = 4√
y = 465
√
z = 14 5 √
√
8 41
17. x = 32
5 , y = √ 5 , z = 2 41
18. x = 9,√
y = 3 34
19. x = 9 20481 , y = 81
40 , z = 40
20. See the following table:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
TABLE 8.2:
Statement
1. 4ABD with AC⊥DB and 6 DAB is a right angle.
2. 6 DCA and 6 ACB are right angles
3. 6 DAB ∼
= 6 DCA ∼
= 6 ACB
∼
6
6
4. D = D
5. 4CAD ∼
= 4ABD
∼
6
6
6. B = B
7. 4CBA ∼
= 4ABD
8. 4CAD ∼
= 4CBA
Reason
Given
Definition of perpendicular lines.
All right angles are congruent.
Reflexive PoC
AA Similarity Postulate
Reflexive PoC
AA Similarity Postulate
Transitive PoC
21. See the following table:
TABLE 8.3:
Statement
1. 4ABD with AC⊥DB and 6 DAB is a right angle.
2. 4ABD ∼ 4CBA ∼ 4CAD
AB
3. BC
AB = DB
Reason
Given
Theorem 8-5
Corresponding sides of similar triangles are proportional.
109
8.3. Geometry - Second Edition, Using Similar Right Triangles, Review Answers
22.
23.
24.
25.
26.
27.
28.
29.
30.
6.1%
10.4%
9.4%
ratios are 31 and 93 , which both reduce to the common ratio 3. Yes, this is true for the next pair of terms since
27
9 also reduces to 3.
geometric mean; geometric mean
10
20
1
See the following table:
TABLE 8.4:
Statement
a
b
1. ae = d+e
and db = d+e
2. a2 = e(d + e) and b2 = d(d + e)
3. a2 + b2 = e(d + e) + d(d + e)
4. a2 + b2 = (e + d)(d + e)
5. c = d + e
6. a2 + b2 = c2
110
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Reason
Theorem 8-7
Cross-Multiplication Property
Combine equations from #2.
Distributive Property
Segment Addition Postulate
Substitution PoE
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Chapter 8. Right Triangle Trigonometry, Answer Key
8.4 Geometry - Second Edition, Special Right
Triangles, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
√
x √2
x √
3, 2x
15 √2
11 2
8 √
90 2√or 127.3 ft.
a = 2 √ 2, b = 2
c = 6 2, d √
= 12
e = f =√13 2
g = 10 3, h√= 20
k = 8, j√= 8 3 √
x = 11 3, y = 22 3
m = 9, n√= 18
√
q = 14 6, p√= 28 3
s = 9,t = 3√ 3
x = w√
=9 2 √
a = 9 √3, b = 18 √3
p = 6 15, q = 6 5
Yes, it’s a 30-60-90 triangle.
No, it is√not even a right triangle.
16 + 6√ 3
8 + 8√ 3
x√
:x 3
4 √2 in
3
3 in
2 √
25
2
4 √3 f t
27
2
3 in
2
12
3960
√ ft
s
3
2
111
8.5. Geometry - Second Edition, Tangent, Sine and Cosine, Review Answers
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8.5 Geometry - Second Edition, Tangent, Sine
and Cosine, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
112
d
f
f
e
f
d
d
e
d
e
f
e
D, D
equal, complement
reciprocals
0.4067
0.7071
28.6363
0.6820
sin A = 45 , cos A = 53 , tan A = 43
√
√
2
sin A = 2 , cos A = 2 2 , tan A = 1
√
√
sin A = 31 , cos A = 2 3 2 , tan A = 4 2
x = 9.37, y = 12.72
x = 14.12, y = 19.42
x = 20.84, y = 22.32
x = 19.32, y = 5.18
x = 5.85, y = 12.46
x = 20.89, y = 13.43
x = 435.86 f t.
x = 56 m
25.3 ft
42.9 ft
94.6 ft
49 ft
14 miles
The hypotenuse is the longest side in a right triangle. Since the sine and cosine ratios are each a leg divided
by the hypotenuse, the denominator is always going to be greater than the numerator. This ensures a ratio that
is less than 1.
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Chapter 8. Right Triangle Trigonometry, Answer Key
8.6 Geometry - Second Edition, Inverse
Trigonometric Ratios, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.7◦
31.0◦
44.7◦
39.4◦
46.6◦
36.9◦
34.6◦
82.9◦
70.2◦
m6 A = √
38◦ , BC = 9.38, AC = 15.23
AB = 4√ 10, m6 A = 18.4◦ , m6 B = 71.6◦
◦
BC = 51, m6 A = 45.6, m6 C =
√ 44.4
◦
m6 A = 60
√ , BC = 12, AC = 12 3
CB = 7 5, m6 A = 48.2◦ , m6 B = 41.8◦
m6 B = 50◦ , AC = 38.14, AB = 49.78
You would use a trig ratio when given a side and an angle and the Pythagorean Theorem if you are given two
sides and no angles.
47.6◦
1.6◦
44.0◦
192
◦
11 f t ≈ 17 f t 5 in; 54
51◦
For problem 20: since the earth tilts on its axis, the position of the sun in the sky varies throughout the year
for most places on earth. Thus, the angle at which the sun hits a particular object will vary at different times
of the year. For problem 21: the water pressure in the hose will affect the path of the water, the more pressure,
the longer the water will travel in a straight path before gravity causes the path of the water to arc and come
back down towards the ground.
Tommy used OA instead of OA for his tangent ratio.
Tommy used the correct ratio in his equation here, but he used the incorrect angle measure he found previously
which caused his answer to be incorrect. This illustrates the benefit of using given information whenever
possible.
Tommy could have used Pythagorean Theorem to find the hypotenuse instead of a trigonometric ratio.
cos 50◦
sin 20◦
As the angle measures increase, the sine value increases.
As the angle measures increase, the cosine value decreases.
The sine and cosine values are between 0 and 1.
tan 85◦ = 11.43, tan 89◦ = 57.29, and tan 89.5◦ = 114.59. As the tangent values get closer to 90◦ , they get
larger and larger. There is no maximum, the values approach infinity.
The sine and cosine ratios will always be less than one because the denominator of the ratios is the hypotenuse
which is always longer than either leg. Thus, the numerator is always less than the denominator in these ratios
resulting in a value less than one.
113
8.7. Geometry - Second Edition, Extension: Laws of Sines and Cosines, Review Answers
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8.7 Geometry - Second Edition, Extension:
Laws of Sines and Cosines, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
114
m6 B = 84◦ , a = 10.9, b = 13.4
m6 B = 47◦ , a = 16.4, c = 11.8
m6 A = 38.8◦ , m6 C = 39.2◦ , c = 16.2
b = 8.5, m6 A = 96.1◦ , m6 C = 55.9◦
m6 A = 25.7◦ , m6 B = 36.6, m6 C = 117.7◦
m6 A = 81◦ , m6 B = 55.4◦ , m6 C = 43.6◦
b = 11.8, m6 A = 42◦ , m6 C = 57◦
b = 8.0, m6 B = 25.2◦ , m6 C = 39.8◦
m6 A = 33.6◦ , m6 B = 50.7◦ , m6 C = 95.7◦
m6 C = 95◦ , AC = 3.2, AB = 16.6
BC = 33.7, m6 C = 39.3◦ , m6 B = 76.7◦
m6 A = 42◦ , BC = 34.9, AC = 22.0
m6 B = 105◦ , m6 C = 55◦ , AC = 14.1
m6 B = 35◦ , AB = 12, BC = 5
Yes, BC would still be 5 units (see isosceles triangle below); the measures of 6 C are supplementary as shown
below.
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Chapter 8. Right Triangle Trigonometry, Answer Key
8.8 Chapter Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
◦
6
BC = 4.4,
√ AC = 10.0, m A = 26
AB = 5 √10, m6 A = 18.4◦ , m6 B = 71.6◦
◦
◦
6
BC = 6 7, m6 A = 41.4
√ , m C = 48.6
◦
m6 A = 30
√ , AC = 25 3, BC = 25
BC = 7 13, m6 A = 31◦ , m6 B √
= 59◦
◦
m6 B = 45 , AC = 32, AB = 32 2
m6 B = 63◦ , BC = 19.1, AB = 8.7
m6 C = √
19◦ , AC = 22.7, AB = 7.8
BC = 4 13, m6 B = 33.7◦ , m6 C = 56.3◦
acute
right, Pythagorean triple
obtuse
right
acute
obtuse
x = 2√
x = 2 √110
x=6 7
2576.5 ft.
x = 29.2◦
AC = 16.1, m6 A = 41.6◦ , m6 C = 63.4◦
m6 A = 123.7◦ , m6 B = 26.3◦ , m6 C = 30◦
115
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C HAPTER
9
Circles, Answer Key
Chapter Outline
116
9.1
G EOMETRY - S ECOND E DITION , PARTS OF C IRCLES AND TANGENT L INES ,
R EVIEW A NSWERS
9.2
G EOMETRY - S ECOND E DITION , P ROPERTIES OF A RCS , R EVIEW A NSWERS
9.3
G EOMETRY - S ECOND E DITION , P ROPERTIES OF C HORDS , R EVIEW A NSWERS
9.4
G EOMETRY - S ECOND E DITION , I NSCRIBED A NGLES , R EVIEW A NSWERS
9.5
G EOMETRY - S ECOND E DITION , A NGLES OF C HORDS , S ECANTS , AND TAN GENTS , R EVIEW A NSWERS
9.6
G EOMETRY - S ECOND E DITION , S EGMENTS OF C HORDS , S ECANTS , AND TAN GENTS , R EVIEW A NSWERS
9.7
G EOMETRY - S ECOND E DITION , E XTENSION : W RITING AND G RAPHING THE
E QUATIONS OF C IRCLES , R EVIEW A NSWERS
9.8
C HAPTER R EVIEW A NSWERS
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Chapter 9. Circles, Answer Key
9.1 Geometry - Second Edition, Parts of Circles
and Tangent Lines, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
diameter
secant
chord
point of tangency
common external tangent
common internal tangent
center
radius
the diameter
4 lines
11. 3 lines
12. none
13.
14.
15.
16.
17.
18.
19.
J
J
J
radius of B = 4, radius of C = 5, radius of D = 2, radius of
J ∼J
D = E because they have the same radius length.
2 common tangents
CE = 7
y = x−2
yes
no
J
E =2
117
9.1. Geometry - Second Edition, Parts of Circles and Tangent Lines, Review Answers
20.
21.
22.
23.
24.
25.
26.
27.
www.ck12.org
yes
√
4 √10
4 11
x=9
x=3
x = 5√
x=8 2
a.
b.
c.
d.
Yes, by AA. m6 CAE = m6 DBE = 90◦ and 6 AEC ∼
= 6 BED by vertical angles.
BC = 37
AD = 35
m6 C = 53.1◦
28. See the following table:
TABLE 9.1:
Statement
1. AB and CB with points of tangency at A and C. AD
and DC are radii.
2. AD ∼
= DC
3. DA⊥AB and DC⊥CB
4. m6 BAD = 90◦ and m6 BCD = 90◦
5. Draw BD.
6. 4ADB and 4DCB are right triangles
7. DB ∼
= DB
8. 4ABD ∼
= 4CBD
9. AB ∼
= CB
Reason
Given
All radii are congruent.
Tangent to a Circle Theorem
Definition of perpendicular lines
Connecting two existing points
Definition of right triangles (Step 4)
Reflexive PoC
HL
CPCTC
29.
a. kite
b. center, bisects
30. AT ∼
= BT ∼
= CT ∼
= DT by theorem 10-2 and the transitive property.
31. 9.23
8
32. √8 ; √
3 3 3
←
→
←→
33. Since AW and W B both share point W and are perpendicular to VW because a tangent is perpendicular to the
radius of the circle. Therefore A, B and W are collinear. V T ∼
= VW because they are tangent segments to circle
A from the same point, V , outside the circle. Similarly, VW ∼
= VU because they are tangent segments to circle
B from V . By the transitive property of congruence, V T ∼
= VU. Therefore, all three segments are congruent.
118
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Chapter 9. Circles, Answer Key
9.2 Geometry - Second Edition, Properties of
Arcs, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
minor
major
semicircle
major
minor
semicircle
c ∼
c
yes, CD
= DE
◦
66
228◦
yes, they are in the same circle with equal central angles
yes, the central angles are vertical angles, so they are equal, making the arcs equal
no, we don’t know the measure of the corresponding central angles.
90◦
49◦
82◦
16◦
188◦
172◦
196◦
270◦
x = 54◦
x = 47◦
x = 25◦
J ∼J
A= B
◦
62
77◦
139◦
118◦
257◦
319◦
75◦
105◦
68◦
105◦
255◦
217◦
119
9.3. Geometry - Second Edition, Properties of Chords, Review Answers
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9.3 Geometry - Second Edition, Properties of
Chords, Review Answers
1. No, see picture. The two chords can be congruent and perpendicular, but will not bisect each other.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
120
AC
c
DF
c
JF
DE
6 HGC
6 AGC
AG, HG,CG, FG, JG, DG
107◦
8◦
118◦
133◦
140◦
120◦
x = 64◦ , y = 4
x = 8,√
y = 10
x = 3 √26, y ≈ 12.3
x=9 5
x = 9, y = 4
x = 4.5
x=3
x = 7√
x = 4 11
c = 121.3◦
mAB
c = 112.9◦
mAB
c∼
c by Theorem 10-5.
BF ∼
= FD and BF
= FD
∼
CA = AF by Theorem 10-6.
QS is a diameter by Theorem 10-4.
a-c shown in the diagram below; d. it is the center; e. shown in the diagram; this construction is not done to
scale and your chords might be in different places but this should give you an idea of what it should look like.
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Chapter 9. Circles, Answer Key
30. for AB:
a.
b.
c.
d.
(1, 5)
m = 0, ⊥m is undefined
x=1
for BC:
a. 92 , 23
b. m = 7, ⊥m = − 71
c. y = − 17 x + 15
7
e. Point of intersection (center of the circle) is (1, 2).
f. radius is 5 units
31.
a. 120◦
b. 60◦
121
9.4. Geometry - Second Edition, Inscribed Angles, Review Answers
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9.4 Geometry - Second Edition, Inscribed Angles, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
48◦
120◦
54◦
45◦
87◦
27◦
100.5◦
95.5◦
76.5◦
84.5◦
51◦
46◦
x = 180◦ , y = 21◦
x = 60◦ , y = 49◦
x = 30◦ , y = 60◦
x = 72◦ , y = 92◦
x = 200◦ , y = 100◦
x = 68◦ , y = 99◦
x = 93◦ , y = 97◦
x = 10◦
x = 24◦
x = 74◦ , y = 106◦
x = 35◦ , y = 35◦
55◦
70◦
110◦
90◦
20◦
90◦
TABLE 9.2:
Statement
1. Inscribed 6 ABC and diameter BD
m6 ABE = x◦ and m6 CBE = y◦
2. x◦ + y◦ = m6 ABC
3. AE ∼
= EB and EB ∼
= EC
4. 4AEB and 4EBC are isosceles
5. m6 EAB = x◦ and m6 ECB = y◦
6. m6 AED = 2x◦ and m6 CED = 2y◦
c = 2x◦ and mDC
c = 2y◦
7. mAD
c + mDC
c = mAC
c
8. mAD
c = 2x◦ + 2y◦
9. mAC
c = 2(x◦ + y◦ )
10. mAC
c = 2m6 ABC
11. mAC
122
Reason
Given
Angle Addition Postulate
All radii are congruent
Definition of an isosceles triangle
Isosceles Triangle Theorem
Exterior Angle Theorem
The measure of an arc is the same as its central angle.
Arc Addition Postulate
Substitution
Distributive PoE
Subsitution
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Chapter 9. Circles, Answer Key
TABLE 9.2: (continued)
Statement
c
12. m6 ABC = 12 mAC
Reason
Division PoE
TABLE 9.3:
Statement
c
1. 6 ACB and 6 ADB intercept AB
c
2. m6 ACB = 12 mAB
1
c
m 6 ADB = 2 mAB
6
6
3. m ACB = m ADB
4. 6 ACB ∼
= 6 ADB
Reason
1. Given
2. Inscribed Angle Theorem
3. Transitive Property
4. Definition of Congruence
←
→
32. Since AC || OD, m6 CAB = m6 DOB by Corresponding Angles Postulate.
c so mDB
c This makes D the midpoint of CB.
c
c and m6 CAB = 1 mCB,
c = 1 mCB.
Also, m6 DOB = mDB
2
2
123
9.5. Geometry - Second Edition, Angles of Chords, Secants, and Tangents, Review Answers
1.
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9.5 Geometry - Second Edition, Angles of
Chords, Secants, and Tangents, Review
Answers
a.
b.
c.
2. No, by definition a tangent line cannot pass through a circle, so it can never intersect with any line inside of
one.
3.
a.
b.
4.
5.
6.
7.
8.
9.
10.
11.
124
center, equal
inside, intercepted
on, half
outside, half
x = 103◦
x = 25◦
x = 100◦
x = 44◦
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12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
Chapter 9. Circles, Answer Key
x = 38◦
x = 54.5◦
x = 63◦ , y = 243◦
x = 216◦
x = 42◦
x = 150◦
x = 66◦
x = 113◦
x = 60, y = 40◦ , z = 80◦
x = 60◦ , y = 25◦
x = 35◦ , y = 55◦
x = 75◦
x = 45◦
x = 35◦ , y = 35◦
x = 60◦
x = 47◦ , y = 78◦
x = 84◦ , y = 156◦
x = 10◦
x = 3◦
See the following table:
TABLE 9.4:
Statement
1. Intersecting chords AC and BD.
2. Draw BC
3.
4.
5.
6.
m6
m6
m6
m6
c
DBC = 12 mDC
1 c
ACB = 2 mAB
a = m6 DBC + m6 ACB
c + 1 mAB
c
a = 1 mDC
2
2
Reason
Given
Construction
Inscribed Angle Theorem
Inscribed Angle Theorem
Exterior Angle Theorem
Substitution
32. See the following table:
TABLE 9.5:
Statement
−
→
−
→
1. Intersecting secants AB and AC.
2. Draw BE.
Reason
Given
Construction
c
3. m6 BEC = 12 mBC
Inscribed Angle Theorem
125
9.5. Geometry - Second Edition, Angles of Chords, Secants, and Tangents, Review Answers
TABLE 9.5: (continued)
Statement
c
4. m6 DBE = 12 mDE
6
6
5. m a + m DBE = m6 BEC
6. m6 a = m6 BEC − m6 DBE
c 1 c
7. m6 a = 12 m
BC − 2 mDE c − mDE
c
8. m6 a = 1 mBC
2
126
Reason
Inscribed Angle Theorem
Exterior Angle Theorem
Subtraction PoE
Substitution
Distributive Property
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Chapter 9. Circles, Answer Key
9.6 Geometry - Second Edition, Segments of
Chords, Secants, and Tangents, Review
Answers
1. x = 12
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
x = 1.5
x = 12
x = 7.5
√
x=6 2
x = 10
x = 10
x=8
x=9
x = 22.4
x = 11
x = 20
x = 120
7√ ≈ 17.14
x = 4 66
x = 6√
x= √
231
x = 4 42
x = 10
The error is in the set up. It should be 10 · 10 = y · (15 + y). The correct answer is y = 5.
10 inches
x=7
x=5
x=3
x=3
x=8
x=6
x=2
x=8
x=2
x = 12, y = 3
127
9.7. Geometry - Second Edition, Extension: Writing and Graphing the Equations of Circles, Reviewwww.ck12.org
Answers
9.7 Geometry - Second Edition, Extension:
Writing and Graphing the Equations of Circles, Review Answers
1. center: (-5, 3), radius = 4
2.
3.
4.
5.
6.
7.
8.
9.
center: (0, -8), radius = 2 √
center: (7, 10), radius = 2√ 5
center: (-2, 0), radius = 2 2
(x − 4)2 + (y + 2)2 = 16
(x + 1)2 + (y − 2)2 = 7
(x − 2)2 + (y − 2)2 = 4
(x + 4)2 + (y + 3)2 = 25
a. yes
b. no
c. yes
(x − 2)2 + (y − 3)2 = 52
(x − 10)2 + y2 = 29
(x + 3)2 + (y − 8)2 = 200
(x − 6)2 + (y + 6)2 = 325
7
37
a-d. ⊥ bisector of AB is y = − 24
x + 24
, ⊥ bisector of BC is y = x + 8 (e) center of circle (-5, 3) (f) radius 25
2
2
(g) (x + 5) + (y − 3) = 625
15. (x − 2)2 + (y − 2)2 = 25
16. (x + 3)2 + (y − 1)2 = 289
10.
11.
12.
13.
14.
128
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Chapter 9. Circles, Answer Key
9.8 Chapter Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
I
A
D
G
C
B
H
E
J
F
129
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C HAPTER
10Perimeter and Area, Answer
Key
Chapter Outline
10.1
G EOMETRY - S ECOND E DITION , T RIANGLES AND PARALLELOGRAMS , R EVIEW
A NSWERS
10.2
G EOMETRY - S ECOND E DITION , T RAPEZOIDS , R HOMBI , AND K ITES , R EVIEW
A NSWERS
10.3
G EOMETRY - S ECOND E DITION , A REAS OF S IMILAR P OLYGONS , R EVIEW A N SWERS
130
10.4
G EOMETRY - S ECOND E DITION , C IRCUMFERENCE AND A RC L ENGTH , R EVIEW
A NSWERS
10.5
G EOMETRY - S ECOND E DITION , A REAS OF C IRCLES AND S ECTORS , R EVIEW
A NSWERS
10.6
G EOMETRY - S ECOND E DITION , A REA AND P ERIMETER OF R EGULAR P OLY GONS , R EVIEW A NSWERS
10.7
C HAPTER R EVIEW A NSWERS
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Chapter 10. Perimeter and Area, Answer Key
10.1 Geometry - Second Edition, Triangles and
Parallelograms, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
A = 144 in2 , P = 48 in
A = 144 cm2 , P = 50 cm
A = 360 m2
A = 112 u2 , P = 44 u
A = 324 f t 2 , P = 72 f t
P = 36 f t
A = 36 in2
A = 210 cm2
6m
Possible answers: 10 × 6, 12 × 4
Possible answers: 9 × 10, 3 × 30
If the areas are congruent, then the figures are congruent. We know this statement is false, #11 would be a
counterexample.
√
8 2 cm
P ≈ 54.9
√cm
A = 96 2 ≈ 135.8 cm2
15 in
P ≈ 74.3 in
A = 180 in2
315 units2
90 units2
14 units2
407.5 units2
560 units2
30 units2
814 units2
72 units2
72 units2
24 acres
6×4
12 × 24
√
√
3
h = 3 √3, A = 9 √
h = 5 √3, A = 25 √ 3
2
h = 2x 3, A = x4 3
x = 20 f t, y = 60 f t
Perimeter is 16 units, Area is 15 square units
131
10.2. Geometry - Second Edition, Trapezoids, Rhombi, and Kites, Review Answers
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10.2 Geometry - Second Edition, Trapezoids,
Rhombi, and Kites, Review Answers
1. If a kite and a rhombus have the same diagonal lengths the areas will be the same. This is because both
formulas are dependent upon the diagonals. If they are the same, the areas will be the same too. This does not
mean the two shapes are congruent, however.
2. h(b1 ) + 24s
1
h(b1 ) + 2 12 · h · b2 −b
2
hb1 + h(b22−b1 )
2hb1 +hb2 −hb1
2
hb1 +hb2
= h2 (b1 + b2 )
2
3. 44s
4 · 21 12 d1 · 12 d2
4
8 d1 · d2
1
2 d1 d2
4. 24s + 24s 2 12 · 12 d1 · x + 2 12 · 12 d1 (d2 − x)
1
1
1
2 d1 · x + 2 d1 d2 − 2 d1 x
1
2 d1 d2
5. 160 units2
6. 315 units2
7. 96 units2
2
8. 77 units
√
9. 100 3 units2
10. 84 units2
11. 1000 units2
12. 63 units2
13. 62.5 units2
14. A = 480 units2
P = 104 units√ 15. A = 36 1 + 3 units2
√ P = 12 2 + 2 units
2
16. A = 108 units√
P = 12 3 + 2 units
√ √ 17. A = 5 3 5 + 77 units2
P = 52 units
√
18. A = 396 3 units2
P = 116 √
units
19. A = 256 5 units2
P = 96 units
20. A = 12 units2
21. 24 units2
22. Any two numbers with a product of 64 would work.
23. Any two numbers with a product of 108 would work.
24. 90 units2
25. kite, 24 units2
132
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26.
27.
28.
29.
30.
31.
Chapter 10. Perimeter and Area, Answer Key
2
2
Trapezoid, 47.5
√ units units
rhombus, 12 5 units2
8, 14
9, 12
192 units2
a. 200 f t 2
b. 400 f t 2
c. 21
32.
a. 300 f t 2
b. 900 f t 2
c. 31
133
10.3. Geometry - Second Edition, Areas of Similar Polygons, Review Answers
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10.3 Geometry - Second Edition, Areas of Similar Polygons, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
134
9
25
1
16
49
4
36
121
1
6
2
9
7
3
5
12
1
4
1
2
5 units2
24 units
100 cm
468.75 cm2
96 units2
198 f t 2
54 in
32 units
4
9
2
3
√
√
Diagonals are 12 and 16. The length of the sides are 12 2 and 16 2.
Because
the diagonals of these rhombi are congruent, the rhombi are actually squares.
√
25 2
2.34 inches
1
Scale: 192
, length of model 5.44 inches
27.5 by 20 cm, yes because the drawing is 10.8 by 7.87 inches
9 by 6 inches
10 by 14 inches
Baby Bella $0.05, Mama Mia $0.046, Big Daddy $0.046, the Mama Mia or Big Daddy are the best deals.
1.5 bottles, so she’ll need to buy 2 bottles.
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Chapter 10. Perimeter and Area, Answer Key
10.4 Geometry - Second Edition, Circumference
and Arc Length, Review Answers
TABLE 10.1:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
diameter
15
8
6
84
18
25
2
36
radius
7.5
4
3
42
9
12.5
1
18
Circumference
15π
8π
6π
84π
18π
25π
2π
36π
r = 44
π in
C = 20 cm
16
The√diameter is the same length as the diagonals of the square.
32 2
16π
9π
80π
15π
r = 108
r = 30
r = 72
120◦
162◦
15◦
40π ≈ 125.7 in.
a. 26π ≈ 81.7 in
b. 775 complete rotations
26. The Little Cheese, 3.59:1; The Big Cheese, 3.49:1; The Cheese Monster, 3.14:1; Michael should buy The
Little Cheese
27. 31 gumdrops
28. 18 in
29. 93 in
30. 30 ft
135
10.5. Geometry - Second Edition, Areas of Circles and Sectors, Review Answers
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10.5 Geometry - Second Edition, Areas of Circles and Sectors, Review Answers
TABLE 10.2:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
136
radius
2
4
5
12
9√
3 10
17.5
Area
4π
16π
25π
144π
81π
90π
306.25π
7
π
30
π
√6
49
π
900
π
π
36
54π
1.0416π
189π √
2.6π − 4 3
33π
20.25π
√ − 40.5
8 3
2
15
120◦
10◦
198◦
123.61
292.25
1033.58
13.73
21.21
54.4
Square ≈ 10, 000 f t 2 ; Circle ≈ 12, 732 f t 2 ; the circle has more area.
18 units
40◦
circumference
4π
8π
10π
24π
18π
√
6 10π
35π
14
60 √
12 π
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Chapter 10. Perimeter and Area, Answer Key
10.6 Geometry - Second Edition, Area and
Perimeter of Regular Polygons, Review
Answers
1. radius
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
apothem
6
equilateral
10√cm
5 3 cm
60 cm
√
150 3 √
A = 384 3
P = 96√
A=8 2
P = 6.12
A = 68.26
A = 72
A = 688.19
P = 100
A = 73.47
P = 15.45
A = 68.26
P = 63
6.5
12
a = 11.01
a = 14.49
93.86, 94.15
30π ≈ 94.25
The perimeter of the 40-gon is closer to the circumference because it is closer in shape to the circle. The more
sides a polygon has, the closer it is to a circle.
695.29, 703.96
225π ≈ 706.86
The area of the 40-gon is closer to the area of the circle because it is closer in shape to the circle than the
20-gon.
Start with 21 asn. n = 6, so all the internal triangles are equilateral triangles with sides s.Therefore the apothem
√
√
3
3 s (s)(6). Reducing this
1
is
s from the 30-60-90 ratio. Plugging this in for n and a, we have A =
2
2
√
we end up with A = 3 2 3 s2 .
2
26.
a.
b.
c.
d.
e.
◦
◦
sin x2 = 2rs ; cos x2 = ar
s = 2r sin 2x
◦
a = r cos x2
1
x◦
x◦
2r
sin
r
cos
= r2 sin
2
2
2
x◦
x◦
2
nr sin 2 cos 2
x◦
2
cos
x◦
2
27. 421.21 cm2
28. 77.25 in2
137
10.6. Geometry - Second Edition, Area and Perimeter of Regular Polygons, Review Answers
www.ck12.org
29. 195.23 cm2
30. 153.44 in2
31. polygon with 30 sides: 254.30 in2 ; circle 254.47 in2 ; They are very close, the more sides a regular polygon
has the closer to a circle it becomes.
2
x
s
s
32. First, take s = 2r sin 2 and solve for r to get = 2 sin x◦ . Next, replace r in the formula to get n 2 sin x◦
sin
(2)
(2)
◦
ns2 cos( x )
We can reduce this to 4 sin x◦2 .
(2)
2
33. 16055.49 cm
34. 4478.46 in2
138
x◦
2
cos
x◦
2
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Chapter 10. Perimeter and Area, Answer Key
10.7 Chapter Review Answers
1. A = 225
P = 60
2. A = 198
P = 58
3. A = 124.71
P = 48
4. A = 139.36
P = 45
5. A = 3000
P = 232
6. A = 403.06
P = 72
7. 72
8. 154 √
9. 162 3
10. C = 34π
A = 289π
11. C = 30π
A = 225π
12. 54 units2
13. 1070.12
14. 1220.39
15. 70.06
139
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C HAPTER
11
Surface Area and Volume,
Answer Key
Chapter Outline
11.1
G EOMETRY - S ECOND E DITION , E XPLORING S OLIDS , R EVIEW A NSWERS
11.2
G EOMETRY - S ECOND E DITION , S URFACE A REA OF P RISMS AND C YLINDERS ,
R EVIEW A NSWERS
11.3
G EOMETRY - S ECOND E DITION , S URFACE A REA OF P YRAMIDS AND C ONES ,
R EVIEW A NSWERS
11.4
G EOMETRY - S ECOND E DITION , VOLUME OF P RISMS AND C YLINDERS , R E VIEW A NSWERS
11.5
G EOMETRY - S ECOND E DITION , VOLUME OF P YRAMIDS AND C ONES , R EVIEW
A NSWERS
11.6
G EOMETRY - S ECOND E DITION , S URFACE A REA AND VOLUME OF S PHERES ,
R EVIEW A NSWERS
11.7
G EOMETRY - S ECOND E DITION , E XPLORING S IMILAR S OLIDS , R EVIEW A N SWERS
11.8
140
C HAPTER R EVIEW A NSWERS
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Chapter 11. Surface Area and Volume, Answer Key
11.1 Geometry - Second Edition, Exploring
Solids, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
V =8
F =9
E = 30
F =6
E =6
V =6
F =9
V =6
Yes, hexagonal pyramid. F = 7,V = 7, E = 12
No, a cone has a curved face.
Yes, hexagonal prism. F = 8,V = 12, E = 18
No a hemisphere has a face.
Yes, trapezoidal prism. F = 6,V = 8, E = 12
Yes, concave decagonal prism. F = 10,V = 16, E = 24
Rectangle
Circle
Trapezoid
18.
19.
20.
21.
22.
23.
24.
Regular Icosahedron
Decagonal Pyramid
Trapezoidal Prism
All 11 nets
141
11.1. Geometry - Second Edition, Exploring Solids, Review Answers
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25. The truncated icosahedron has 60 vertices, by Euler’s Theorem.
F +V = E + 2
32 +V = 90 + 2
V = 60
26. regular tetrahedron
27. Use the construction directions from problem 26 to make an equilateral triangle with midsegments. Using
one of the midpoints of the equilateral triangle as a vertex, construct another adjacent equilateral triangle with
midsegments. Your result should look like the picture below.
28. regular dodecahedron, 31
29. 19
30. 1 red face, 8 yellow faces, 7 blue faces and 4 green faces
142
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Chapter 11. Surface Area and Volume, Answer Key
11.2 Geometry - Second Edition, Surface Area
of Prisms and Cylinders, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
9 f t.2
10, 000 cm2
triangles, A = 6
The rectangles are 3 × 6, 4 × 6, and 6 × 5. Their areas are 18, 24, and 30.
72
84
Lateral surface area is the area of all the sides, total surface area includes the bases.
rectangle, 2πrh
a. 96 in2
b. 192 in2
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
350π cm2
1606.4
390.2
486π
182
34π
2808
x=8
x = 40
x = 25
60π in2
4100π cm2
The height could be 1, 3, 5, or 15.
4060 f t 2
2940 f t 2
5320 f t 2
22 gal
$341
5 in by 4π + 1 in, 20π + 5 in2 ≈ 67.83 in2
x2 − 16 in2 , x = 25 in
5 2
2 x π, x = 8
143
11.3. Geometry - Second Edition, Surface Area of Pyramids and Cones, Review Answers
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11.3 Geometry - Second Edition, Surface Area
of Pyramids and Cones, Review Answers
1.
2.
3.
4.
5.
6.
vertex
y
lateral edge
w
z
t
7.
√
8. 5 10 cm
9. 15 in
10. To find the slant√
height, we need to find the distance from the center of the edge of the equilateral triangle.
This distance is 3.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
144
This is a picture of the base. The slant height is 62 +
671
135
64
1413.72
360
422.35
1847.26
896
1507.96
3, the
√ lateral faces
36√ 3
s2 3
576; 321.53
1159.25
1152.23
1473.76
100.8◦
7
√ 2
√
3 = l 2 → l = 39
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29.
30.
31.
32.
33.
Chapter 11. Surface Area and Volume, Answer Key
24
175π
10 in
13 in
360 in2
145
11.4. Geometry - Second Edition, Volume of Prisms and Cylinders, Review Answers
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11.4 Geometry - Second Edition, Volume of
Prisms and Cylinders, Review Answers
1. No, the volumes do not have to be the same. One cylinder could have a height of 8 and a radius of 4, while
another could have a height of 22 and a radius of 2. Both have a surface area of 96π, but the volumes are not
the same.
2. 960 cubes, yes this is the same as the volume.
3. 280 in3
4. 4π in3
5. 6 in
6. r = 9
7. 5
8. 36 units3
9.
a. 64 in3
b. 128 in3
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
146
882π cm3
3960
902.54
4580.44
147
50.27
7776
x=7
x = 24
x = 32
294π in3
24000π cm3
75π m3
330, 000 f t.3
165, 000 f t.3
495, 000 f t.3
36891.56 cm3
15901.92 cm3
r = 3 cm, h = 12 cm
11 cm
300.44 in3
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Chapter 11. Surface Area and Volume, Answer Key
11.5 Geometry - Second Edition, Volume of
Pyramids and Cones, Review Answers
Unless otherwise specified, all units are units3 .
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
9680
1280
778.71
3392.92
400
396.55
5277.88
128
1884.96
100.53
113.10
188.50
42
200
1066.67
√
9 √3
2 √
6
18 √
2
1 3
s
2
12
Find
√ the volume of one square pyramid then multiply it by 2.
72 √3
1 3
2
3s
h = 13.5 in
h = 3.6 cm
r = 3 cm
112 in3
190.87 cm3
471.24 cm3
h = 9 m, r = 6 m
15 ft
147
11.6. Geometry - Second Edition, Surface Area and Volume of Spheres, Review Answers
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11.6 Geometry - Second Edition, Surface Area
and Volume of Spheres, Review Answers
1. No, all the cross sections must be circles because there are no edges.
2. SA = 256π in2
3
V = 2048
3 π in
3. SA = 324π cm2
V = 972π cm3
4. SA = 1600π f t 2
3
V = 32000
3 π ft
2
5. SA = 16π m
2
V = 32
3πm
6. SA = 900π f t 2
V = 4500π f t 3
7. SA = 1024π in2
3
V = 16384
3 π in
8. SA = 676π cm2
3
V = 8788
3 π cm
9. SA = 2500π yd 2
3
V = 62500
3 π yd
10. r = 5.5 in
11. r = 33 m
12. V = 43 π f t 3
13. SA = 36π mi2
14. r = 4.31 cm
15. r = 7.5 f t.
16. 2025π cm2
17. 1900π units2
18. 4680 f t 2
19. 91.875π units2
20. 381703.51 cm3
21. 7120.94 units3
22. 191134.50 f t 3
23. 121.86 units3
350π
2
24. h = 20
3 cm, SA = 3 cm
25. 21.21 in3
26. 12π cm3 , 19 minutes
27.
a.
b.
c.
d.
28.
148
SA = 2πr2 + 2πrh
SA = 4πr2
SA = 4πr2
They are the same. Think back to the explanation for the formula for the surface area of a sphere using
the baseball-it is really the sum of the area of four circles. For the cylinder, the SA is the sum of the areas
of the two circular bases and the lateral area. The lateral area is 2πrh, when we replace h with r this part
of the formula becomes the area of two more circles. That makes the total surface area of the cylinder
equal to the area of four circles, just like the sphere.
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Chapter 11. Surface Area and Volume, Answer Key
a. 24429 in3
b. 732.87 lbs
c. 50 in
29. 25,132.74 miles
30. 201 million square miles
31. 268 billion cubic miles
149
11.7. Geometry - Second Edition, Exploring Similar Solids, Review Answers
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11.7 Geometry - Second Edition, Exploring Similar Solids, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
150
42
No, 14
10 6= 35
Yes, the scale factor is 4:3.
Yes, the scale factor is 3:5.
No, the top base is not in the same proportion as the rest of the given lengths.
Yes, cubes have the same length for each side. So, comparing two cubes, the scale factor is just the ratio of
the sides.
1:16
8:343
125:729
8:11
5:12
87.48π
The volume would be 43 or 64 times larger.
4:9
60 cm
91125 m3
2:3
4:9
y = 8, x√= h = 12√
w = 4 5, z = 6 5
Vs = 170.67,V
√ l = 576 √
LAs = 16 5, LAl = 36 5
Yes, just like the cubes spheres and hemispheres only have a radius to compare. So, all spheres and hemispheres are similar.
49:144, 343:1728
98π, 288π
The ratio of the lateral areas is 49:144, which is the same as the ratio of the total surface area.
9:25, about 2.78 times as strong
27:125
Animal A, Animal B’s weight is about 4.63 times the weight of animal A but his bones are only 2.78 times as
strong.
81 sq in
small $0.216, large $0.486
8:27
The larger can for $2.50 is a better deal. Using the cost of the canning material and the ratio of the volume of
beans, the “equivalent” cost of producing the larger can is $2.62. If we just use the volume of bean ratio (as a
consumer would) the cost should be $2.87. Both of these are higher than the $2.50 price.
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Chapter 11. Surface Area and Volume, Answer Key
11.8 Chapter Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
F
K
G
A
E
D
J
B
L
C
H
I
H
G
A
B
D
J
I
E
F
C
151
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C HAPTER
12
Rigid Transformations,
Answer Key
Chapter Outline
12.1
G EOMETRY - S ECOND E DITION , E XPLORING S YMMETRY, R EVIEW A NSWERS
12.2
G EOMETRY - S ECOND E DITION , T RANSLATIONS AND V ECTORS , R EVIEW A N SWERS
12.3
G EOMETRY - S ECOND E DITION , R EFLECTIONS , R EVIEW A NSWERS
12.4
G EOMETRY - S ECOND E DITION , R OTATIONS , R EVIEW A NSWERS
12.5
G EOMETRY - S ECOND E DITION , C OMPOSITION OF T RANSFORMATIONS , R E VIEW A NSWERS
12.6
G EOMETRY - S ECOND E DITION , E XTENSION : T ESSELLATIONS , R EVIEW A N SWERS
12.7
152
C HAPTER R EVIEW A NSWERS
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Chapter 12. Rigid Transformations, Answer Key
12.1 Geometry - Second Edition, Exploring
Symmetry, Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
sometimes
always
always
never
sometimes
never
never
always
always
sometimes
a kite that is not a rhombus
a circle
an isosceles trapezoid
n
15.
16.
17.
18. none
20.
21.
22.
23.
24.
25.
19.
H is the only one with rotational symmetry, 180◦ .
line symmetry
rotational symmetry
line symmetry
line symmetry (horizontal)
rotational symmetry
153
12.1. Geometry - Second Edition, Exploring Symmetry, Review Answers
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
154
2 lines
6 lines
4 lines
180◦
60◦ , 120◦ , 180◦ , 240◦ , 300◦
90◦ , 180◦ , 270◦
none
120◦ , 240◦
40◦ , 80◦ , 120◦ , 160◦ , 200◦ , 240◦ , 280◦ , 320◦
8 lines of symmetry; angles of rotation: 45◦ , 90◦ , 135◦ , 180◦ , 225◦ , 270◦ , and 315◦
3 line of symmetry; angles of rotation: 120◦ , 240◦
1 line of symmetry; no rotational symmetry
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Chapter 12. Rigid Transformations, Answer Key
12.2 Geometry - Second Edition, Translations
and Vectors , Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
A vector has direction and size, a ray is part of a line, so it has direction, but no size.
A0 (−1, −6)
B0 (9, −1)
C(0, 6)
A00 (4, −15)
D(7, 16)
A000 (9, −24)
All four points are collinear.
A0 (−8, −14), B0 (−5, −17),C0 (−7, −5)
A0 (5, −3), B0 (8, −6),C0 (6, 6)
A0 (−6, −10), B0 (−3, −13),C0 (−5, −1)
A0 (−11, 1), B0 (−8, −2),C0 (−10, 10)
(x, y) → (x − 6, y + 2)
(x, y) → (x + 9, y − 7)
(x, y) → (x − 3, y − 5)
(x, y) → (x + 8, y + 4)
√
√
√
Using the distance formula, AB = A0 B0 = 5, BC = B0C0 = 3 5, and AC = A0C0 = 5 2.
(x, y) → (x − 8, y − 4)
*
19. GH= h6, 3i
*
20. KJ= h−2, 4i
*
21. LM= h3, −1i
22.
23.
25.
26.
27.
28.
29.
30.
31.
32.
24.
D0 (9, −9), E 0 (12, 7), F 0 (10, 14)
Q0 (−9, −6),U 0 (−6, 0), A0 (1, −9), D0 (−2, −15)
h−3, 8i
h9, −12i
h0, −7i
(x, y) → (x − 7, y + 2)
(x, y) → (x + 11, y + 25)
(x, y) → (x + 15, y − 9)
155
12.3. Geometry - Second Edition, Reflections, Review Answers
www.ck12.org
12.3 Geometry - Second Edition, Reflections,
Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
d
p
(-3, 2), (-8, 4), (-6, 7), (-4, 7)
(-6, 4), (-2, 6), (-8, 8)
(2, 2), (8, 3), (6, -3)
(2, 6), (-6, 2), (4, -2)
(2, -2), (8, -6)
(2, -4), (-4, 2), (-2, -6)
(2, 3), (4, 8), (7, 6), (7, 4)
(4, 6), (6, 2), (8, 8)
(2, 4), (-4, 3), (-2, 9)
(-4, -14), (4, -10), (-6, -6)
(-2, -2), (-6, -8)
(-4, 2), (2, -4), (-6, -2)
y = −2
y−axis
y=x
18-20.
21. It is the same as a translation of 8 units down.
22-24.
156
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Chapter 12. Rigid Transformations, Answer Key
25. It is the same as a translation of 12 units to the left.
26-28.
29. A rotation of 180◦ .
30.
31. Perpendicular Bisector
32.
157
12.4. Geometry - Second Edition, Rotations, Review Answers
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12.4 Geometry - Second Edition, Rotations, Review Answers
1.
2.
4.
5.
6.
7.
8.
9.
10.
11.
158
3.
d
d, they are the same because the direction of the rotation does not matter.
270◦
90◦
Not rotating the figure at all; 0◦
(-6, -2)
(-6, -4)
(2, -2) and (6, 4)
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Chapter 12. Rigid Transformations, Answer Key
12.
13.
14.
15.
159
12.4. Geometry - Second Edition, Rotations, Review Answers
16.
17.
18.
19.
20.
21.
22.
23.
x=3
x = 4.5
x = 21
90◦
180◦
180◦
24-26.
27. A rotation of 180◦ .
28-30.
160
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Chapter 12. Rigid Transformations, Answer Key
31. Angle of rotation is double the angle between the lines.
161
12.5. Geometry - Second Edition, Composition of Transformations, Review Answers
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12.5 Geometry - Second Edition, Composition
of Transformations, Review Answers
1. Every isometry produces a congruent figure to the original. If you compose transformations, each image will
still be congruent to the original.
2. a translation
3. a rotation
4. (2, 2), (-2, -4), (0, -8), (4, -6)
5. (x, y) → (x + 6, −y)
6. (x, y) → (x − 6, −y)
7. No, because order does not matter.
8. (-2, -3), (-4, 2), (-9, -3)
9. (x, y) → (−x, y − 5)
10. (x, y) → (−x, y + 5)
11. (2, -10), (10, -6), (8, -4)
12. A translation of 12 units down.
13. (x, y) → (x, y + 12)
14. This image is 12 units above the original.
15. #11 → (x, y) → (x, y − 12), #14 → (x, y) → (x, y + 12), the 12’s are in the opposite direction.
16. (-8, 2), (-6, 10), (-2, 8), (-3, 4)
17. A rotation of 270◦
18. A rotation of 90◦
19. It is in the 4th quadrant and are 180◦ apart.
20. #16 → (x, y) → (y, −x), #19 → (x, y) → (−y, x), the values have the opposite sign.
21. 14 units
22. 14 units
23. rotation, 180◦
24. the origin
25. 166◦
26. 122◦
27. 315◦
28. 31 units
29. 2(b − a), right
30. 2(b − a), left
162
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Chapter 12. Rigid Transformations, Answer Key
12.6 Geometry - Second Edition, Extension:
Tessellations, Review Answers
1-7. Yes, all quadrilaterals will tessellate.
8. Equilateral triangle, square, and regular hexagon.
9. Here is one possibility.
10. The figure is an equilateral concave hexagon.
11.
12.
13. Answers will vary.
163
12.7. Chapter Review Answers
12.7 Chapter Review Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
164
C
E
A
F
J
B
H
D
I
G
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