Chapter 5 ● Assignments 97
© 2009 Carnegie Learning, Inc.
5
Assignment for Lesson 5.1
Riding a Ferris Wheel
Introduction to Circles
1. For each term, name all of the components of circle Y that are examples of the term.
a. Chord
b. Radius
c. Diameter
d. Tangent
e. Arc
f. Major arc
g. Minor arc
h. Semicircle
i. Secant
R
Y
O
M
T
G
E
Assignment
Name _____________________________________________ Date _____________________
GM, RE, RO
YR, YO
RO
↔
RT
Sample Answer:
៣
RG,
៣
GE,
៣
EO
Sample Answer:
៣
GRM,
៣
GRO,
៣
GRE
Sample Answer:
៣
RG,
៣
GE,
៣
EO,
៣
OM
↔
RE
semicircle REO, semicircle RMO
98 Chapter 5 ● Assignments
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5
2. On circle R, draw each item.
a. tangent JB
b. radius RB
c. diameter BH
d. chord HI
e. secant HJ
3. Use circle R from Question 2. Name all of the examples of each term.
a. Major arc
b. Minor arc
c. Semicircle
J
B
R
H
I
៣
HBI
៣
HI,
៣
BI
semicircle
៣
BIH
Sample Answer:
Sample answers are based on the answer for
Question 2.
Holding the Wheel
Central Angles, Inscribed Angles, and Intercepted Arcs
Use circle S to complete Questions 1 through 5.
1. Suppose that ⫽ 59º. Find .
2. Suppose that m ⫽ 124º. Find .
3. Suppose that m ⫽ 55º. Find m .
4. Suppose that m ⫽ 71º. Find .
5. What is ?
6. In circle E shown below, . Find and .
N
A
E
G
m
៣
ANGm⬔AEGm⬔ANG ⫽ 74⬚
m
៣
FEC
m
៣
IC⬔FSI
⬔EFC
៣
CE
m
៣
FI⬔CSI
m
៣
CFEm
៣
CE
F
I
S
E
H
C
R
Chapter 5 ● Assignments 99
© 2009 Carnegie Learning, Inc.
5
Assignment
Name _____________________________________________ Date _____________________
Assignment for Lesson 5.2
180º
301º
56º
27.5º
109º
m
៣
ANG ⴝ 360º ⴚ 148º ⴝ 212º
m⬔AEG ⴝ 2
ⴢ
74º ⴝ 148º
100 Chapter 5 ● Assignments
© 2009 Carnegie Learning, Inc.
5
7. In circle H shown below, is 105º, is 47º, and is 100º. Find ,
, , and .
T
C
H
E
A
m⬔TEAm⬔CAEm⬔TCE
m⬔ETCm
៣
TEm
៣
EAm
៣
CA
m⬔TEA ⴝ
1
2
(108º ⴙ 105º) ⴝ 106.5º
m⬔CAE ⴝ
1
2
(100º ⴙ 108º) ⴝ 104º
m⬔TCE ⴝ
1
2
ⴢ
100º ⴝ 50º
m⬔ETC ⴝ
1
2
(47º ⴙ 105º) ⴝ 76º
Chapter 5 ● Assignments 101
© 2009 Carnegie Learning, Inc.
Manhole Covers
Measuring Angles Inside and Outside of Circles
1. In circle P shown below, and . Find the value of x.
2. In circle P shown below, and Find .
3. In circle O shown below, and . Find .
S
N
A
H
O
C
m⬔SCHm
៣
HA ⫽ 35⬚m
៣
SN ⫽ 55⬚
D
E
A
N
P
C
m
៣
EAm⬔NCA ⫽ 68⬚.m
៣
DN ⫽ 144⬚
D
N
A
E
x
P
m
៣
NA ⫽ 49⬚m
៣
DE ⫽ 75⬚
Assignment for Lesson 5.3
Assignment
Name _____________________________________________ Date _____________________
5
80º ⴝ m
៣
EA
224º ⴝ 144º ⴙ m
៣
EA
112º ⴝ
1
2
(144 º ⴙ m
៣
EA)
m⬔DCN ⴝ
1
2
(m
៣
DN ⴙ m
៣
EA)
m⬔DCN ⴝ 180º ⴚ 68º ⴝ 112º
ⴝ 62º
ⴝ
1
2
ⴢ
124º
x ⴝ
1
2
(75º ⴙ 49º)
Because is a semicircle,
So, .
Because is a semicircle,
So,
So, is 135º.m⬔SCH
ⴝ 135º
ⴝ
1
2
(270º)
ⴝ
1
2
(125º ⴙ 145º)
m⬔SCH ⴝ
1
2
(m
៣
SH ⴙ m
៣
NA)
m
៣
NA ⴝ 180º ⴚ 35º ⴝ 145º.
m
៣
HA ⴙ m
៣
NA ⴝ 180º.
៣
HAN
m
៣
SH ⴝ 180º ⴚ 55º ⴝ 125º
m
៣
SH ⴙ m
៣
SN ⴝ 180º.
៣
HSN
102 Chapter 5 ● Assignments
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5
4. In circle O shown below, and . Find .
5. In circle S shown below, and . Find
U
E
R
O
T
S
m⬔OUT.m
៣
OT ⫽ 121⬚m
៣
ER ⫽ 38⬚
A
S
D
M
O
C
m⬔DCMm
៣
MS ⫽ 104⬚m
៣
AS ⫽ 11⬚
Because is a semicircle, .
So,
So, is 43.5º.m⬔DCM
ⴝ
1
2
(87º) ⴝ 43.5º
ⴝ
1
2
(76º ⴙ 11º)
m⬔DCM ⴝ
1
2
(m
៣
DM ⴙ m
៣
AS)
m
៣
DM ⴝ 180º ⴚ 104º ⴝ 76º.
m
៣
DM ⴙ m
៣
MS ⴝ 180º
៣
DMS
So, is 41.5º.m⬔OUT
ⴝ 41.5º
ⴝ
1
2
(83º)
ⴝ
1
2
(121º ⴚ 38º)
m⬔OUT ⴝ
1
2
(m
៣
OT ⴚ m
៣
ER)
Name _____________________________________________ Date ______________________
Chapter 5 ● Assignments 103
© 2009 Carnegie Learning, Inc.
5
6. In circle S shown below, is the diameter of the circle and . Find
7. In circle G shown below, and . Find .
U
O
S
E
H
G
41
⬚
41
⬚
171
⬚
m⬔EUHm
៣
HE ⫽ 171⬚OH ⫽ ES, m
៣
OH ⫽ 41⬚
U
E
S
O
T
P
m⬔OUT.
m
៣
OT ⫽ 132⬚OE
Draw chord OT. Find the measure of exterior angle OTP.
Next, find Because is a diameter,
So, m
Because is an exterior angle of
So, is 42º.m⬔OUT
42º ⴝ m⬔OUT
66º ⴝ m⬔OUT ⴙ 24º
䉭OUT, m⬔OTP ⴝ m⬔OUT ⴙ m⬔UOT.⬔OTP
ⴝ 24º
ⴝ
1
2
(48º)
m⬔UOT ⴝ
1
2
(m
៣
ET)
៣
ET ⴝ 180º ⴚ 132º ⴝ 48º.
m
៣
OT ⴙ m
៣
ET ⴝ 180º.OEm⬔UOT.
ⴝ 66º
ⴝ
1
2
(132º)
m⬔OTP ⴝ
1
2
(m
៣
OT)
Because and .
So, is 32º.m⬔EUH
ⴝ 32º
ⴝ
1
2
(171º ⴚ 107º)
m⬔EUH ⴝ
1
2
(m
៣
HE ⴚ m
៣
OS)
m
៣
OS ⴝ 360º ⴚ 41º ⴚ 41º ⴚ 171º ⴝ 107º
m
៣
ES ⴝ 41ºm
៣
OH ⴝ 41º,OH ⴝ ES
104 Chapter 5 ● Assignments
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5
8. In circle G shown below, . Find .
9. In circle T shown below, is 57º and is 141º. Find .
C
L
B
R
T
E
m
៣
BLm
៣
REm⬔RCE
R
H
G
U
E
m⬔HUEm
៣
HE ⫽ 99⬚
Because
So, is 81º.m⬔HUE
ⴝ 81º
ⴝ
1
2
(261º ⴚ 99º)
m⬔HUE ⴝ
1
2
(m
៣
HRE ⴚ m
៣
HE)
m
៣
HE ⴝ 99º, m
៣
HRE ⴝ 360º ⴚ 99º ⴝ 261º.
The measure of is equal to the difference of
the measure of the arcs that are intercepted by the
angle. So,
So, m is 27º.
៣
BL
m
៣
BL ⴝ 27 º
57º ⴝ
1
2
(141º ⴚ m
៣
BL)
m⬔RCE ⴝ
1
2
(m
៣
RE ⴚ m
៣
BL)
1
2
⬔RCE
Assignment
Name _____________________________________________ Date _____________________
Assignment for Lesson 5.4
Color Theory
Chords and Circles
Use circle T below to complete Questions 1 through 7.
1. Draw an inscribed right angle in circle T. Label each point where the angle intersects
the circle. What is the name of the right angle?
2. Draw the chord determined by the inscribed right angle. What is the name of the chord?
3. What else do you know about the chord determined by an inscribed right angle?
4. Draw a second inscribed right angle in circle T. Label each point where the angle
intersects the circle. What is the name of the second right angle?
5. Draw the chord determined by the second inscribed right angle. What is the name
of the chord?
6. What else do you know about the chord determined by the second inscribed right angle.
7. Do you think every inscribed right angle will determine the longest chord of the circle,
which is the diameter of the circle?
T
Chapter 5 ● Assignments 105
© 2009 Carnegie Learning, Inc.
5
R
I
G
A
E
N
⬔RIG
RG
The chord determined by an inscribed right angle is a diameter of the circle.
.⬔ENA
AE
The chord determined by the second inscribed right angle is also a diameter of the
circle.
Yes. The arc of a semicircle is 180º, so the measure of an angle inscribed in a
semicircle is half of 180º, or 90º.
106 Chapter 5 ● Assignments
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5
8. The figure below shows a section of a circle. Draw two chords and construct their
perpendicular bisectors to locate the center of the circle.
9. In circle G shown below, = 1.84 centimeters, = 1.98 centimeters, = 90°,
and . Determine which chord is longer, or .
G
I
L
H
J
M
K
៣
JK
៣
IHm⬔GMK ⫽ 90⬚
m⬔GLH
៣
GL
៣
MG
Because point M is closer to the center of the circle than point L, chord JK is
longer than chord IH.
A
B
C
D
E
F
G
H
O
The intersection of the perpendicular bisectors of two chords is the center of the
circle. Draw chords AB and CD. Construct line EF, which is the perpendicular
bisector of AB. Construct line GH, which is the perpendicular bisector of CD. Point
O, which is the intersection of lines EF and GH, is the center of the circle.
Solar Eclipses
Tangents and Circles
1. In the space below, draw circle O with a tangent line drawn. Label the point of tangency
as point A.
2. Label another point on the tangent as point B.
3. Draw a second tangent line to the circle that passes through point B. Label this second
point of tangency as point C.
4. Draw the radii and .
5. What is ? Explain how you found your answer.
6. What is ? Explain how you found your answer.
7. Use a protractor to measure .
8. What is ? Explain how you found this measure.m⬔ABC
⬔AOC
m⬔OCB
m⬔OAB
OC
OA
Chapter 5 ● Assignments 107
© 2009 Carnegie Learning, Inc.
Assignment
Name _____________________________________________ Date _____________________
Assignment for Lesson 5.5
5
The measure of is 90º, because a tangent to a circle is perpendicular to the
radius that is drawn to the point of tangency.
⬔OAB
360º – 90º – 90º – 114º = 66º
The sum of the angles in any quadrilateral is 360º. Subtract the three known angles
from 360º.
A
O
B
C
The measure of is 90º, because a tangent to a circle is perpendicular to the
radius that is drawn to the point of tangency.
⬔OCB
m⬔AOC ⴝ 114º
108 Chapter 5 ● Assignments
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5
Chapter 5 ● Assignments 109
© 2009 Carnegie Learning, Inc.
Gears
Arc Length
1. In circle A shown below, describe the difference between the measure of minor arc
BC and the length of minor arc BC.
2. In circle E shown below, the radius of the circle is 16 centimeters and is
40°. Find the arc length of .
E
S
B
J
៣
JB
m⬔JSB
A
B
C
Assignment
Name _____________________________________________ Date _____________________
Assignment for Lesson 5.6
5
The radius of circle A is 3 centimeters. The circumference of circle A is , or about
18.84 centimeters. The measure of minor arc BC is determined by the measure of its
central angle, So, is 140º and is 140º. The length of arc BC is
determined by the portion of the circumference that lies between points B and C.
Arc BC is of the circle. So, the length of is centimeters.
140º
360º
ⴢ
18.84 ⬇ 7.33
៣
BC
140º
360º
m
៣
BCm⬔BAC⬔BAC.
2(3)
The circumference of circle E is , or about 100.48 centimeters. The measure
of is So,
40º
; So, the arc length of is centimeters.
80º
360º
ⴢ
100.48 ⬇ 22.33
៣
JB
៣
JB80º ⴝ m
៣
JBⴝ
1
2
m
៣
JBm⬔JSB ⴝ
1
2
m
៣
JB.
1
2
m⬔JSB
2(16)
110 Chapter 5 ● Assignments
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5
3. In circle I shown below, the radius is 6 millimeters and m is 80°. Find the arc
length of .
4. In circle H shown below, the arc length of is centimeters and is
80°. Find the length of a diameter of circle H.
S
O
J
H
m⬔JOS24
៣
SJ
S
R
I
C
H
D
80
⬚
៣
SC
៣
HC
The circumference of circle I is , or about 37.68 millimeters. Because is a
diameter, is a semicircle. Because is 80º,
. The arc length of is
millimeters.
m
៣
SC
360º
ⴢ
37.68 ⴝ
100º
360º
ⴢ
37.68 ⬇ 10.47
៣
SCm
៣
SC ⴝ 180º ⴚ 80º ⴝ 100º
m
៣
HC
៣
SCH
SH2(6)
Because is 80°, . Because the length of an arc is
equal to the ratio of the measure of the arc and 360° times the circumference of the
circle,
So, the radius of circle H is 27 centimeters and the diameter of circle H is
54 centimeters.
27 ⴝ r.
8640 ⴝ 320ºr
8640 ⴝ 160º
ⴢ
2r
24 ⴝ
160º
360º
ⴢ
2r
m
៣
SJ ⴝ 2(80º) ⴝ 160ºm⬔JOS
Chapter 5 ● Assignments 111
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5
Playing Darts
Areas of Parts of Circles
In circle C shown below, is equilateral, AC = 10 inches, and m ACB
is 60º. Use the figure to complete Questions 1 through 3.
1. Find the area of the sector ACB.
2. The height of is about 8.66 inches. Find the area of .
3. What is the shaded region called? Find the area of the shaded region.
In circle A shown below, the radius is 18 centimeters, is equilateral,
and is 60°. Use the figure to complete Questions 4 and 5.
4. Find the area of the sector of the circle determined by radii AB and AC.
A
B
C
60
⬚
18 cm
m⬔BAC
⌬ABC
䉭ABC䉭ABC
A
B
C
10 in.
⬔⌬ABC
Assignment
Name _____________________________________________ Date _____________________
Assignment for Lesson 5.7
The area of circle C is ( ) or about 314 square inches. Because is 60°,
is 60°. So, the area of sector ACB is square inches, or about
square inches.
52.3
60º
360º
ⴢ
314m
៣
AB
m⬔ACB10
2
The area of ΔABC is square inches.
1
2
(8.66)10 ⴝ 43.3
The shaded region is called a segment of the circle.
The area of the segment of the circle is equal to the area of the sector minus the area
of the triangle. So, the area of the segment is , or about 9.03 square inches.52.3 ⴚ 43.3
The area of the sector is square centimeters.
60º
360º
ⴢ
(18
2
) ⬇ 169.56
112 Chapter 5 ● Assignments
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5
5. Find the area of the segment of the circle bounded by chord BC.
In circle A shown below, the radius is 22 centimeters and is 90°.
Use the figure to complete Questions 6 and 7.
6. Find the area of the sector of the circle determined by radii AR and AT.
7. Find the area of the segment of the circle bounded by chord RT.
22 cm
R
A
T
m⬔RAT
The area of the segment is the difference between the area of the sector and the
area of the triangle.
Because is equilateral, BC is 18 centimeters. The height of is
centimeters. The area of is square centimeters. So, the
area of the segment is 169.56 – 140.30 = 29.26 square centimeters.
1
2
(18)(9
冪
3) ⬇ 140.30䉭ABC
9
冪
3䉭ABC䉭ABC
The area of the sector is square centimeters.
90º
360º
ⴢ
(22
2
) ⬇ 379.94
The area of the segment is the difference between the area of the sector and the
area of the triangle.
The area of is square centimeters. So, the area of the segment
is square centimeters.379.94 ⴚ 242 ⴝ 137.94
1
2
(22)(22) ⴝ 242䉭 RAT
Chapter 5 ● Assignments 113
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5
Crop Circles
Circle Measurements and Relationships
Use the diagram and the given information to answer Questions 1 through 5.
Explain your reasoning.
The radius of circle P in the diagram below is 8 inches and segment GF is a diameter of the circle.
Segments HK and GK are tangents to circle P. The measure of angle HGK is 43 degrees.
Assignment
Name _____________________________________________ Date _____________________
Assignment for Lesson 5.8
H
L
F
P
G
K
J
1. Find .
2. Find m⬔GKH.
3. Find m⬔HFG.
GF
The measure of angle FGK is 90 degrees because is tangent to circle P, and you
are given that the measure of angle HGK is 43 degrees. So, the measure of angle
HGF is 90ⴗⴚ43ⴗ, or 47ⴗ. The base of triangle GHF is the diameter of circle P, so
triangle GHF is a right triangle, and the measure of angle GHF is 90 degrees. So, the
measure of angle HFG is 180ⴗⴚ(90 ⴙ 47ⴗ), or 43ⴗ.
GK
Tangent segments to the same circle are equal in length. So, triangle KHG is an
isosceles triangle. The measure of angle HGK is given as 43 degrees. So, the
measure of GKH is equal to 180ⴗⴚ(2 ⴛ 43ⴗ), or 94ⴗ.
The radius of circle P is given as 8 inches. Segment GF is a diameter, so the length
of segment GF is 2 8 ⴝ 16 inches.
ⴢ
114 Chapter 5 ● Assignments
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5
4. Find .
5. Find m .
Use the diagram and the given information to answer Questions 6 through 11.
Explain your reasoning.
In the diagram below, line SU is tangent to circle A at point U. Segment VT is a diameter of the
circle. The measure of arc VW is 33⬚ and the measure of arc WX is 28⬚.
៣
GH
៣
HG
S
U
A
T
V
R
X
W
Y
Z
6. Find m⬔VXT.
Triangle GHF is a right triangle. The length of the hypotenuse is 16 inches, and you
found in Question 3 that the measure of angle HFG is 43ⴗ. Use a trigonometric ratio
to find the length HG.
HG ⬇ 10.9 inches
HG ⴝ 16 sin 43º
sin 43º ⴝ
HG
16
sin ⬔HFG ⴝ
HG
hypotenuse
The measure of an intercepted arc is twice the measure of its inscribed angle.
From Question 3, the measure of angle HFG is 43ⴗ. So, the measure of arc GH is
2 ⴛ 43ⴗ, or 94ⴗ.
The base of triangle VXT is the diameter of circle A, so triangle VXT is a right
triangle. The measure of angle VXT is 90ⴗ.
Chapter 5 ● Assignments 115
© 2009 Carnegie Learning, Inc.
5
Name _____________________________________________ Date ______________________
7. Find m⬔VTX.
8. Find m .
9. Find m⬔VYZ.
10. Find m⬔ZYX.
11. Find m⬔ZUS.
៣
TX
The measure of angle VTX is equal to half the measure of its intercepted arc, which is
arc VX. The measure of arc VW is 33ⴗ and the measure of arc WX is 28ⴗ, so the
measure of arc VX is 33ⴗⴙ28ⴗ, or 61ⴗ. So, the measure of angle VTX is 61ⴗ, or 30.5ⴗ.
ⴢ
1
2
The measure of arc TX is equal to twice the measure of its intercepted angle,
which is angle TVX. In Questions 6 and 7 you found that the measure of angle VXT
is 90ⴗ and the measure of angle VTX is 30.5ⴗ, so the measure of angle TVX is 180ⴗⴚ
(90ⴗⴙ30.5ⴗ), or 59.5ⴗ. So, the measure of arc TX is 2 ⴛ 59.5ⴗ, or 119ⴗ.
First, find the measure of angle VZY. The measure of angle VZY is equal to half the
sum of the measures of the two intercepted arcs, VW and TU. The measure of arc
VW is 33ⴗ. Angle TAU is a right angle, so the measure of arc TU is 90ⴗ. The measure
of angle VZY is (33ⴗⴙ90ⴗ), or 61.5ⴗ. In Question 8 you found that the measure of
angle TVX is 59.5ⴗ. So, the measure of angle VYZ is 180ⴗⴚ(61.5ⴗⴙ59.5ⴗ), or 59ⴗ.
1
2
Angles ZYX and VYZ form a linear pair. In Question 9 you found that the measure of
angle VYZ is 59ⴗ, so the measure of angle ZYX is 180ⴗⴚ59ⴗ, or 121ⴗ.
Because line SU is tangent to circle A at point U, angle AUS is a right angle. So, the
measure of angle ZUS is equal to 90ⴗ minus the measure of angle AUZ. The measure
of angle ZAU is 90ⴗ and the measure of angle AZU is 61.5ⴗ (angles AZU and VZY
are vertical angles and thus congruent). So, the measure of angle AUZ is 180ⴗⴚ
(90ⴗⴙ61.5ⴗ), or 28.5ⴗ. This means that the measure of angle ZUS is 90ⴗⴚ28.5ⴗ,
or 61.5ⴗ.
116 Chapter 5 ● Assignments
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5
