Ch 6 Worksheet L1 Shorten Key
Lesson 6.1 Tangent Properties
Name ___________________________
Investigation 1
Converse of the Tangent Theorem
Tangent Conjecture
“If you draw a tangent to a circle, then…”
Draw a line perpendicular to
it
Draw a radius to the point of tangency.
What do you notice? perpendicular
Would this be true for all tangent lines? Yes
OT at point T, call
TA .
What type of line is TA ? tangent
Would this work for any radius? Yes
T
O
A
O
T
Write the Tangent
Conjecture in your
notes.
Write the Converse of the Tangent Conjecture in
your notes.
N
Investigation 2
Tangent Segments Conjecture
Draw tangent segments to circle E from point N.
What do you notice about these segments? They’re congruent.
Measure them.
Write the Tangent Segments Conjecture in your notes.
4.5 cm
A
4.5 cm
E
G
Draw Kite ANGE. Do you know any of the angles of this kite? What relationships can you make
between the angles of this kite? Make sure you can justify your answers with properties!
mA  90 and mG  90 because tangents are perpendicular to the radii at the point of tangency.
Sum of the angles of a quadrilateral are 360.
So 360  90  90  mAEG  mN and 180  mAEG  mN
S. Stirling
Always? Yes
Page 1 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
EXERCISES Lesson 6.1 Page 313-314 #1 – 5, 8 – 10.
Show how you are finding your answers! State the properties you are applying and show calculations!!.
w = 180 – 130 = 50
50
Tangents from a point
outside a  =.
Isos.  base angles =
Or Tangent  radius
Quad  sum – 360
w = 360 – 90 – 90 – 130 = 50
Linear pair supplementary.
Tangent  radius
 sum = 180
y = 180 – 60 – 90 = 30
and  sum = 180
x = (180 – 70)/2 = 55
Tangent  radius
Quad. sum = 360
z = 360 – 180 – 75 = 105
60
13
13
13
6
6
S. Stirling
Tangents from a point
outside a  =.
13
OR = OA = AP = PC = 13
TC = TD = DS = SR
and TD = ½ of 12 = 6
6
Perim = 4 * 13 + 4 * 6 = 76
6
Page 2 of 15
Ch 6 Worksheet L1 Shorten Key
r
Name ___________________________
t
diameter
Various lines.
Tangents must be  radii!
r
X
t
Y
Z
10. Draw an obtuse triangle ABC inscribed in the circle given below.
Is the longest side of triangle ABC longer or shorter than the diameter?
B
Various triangles.
A
Shorter
C
S. Stirling
Page 3 of 15
Ch 6 Worksheet L1 Shorten Key
Lesson 6.2 Chord Properties
Investigation 3
Chord Properties
If two chords in a circle are congruent, then…
Investigate the following:
 the central angles associated with those chords
 the intercepted arcs associated with those chords
If AB  CD , then
Name ___________________________
What if the chords are not congruent?
EF  GH
F
B
D
O
116
116
P
116
H
116
A
Write your observations:
C
E
mBOA  mDOC  116
G
None of the measures are equal.
equal central angles
mAB  mCD  116
equal intercepted arcs
Write the Chord Conjectures in your notes.
EXERCISES Lesson 6.2 Pages 320 – 321 # 1 – 3, 5, 6, 8 – 11
Write the properties you are using as you are finding the missing measures. (You don’t need to name
them, you just need to state them.)
165
= chords cut =
arcs and =
Central angles.
w = 70
= chords cut = arcs.
Circle’s arcs = 360.
z = 360 – 276 = 84
Central
angle =
intercepted
arc.
x = 165
70
128
70
70
84
S. Stirling
Page 4 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
Various:
Central angle =
intercepted arc.
mAOI  65
65
mAC  68
34
34
Various:
Linear pair supplementary.
115
Radii = so COB
112
isos. & base  =
68
65
65
 sum = 180
(180 – 112)/2 = 34
mB  34
Central angle = intercepted
arc. w = 115
= chords cut = arcs and =
central angles.
x = 115 and y = 65
115
mAC  130 so
mAB  130  48  82
Various:
Circle’s arcs add to 360.
= chords cut = arcs and =
central angles.
Central angle =
intercepted arc.
x = 48, y = 82, w = 110
mFAT  360  72  288
288 ÷ 3 = 96 = x
Various:
110
120
82
48
82
120 Circle’s arcs add to 360.
96
360 – 48 – 82 – 110 =
120
z = 120
y = 96
Radii = so FOE isos. &
42
96
42
96
96
base  = ,  sum = 180
(180 – 96)/2 = 42 = z
96
66
66 66
66 48
66
S. Stirling
66
Various:
Central angle = intercepted
arc.
||, so corresponding angles
=. x = 66
Since radii of a circle =,
AOB isos. & base angles
=.  sum = 180, so
180 – 66 – 66 = 48 = y
mAOC  180 114  66
and z = 66.
Various:
Radius = 18 so the
diameter = 36.
The diameter
would have to be
the longest chord of
the circle, so the
chord can’t be
greater than 36.
Page 5 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
Lesson 6.3 Arcs and Angles
Investigation 4
The Big Question: What is the measure of an inscribed angle?
142
A
C
What is the measure of mAB ?
76
76
142
Draw an inscribed angle, AXB .
What is mAXB ? 38
What is the measure of mCD ?
38
X
O
D
P
76
B
38
71
71
Y
142
Draw an inscribed angle, CYD .
What is mCYD ? 71
What is the relationship between an inscribed angle and its intercepted arc?
Write the Inscribed Angle Conjecture in your notes.
Investigation 5: Inscribed Quadrilaterals
Use your notes and draw a cyclic quadrilateral in
Remember each angle must be an inscribed angle
and each side must be a chord.
Label your quadrilateral ABCD.
inscribed angle = ½
intercepted arc
P.
P
Measure all of the angles of your quadrilateral.
Are there any relationships between the angles?
The opposite angles are supplementary in a
cyclic quadrilateral.
Write the Cyclic Quadrilateral Conjecture in your notes.
Try to draw a cyclic parallelogram in circle O.
What type of parallelogram can be inscribed in a circle?
O
Only rectangles (and squares) can be
inscribed in a circle.
.
Write the Cyclic Parallelogram Conjecture in your notes.
S. Stirling
Page 6 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
Investigation 6
Given AB CF ED .
Examine the measures of the arcs. What could
you conclude about the intercepted arcs?
A
F
B
22
22
C
22
mAF  mBC
54
54
mEF  mDC
22
P
54
54
.
E
Write the Parallel Lines (Secants) Intercepted
Acrs Conjecture in your notes.
D
EXERCISES Lesson 6.3 Pages 327 – 328 # 1 – 14
Write the properties you are using as you are finding the missing measures. (You don’t need to name
them, you just need to state them.)
Various:
Inscribed angle =
60 ½ intercepted arc.
Semi circle
measures 180 
180 – 120 = 60
60 ÷ 2 = 30
120
Inscribed
angle = ½
intercepted
arc.
65
Inscribed
angle = ½
intercepted
arc.
95 * 2 = 190
c = 190 – 120
= 70
70
Various:
Inscribed angle =
½ intercepted arc &
Semi circle = 180 
40
42
140
84
S. Stirling
20 * 2 = 40
d = 180 – 40 = 140
180 – 96 = 84
e = 84 ÷ 2 = 42
30
Various:
Radius  tangent.
50
 sum = 180
180 – 90 – 40 = 50
Central angle =
intercepted arc.
h = 50
Various:
Inscribed angle = ½
intercepted arc
75 * 2 = 150
100
Circle’s arcs = 360
90
g = 360 – 150 – 110 = 100
x = (110 + 100)/2 = 105
Quad. sum = 360
105
x
150 f = 360 – 75 – 105 – 90 = 90
Page 7 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
Various:
Central  = arc &
vertical s =.
50
32
Radius  tangent.
130
2 x  64  360
Quad. sum = 360 
w = 360 – 180 – 130
= 50
130
2 x  296
x
Various:
NDO is a
semicircle.
44
180 – 136 = 44
Kite, so = chords
make = arcs so
Various:
Parallel secants cut
= arcs.
Circle = 360º
142
142
y = 44
x  148
Various:
Inscribed angle = ½
intercepted arc.
38 * 2 = 76
Circle = 360 and =
chords cut = arcs
k = (360 – 76)/2
= 142
76
60
60
60
60
Various:
Circle = 360 & =
chords cut = arcs
s = 360/6 = 60
Inscribed angle = ½
intercepted arc.
r = ½ (60 * 4) = 120
Various:
Inscribed angle = ½
intercepted arc.
90 = ½ (98 + x)
180 = 98 + x, x = 82
m = 360 – 220 = 140
140
x
82
n = ½ (140 + 82) = 111
Various:
Isos Δ, base s =.
q
120
p
41
41
98
S. Stirling
180  98
 41
2
Central  =
intercepted arc.
Parallel secants cut
= arcs.
Circle = 360º.
2 p  218  360
2 p  142
p  71
2c
Various:
Inscribed angle = ½
intercepted arc.
Circle = 360º.
2d
2b
2a  2b  2c  2d  2e  360
2e
2  a  b  c  d  e   360
a  b  c  d  e  180
2a
Sum = 180º
Page 8 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
EXERCISES Lesson 6.5
Pages 337 – 340 # 1 – 13, 15, 19.
On all problems, show algebraic procedures: write the formula, substitute in known information, then
solve. On #1 – 6, leave your answers in terms of π. On #7 – 9, use the π approximation on the calculator
and round final answers to 3 decimal places. For #10 – 15, see your book for the problem statement.
1. If C = 5π cm, find d.
C  d
5  d 
5d
4. If d = 5.5 cm, find C.
2. If r = 5 cm, find C.
C  2 r
C  2 r
C  2  5 
24  2 r
24 2 r

2 2
12
r

C  10
5. If a circle has a diameter of 12
cm, what is its circumference?
C  d
C  5.5
3. If C = 24 cm, find r.
6. If a circle has a circumference of
46π, what is its diameter?
C  d
C  d
46  d 
C  12
46  d
7. If d = 5 cm, find C.
8. If r = 4 cm, find C.
C  d
C  2 r
C  5
C  2  4 
C  15.708
C  8
9. If C = 44 m, find r.
C  25.133
C  2 r
44  2 r
44 2 r

2 2
22
 r  7.003

10. A bicycle tire with a 27 inch
diameter, find C.
11. Ferris wheel with r = 24 cm, find distance
traveled by a seat in one revolution.
C  d
C  2 r
C  27
C  2  24 
C  84.823 in
C  48
C  150.796
S. Stirling
Page 9 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
12. Circle inscribed in a square with perimeter 24 cm,
6
p  4s
C  d
24  4 s
C  6
6s
13. Circle with C = 16π inches is circumscribed about a
square, find length of the diagonal.
find C.
C  d
16  d 
6
P
C  18.850
16
P
16  d
15. Find number of 1 inch tiles to put around the edge of the pool.
The circular ends:
C  d
C  18
C  56.549
Sides of the rectangle are =.
perim  56.549  2  30   116.549 ft
116.549 * 12 = 1398.6 one-inch tiles
So need 1399 one-inch tiles
18
30
#19 Hint: Start with the 42 degree angles!
b = 90
c = 42
d = 70
e = 48
f = 132
g = 52
180 – 42 – 90 = 48
84
H
K
180 – 42 – 48 = 90
90
R
48
48
84/2 = 42
132
P
52
M
70
360 – 90 – 68 – 132 = 70
52
N
S. Stirling
S
(180 – 76)/2 = 52
Page 10 of 15
Ch 6 Worksheet L1 Shorten Key
Investigation 10: Arc Length
Name ___________________________
So far the measure of an arc = the measure of its central angle (in degrees).
In the diagram, mAB  mCD  120 .
If you are thinking in terms of “turn” or
degrees, it makes sense that if you are
standing at point O you will turn 120 to get
from A to B and you would turn the same
amount of degrees to turn from C to D.
But if you are on the circle itself, and if you
are traveling from point A to point B did you
travel the same distance from point C to point
D?
OC = 12 cm
D
B
120
O
4 cm
A
8 cm
C
NO! The distance from C to D is
longer than the distance from A to B.
How can you explain this? The distance
would be part of the circumference, but what part? What part (fraction) of the circle are we talking about?
Fraction =
120 1

360 3
If OA  4 cm and OC
of the circumference!
 12 cm , how far is it from A to B? How far is it from C to D? Think part
1
8
2  4  
 8.378 cm
3
3
1
length of CD = 2 12   8  25.133 cm
3
length of AB =
So if you are looking at the length of the arc, and not the amount of turn (or degree of the arc), then it
makes complete sense.
S. Stirling
Page 11 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
EXERCISES Lesson 6.7 Pages 351 # 1 – 9, 13 – 14
On all problems, show algebraic procedures: write the formula, substitute in known information, then
solve. Leave your answers in terms of π!!
80
2  3
360
2
  6
9
4
 
3
 4.189
length 
120
2 12 
360
1
  24
3
 8
length 
 25.133
120
2 r
360
 3 
 3  2
r

 6  

 2 
 2  3
9r
210
2 12 
360
7
  24
12
 14
6 
length 
210
 43.982
60
80
 12.566
 18.850
160
80
2  9 
360
2
  18
9
 4
length 
60
length 
2 18 
360
1
  36
6
 6
160
d
360
 9 
 9  4
d
 12   
 4 
 4  9
27  r
12 
72
72
2 r
360
 5  40  5  2
 
r
 
 2  1
 2  5
100  r
40 
72
S. Stirling
Page 12 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
Read the problems from the book pages 352 – 353.
9. Completes 4 laps in 6 minutes. Calculate average speed in meters per minute. Round to two decimal
places! Make a drawing.
P  d  2 100
P  40  200
100 meters
325.66 m
 13.57 m
min
24 min
P  325.66 meters
13. 1 revolution in 20 seconds, what is the angular
velocity?
360
 18 deg
sec
20 sec
Since all of the horses rotate 360º in one
revolution, they all have the same angular
velocity.
40
meters
14. 2 horses complete 1 revolution in 20 seconds.
The horses are 8 m and 6 m from the center.
What are the tangential velocities of the two
horses? Round to two decimal places!
Horse #1:
1 rev  2  8  16 meters
16 m
 2.51 m
sec
20 sec
Horse #2:
1 rev  2  6   12 meters
12 m
 1.88 m
sec
20 sec
The horse on the outside is moving faster
because he has to travel further to make one
revolution in the same amount of time (20
seconds).
S. Stirling
Page 13 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
EXERCISES Chapter 6 Review Pages 359 – 360 # 4 – 19, 21, 22
Write the properties you are using as you are finding the missing measures. (You don’t need to name
them. You just need to state them.) Mark diagrams with the information as you go!
The degree measure describes the amount of turn, based on the central angle.
The arc length is part of the circumference. Measured in a unit of length, like inches.
90
Various:
Circle = 360
= chords cut =
arcs.
Various:
Tangent  Radius
Central angle =
intercepted arc.
 sum = 180
180 – 90 – 35 = 55
b = 55
b
Various:
Inscribed angle
= ½ intercepted
arc.
110 * 2 = 220
a = 220 – 155
= 65
Various:
Vertex inside so
2
x  62
x 1
60  64 
180
Linear pair supp
x
e  180  62  118
92
Various:
Linear pair supp
Vertex inside so
92  1 118  f 
2
184  118  f
C  2  20 
132  d 
C  40
132
C  125.664
66  f
100
S. Stirling
Various:
Equal chords cut = arcs.
100
length 
2  27 
360
5
  54
18
 15
 47.124
Various:
Inscribed angle = ½
intercepted arc.
90 * 2 = 180
Circle = 360
d = 360 – 180 – 89
= 91
C  2 r
180  88  92
92
c = (360 – 104)/2
= 128
70
160
d

d  42.017
Various:
Equal chords cut = arcs.
(360 – 220)/2 = 70
70
length 
2  36 
360
7
  72
36
 14
 43.982
Page 14 of 15
Ch 6 Worksheet L1 Shorten Key
Name ___________________________
Various:
 sum = 180.
x = 180 – 35 – 57 = 108
Inscribed angle = ½
intercepted arc.
108 * 2 = 216 but the
angle intercepts a semicircle which = 180. x
should = 90.
x
Various:
Parallel secants cut = arcs, so
x = 56.
x =56
s arcs add to 360º, but
72
84 + 56 + 56 + 158 = 354,
not 360.
Various:
Semi- = 180. 180 – 108 = 72
Inscribed angle = ½ intercepted
arc. 72 ÷ 2 = 36
72
36
Various:
Circle’s arcs = 360.
360 – 152 – 56 = 152
= chords cut = arcs.
So JI  IM and
152
JIM is isos.
Alternate interior s =,
so lines ||.
Various:
Inscribed angle = ½ intercepted arc.
70
mKIM  140 &
mKI  140  70  70
= chords cut = arcs.
so KIM is isos.
(Place answers below.)
Need at least 2 perpendicular
bisectors. Point V is equidistant from
the triangle’s vertices.
Need at least 2 angle bisectors and a radius
drawn perpendicular to a side (for the
radius). Point S is equidistant from the
triangle’s sides.
B
B
V
S
A
S. Stirling
C
A
C
Page 15 of 15