Ch 6 Worksheets L2 Shortened Key
Name ___________________________
Worksheets Chapter 6:
Discovering and Proving Circle Properties
Lesson 6.1 Tangent Properties
Investigation 1
Tangent Conjecture
“If you draw a tangent to a circle, then…”
Converse of the Tangent Theorem
Draw a radius to the point of tangency.
What do you notice? perpendicular
Would this be true for all tangent lines? Yes
it
Draw a line perpendicular to
OT at point T, call
TA .
What type of line is TA ? tangent line
Would this work for any radius? Yes
T
O
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.
A
x
E
x
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? The intercepted arc? 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°. Let m∠AEG = x
So 360 = 90 + 90 + x + m∠N and 180 = x + m∠N , always.
Also since x = m∠AEG = mAG , since the central angle = it’s intercepted arc.
S. Stirling
Page 1 of 13
Ch 6 Worksheets L2 Shortened Key
Name ___________________________
EXERCISES Lesson 6.1 Page 313-314 #1 – 5, 8, 9.
Show how you are finding your answers! State the properties you are applying when possible.
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
6
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
Page 2 of 13
Ch 6 Worksheets L2 Shortened Key
r
Name ___________________________
t
diameter
112
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.
Shorter
A
C
S. Stirling
Page 3 of 13
Ch 6 Worksheets L2 Shortened Key
Name ___________________________
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
What if the chords are not congruent?
EF ≅ GH
If AB ≅ CD , then
F
B
D
O
116
116
P
116
H
116
A
C
E
G
Write your observations:
none of the measures are equal
m∠BOA = m∠DOC = 116
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 13
Ch 6 Worksheets L2 Shortened 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
66 48
66
S. Stirling
Various:
Central angle = intercepted
arc.
||, so corresponding angles
=. x = 66
Since radii of a circle =,
ΔAOB 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 13
Ch 6 Worksheets L2 Shortened Key
Name ___________________________
Lesson 6.3 Arcs and Angles
Investigation 4
The Big Question: What is the measure of an inscribed angle?
A
What is the measure of mAB ?
76°
Draw an inscribed angle, ∠AXB .
What is m∠AXB ? 38°
No matter where you draw the
inscribed angle it = 38°
76
38
X
O
76
B
142
C
38
142
What is the measure of mCD ?
P
142°
71
Draw an inscribed angle, ∠CYD .
What is m∠CYD ? 71°
D
71
Y
What is the relationship between an inscribed angle and its intercepted arc? inscribed angle = ½ arc
Write the Inscribed Angle Conjecture in your notes.
EXERCISES Lesson 6.3 Pages 327 – 328 # 1 – 7, 9 – 11, 16
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
65
70
S. Stirling
Inscribed
angle = ½
intercepted
arc.
Inscribed
angle = ½
intercepted
arc.
95 * 2 = 190
c = 190 – 120
= 70
30
Various:
Radius ⊥ tangent.
50
+ sum = 180
180 – 90 – 40 = 50
Central angle =
intercepted arc.
h = 50
Page 6 of 13
Ch 6 Worksheets L2 Shortened Key
Name ___________________________
Various:
Inscribed angle =
½ intercepted arc &
Semi circle = 180 °
40
42
84
Various:
Inscribed angle = ½
intercepted arc
75 * 2 = 150
100
90
20 * 2 = 40
d = 180 – 40 = 140
180 – 96 = 84
e = 84 ÷ 2 = 42
Circle = 360°
g = 360 – 150 – 110 = 100
(110 + 100)/2 = 105
150 Quad. sum = 360°
f = 360 – 75 – 105 – 90 = 90
Various:
Various:
Central ∠ = arc &
44
vertical ∠s =.
50
180 – 136 = 44
Kite, so = chords
make = arcs so
Radius ⊥ tangent.
130
130
142
142
Quad. sum = 360 °
w = 360 – 180 – 130
= 50
Various:
Inscribed angle = ½
intercepted arc.
38 * 2 = 76
Circle = 360° & =
chords cut = arcs
k = (360 – 76)/2
= 142
NDO is a
semicircle.
y = 44
60
60
60
60
Various:
Circle = 360° & =
chords cut = arcs
s = 360/6 = 60
Inscribed angle = ½
intercepted arc.
r = ½ (60 * 4) = 120
76
74
S. Stirling
Various:
Inscribed angle = ½
intercepted arc.
37 * 2 = 74
But 35 ≠ ½ * 74
Page 7 of 13
Ch 6 Worksheets L2 Shortened 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 = dπ
5π = dπ
5=d
2. If r = 5 cm, find C.
3. If C = 24 cm, find r.
C = 2π r
C = 2π r
C = 2π ( 5)
24 = 2π r
24 2π r
=
2π 2π
12
=r
C = 10π
π
4. If d = 5.5 cm, find C.
5. If a circle has a diameter of 12
cm, what is its circumference?
C = dπ
C = 5.5π
7. If d = 5 cm, find C.
6. If a circle has a circumference of
46π, what is its diameter?
C = dπ
C = dπ
C = 12π
8. If r = 4 cm, find C.
C = dπ
C = 2π r
C = 5π
C ≈ 15.708
C = 2π ( 4 )
46π = dπ
46 = d
9. If C = 44 m, find r.
C = 8π
C ≈ 25.133
C = 2π r
44 = 2π r
44 2π r
=
2π 2π
22
= r ≈ 7.003
π
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 = dπ
C = 2π r
C = 27π
C ≈ 84.823 in
C = 2π ( 24 )
S. Stirling
C = 48π
C ≈ 150.796
Page 8 of 13
Ch 6 Worksheets L2 Shortened Key
Name ___________________________
12. Circle inscribed in a square with perimeter 24 cm,
find C.
p = 4s
24 = 4s
6=s
13. Circle with C = 16π inches is circumscribed about a
square, find length of the diagonal.
6
C = dπ
C = dπ
C = 6π
C ≈ 18.850
16π = dπ
16 = d
6
P
16
P
15. Find number of 1 inch tiles to put around the edge of the pool.
The circular ends:
C = dπ
C = 18π
C ≈ 56.549
Sides of the rectangle are =.
perim = 56.549 + 2 ( 30 ) = 116.549 ft
18
116.549 * 12 = 1398.6 one-inch tiles
So need 1399 one-inch tiles
30
#19
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 9 of 13
Ch 6 Worksheets L2 Shortened Key
Name ___________________________
Investigation 5: Arc Length
So far, the measure of an arc = the measure of its central angle (in degrees).
In the diagram, m AB = 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.
D
But if you are on the circle itself, and if you
are traveling from point A to point B (on the
circle) did you travel the same distance as you
would from point C to point D?
B
120
O
A
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 of the circle, 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
16
2
iπ ( 4 ) = π ≈ 16.755 cm
3
3
1
2
length of CD = iπ (12 ) = 48π ≈ 150.796 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 10 of 13
Ch 6 Worksheets L2 Shortened Key
Name ___________________________
EXERCISES Lesson 6.7 Pages 351 # 1 – 7
On all problems, show algebraic procedures: write the formula, substitute in known information, then
solve. Leave your answers in terms of π.
120
2π (12 )
360
1
= • 24π
3
= 8π
80
length =
2π ( 3)
360
2
= • 6π
9
4
= π
3
≈ 4.189
length =
≈ 25.133
210
2π (12 )
360
7
= • 24π
12
= 14π
≈ 43.982
120
2π r
360
⎛ 3 ⎞
⎛ 3 ⎞ 2π
r
⎜ ⎟ 6π = ⎜ ⎟
⎝ 2π ⎠
⎝ 2π ⎠ 3
9=r
length =
210
60
2π (18)
360
1
= • 36π
6
= 6π
length =
60
6π =
80
2π ( 9 )
360
2
= •18π
9
= 4π
length =
80
≈ 12.566
≈ 18.850
160
dπ
360
⎛ 9 ⎞
⎛ 9 ⎞ 4π
d
⎜ ⎟12π = ⎜ ⎟
⎝ 4π ⎠
⎝ 4π ⎠ 9
27 = r
12π =
160
S. Stirling
Page 11 of 13
Ch 6 Worksheets L2 Shortened Key
Name ___________________________
EXERCISES Chapter 6 Review Pages 359 – 360 # 4 – 19
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:
Tangent ⊥ Radius
Central angle =
intercepted arc.
+ sum = 180
180 – 90 – 35 = 55
b = 55
b
180
100
Various:
Circle = 360°
= chords cut =
arcs.
Various:
Inscribed angle
= ½ intercepted
arc.
110 * 2 = 220
a = 220 – 155
= 65
Various:
Inscribed angle = ½
intercepted arc.
90 * 2 = 180
Circle = 360°
d = 360 – 180 – 89
= 91
Various:
Equal chords cut = arcs.
100
length =
2π ( 27 )
360
5
= • 54π
18
= 15π
C = 2π r
108
S. Stirling
132 = dπ
132
=d
C = 2π ( 20 )
C = 40π
C ≈ 125.664
π
d ≈ 42.017
Various:
Equal chords cut = arcs.
(360 – 220)/2 = 70
70
length =
2π ( 36 )
360
7
= • 72π
36
= 14π
160
≈ 47.124
Various:
+ sum = 180.
180 – 35 – 57 = 108
Inscribed angle = ½
intercepted arc.
108 * 2 = 216 but the
angle intercepts a semicircle which = 180. It
should = 90.
c = (360 – 104)/2
= 128
≈ 43.982
72
36
Various:
Semi-: = 180. 180 – 108 = 72
Inscribed angle = ½ intercepted
arc. 72 ÷ 2 = 36
Alternate interior ∠s =,
so lines ||.
Page 12 of 13
Ch 6 Worksheets L2 Shortened Key
152
Name ___________________________
Various:
Circle = 360.
360 – 152 – 56 = 152
= chords cut = arcs.
So JI = IM and
Various:
Inscribed angle = ½ intercepted arc.
70
mKIM = 140 &
mKI = 140 − 70 = 70
= chords cut = arcs.
so ΔKIM is isos.
ΔJIM is isos.
B
Need perpendicular
bisectors.
C
A
B
Need angle bisectors.
And a radius drawn
perpendicular to a
side.
A
S. Stirling
C
Page 13 of 13