Download radii: AP , PR,PB diameter: AB chords: AB , CD, AF secant: AG or AG

Ch 6 Note Sheet L1 Key
Review: Circles Vocabulary
Name ___________________________
If you are having problems recalling the vocabulary, look back at your notes for Lesson 1.7 and/or page 69 – 71 of
your book. Also, pay close attention to the geometry notation you need to use to name the parts!!
Circles
G
Circle is a set of points in a plane a given distance
(radius) from a given point (center).
F
R
Congruent circles are two or more circles with the
same radius measure.
Concentric circles are two or more circles with the
same center point.
A
C
P
B
D
Types of Lines [Segments]
Radius [of a circle] is
• A segment that goes from the center to any point
on the circle.
• the distance from the center to any point on the
circle.
Diameter [of a circle] is
• A chord that goes through the center of a circle.
• the length of the diameter. d = 2r or ½ d = r.
Chord is a segment connecting any two points on the
circle.
Secant A line that intersects a circle in two points.
Tangent [to a circle] is a line that intersects a circle in
only one point. The point of intersection is called the
point of tangency.
Arcs & Angles
Arc [of a circle] is formed by two points on a circle and a
continuous part of the circle between them. The two
points are called endpoints.
Semicircle is an arc whose endpoints are the endpoints of
the diameter.
Minor arc is an arc that is smaller than a semicircle.
Major arc is an arc that is larger than a semicircle.
Intercepted Arc An arc that lies in the interior of an
angle with endpoints on the sides of the angle.
Central angle An angle whose vertex lies on the center
of a circle and whose sides are radii of the circle.
Measure of a central angle determines the measure of
an intercepted minor arc.
S. Stirling
E
radii: AP , PR , PB
diameter: AB
chords: AB , CD , AF
secant: AG or AG
tangent: EB
semicircles: ACB , ARB
minor arcs: AC , AR , RD , BC ,…
major arcs: ARC , ARD , FRD ,
BAC ,…
central angles with their intercepted
arcs:
m∠APR = mAR
m∠RPB = mRB
Page 1 of 11
Ch 6 Note Sheet L1 Key
Lesson 6.1 Tangent Properties
Investigation 1: See Worksheet page 1.
O
T
Name ___________________________
Example 1
AB is tangent to both circles and
mAMC = 295° . m∠BQX = ?
Tangent Conjecture
A tangent to a circle is
perpendicular to
the radius drawn to the
point of tangency.
Converse of the Tangent Theorem
A line that is perpendicular to a radius at its
endpoint on the circle is tangent to the circle.
mAC = 360 − 295 = 65 so central angle
m∠APC = 65 .
Tangents are perpendicular to radii so
m∠APX = 90 = m∠XBQ .
N
Investigation 2: See Worksheet page 1.
Tangent Segments Conjecture
Tangent segments to a circle
from a point outside the circle
are congruent.
Since vertical angles are equal
m∠PXA = m∠BXQ
And the third angles of the triangles are
equal m∠APC = m∠BQX = 65
A
E
G
Vocabulary:
Intercepted Arc An arc that lies in the interior of an angle with
endpoints on the sides of the angle.
G
Central angle
∠GNA determines the measure of minor arc GA .
∠GNA is said to intercept GA because the arc GA is within
the angle.
N
T
Are there any relationships between the angles formed when you
have tangent segments from a point outside a circle?
A
M
If mGA = 120° …
The measure of a minor arc is defined as the measure of its central angle, m∠GNA = mGA = 140° .
If TA and TG are both tangent to circle N, so m∠A and m∠G must both = 90°.
The sum of the angles of quadrilateral TANG must be 360°. So m∠T + 90 + 140 + 90 = 360° , which
means m∠T = 40° .
What is the relationship between ∠T and the intercepted arc, GA ? They add to 180°
S. Stirling
Page 2 of 11
Ch 6 Note Sheet L1 Key
Name ___________________________
Intersecting Tangents Conjecture
The measure of an angle formed by two intersecting tangents to a circle is
180 minus the measure of the intercepted arc, m∠ = 180 − small arc .
Mini-proof:
x
a
If radii are drawn, the quadrilateral makes a kite
with two 90º angles.
Sum of the other two equal 180.
So 180 = x + a and since x = 180 − a .
b
Example 2
Rays r as s are tangents. w = ?
180 – 54 = w
OR
so central angle = w
360 – 90 – 90 – 54 = w
126 = w
Vocabulary:
“Tangent” means to intersect in one point.
Tangent Circles Circles that are tangent to the same line at the same point.
Internally Tangent
Circles
Two tangent circles
having centers on the
same side of their
common tangent.
Externally Tangent
Circles
Two tangent circles
having centers on
opposite sides of their
common tangent.
Polygons and Circles
The triangle is inscribed in the circle, or
the circle is circumscribed about the triangle.
The triangle is circumscribed about the circle, or
the circle is inscribed in the triangle.
B
All of the vertices
of the polygon are
on the circle.
The sides are all
chords of the circle.
P
C
All of the sides of the polygon
are tangent to the circle.
O
C
A
A
S. Stirling
B
Page 3 of 11
Ch 6 Note Sheet L1 Key
Name ___________________________
Lesson 6.2 Chord Properties
Read top of page 317. Then use the diagrams to define the following.
Central angle
An angle whose vertex lies on the
center of a circle and whose sides
are radii of the circle.
Examples:
Non-Examples:
R
D
P
O
Q
A
S
T
B
Measure of a central angle =
measure of its intercepted arc.
Inscribed angle
An angle whose vertex lies on a
circle and whose sides are chords
of the circle.
∠PQR , ∠PQS , ∠RST ,
∠QST and ∠QSR are NOT
∠AOB , ∠DOA and
∠DOB are central
angles of circle O.
central angles of circle O.
Examples:
Non-Examples:
Q
A
C
V
R
P
B
E
T
W
X
D
∠ABC , ∠BCD and
∠CDE are inscribed
U
∠PQR , ∠STU , and ∠VWX
are NOT inscribed angles.
angles.
Investigation 3: See Worksheet page 2.
S
B
D
O
Chord Central Angles Conjecture
If two chords in a circle are congruent, then they
determine two central angles that are congruent.
A
C
If AB ≅ CD , then ∠BOA ≅ ∠COD .
Chord Arcs Conjecture
If two chords in a circle are congruent, then their
intercepted arcs are congruent.
If AB ≅ CD , then AB ≅ CD .
B
D
Chord Distance to Center Conjecture
Two congruent chords in a circle are equidistant
from the center of the circle.
(Remember, that distance from a point to a line is defined as
the length of the perpendicular from the point to the line.)
S. Stirling
O
M
A
N
C
If AB ≅ CD , then OM = ON .
Page 4 of 11
Ch 6 Note Sheet L1 Key
Investigation 4: See Worksheet page 2.
Name ___________________________
F
If OM ⊥ EF , then
Perpendicular to a Chord Conjecture
The perpendicular from the center of a circle
to a chord bisects the chord and its
intercepted arc.
OM bisects EF
or EM ≅ MF
and EQ ≅ QF
A
M
E
j
Perpendicular Bisector of a Chord
Conjecture
The perpendicular bisector of a chord passes
through the center of the circle.
O
Q
D
B
P
k
C
Note: This property is used to find the center of any
circle. Just draw two nonparallel chords, of any
length, then construct the perpendicular bisectors of
both of them. By the Perpendicular Bisector of a
Chord Theorem, the intersection must be the center
of the circle.
If line j is the perpendicular bisector of BD ,
then line j passes through point P.
If line k is the perpendicular bisector of AC ,
then line k passes through point P.
Lesson 6.3 Arcs and Angles Read top of page 324.
Investigation 5: See Worksheet page 5.
Inscribed Angle Conjecture
The measure of an angle inscribed in a circle equals
half the measure of its intercepted arc.
If m∠COR = 92° ,
C
then mCR = 92° .
The measure of the
central angle equals
the measure of its
intercepted arc.
92
O
A
R
If mCR = 92° , then m∠CAR = 46° .
Investigation 6: See Worksheet page 5.
Inscribed Angles Intercepting Arcs Conjecture
Inscribed angles that intercept the same arc [or congruent arcs] are congruent.
If mAB = 170° , then
If m AB = mCD = 42° ,
P
m∠APB = 1 (170 ) ° = 85°
2
and m∠AQB = 1 (170) ° = 85°
2
so m∠APB = m∠AQB .
A
then
m∠APB = m∠CQD = 21° .
170
Q
42
B
O
C
42
B
The inscribed angles
intercept the same arcs, so they’re congruent.
S. Stirling
A
P
D
Q
The inscribed angles intercept congruent arcs
so they’re congruent.
Page 5 of 11
Ch 6 Note Sheet L1 Key
Name ___________________________
Investigation 7: See Worksheet page 6.
If ∠ACD , ∠ABD and/or
∠AED intercept semicircle
Angles Inscribed in a Semicircle Conjecture
Angles inscribed in a semicircle are right angles.
R
A
ARD , then
O
C
D
B
E
m∠ACD = m∠ABD = m∠AED = 90° .
Investigation 8: See Worksheet page 6.
Cyclic Quadrilateral A quadrilateral that
can be inscribed in a circle. (Each of its
angles are inscribed angles and each of its
sides is a chord of the circle.)
If ABDC is a cyclic quadrilateral
m∠A + m∠D = 180° and m∠B + m∠C = 180°
A
Cyclic Quadrilateral Conjecture
The opposite angles of a cyclic
quadrilateral are supplementary.
C
m ∠BAC = 106.61 °
m ∠CDB = 73.39°
m ∠DBA = 85.99°
B
m ∠ACD = 94.01°
D
Cyclic Parallelogram Conjecture
If a parallelogram is inscribed within a circle, then the
parallelogram is a rectangle.
G
O
Y
D
L
Secant
A line that intersects a circle in two
points.
Secants: EC , TN
L
Secant Segments: LN , SA ,
T
SC , EA .
Secant Segment
A segment that intersects a circle in two
points and it contains at least one point
on the exterior of the circle.
Investigation 9: See Worksheet page 7.
Parallel Lines [Secants] Intercepted Arcs
Conjecture
Parallel lines [secants] intercept congruent
arcs on a circle.
N
Chords: EC , TN because they
begin and end on the circle.
S
If AB DC ,
then mAD = mBC = 64° .
D
64
A
Note: mAB ≠ mDC .
S. Stirling
A
C
E
Page 6 of 11
C
B
64
Ch 6 Note Sheet L1 Key
Name ___________________________
Are there any short cuts for finding the measures of angles and their intercepted arcs?
Quick Formulas for Angles and Arcs:
Diagram:
Where is the What is the
vertex of the figure formed
angle?
by?
Center of the 2 radii
A
circle
“Central
O 1
angle”
Measure of the angle formed?
angle = intercepted arc
m∠1 = m AB
B
On the circle
2 chords
“Inscribed
angle”
angle = ½ (intercepted arc)
C
E
m∠ 2 = 1 i mCE
2
2
D
1 chord &
1 tangent
m∠ 2 = 1 i mFG
2
F
2
H
Inside the
circle
G
2 chords
angle = ½ (sum of intercepted arcs)
L
m∠3 = 1
M
3
V
2
( m JK + mLM )
J
K
Outside the
circle
2 secants
angle
= ½ (difference of intercepted arcs)
N
O
4
P
1 secant &
1 tangent
Q
W
U
T
R
4
( m NP − mOQ )
m∠ 4 = 1 ( mWS − mSU )
2
m∠ 4 = 1
2
S
2 tangents
X
4
Y
Summary:
Vertex center
m∠ 4 = 1
Z
2
( mYAZ − mYZ )
A
○ → angle = arc
○ → angle = ½ arc
Vertex inside ○ → angle = ½ sum arcs
Vertex outside ○ → angle = ½ difference arcs
Vertex on
S. Stirling
Page 7 of 11
Ch 6 Note Sheet L1 Key
Name ___________________________
Lesson 6.5 The Circumference/Diameter Ratio Read top of page 335.
Pi is a number, just like 3 is a number. It represents the number you get when you take the circumference of a
circle and divide it by the diameter. There is no exact value for pi, so you use the symbol π. You will leave π in the
answers to your problems unless they ask for an approximate answer. If they do, use 3.14 or the π key on your
calculator.
Circumference
The perimeter of a circle, which is the distance around the circle. Also, the curved path of
the circle itself.
Circumference Conjecture
If C is the circumference and d is the diameter of a circle, then there is a number π such
that C = dπ
Since d = 2r, where r is the radius, then C = 2rπ or 2πr
Example A: If a circle has a diameter of 3 meters,
what is it’s circumference?
Example B: If a circle has a circumference
of 12π meters, what is it’s radius?
C =πd
C = π •3
C = 3π
C ≈ 9.4 m
C = 2π r
12π = 2π r
12π 2π r
=
2π
2π
r = 6m
write the formula
substitute
simplify (pi written last, like a variable)
Only an approximate answer!
write the formula
substitute
divide to get r.
simplify
Intersecting Chords Conjecture Page 339 #17 (Optional)
The measure of an angle formed by two intersecting chords is half
the sum of the measures of the two intercepted arcs.
Given: Circle O with chords AG and LN
Show: m∠AEN = 1
LG + AN )
2(
Circle O with chords AG and LN Given
Draw auxiliary line NG .
m∠N = 1 mLG and m∠G = 1 mAN Inscribed Angle Conjecture
2
2
L
A
E
O
N
m∠AEN = m∠N + m∠G Triangle Exterior Angle Conjecture
m∠AEN = 1 LG + 1 AN Substitution
2
2
m∠AEN = 1 LG + AN Distributive Property
2
(
S. Stirling
)
Page 8 of 11
G
Ch 6 Note Sheet L1 Key
Lesson 6.6 Around the World
Name ___________________________
Read top of page 335.
You will need to understand units, fractions and multiplication of fractions to get a grip on this section.
(P.S. You’ll thank me later when you take Chemistry too!!)
Concept 1: You may divide out common factors when multiplying fractions…before actually multiplying
the numerators and then the denominators to get your answer. Warning! You must divide
out the same factor from the top and the bottom. (Your making 1s.)
Example A: Multiply.
2 7 15
• • =
5 8 21
Example B: Multiply.
1
4
3 7
16 4
• •5• • =
4 8
14 3
5
Concept 2: You may use the technique of dividing out common factors on units as well. This is called
“dimensional analysis”. Here are the key ideas:
Example C: How many yards are there in 182.54 centimeters?
Idea 1: A fraction like 5 = 1
5
So since
1 yard 3 feet
=
=1
1 yard = 3 feet, so
3 feet 1 yard
12 inches
1 foot
=
=1
12 inches = 1 foot, so
1 foot
12 inches
1 inch
2.54 cm
=
=1
1 inch = 2.54 cm, so
2.54 cm 1 inch
Idea 2: If you multiply a value by 1, you don’t
change it’s value. So if I multiply by
fractions that equal 1, I don’t change the
quantity.
Idea 3: You may divide out common factors
…even the units!
Make sure you set up your fractions so that the
units will divide out.
182.54 cm
1 in
1 ft 1 yd
•
•
•
=
1
2.54 cm 12 in 3 ft
Now multiply the numerators and then the
denominators together, then divide the top by the
bottom.
182.54
yd = 1.996 yd
91.44
Example D: You traveled 260 km in 2 hours, what is your mph? How many miles would you go in a
minute? (Note: 1 mile = 1.609 km and the word “per” indicates division)
260 km
1 mi
•
= 80.8 mph
2 hr 1.609 km
S. Stirling
80.8 mi 1 hr
mi
•
= 1.347
1 hr
60 min
min
Page 9 of 11
Ch 6 Note Sheet L1 Key
Name ___________________________
EXAMPLE page 341.
If the diameter of earth is 8000 miles, find the average speed in miles per hour Phileas Fogg needs to
circumnavigate Earth about the equator in 80 days. (Remember distance = rate times time.)
Summarize Problem:
d = 8000 mi, 80 days, around the equator.
Find speed in mph.
C =πd
C = 8000π miles
average speed = mph
24 hr
80 day •
= 1920 hrs
1 day
8000π mi
≈ 13.09 mph
1920 hrs
OR
80 day •
24 hr
= 1920 hrs
1 day
8000π mi 1 day
mi
•
≈ 13.09
80 day 24 hr
hr
OR for an exact answer…
8000π mi 1 day
•
=
80 day 24 hr
100π mi 1
•
=
1
24 hr
25π mi
6 hr
Speedy Ms. A was running the track at the high school. The track is pictured below. If she makes it
around the track 6 times in 6.5 minutes, what was her average speed in meters per second. Keep two
decimal places!
112 m
C = 2i112 + 2π ( 28)
C ≈ 399.93
28 m
399.93 • 6 = 2399.58 meters
2399.58 m 1 min
m
•
= 6.15
6.5 min 60 sec
sec
S. Stirling
Page 10 of 11
Ch 6 Note Sheet L1 Key
Lesson 6.7 Arc Length
Investigation 10: See Worksheet page 11.
Name ___________________________
Measure of an Arc: The measure of an arc equals the measure of its central angle,
measured in degrees.
Arc length: The portion of (or fraction of) the circumference of the circle described by an
arc, measured in units of length.
Arc Length Conjecture
The length of an arc equals the measure of the arc divided by 360° times the
circumference.
It is a fraction of the circle! So…….
Use the formula every time!! Arc Length =
arc degrees
• Circumference
360
Example (a):
Given central ∠ATB = 90°
with radius 12.
Example (b):
Given central ∠COD = 180°
with diameter 15 cm.
mAB = 90°
mCED = 180°
90 1
=
fraction of circle =
360 4
90
length of AB =
( 2π i12)
360
1
= ( 24π ) = 6π
4
180 1
=
fraction of circle =
360 2
180
length of CED =
(15π )
360
15
= π cm
2
EXAMPLE B:
If the radius of a circle is 24 cm and
m∠BTA = 60° , what is the length of AB ?
mAB = 120°
T
S. Stirling
fraction of circle =
120 1
=
360 3
1
( 2π i9)
3
= 6π ft
length of EF =
The length of ROT is 116π, what is the radius of
the circle?
length of AB =
60
Given mEF = 120° with
radius = 9 ft.
EXAMPLE C:
B
120
( 2π i24)
360
= 16π ≈ 50.3 cm
Example (c):
A
mROT = 360 − 120° = 240
240
116π =
( 2π ir )
360
4π
116π =
r
3
116π 3
i = r = 87 .
then
1 4π
O
R
Page 11 of 11
120
T