Download Ch 6 Note Sheets Key - Palisades School District

Ch 6 Note Sheet L1 Shortened Key.doc
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.
E
radii: AP , PR , PB
diameter: AB
chords: AB , CD , AF
secant: AG or AG
tangent: EB or EB or BE …
Arcs & Angles
semicircles: ACB , ARB
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.
minor arcs: AC , AR , RD , BC ,…
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
major arcs: ARC , ARD , FRD ,
BAC ,…
central angles with their intercepted
arcs:
mAPR  mAR
mRPB  mRB
Page 1 of 12
Ch 6 Note Sheet L1 Shortened Key.doc
Lesson 6.1 Tangent Properties
Investigation 1: See Worksheet page 1.
Tangent Conjecture
A tangent to a circle is
perpendicular to
the radius drawn to the
point of tangency.
Name ___________________________
Converse of the Tangent
Conjecture
A line that is perpendicular
to a radius at its endpoint
on the circle is tangent to
the circle.
O
T
N
Investigation 2: See Worksheet page 1.
Tangent Segments Conjecture
Tangent segments to a circle from a point outside
the circle are congruent.
A
E
G
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,
x  180  a .
x
a
a
b
“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 2 of 12
Ch 6 Note Sheet L1 Shortened Key.doc
Name ___________________________
Example 1
Example 2
Rays r as s are tangents. w = ?
70
5 cm
70
110
126
250
180 – 54 = w using formula
OR
Tangent  radius, so mONM  mOPM  90
Quad. sum = 360 so 360  180  110  x , x  70
y  5 cm Tangents to a circle from a point are congruent.
mNP  110 Measure of a central angle equals its intercepted
arc.
central angle = w
now use the kite formed by =
tangent segments and = radii
360 – 90 – 90 – 54 = w
126 = w
mPQN  360  110  250 Total degrees in a circle = 360
Example 3
AD is tangent to both circle B and circle C.
w  100
mAXT  260
100
260
mA  90 and mD  90 Tangent  radius, Quadrilateral sum = 360 so w  360  90  90  80  100
w  mAT  100 Measure of central angle = intercepted arc and Circle’s degree = 360 so
mAXT  360  100  260
S. Stirling
Page 3 of 12
Ch 6 Note Sheet L1 Shortened Key.doc
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.
Measure of a central angle =
measure of its intercepted arc.
i.e. mBOA  mAB
Inscribed angle
An angle whose vertex lies on a
circle and whose sides are chords of
the circle.
Name ___________________________
Examples:
Non-Examples:
R
D
P
O
Q
A
S
T
B
AOB , DOA and
DOB are central
PQR , PQS , RST ,
QST and QSR are NOT
angles of circle O.
central angles of circle O.
Examples:
Non-Examples:
Q
A
C
V
R
P
B
E
T
W
X
D
U
S
ABC , BCD and
CDE are inscribed
PQR , STU , and VWX
angles.
are NOT inscribed angles.
Investigation 3: See Worksheet page 4 top.
Example 1
Chord Conjectures
If two chords in a circle are congruent, then
 they determine two central angles that
are congruent.
 their intercepted arcs are congruent.
S. Stirling
B
D
If AB  CD , then
O
BOA  COD .
If AB  CD , then
A
AB  CD
Page 4 of 12
C
Ch 6 Note Sheet L1 Shortened Key.doc
Lesson 6.3 Arcs and Angles
Read top of page 324.
Review (from Chapter 1.7):
A circle measures 360.
Name ___________________________
C
Example 1
If mCOR  92 ,
then mCAR  ? .
92
92
O 46
A
R
A semicircle measures 180.
mCR  92 .
The central angle equals its intercepted arc.
Investigation 4: See Worksheet page 6.
If mCR  92 , then mCAR  46 .
The inscribed angle equals half its
intercepted arc.
Inscribed Angle Conjecture
The measure of an angle inscribed in a circle
equals half the measure of its intercepted arc.
Example 2
mAB  170 , find mAPB and mAQB .
85
mAPB  1 170    85
2
mAQB  1 170    85
2
A
P
85
Q
170
An inscribed angle measures ½ the intercepted arc.
B
Example 3
mAB  mCD  42 , find mAPB and mCQD
mAPB  1 42  21
2
1
mCQD 
42  21
2
The inscribed angle equals half its intercepted arc.
A
P
42
21
B
O
C
42
21
Q
D
Example 4
ACD , ABD and AED intercept semicircle ARD .
R
A
180
Find the measure of each angle.
All are 1 180  90
2
mACD  mABD  mAED  90
.
S. Stirling
C
O
90
90
B
Page 5 of 12
90
E
D
Ch 6 Note Sheet L1 Shortened Key.doc
Name ___________________________
Investigation 5: 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.
N
Chords: EC , TN because they
begin and end on the circle.
S
A
C
E
Investigation 6: See Worksheet page 7.
Parallel Lines [Secants] Intercepted Arcs
Conjecture
Parallel lines [secants] intercept congruent
arcs on a circle.
If AB DC ,
then mAD  mBC  64 .
D
64
A
Note: mAB  mDC .
S. Stirling
Page 6 of 12
C
B
64
Ch 6 Note Sheet L1 Shortened Key.doc
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  mAB
B
On the circle
2 chords
“Inscribed
angle”
angle = ½ (intercepted arc)
C
E
m2  1 mCE
2
2
D
1 chord &
1 tangent
m2  1 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
 mJK  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
 mNP  mOQ 
m4  1  mWS  mSU 
2
m4  1
2
S
2 tangents
m4  1
Z
X
4
Y
2
 mYAZ  mYZ 
A
Summary:
Vertex at center
→ angle = arc
Vertex on
→ angle = ½ arc
Vertex inside
→ angle = ½ sum of arcs
Vertex outside
→ angle = ½ difference between arcs
S. Stirling
Page 7 of 12
Ch 6 Note Sheet L1 Shortened Key.doc
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
ALL problems must be completed by the process shown in Example A!!
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 d
C   3
C  3
C  9.4 m
C  2 r
12  2 r
12 2 r

2
2
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
Example C:
If a circle has a circumference of 20 meters, what
is its diameter?
C d
20   d
20
d

write the formula
substitute
accurate answer
d  6.366 m approximate answer
S. Stirling
Page 8 of 12
Ch 6 Note Sheet L1 Shortened Key.doc
Lesson 6.6 Around the World Read top of page 335.
Name ___________________________
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 or Physics 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. (You’re 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 its value. So if I multiply by
fractions that equal 1, I don’t change the
quantity just the look of it.
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 together and then
the denominators together, then divide the
numerator by the denominator.
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 12
Ch 6 Note Sheet L1 Shortened Key.doc
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 d
C  8000 miles
average speed = mph
24 hr
80 day 
 1920 hrs
1 day
8000 mi
 13.09 mph
1920 hrs
OR
80 day 
24 hr
 1920 hrs
1 day
8000 mi 1 day
mi

 13.09
80 day 24 hr
hr
OR for an exact answer…
8000 mi 1 day

=
80 day 24 hr
100 mi 1

=
1
24 hr
25 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  2 112  2  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 12
Ch 6 Note Sheet L1 Shortened Key.doc
Lesson 6.7 Arc Length
Investigation 10: See Worksheet page 14.
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
Example (b): Given central COD  180 with
radius 12. Find the length of AB .
diameter 15 cm. Find the length of CED .
B
90
mAB  90
fraction of circle =
length of AB 

90 1

360 4
90
T
12
mCED  180
A
90
 2 12 
360
1
 24   6
4
180 1

fraction of circle =
360 2
180
length of CED 
15 
360
15
  cm
2
D
Example (c): Given mEF  120 with radius =
9 ft. Find the length of EF .
120
F
fraction of circle =
120 1

360 3
1
 2 9 
3
 6 ft
O
9
E
length of EF 
S. Stirling
Page 11 of 12
C
O
E
Ch 6 Note Sheet L1 Shortened Key.doc
Name ___________________________
Example (d): If the radius of a circle is 24 cm
and mBTA  60 , what is the length of AB ?
mAB  120
Example (e): The length of ROT is 116π, what
is the radius of the circle?
O
B
120
60
mROT  360  120  240
R
A
120
T
degree
 Circumference
360
120
length of AB 
 2 24 
360
 16  50.3 cm
arc length 
S. Stirling
arc length 
degree
 Circumference
360
240
 2 r 
360
4
116 
r
3
116 3
r
then
1 4
r  87
116 
Page 12 of 12
T