Transcript Geometry 10.2 Arcs and Chords Geometry Mrs. Spitz Spring 2005 Geometry Objectives/Assignment • Use properties of arcs of circles, as applied. • Use properties of chords of circles. •

Geometry
10.2 Arcs and Chords
Geometry
Mrs. Spitz
Spring 2005
Geometry
Objectives/Assignment
• Use properties of arcs of circles, as
applied.
• Use properties of chords of circles.
• Assignment: pp. 607-608 #3-47
• Reminder Quiz after 10.3 and 10.5
Geometry
Using Arcs of Circles
• In a plane, an angle
whose vertex is the
center of a circle is
a central angle of
the circle. If the
measure of a
central angle, APB
is less than 180°,
then A and B and the
points of P
central angle
A
major
arc
minor
arc
P
B
C
Geometry
Using Arcs of Circles
• in the interior of APB
form a minor arc of the
circle. The points A and
B and the points of
P
in the exterior of APB
form a major arc of the
circle. If the endpoints
of an arc are the
endpoints of a diameter,
then the arc is a
semicircle.
central angle
A
major
arc
minor
arc
P
B
C
Naming Arcs
Geometry
G
• Arcs are named by
their endpoints. For
example, the minor
arc associated with
APB above is AB.
Major arcs and
semicircles are
named by their
endpoints and by a
point on the arc.

60°
60°
E
H
F
E
180°
Naming Arcs
Geometry
G
• For example, the
major arc
associated with
APB is ACB .
E GFhere on the
right is a semicircle.
The measure of a
minor arc is defined
to be the measure of
its central angle.


60°
60°
E
H
F
E
180°
Geometry
Naming Arcs

• For instance, m GF=
mGHF = 60°.
• m GF is read “the
measure of arc GF.”
You can write the
measure of an arc next
to the arc. The
measure of a
semicircle is always
180°.

G
60°
60°
E
H
F
E
180°
Geometry
Naming Arcs

G
• The measure of a GF
major arc is defined as
the difference between E
360° and the measure
of its associated minor
arc. For example, m GEF
= 360° - 60° = 300°.
The measure of the
whole circle is 360°.
60°
60°

H
F
E
180°
Geometry
Ex. 1: Finding Measures of Arcs
•
Find the measure
of each arc of R.

a. MN
b. MPN
c. PMN

N
80°
R
M
P
Geometry
Ex. 1: Finding Measures of Arcs
•
Find the measure
of each arc of R.

a. MN
b. MPN
c. PMN
Solution:
MN is a minor arc, so
m MN = mMRN
= 80°

 
N
80°
R
M
P
Geometry
Ex. 1: Finding Measures of Arcs
•
Find the measure
of each arc of R.

a. MN
b. MPN
c. PMN
Solution:
MPN is a major arc, so
m MPN = 360° – 80°
= 280°

 
N
80°
R
M
P
Geometry
Ex. 1: Finding Measures of Arcs
•
Find the measure
of each arc of R.

a. MN
b. MPN
c. PMN
Solution:
PMN is a semicircle, so
m PMN = 180°

 
N
80°
R
M
P
Note:
C
Geometry
A
• Two arcs of the same
circle are adjacent if
they intersect at exactly
one point. You can add
the measures of
adjacent areas.
• Postulate 26—Arc
Addition Postulate. The
measure of an arc
formed by two adjacent
arcs is the sum of the
measures of the two
arcs.

 
B
m ABC = m AB + m BC
Geometry
Ex. 2: Finding Measures of Arcs
•
G
Find the measure of
each arc.

  
a. GE
b. GEF
c. GF
m GE = m GH + m HE =
40° + 80° = 120°
H
40°
80°
R
110°
F
E
Geometry
Ex. 2: Finding Measures of Arcs
•
G
Find the measure of
each arc.

  
a. GE
b. GEF
c. GF
m GEF = m GE + m EF
120° + 110° = 230°
H
40°
80°
R
110°
=
F
E
Geometry
Ex. 2: Finding Measures of Arcs
•
G
Find the measure of
each arc.


a. GE
b. GEF
c. GF
m GF = 360° - m GEF =
360° - 230° = 130°
H
40°

80°
R
110°
F
E
Geometry
Ex. 3: Identifying Congruent Arcs
• Find the measures
of the blue arcs.
Are the arcs
congruent?
•
 
 
AB and DC are in
the same circle and
m AB = m DC = 45°.
So, AB  DC
A
D
B
45°
45°
C
Geometry
Ex. 3: Identifying Congruent Arcs
• Find the measures
of the blue arcs.
Are the arcs
congruent?
•
 
 
P Q and RS are in
congruent circles and
m P Q = m RS = 80°.
So, P Q  RS
80°
Q
P
80°
S
R
Geometry
Ex. 3: Identifying Congruent Arcs
• Find the measures of
the blue arcs. Are the
arcs congruent?


 

Z
X
• m XY = m ZW = 65°, but
XYand ZW are not arcs of the
same circle or of
congruent circles, so XY
and ZW are NOT
congruent.

65°
Y
W
Geometry
Using Chords of Circles
• A point Y is called
the midpoint of
if XY  YZ . Any
line, segment, or ray
that contains Y
bisects X YZ .
 

Geometry
Theorem 10.4
• In the same circle, or in
congruent circles, two
minor arcs are
congruent if and only if
their corresponding
chords are congruent.

AB  BC if and only if
AB  BC
A
C
B
Geometry
Theorem 10.5
• If a diameter of a
circle is
perpendicular to a
chord, then the
diameter bisects the
chord and its arc.
F
E
 
G
DE  EF ,
DG  GF
D
Geometry
Theorem 10.5
• If one chord is a
perpendicular
bisector of another
chord, then the first
chord is a diameter.
J
M
JK is a diameter of
the circle.
K
L
Ex. 4: Using Theorem 10.4
Geometry
(x + 40)°
• You can use
Theorem 10.4 to
find m AD .



 
C
A
• Because AD  DC,
and AD  DC . So,
m AD = m DC
2x = x + 40
x = 40
2x°
B
Substitute
Subtract x from each
side.
Geometry
Ex. 5: Finding the Center of a
Circle
• Theorem 10.6 can
be used to locate a
circle’s center as
shown in the next
few slides.
• Step 1: Draw any
two chords that are
not parallel to each
other.
Geometry
Ex. 5: Finding the Center of a
Circle
• Step 2: Draw the
perpendicular
bisector of each
chord. These are
the diameters.
Geometry
Ex. 5: Finding the Center of a
Circle
• Step 3: The
perpendicular
bisectors intersect
at the circle’s center.
center
Geometry
Ex. 6: Using Properties of Chords
• Masonry Hammer. A
masonry hammer has
a hammer on one end
and a curved pick on
the other. The pick
works best if you
swing it along a
circular curve that
matches the shape of
the pick. Find the
center of the circular
swing.
Geometry
Ex. 6: Using Properties of Chords
• Draw a segment AB,
from the top of the
masonry hammer to
the end of the pick.
Find the midpoint C,
and draw
perpendicular bisector
CD. Find the
intersection of CD with
the line formed by the
handle. So, the center
of the swing lies at E.
Geometry
Theorem 10.7
• In the same circle,
or in congruent
circles, two chords
are congruent if and
only if they are
equidistant from the
center.
• AB  CD if and only
if EF  EG.
C
G
D
E
B
F
A
Geometry
Ex. 7: Using Theorem 10.7
AB = 8; DE = 8, and
CD = 5. Find CF.
A
8 F
B
C
E
5
8
G
D
Geometry
Ex. 7: Using Theorem 10.7
Because AB and DE
are congruent
chords, they are
equidistant from the
center. So CF 
CG. To find CG,
first find DG.
CG  DE, so CG
bisects DE.
Because
DE = 8,
8
DG = 2 =4.
A
8 F
B
C
E
5
8
G
D
Geometry
Ex. 7: Using Theorem 10.7
Then use DG to find
CG. DG = 4 and
CD = 5, so ∆CGD is
a 3-4-5 right
triangle. So CG = 3.
Finally, use CG to
find CF. Because
CF  CG, CF = CG
=3
A
8 F
B
C
E
5
8
G
D
Geometry
Reminders:
• Quiz after 10.3
• Last day to check Vocabulary from
Chapter 10 and Postulates/Theorems
from Chapter 10.