Geometry
9.3 Arcs and Chords
Geometry
Objectives/Assignment
• Use properties of arcs of circles.
• Use properties of chords of circles.
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.
central angle
A
major
arc
minor
arc
P
B
C
Geometry
Using Arcs of Circles
• minor arc of the circle is
less than 180
• major arc of the circle is
greater than 180
• semicircle if the
endpoints of an arc are
the endpoints of a
diameter.
central angle
A
major
arc
minor
arc
P
B
C
Naming Arcs
Geometry
G
The measure of a
minor arc is defined
to be the measure of
its central angle.

60°
60°
E
H
F
m GF = m  GHF = 60°.
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
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
Geometry
Theorem 9.1
• 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 9.2
• 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
• 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
Using thisTheorem
(x + 40)°
Geometry
D
• You can use
Theorem 9.3 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
Theorem 9.3
• 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
Using Theorem 9.4
AB = 8; DE = 8, and
CD = 5. Find CF.
A
8 F
B
C
E
5
8
G
D
Geometry
Using Theorem 9.4
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
Using Theorem 9.4
Then use DG to find
CG. DG = 4 and
CD = 5, so ∆CGD is
right triangle.
A
8 F
B
C
E
So CG = 3
CF = CG = 3
5
8
G
D
Geometry
Practice
•360 – 120 =240
•240/3 = 80
Geometry
Practice
•12 * 2 = 24
Geometry
Practice
•BY² = 20² - 12²
•BY = 16
•XY = 16*2 = 32
Geometry
•BC = 360 / 5 = 72
Geometry
• MX = 12
•XN = 12
•MN = PQ
•YQ = 12
Geometry
• Radius² = 8² + 6²
•R² = 100
•R = 10 inches
16
6
•
Geometry
• Chord ² =13² - 5²
•Chord ² = 144
•C= 12 + 12
•C= 24 inches
5
13
•
Geometry
• Using Pythagorean Formula
•Distance ² = 17 ² - 15 ²
•D ² = 64
•Distance = 8 inches
30
17
•
Geometry
• Using Pythagorean Formula
•Radius = 6 inches
• APB = 120°
•PAB isosceles triangle
• A= (180 ° - 120 °)/2
• A = 30 °
•AB = 5.2 +5.2 = 10.4
•Cos 30 = x/ 6
•X = 6*Cos 30 = 5.2
Geometry
• PD = PC = PB = Radius
•CE = 2
•EP = 5 -2 = 3
•Using Pythagorean Formula for EPB
•EB ² = PB ² - PE ²
•EB ² = 5 ² - 3 ²
•EB ²= 16
•EB = 4
•AB = AE + EB = 4+ 4 = 8