lesson 6.2 properties of chords
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Objective : Discover properties of chords of a circle. Lesson 6.2 Properties of Chords. Homework: Lesson 6.2/1-12. What is a chord? A chord is a segment with endpoints on a circle. Any chord divides the circle into two arcs. A diameter divides a circle into two semicircles. - PowerPoint PPT PresentationTRANSCRIPT
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Lesson 6.2 Properties of Chords
Homework: Lesson 6.2/1-12
Objective: Discover properties of chords of a circle
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What is a chord? A chord is a segment with endpoints on a circle.
Any chord divides the circle into two arcs. A diameter divides a circle into two semicircles. Any other chord divides a circle into a minor arc and a major arc.
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Chord Arcs Conjecture
In the same circle, two minor arcs are congruent if and only if their corresponding chords are congruent.
CB
A
IFF
IFF
and
G
and
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Perpendicular Bisector of a Chord Conjecture
If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
E
D
G
FH
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JK is a diameter of the circle.
J
L
K
M
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Perpendicular Bisector to a Chord ConjectureIf one chord is a perpendicular bisector of another chord, then the first chord passes through the center of the circle and is a diameter.
JK is a diameter of the circle.
J
L
K
M
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If one chord is a perpendicular bisector of another chord, then the
first chord is a diameter.
E
D
G
FDG
GF
, DE EF
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Ex. 4: Using Chord Arcs Conjecture
2x = x + 40 x = 40
BA
C
2x°(x + 40)°D
A
C
B
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Ex. 5: Finding the Center of a Circle
• Perpendicular bisector to a chord 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.
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Ex. 5: Finding the Center of a Circle
Step 2: Draw the perpendicular bisector of each chord. These are the diameters.
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Ex. 5: Finding the Center of a Circle
Step 3: The perpendicular bisectors intersect at the circle’s center.
center
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Chord Distance to the Center Conjecture
F
G
E
B
A
C
D
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F
G
E
B
A
C
D
AB CD if and only if EF EG.
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Ex. 7: AB = 8; DE = 8, and CD = 5. Find CF.
58
8 F
G
C
E
D
A
B
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2x = x + 40 x = 40
BA
C
2x°
(x + 40)°D
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Ex.4: Solve for the missing sides.
A
B
D C
7m
3m
BC =
AB =
AD ≈
7m
14m
7.6m
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.AC
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Ex.6: QR = ST = 16. Find CU.
x = 3
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Ex 7: AB = 8; DE = 8, and CD = 5. Find CF.
58
8 F
G
C
E
D
A
B
CG = CF
CG = 3 = CF
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Ex.8: Find the length of Tell what theorem you used. .BF
BF = 10
Diameter is the perpendicular bisector of
the chordTherefore, DF = BF
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Ex.9: PV = PW, QR = 2x + 6, and ST = 3x – 1. Find QR.
Congruent chords are equidistant from the center.
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Congruent chords intercept congruent arcs
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Ex.11:
Congruent chords are equidistant from the center.
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