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Properties of a Chord

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Circle Geometry Properties of a Chord Homework: Lesson 6.2/1-12, 18 Quiz Friday Lessons 6.1 6.2 Ying Yang Project Due Friday – PowerPoint PPT presentation

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Title: Properties of a Chord


1
Properties of a Chord
  • Circle Geometry

Homework Lesson 6.2/1-12, 18 Quiz Friday Lessons
6.1 6.2 Ying Yang Project Due Friday
2
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.

3
What is a chord?
  • The diameter of a circle is the longest chord of
    any circle since it passes through the center.
  • A diameter satisfies the definition of a chord,
    however, a chord is not necessarily a diameter.

Therefore, every diameter is a chord, but not
every chord is a diameter.
4
Chord Property 1, 2 and 3
  • A perpendicular line from the center of a chord
    to the center of a circle
  • 1 Makes a 90 angle with the chord
  • 2 Creates two equal line segments RS and QR
  • 3 Must pass through the center of the circle O

5
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6
is a diameter of the circle.
7
If one chord is a perpendicular bisector of
another chord, then the first chord is a
diameter.
8
AB ? CD if and only if EF ? EG.
9
Chord Arcs Conjecture
  • In the same circle, or in congruent circles, two
    minor arcs are congruent if and only if their
    corresponding chords are congruent.

10
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.

11
Perpendicular Bisector to a Chord Conjecture
  • If one chord is a perpendicular bisector of
    another chord, then the first chord passes
    through the center of the circle and is a
    diameter.

is a diameter of the circle.
12
Ex Using Chord Arcs Conj.
(x 40)
2x
2x x 40
Substitute
Subtract x from each side.
x 40
13
Ex Finding the Center of a Circle
  • Perpendicular Bisector to a Chord Conjecture can
    be used to locate a circles center as shown in
    the next few slides.
  • Step 1 Draw any two chords that are not
    parallel to each other.

14
Ex Finding the Center of a Circle cont
  • Step 2 Draw the perpendicular bisector of each
    chord. These are the diameters.

15
Ex Finding the Center of a Circle cont
  • Step 3 The perpendicular bisectors intersect at
    the circles center.

16
Chord Distance to the Center Conjecture
  • 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.

17
Ex Find CF
  • AB 8 DE 8, and CD 5. Find CF.

18
Ex Find CF cont
  • 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, DG
    4.

19
Ex Find CF cont
  • 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

20
 
 
21
An angle whose vertex is on the center of the
circle and whose sides are radii of the circle
An angle whose vertex is ON the circle and whose
sides are chords of the circle
22
Naming Arcs
Arcs are defined by their endpoints
Minor Arcs require the 2 endpoints
 
Major Arcs require the 2 endpoints AND a point
the arc passes through
 
Semicircles also require the 2 endpoints
(endpoints of the diameter) AND a point the arc
passes through
K
 
23
  •  

24
  •  

(x 40)
D
2x
 
2x x 40
x 40
25
Ex.3 Solve for the missing sides.
 
A
7m
C
3m
D
BC AB AD
7m 14m 7.6m
B
26
  •  

27
What can you tell me about segment AC if you know
it is the perpendicular bisectors of segments DB?
D
Its the DIAMETER!!!
A
C
B
28
Finding the Center of a Circle
  • If one chord is a perpendicular bisector of
    another chord, then the first chord is a
    diameter.
  • This can be used to locate a circles center as
    shown in the next few slides.
  • Step 1 Draw any two chords that are not
    parallel to each other.

29
Finding the Center of a Circle
  • Step 2 Draw the perpendicular bisector of each
    chord.
  • These are diameters of the circle.

30
Finding the Center of a Circle
  • Step 3 The perpendicular bisectors intersect at
    the circles center.

31
Ex.6 QR ST 16. Find CU.
 
 
x 3
32
Ex 7 AB 8 DE 8, and CD 5. Find CF.
CG CF
 
 
 
 
CG 3 CF
33
  • Ex.8 Find the length of
  • Tell what theorem you used.

BF 10
Diameter is the perpendicular bisector of the
chord Therefore, DF BF
34
Ex.9 PV PW, QR 2x 6, and ST 3x 1. Find
QR.
 
 
 
 
 
Congruent chords are equidistant from the center.
35
  •  

Congruent chords intercept congruent arcs
 
36
Ex.11
Congruent chords are equidistant from the center.
 
 
 
 
 
37
Lets get crazy
Find c.
3
8
c
12
38
Lets get crazy
Find c.
3
8
c
12
Step 1 What do we need to find?
We need a radius to complete this big triangle.
39
Lets get crazy
Find c.
3
8
c
12
How do we find a radius?
We can draw multiple radii (radiuses).
40
Lets get crazy
Find c.
5
4
3
8
c
12
How do we find a radius?
Now what do we have and what will we do?
We create this triangle, use the chord property
to find on side the Pythagorean theorem find
the missing side.
41
Lets get crazy
Find c.
5
4
3
13
c
12
!!! Pythagorean Theorem !!!
c2a2b2 c252122 c2169 c 13.
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