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Circle Segments and Volume. Chords of Circles Theorem 1. In the same circle, or in congruent circles two minor arcs are congruent if and only if their corresponding chords are congruent. Chord Arcs Conjecture.

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Circle segments and volume

Circle Segments and Volume



In the same circle, or in congruent circles two minor arcs are congruent if and only if their corresponding chords are congruent.


Chord arcs conjecture
Chord Arcs Conjecture are congruent if and only if their corresponding chords are congruent.

In the same circle, two minor arcs are congruent if and only if their corresponding chords are congruent.

IFF

G

and

IFF

and


Solve for x
Solve for x. are congruent if and only if their corresponding chords are congruent.

8x – 7

3x + 3

8x – 7 = 3x + 3

x = 2


Example
Example are congruent if and only if their corresponding chords are congruent.

Find WX.


Example1
Example are congruent if and only if their corresponding chords are congruent.

Find

130º


Chords of circles theorem 2
Chords of Circles Theorem 2 are congruent if and only if their corresponding chords are congruent.


If a diameter is perpendicular to a chord, then it also bisects the chord and its arc.This results in congruent arcs too.Sometimes, this creates a right triangle & you’ll use Pythagorean Theorem.


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.

Perpendicular Bisector of a Chord Conjecture

H


IN the diameter bisects the chord and its arc.Q, KL  LZ. If CK = 2x + 3 and CZ = 4x, find x.

x = 1.5

Q

C

Z

K

L


In the diameter bisects the chord and its arc.P, if PM  AT, PT = 10, and PM = 8, find AT.

P

A

M

MT = 6

T

AT = 12


Chords of circles theorem 3
Chords of Circles Theorem 3 the diameter bisects the chord and its arc.


Perpendicular bisector to a chord conjecture

If one chord is a the diameter bisects the chord and its arc.perpendicular bisector of another chord, then the bisecting chord is a diameter .

Perpendicular Bisector to a Chord Conjecture

  • is a diameter of the circle.


  • the diameter bisects the chord and its arc.

  • ,

If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.


Chords of circles theorem 4
Chords of Circles Theorem 4 the diameter bisects the chord and its arc.


In the same circle or in congruent circles two chords are congruent when they are equidistant from the center.


Chord distance to the center conjecture
Chord Distance to the Center Conjecture congruent when they are equidistant from the center.


In k k is the midpoint of re if ty 3x 56 and us 4x find the length of ty
In congruent when they are equidistant from the center.K, K is the midpoint of RE. If TY = -3x + 56 and US = 4x, find the length of TY.

U

T

K

E

x = 8

R

S

Y

TY = 32


Example2
Example congruent when they are equidistant from the center.

30

CE =


Example3
Example congruent when they are equidistant from the center.

LN =

96


Segment congruent when they are equidistant from the center.Lengths in Circles


Two chords intersect congruent when they are equidistant from the center.

INSIDE the circle

Type 1:

part

part

part

part

Go down the chord and multiply


Solve for x. congruent when they are equidistant from the center.

9

6

x

2

x = 3


Find the length of db

12 congruent when they are equidistant from the center.

2x

8

3x

Find the length of DB.

A

D

x = 4

C

DB = 20

B


Find the length of ac and db
Find the length of AC and DB. congruent when they are equidistant from the center.

D

A

x – 4

x

C

5

10

x = 8

AC = 13

DB = 14

B


Two secants intersect congruent when they are equidistant from the center.

OUTSIDE the circle

Type 2:


Ex: 3 Solve for x. congruent when they are equidistant from the center.

B

13

A

7

E

4

C

x

D

7

(7 + 13)

=

4

(4 + x)

x = 31

140 = 16 + 4x

124 = 4x


Ex: 4 Solve for x. congruent when they are equidistant from the center.

B

x

A

5

D

8

6

C

E

6

(6 + 8)

=

5

(5 + x)

84 = 25 + 5x

x = 11.8

59 = 5x


Ex: 5 Solve for x. congruent when they are equidistant from the center.

B

10

A

x

D

4

8

C

E

x

(x + 10)

=

(8 + 4)

8

x2+10x = 96

x = 6

x2 +10x – 96 = 0


Type 2 (with a twist): congruent when they are equidistant from the center.

Secant and Tangent


Ex: 5 Solve for x. congruent when they are equidistant from the center.

x

12

24

(12 + x)

242

=

12

x = 36

576 = 144 + 12x


Ex: 6 congruent when they are equidistant from the center.

5

15

x

(5 + 15)

x2

=

5

x2 = 100

x = 10


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