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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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Presentation Transcript
slide3
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

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.

8x – 7

3x + 3

8x – 7 = 3x + 3

x = 2

example
Example

Find WX.

example1
Example

Find

130º

slide9

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 to a chord conjecture
If one chord is a perpendicular bisector of another chord, then the bisecting chord is a diameter .Perpendicular Bisector to a Chord Conjecture
  • is a diameter of the circle.
slide15

  • ,

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

slide17
In the same circle or in congruent circles two chords are 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 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

30

CE =

example3
Example

LN =

96

slide24

Two chords intersect

INSIDE the circle

Type 1:

part

part

part

part

Go down the chord and multiply

slide25

Solve for x.

9

6

x

2

x = 3

find the length of db

12

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.

D

A

x – 4

x

C

5

10

x = 8

AC = 13

DB = 14

B

slide28

Two secants intersect

OUTSIDE the circle

Type 2:

slide29

Ex: 3 Solve for x.

B

13

A

7

E

4

C

x

D

7

(7 + 13)

=

4

(4 + x)

x = 31

140 = 16 + 4x

124 = 4x

slide30

Ex: 4 Solve for x.

B

x

A

5

D

8

6

C

E

6

(6 + 8)

=

5

(5 + x)

84 = 25 + 5x

x = 11.8

59 = 5x

slide31

Ex: 5 Solve for x.

B

10

A

x

D

4

8

C

E

x

(x + 10)

=

(8 + 4)

8

x2+10x = 96

x = 6

x2 +10x – 96 = 0

slide32

Type 2 (with a twist):

Secant and Tangent

slide33

Ex: 5 Solve for x.

x

12

24

(12 + x)

242

=

12

x = 36

576 = 144 + 12x

slide34

Ex: 6

5

15

x

(5 + 15)

x2

=

5

x2 = 100

x = 10

ad