# Circle Segments and Volume - PowerPoint PPT Presentation

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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

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#### Presentation Transcript

Circle Segments and Volume

### 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

8x – 7

3x + 3

8x – 7 = 3x + 3

x = 2

Find WX.

Find

130º

### Chords of Circles Theorem 2

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.

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 Q, KL  LZ. If CK = 2x + 3 and CZ = 4x, find x.

x = 1.5

Q

C

Z

K

L

In P, if PM  AT, PT = 10, and PM = 8, find AT.

P

A

M

MT = 6

T

AT = 12

### Chords of Circles Theorem 3

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.

• ,

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

U

T

K

E

x = 8

R

S

Y

TY = 32

30

CE =

### Example

LN =

96

Segment Lengths in Circles

Two chords intersect

INSIDE the circle

Type 1:

part

part

part

part

Go down the chord and multiply

Solve for x.

9

6

x

2

x = 3

12

2x

8

3x

A

D

x = 4

C

DB = 20

B

### Find the length of AC and DB.

D

A

x – 4

x

C

5

10

x = 8

AC = 13

DB = 14

B

Two secants intersect

OUTSIDE the circle

Type 2:

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

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

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

Type 2 (with a twist):

Secant and Tangent

Ex: 5 Solve for x.

x

12

24

(12 + x)

242

=

12

x = 36

576 = 144 + 12x

Ex: 6

5

15

x

(5 + 15)

x2

=

5

x2 = 100

x = 10