Circle Segments and Volume

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# Circle Segments and Volume - PowerPoint PPT Presentation

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
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

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.

8x – 7

3x + 3

8x – 7 = 3x + 3

x = 2

Example

Find WX.

Example

Find

130º

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 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.

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.

U

T

K

E

x = 8

R

S

Y

TY = 32

Example

30

CE =

Example

LN =

96

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

Find the length of DB.

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