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

CB

A

   IFF

 

IFF

andG

and

   

8x – 7 3x + 3

8x – 7 = 3x + 3

Solve for x.

x = 2

Find WX.4 2 3y y

4 3y

7y11WX cm

Example

Find mAB

130º

Example

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.

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. E

D

G

FH

 

 

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

K

Q

C

L

Z

x = 1.52 3 4x x

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

T

AM

P

MT = 6AT = 12

22 28 10MT

Chords of Circles Theorem 3

Perpendicular Bisector to a Chord Conjecture

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

JK is a diameter of the circle.

J

L

K

M

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

first chord is a diameter.

E

D

G

FDG

GF

, DE EF

Chords of Circles Theorem 4

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

F

G

E

B

A

C

D

In K, K is the midpoint of RE. If TY = -3x + 56 and US = 4x, find the length of TY.

Y

T

S

Kx = 8

U

RE

3 56 4x x 56 7x

TY = 32

Example

CE =302 2 220 25x

15x

Example

LN = 962 2 214 50x

48x

Segment Lengths

in Circles

partpart

partpart

part part = part part Go down the chord and multiply

9

2

6x

x = 3

Solve for x.

9 2 6x 18 6x

Find the length of DB.

8

122x

3x x = 4

A

B

C

D

12 8 3 2x x 296 6x

216 x

DB = 20

Find the length of AC and DB.

x = 8

x5

x – 4

10

A

B

C

D 5 10 4x x 5 10 40x x

5 40x

AC = 13

DB = 14

outside whole = outside whole

EA

B

C

D

7 13

4

x

7(7 + 13) 4 (4 + x)=

Ex: 3 Solve for x.

140 = 16 + 4x124 =

4xx = 31

E

A

B

CD 8

5

6

x

6 (6 + 8)

5(5 + x)=

Ex: 4 Solve for x.

84 = 25 + 5x59 = 5x x =

11.8

E

A

B

CD 4

x

8

10

x (x + 10)

8(8 + 4)=

Ex: 5 Solve for x.

x2+10x = 96x2 +10x – 96 =

0x = 6

2tan = outside whole

24

12 x

242 = 12 (12 + x)576 = 144 + 12x x = 36

Ex: 5 Solve for x.

155

x

x2 = 5 (5 + 15)x2 = 100x = 10

Ex: 6