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