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
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.
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 ConjectureH
IN the diameter bisects the chord and its arc.Q, KL LZ. If CK = 2x + 3 and CZ = 4x, find x.
x = 1.5
Q
C
Z
K
L
In the diameter bisects the chord and its arc.P, if PM AT, PT = 10, and PM = 8, find AT.
P
A
M
MT = 6
T
AT = 12
If one chord is a the diameter bisects the chord and its arc.perpendicular bisector of another chord, then the bisecting chord is a diameter .
Perpendicular Bisector to a Chord ConjectureIf 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
Segment congruent when they are equidistant from the center.Lengths in Circles
Two chords intersect congruent when they are equidistant from the center.
INSIDE the circle
Type 1:
part
part
part
part
Go down the chord and multiply
12 congruent when they are equidistant from the center.
2x
8
3x
Find the length of DB.A
D
x = 4
C
DB = 20
B
D
A
x – 4
x
C
5
10
x = 8
AC = 13
DB = 14
B
Two secants intersect congruent when they are equidistant from the center.
OUTSIDE the circle
Type 2:
Ex: 3 Solve for x. congruent when they are equidistant from the center.
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. congruent when they are equidistant from the center.
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. congruent when they are equidistant from the center.
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): congruent when they are equidistant from the center.
Secant and Tangent
Ex: 5 Solve for x. congruent when they are equidistant from the center.
x
12
24
(12 + x)
242
=
12
x = 36
576 = 144 + 12x