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.
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 a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
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
If one chord is a perpendicular bisector of another chord, then the bisecting chord is a diameter .
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 =
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
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