Circle Segments and Volume
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Transcript of Circle Segments and Volume
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Circle Segments
and Volume
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Chords of Circles Theorem 1
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In the same circle, or in congruent circles two minor arcs are congruent if and only if their corresponding chords are congruent.
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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
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8x – 7 3x + 3
8x – 7 = 3x + 3
Solve for x.
x = 2
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Find WX.4 2 3y y
4 3y
7y11WX cm
Example
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Find mAB
130º
Example
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Chords of Circles Theorem 2
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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.
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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
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IN Q, KL LZ. If CK = 2x + 3 and CZ = 4x, find x.
K
Q
C
L
Z
x = 1.52 3 4x x
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In P, if PM AT, PT = 10, and PM = 8, find AT.
T
AM
P
MT = 6AT = 12
22 28 10MT
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Chords of Circles Theorem 3
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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
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If one chord is a perpendicular bisector of another chord, then the
first chord is a diameter.
E
D
G
FDG
GF
, DE EF
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Chords of Circles Theorem 4
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In the same circle or in congruent circles two chords are congruent when they are equidistant from the center.
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•
Chord Distance to the Center Conjecture
F
G
E
B
A
C
D
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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
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Example
CE =302 2 220 25x
15x
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Example
LN = 962 2 214 50x
48x
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Segment Lengths
in Circles
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partpart
partpart
part part = part part Go down the chord and multiply
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9
2
6x
x = 3
Solve for x.
9 2 6x 18 6x
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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
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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
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outside whole = outside whole
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EA
B
C
D
7 13
4
x
7(7 + 13) 4 (4 + x)=
Ex: 3 Solve for x.
140 = 16 + 4x124 =
4xx = 31
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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
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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
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2tan = outside whole
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24
12 x
242 = 12 (12 + x)576 = 144 + 12x x = 36
Ex: 5 Solve for x.
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155
x
x2 = 5 (5 + 15)x2 = 100x = 10
Ex: 6