geometrygeometry arcs and chords geometry. geometry geometry objective use properties of arcs of...
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Arcs and Chords
Geometry
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Objective
• Use properties of arcs of circles to find missing measurements and centers of circles.
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Using Arcs of Circles
• In a plane, an angle whose vertex is the center of a circle is a central angle of the circle. If the measure of a central angle, APB is less than 180°, then A and B and the points of P
central angle
minorarcmajor
arcP
B
A
C
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Inscribed Angles• The measure of an inscribed angle is half the
measure of its intercepted arc
• A central angle has the same measure as its intercepted arc.
• Two inscribed angles that intercept the same arc are congruent.
• An angle inscribed in a semicircle is a right angle.
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Ex. 1: Finding Measures of Arcs
• Find the measure of each arc of R.
a.
b.
c.
Solution:
is a minor arc, so m = mMRN = 80°
MNMPN
PMN PR
M
N80°
MN
MN
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Ex. 1: Finding Measures of Arcs
• Find the measure of each arc of R.
a.
b.
c.
Solution:
is a major arc, so m = 360° – 80° = 280°
MNMPN
PMN PR
M
N80°
MPN
MPN
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Ex. 2: Finding Measures of Arcs
• Find the measure of each arc.
a.
b.
c.
m = m + m =
40° + 80° = 120°
GE
R
EF
G
H
GEFGFGE
GH
HE
40°
80°
110°
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Ex. 3: Identifying Congruent Arcs• Find the measures
of the blue arcs. Are the arcs congruent?
C
D
A
BAB• and are in the same circle and
m = m = 45°. So,
DC
ABDCDC
AB
45°
45°
•Theorem•Within a circle congruent chords have congruent arcs and congruent arcs have congruent central angles.
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Q
S
P
R
Ex. 3: Identifying Congruent Arcs
• Find the measures of the blue arcs. Are the arcs congruent?
RSPQ• and are in
congruent circles and
m = m = 80°. So,
PQ
RSRS
PQ
80°
80°
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Theorem 10.4
• In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.
ABBCif and only if
AB BC
C
B
A
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Ex. 4: Using Theorem 10.4
• You can use Theorem 10.4 to find m .
AD
• Because AD DC, and . So, m = m
AD
DC
ADDC
2x = x + 40 Substitute x = 40 Subtract x from each
side.
BA
C
2x°
(x + 40)°
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Theorem 10.5
• If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
E
D
G
F
DG
GF
, DE EF
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Theorem 10.5
• If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
JK is a diameter of the circle.
J
L
K
M
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Ex. 5: Finding the Center of a Circle
• Theorem 10.6 can be used to locate a circle’s center as shown in the next few slides.
• Step 1: Draw any two chords that are not parallel to each other.
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Ex. 5: Finding the Center of a Circle
• Step 2: Draw the perpendicular bisector of each chord. These are the diameters.
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Ex. 5: Finding the Center of a Circle
• Step 3: The perpendicular bisectors intersect at the circle’s center.
center
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Ex. 6: Using Properties of Chords
• Masonry Hammer. A masonry hammer has a hammer on one end and a curved pick on the other. The pick works best if you swing it along a circular curve that matches the shape of the pick. Find the center of the circular swing.
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Ex. 6: Using Properties of Chords
• Draw a segment AB, from the top of the masonry hammer to the end of the pick. Find the midpoint C, and draw perpendicular bisector CD. Find the intersection of CD with the line formed by the handle. So, the center of the swing lies at E.
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• In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
• AB CD if and only if EF EG.
Theorem 10.7
F
G
E
B
A
C
D
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Ex. 7: Using Theorem 10.7
AB = 8; DE = 8, and CD = 5. Find CF.
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8 F
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Ex. 7: Using Theorem 10.7
Because AB and DE are congruent chords, they are equidistant from the center. So CF CG. To find CG, first find DG.
CG DE, so CG bisects DE. Because DE = 8, DG = =4.
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C
E
D
A
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2
8
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Ex. 7: Using Theorem 10.7
Then use DG to find CG. DG = 4 and CD = 5, so ∆CGD is a 3-4-5 right triangle. So CG = 3. Finally, use CG to find CF. Because CF CG, CF = CG = 3
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