10.2 Arcs and Chords

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10.2 Arcs and Chords

Geometry

Mrs. Spitz

Spring 2005

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Objectives/Assignment

• Use properties of arcs of circles, as applied.

• Use properties of chords of circles.

• Assignment: pp. 607-608 #3-47

• Reminder Quiz after 10.3 and 10.5

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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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Using Arcs of Circles

• in the interior of APB form a minor arc of the circle. The points A and B and the points of P in the exterior of APB form a major arc of the circle. If the endpoints of an arc are the endpoints of a diameter, then the arc is a semicircle.

central angle

minorarcmajor

arcP

B

A

C

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Naming Arcs

• Arcs are named by their endpoints. For example, the minor arc associated with APB above is . Major arcs and semicircles are named by their endpoints and by a point on the arc.

AB

EH F

G

E

60°

60°

180°

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Naming Arcs

• For example, the major arc associated with APB is .

here on the right is a semicircle.

The measure of a minor arc is defined to be the measure of its central angle.

EH F

G

E

60°

60°

180°

ACB

EGF

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Naming Arcs

• For instance, m = mGHF = 60°.

• m is read “the measure of arc GF.” You can write the measure of an arc next to the arc. The measure of a semicircle is always 180°.

EH F

G

E

60°

60°

180°

GF

GF

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Naming Arcs

• The measure of a major arc is defined as the difference between 360° and the measure of its associated minor arc. For example, m = 360° - 60° = 300°. The measure of the whole circle is 360°.

EH F

G

E

60°

60°

180°

GEF

GF

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Ex. 1: Finding Measures of Arcs

• Find the measure of each arc of R.

a.

b.

c.

MNMPN

PMN PR

M

N80°

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a.

b.

c.

Solution:

is a minor arc, so m = mMRN = 80°

MNMPN

PMN PR

M

N80°

MN

MN

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a.

b.

c.

Solution:

is a major arc, so m = 360° – 80° = 280°

MNMPN

PMN PR

M

N80°

MPN

MPN

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a.

b.

c.

Solution:

is a semicircle, so m = 180°

MNMPN

PMN PR

M

N80°

PMN

PMN

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Note:

• Two arcs of the same circle are adjacent if they intersect at exactly one point. You can add the measures of adjacent areas.

• Postulate 26—Arc Addition Postulate. The measure of an arc formed by two adjacent arcs is the sum of the measures of the two arcs.

ABC

B

A

C

m = m + m AB

BC

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• 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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a.

b.

c.

m = m + m =

120° + 110° = 230°

GE

R

EF

G

H

GEFGF

EF

40°

80°

110°GEF

GE

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a.

b.

c.

m = 360° - m =

360° - 230° = 130°

GE

R

EF

G

H

GEFGF

40°

80°

110°GF

GEF

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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°

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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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X

W

Y

Z

• m = m = 65°, but and are not arcs of the same circle or of congruent circles, so and are NOT congruent.

Ex. 3: Identifying Congruent Arcs• Find the measures of

the blue arcs. Are the arcs congruent?

65°XY

ZW

XY

ZW

XY

ZW

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Using Chords of Circles

• A point Y is called the midpoint of if . Any line, segment, or ray that contains Y bisects .

XYZ

XY

YZ

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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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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. 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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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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• Step 2: Draw the perpendicular bisector of each chord. These are the diameters.

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• 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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AB = 8; DE = 8, and CD = 5. Find CF.

58

8 F

G

C

E

D

A

B

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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.

58

8 F

G

C

E

D

A

B

2

8

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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

58

8 F

G

C

E

D

A

B

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Reminders:

• Quiz after 10.3

• Last day to check Vocabulary from Chapter 10 and Postulates/Theorems from Chapter 10.

10.2 Arcs and Chords

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Transcript of 10.2 Arcs and Chords