10.3 Arcs and Chords

10.3 ARCS AND CHORDS

GEOMETRY

OBJECTIVES/ASSIGNMENTUse properties of arcs of circles, as applied.Use properties of chords of circles.Assignment: pp. 607-608 #3-47Reminder Quiz after 10.3 and 10.5

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

USING ARCS OF CIRCLES

central angle

minorarcmajor

arcP

B

A

C

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.

USING ARCS OF CIRCLES

central angle

minorarcmajor

arcP

B

A

C

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°

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.

EGF

EH F

G

E

60°

60°

180°

ACB

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

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°.

GEF

EH F

G

E

60°

60°

180°

GF

EX. 1: FINDING MEASURES OF ARCSFind the measure

of each arc of R.

a. b. c.

MNMPN

PMN PR

M

N 80°


of each arc of R.

a. b. c. Solution:

is a minor arc, so m = mMRN = 80°

MNMPN

PMN PR

M

N 80°

MN

MN


of each arc of R.

a. b. c. Solution:

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

MNMPN

PMN PR

M

N 80°

MPN

MPN


of each arc of R.

a. b. c. Solution:

is a semicircle, so m = 180°

MNMPN

PMN PR

M

N 80°

PMN

PMN

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

AB

B

A

C

m = m + m BC

EX. 2: FINDING MEASURES OF ARCSFind the measure of

each arc.a. b. c. m = m + m

=40° + 80° = 120°

GE

GEFR

EF

G

H

GFGE

GH

HE

40°

80°

110°


each arc.a. b. c. m = m + m

=120° + 110° = 230°

GE

GEFR

EF

G

H

GF

EF

40°

80°

110°GEF

GE


each arc.a. b. c. m = 360° - m

=360° - 230° = 130°

GE

GEFR

EF

G

H

GF

40°

80°

110°GF

GEF

EX. 3: IDENTIFYING CONGRUENT ARCSFind 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°

Q

S

P

R

EX. 3: IDENTIFYING CONGRUENT ARCSFind 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°

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 ARCSFind the measures of

the blue arcs. Are the arcs congruent?

XY

65°ZW

XY

ZW

XY

ZW

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

THEOREM 10.3

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

CB

A

If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.

THEOREM 10.5

E

D

G

F

DG

GF

, DE EF

If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.

THEOREM 10.4

JK is a diameter of the circle.

J

L

K

M

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

AD DC

2x = x + 40 Substitute x = 40 Subtract x from each

side.

BA

C

2x°

(x + 40)°

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.

EX. 5: FINDING THE CENTER OF A CIRCLE

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


Step 3: The perpendicular bisectors intersect at the circle’s center.


center

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.

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.

EX. 6: USING PROPERTIES OF CHORDS

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

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


58

8 F

G

C

E

D

A

B


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.

28

58

8 F

G

C

E

D

A

B


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

REMINDERS:Quiz after 10.3 Last day to check Vocabulary from Chapter 10 and

Postulates/Theorems from Chapter 10.

Documents

10.3 Arcs and Chords