Unit 4:Arcs and Chords

Keystone Geometry

Measures of an ArcMeasures of an Arc:1. The measure of a minor arc is the measure of its central angle. 2. The measure of a major arc is 360 - (measure of its minor arc). 3. The measure of any semicircle is 180.

Adjacent Arcs:Arcs in a circle with exactly one point in common.

A

B

C

T

P

List: Major Arcs Minor Arcs Semicircles Adjacent Arcs

Example: Find the measure of each arc.4x + 3x + (3x +10) + 2x + (2x-14) = 360°

14x – 4 = 360° so 14x = 364x = 26°

4x = 4(26) = 104°

3x = 3(26) = 78°

3x +10 = 3(26) +10= 88°

2x = 2(26) = 52°

2x – 14 = 2(26) – 14 = 38°

3x+10

2x-14

2x4x

3x

B

D

C

E

A

4

Theorem• In the same circle or in congruent circles, two

minor arcs are congruent if and only if their central angles are congruent.

M

N

R

S

O

5

Arc Addition Postulate• The measure of the arc formed by two

adjacent arcs is the sum of the measures of these two arcs.

H

J G

K

F

6

Example• In circle J, find the measures

of the angle or arc named with the given information:

• Find:

H

J G

K

F

7

In Circle C, find the measure of each arc or angle named.• Given: SP is a diameter of

the circle. Arc ST = 80 and Arc QP=60.

• Find:

T

C

P

S

Q

8

Theorem #1:In a circle, if two chords are congruent then their corresponding minor arcs are congruent.

E

A B

C D

Example:

9

Theorem #2:In a circle, if a diameter (or radius) is perpendicular to a chord, then it bisects the chord and its arc.

E

D

A

C

BExample: If AB = 5 cm, find AE.

10

Theorem #3:In a circle, two chords are congruent if and only if they are equidistant from the center.

O

A B

C

DF

E

Example: If AB = 5 cm, find CD.

Since AB = CD, CD = 5 cm.

CD AB iff OF OE

11

Try Some:• Draw a circle with a chord that is 15 inches long and 8

inches from the center of the circle. • Draw a radius so that it forms a right triangle. • How could you find the length of the radius?

8cm

15cm

O

A BD

∆ODB is a right triangle and OD bisects AB

2 2 2

2 2 2

AB 15DB= = =7.5 cm2 2

OD=8 cmOB =OD +DBOB =8 +(7.5) =64+56.25=120.25OB= 120.25 11 cm

Solution:

x

12

Try Some Sketches:• Draw a circle with a diameter that is 20 cm long.

• Draw another chord (parallel to the diameter) that is 14cm long. • Find the distance from the small chord to the center of Circle O.

Solution:

∆EOB is a right triangle. OB (radius) = 10 cm2 2 2

2 2 2

2

10 7

100 49 51

51

OB OE EB

X

X

X

10 cm10 cm

20cm O

A B

DC

7.1 cm

14 cm

Ex