Arc length, area of a sector and segments of a circle

Objectives

Find the circumference of a circle and

the length of a circular arc.

Use circumference and arc length to

solve real-life problems.

Find the area of a sector and a segment

in a circle.

Find the lengths of segments of chords.

Find the lengths of segments of

tangents and secants.

Finding circumference and arc

length

• The circumference of a circle is the

distance around the circle. For all

circles, the ratio of the circumference

to the diameter is the same. This

ratio is known as or pi.

Theorem:

Circumference of a Circle

• The circumference

C of a circle is C =

d or C = 2r,

where d is the

diameter of the

circle and r is the

radius of the circle. diameter d

Ex. 1: Using circumference

• Find (a) the circumference of a circle

with radius 6 centimeters and (b) the

radius of a circle with circumference

31 meters. Round decimal answers

to two decimal places.

Solution: C = 2r

= 2 • • 6

= 12

37.70

So, the

circumference is

about 37.70 cm.

C = 2r

31 = 2r

31 = r

4.93 r

So, the radius is

about 4.93 cm.

2

a.

b.

And . . .

• An arc length is a portion of the

circumference of a circle. You can

use the measure of an arc (in

degrees) to find its length (in linear

units).

Ex. 2: Finding Arc Lengths

• Find the length of each arc.

5 cm

B

A

50°

a. 7 cm

D

C

50°

b. 7 cm

F

E

100°

c.

Arc Length Corollary

• In a circle, the ratio

of the length of a

given arc to the

circumference is

equal to the ratio of

the measure of the

arc to 360°. AB

Arc length of

2r =

360°

or Arc length of = 360°

• 2r

m AB

AB

m

More . . .

• The length of a

semicircle is half the

circumference, and

the length of a 90°

arc is one quarter of

the circumference.

½ • 2r

r¼ • 2r

r r

Ex. 2: Finding Arc Lengths • Find the length of each arc.

5 cm

B

A

50°

a.

a. Arc length of = AB 50°

360° • 2(5)

a. Arc length of = AB # of °

360° • 2r

4.36 centimeters

Ex. 2: Finding Arc Lengths • Find the length of each arc.

7 cm

D

C

50°

b. b. Arc length of = CD # of °

360° • 2r

b. Arc length of = CD 50°

360° • 2(7)

6.11 centimeters

Ex. 2: Finding Arc Lengths • Find the length of each arc.

7 cm

F

E

100°

c. c. Arc length of = # of °

360° • 2r

c. Arc length of = EF 100°

360° • 2(7)

EF

12.22 centimeters

In parts (a) and (b) in Example 2, note that the

arcs have the same measure but different

lengths because the circumferences of the

circles are not equal.

Ex. 3: Using Arc Lengths • Find the indicated measure.

3.82 m

R

Q

P

60°

a. circumference PQ

Arc length of

2r = PQ

m

360°

3.82

2r 6

1 =

3.82

2r 360°

60° =

3.82(6) = 2r

22.92 = 2r

C = 2r; so using substitution, C = 22.92

meters.

Ex. 3: Using Arc Lengths • Find the indicated measure.

XY

b. m XY

Arc length of

2r = 360°

18

2(7.64) 360° =

135° m

XY

m

XY

m

• 360° 360° •

XY

So the m

135° XY7.64 in.

18 in.

Z

Y

X

Ex. 4: Comparing Circumferences

• Tire Revolutions: Tires

from two different

automobiles are shown

on the next slide. How

many revolutions does

each tire make while

traveling 100 feet?

Round decimal answers

to one decimal place.

• Reminder: C = d or

2r.

• Tire A has a diameter

of 14 + 2(5.1), or 24.2

inches.

• Its circumference is

(24.2), or about 76.03

inches.

Ex. 4: Comparing Circumferences

• Reminder: C = d or

2r.

• Tire B has a diameter

of 15 + 2(5.25), or 25.5

inches.

• Its circumference is

(25.5), or about 80.11

inches.

Ex. 4: Comparing Circumferences

Ex. 5: Finding Arc Length a. Find the distance around Lane 1.

The track is made up of two semicircles and two straight sections with length s. To find the total distance around each lane, find the sum of the lengths of each part. Round decimal answers to one decimal place.

• Divide the distance traveled by the tire

circumference to find the number of revolutions

made. First, convert 100 feet to 1200 inches.

TIRE A: 100 ft.

76.03 in.

1200 in.

76.03 in. = 100 ft.

80.11 in.

1200 in.

80.11 in. =

15.8 revolutions

TIRE B:

15.0 revolutions

Ex. 4: Comparing Circumferences

Ex. 5: Finding Arc Length

• Track. The track shown has six lanes. Each lane is 1.25 meters wide. There is 180° arc at

the end of each track. The radii for the arcs in the first two lanes are given.

a. Find the distance around Lane 1.

b. Find the distance around Lane 2.

• Distance = 2s + 2r1

= 2(108.9) + 2(29.00)

400.0 meters

• Distance = 2s + 2r2

= 2(108.9) + 2(30.25)

407.9 meters

Ex. 5: Lane 1

Ex. 5: Lane 2

A = r2

Definition of the Area of a Sector: a

region bound by 2 radii and an arc.

O

As you remember,

the area of a circle is

radius

Theorem: A sec = (mHP) r2

360

Where r is the radius and the arc

HP is measured in degrees.

Find the area, leave in terms of .

12m

60º A = 60π(122)

360

A = 24π m2

Area of a segment: a segment is a

region bound by a chord and its

corresponding arc.

The area of a

segment is equal to

the area of the sector

minus the area of the

triangle.

radius

The red piece is the

segment.

X

Y Z

Given arc XY is 90º and ZX = 8

Find the shaded area.

Segment = sector – triangle

= 90π(82) – ½(8)(8)

360

= 16π – 32 units2

..\Areas of Circles, Sectors & Segments.mp4

What is the length of the arc shown in red? Leave your answer in terms of .

360

mXYlengthof XY d

90(16)

360

4 .in

2360

mXPYlengthof XPY r

2402 (15)

360

20 cm

1) 2)

J F

V

Find the area of the shaded region. Activity:

3) 4)

5)

Finding the Lengths of Chords

• When two chords intersect in the

interior of a circle, each chord is

divided into two segments which are

called segments of a chord. The

following theorem gives a relationship

between the lengths of the four

segments that are formed.

Chord Product Theorem

• If two chords

intersect in the

interior of a circle,

then the product of

the lengths of the

segments of one

chord is equal to

the product of the

lengths of the

segments of the

other chord.

E

A

B

C

D

EA • EB = EC • ED

Proving the Chord Product

Theorem

• You can use similar

triangles to prove Theorem

10.15.

• Given: , are chords

that intersect at E.

• Prove: EA • EB = EC • ED

AB CD

E

D

B

C

A

Paragraph proof: Draw and . Because C and B intercept the same arc, C B. Likewise, A D. By the AA Similarity Postulate, ∆AEC ∆DEB. So the lengths of corresponding sides are proportional.

AC

DB

E

D

B

C

A

ED

EA=

EB

EC

EA • EB = EC • ED

Lengths of sides are

proportional.

Cross Product Property

Proving the Chord Product

Theorem

Ex. 1: Finding Segment

Lengths • Chords ST and PQ

intersect inside the

circle. Find the

value of x.

6

93

XR

T

S

Q P

RQ • RP = RS • RT Use Chord Product Theorem

Substitute values. 9 • x = 3 • 6

9x = 18

x = 2

Simplify.

Divide each side by 9.

Using Segments of Tangents

and Secants • In the figure shown,

PS is called a

tangent segment

because it is tangent

to the circle at an

end point. Similarly,

PR is a secant

segment and PQ is

the external segment

of PR.

Q

S

P

R

Secant – Secant Theorem

• If two secant segments share the same endpoint outside a circle, then the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment.

C

A

D

E

B

EA • EB = EC • ED

• If a secant segment

and a tangent

segment share an

endpoint outside a

circle, then the

product of the length

of the secant segment

and the length of its

external segment

equal the square of

the length of the

tangent segment.

(EA)2 = EC • ED

C

D

E

A

Secant – Tangent Theorem

Ex. 2: Finding Segment

Lengths

• Find the value of x. x

10

11

9

S

P

T

R

Q

RP • RQ = RS • RT Use Secant-Secant Theorem

Substitute values. 9•(11 + 9)=10•(x + 10)

180 = 10x + 100

80 = 10x

Simplify.

Subtract 100 from each side.

8 = x Divide each side by 10.

Ex. 3: Estimating the radius of

a circle • Aquarium Tank.

You are standing

at point C, about 8

feet from a circular

aquarium tank.

The distance from

you to a point of

tangency is about

20 feet. Estimate

the radius of the

tank.

(CB)2 = CE • CD Use Secant-Tangent Theorem

Substitute values.

400 16r + 64

336 16r

Simplify.

21 r Divide each side by 16.

(20)2 8 • (2r + 8)

Subtract 64 from each side.

So, the radius of the tank is about 21 feet.

Solution:

(BA)2 = BC • BD Use Secant-Tangent Theorem

Substitute values.

25 = x2 + 4x

0 = x2 + 4x - 25

Simplify.

Use Quadratic Formula.

(5)2 = x • (x + 4)

Write in standard form.

Simplify.

Use the positive solution because lengths cannot be

negative. So,

Ex. 4: Finding Segment

Lengths

Try This! # 1

Find the value of x.

x 9

18

12

E

B

D

A

C

9(12) = 18x

108 = 18x

x = 6

Try This! # 2

Find the value of x.

x

1012

11

H

G

F

E

D

DE DF = DG DH

11(21) = 12(12 + x)

231 = 144 + 12x

87 = 12x

x = 7.25

Try This! # 3

Find the value of x.

3x5

10

Y

W

X Z

WX2 = XY(XZ)

102 = 5(5 + 3x)

100 = 25 + 15x

75 = 15x

x = 5