MA1700
Practice Exercises: Arc Length, Sector Area, Angular/Linear Velocity
Part 1: Arc Length Exercises
1. In each of the following determine the length of the indicated arc.
b) Length of the major arc?
a) Length of AB  ?
c) Length of AB  ?
A
A
A
r
9
6
115
C
D
B
C
50
D
270
C
B
B
31
d) Given AB  20
 cm and
r  3 cm , find length of
D
e) What is the circumference
of the circle below?
ADB .
D
f) What is the length of
an arc with central angle
64o if AB  15 cm ?
A
16.5 m
C
B
A
B
2. For a circle with radius 4.5m, find the length of an arc with central angle of 36 o.
3. If the circumference of a circle is 26 , what is the radius?
4. Sara ran 5 times around a circular that has a diameter of 30m. Approximately how
far did she run?
5. A motorcycle mechanic wants to put a strip of rubber between the tube and tire of a
22 inch rim. To the nearest inch, how long should the strip of rubber be?
Part II: Sector Area
1. Find the area of the shaded region of each of the following circles.
a)
b)
3 cm
60
100
5 cm
c)
d)
75
16 cm
e)
12 in
135
f)
18mm
210
80
90 mm
2. A sector has an area of 36 cm2 . The radius of the sector is 6cm. What is the
measure of the central angle of the intercepted arc?
3. A sector has an area of 12 in2 . The measure of the intercepted arc is 120o. What is
the radius of the circle?
Page 2 of 7
Part III: Angular / Linear Velocity
1. The second hand on a clock makes one revolution in one minute. What is the linear
velocity, in cm/s, of a point on the tip of the second hand, if the second hand is 9cm
from the centre to the tip? (Note: 1 revolution = 2 )
2. A Merry-Go-Round travels at an angular velocity of 6 revolutions per minute. What
is the linear velocity, in m/min, of a person sitting 5 meters from the centre?
3. The earth makes one rotation every 24 hours. What is the linear velocity, in km/h, of
a person standing on the equator, if the radius of the earth at the equator is
approximately 6438 km?
4. A bicycle wheel is 61cm in diameter. If the bicycle is traveling at 25 km/h, what is the
angular velocity, in revolutions per minute (rpm), of the wheel?
5. A figure skater is spinning with her arms outstretched at an angular velocity of 8
radians per second.
a. What is the angular velocity of the spin, in revolutions per minute?
b. Her fingertips travel in circular motion with a radius of 90cm. What is the
linear velocity of the fingertip, in m/s?
Page 3 of 7
Solutions
Part I: Arc Length
1.
a)
b)
 
o
 180
 23 
S  r  
 26 
 23
 
 26
  115o 
c)
   

o 
 180  2
   360o  270o  
   9
S  9  
2 2
e)
C  2 r   d
AB  15cm  r 
15
cm
2
8
 15 
  
S     64o  
cm 
cm  8.38cm
o 
3
2
 180 
 51.9m
36


180
5
r  4.5m

3.
d)
Circumference = 2 r  6
31
89
ADB  6   

20
20
f)
 16.6 m
2.
   31
 
o 
 180  18
 31  31
S  6    
 18  3
   360o  50o  
9
 
S   4.5    m 
m  2.8m
10
5
C  2 r
26   2  r
r  13
4. d  30m
C d
 30 m
d  5  30  m
 150 m
 471.2 m
Page 4 of 7
5. d  22 in
C d
 22 in
 69.1in
Part II: Sector Area
1.
a)
A
A

1
 r2    r2

2
2
1  o   
3
2
60 
3cm  
cm2  4.7cm2

o  
2
2
 180  
b)
A
1 o 
100 
o
2 
 180
125
2

cm 2  21.8cm 2
   5cm  
18

A
1  o   
160
2
75 
16cm  
cm 2  167cm 2

o  
2
3
 180  
A
1  o   
2
135 
12in   54 in 2  169.6 in 2

o  
2
 180  
A
1  o   
2
80 
90mm   1800 mm2  5654.9 mm2

o  
2
 180  
A
1
2
  
210o 
18mm   189 mm2  593.8 mm2

o  
2
 180  
c)
d)
e)
f)
Page 5 of 7
2.
1
A   r2
2
2 A 2  36 
 2 
 2
r
62
A  36 cm 2
r  6 cm
 ?
3.
1
  2
12  120o
r
2
180o 
 180  2
24 
r
 120 
A  12 in 2
  120o
r ?
36  r 2
r 6
Part III: Angular/Linear Velocity
1.


t
V  r 
2.

t
r
2
3
 9cm  
60s
10
cm
s
 0.94 cm s
6 rev 6  2 

 12 radian min
min
min
m
V   r  12  5 
 60 m min  188.5 m min
min

3.
t  25h
r  6438km
V

t
r
2
1073
 6438km  
24h
2
km
h
 1685.5 km h
4.
61  1km 
4
r  cm 
  3.05 10 km
2
 100000cm 
V  25 km h
V
25 km h

 81967.213 rad h
r 3.05 104 km
rad  1 rev  1 h 
 81967.213



h  2 rad  60 min 

 ?
 217.4 rev min
Page 6 of 7
5.
(a)
 8
rad  1 rev  4 rev  60s  240 rev
 76.4 rpm




s  2 rad   s  1min 
 min
(b)
  8 rad s
r  90 cm
V  r  8 rads  90cm   720 cms  1001mcm   7.2 ms
Page 7 of 7