Basic Engineering Mathematics, Fifth Edition

The circle 233

3. Convert to degrees:
   (a)

7 π

6

rad (b)

4 π

9

rad (c)

7 π

12

rad

4. Convert to degrees and minutes:
   (a) 0.0125rad (b) 2.69rad
   (c) 7.241rad

26.4 Arc length and area of circles and sectors

26.4.1 Arc length

From the definition of the radian in the previous section
and Figure 26.7,

arc length,s=rθ whereθis in radians

26.4.2 Area of a circle

From Chapter 25, for any circle,area=π×(radius)^2

i.e. area=πr^2

Sincer=

d

2

, area=πr^2 or

πd^2

4

26.4.3 Area of a sector

Area of a sector=

θ

360

(πr^2 )whenθis in degrees

=

θ

2 π

(

πr^2

)

=

1

2

r^2 θ whenθis in radians

Problem 8. A hockey pitch has a semicircle of

radius 14.63m aroundeach goal net. Find the area

enclosed by the semicircle, correct to the nearest

square metre

Area of a semicircle=

1

2

πr^2

Whenr= 14 .63m, area=

1

2

π( 14. 63 )^2

i.e. area of semicircle=336m^2

Problem 9. Find the area of a circular metal plate

having a diameter of 35.0mm, correct to the nearest

square millimetre

Area of a circle=πr^2 =

πd^2

4

Whend= 35 .0mm, area=

π( 35. 0 )^2

4

i.e. area of circular plate=962mm^2

Problem 10. Find the area of a circle having a

circumference of 60.0mm

Circumference,c= 2 πr

from which radius,r=

c

2 π

=

60. 0

2 π

=

30. 0

π

Area of a circle=πr^2

i.e. area=π

(

30. 0

π

) 2

= 286 .5mm^2

Problem 11. Find the length of the arc of a circle

of radius 5.5cm when the angle subtended at the

centre is 1.20 radians

Length of arc,s=rθ,whereθis in radians.

Hence,arc length,s=( 5. 5 )( 1. 20 )= 6 .60cm.

Problem 12. Determine the diameter and

circumference of a circle if an arc of length 4.75cm

subtends an angle of 0.91 radians

Since arc length,s=rθthen

radius,r=

s

θ

=

4. 75

0. 91

= 5 .22cm

Diameter= 2 ×radius= 2 × 5. 22 =10.44cm

Circumference,c=πd=π( 10. 44 )=32.80cm

Problem 13. If an angle of 125◦is subtended by

an arc of a circle of radius 8.4cm, find the length of

(a) the minor arc and (b) the major arc, correct to

3 significant figures

Since 180◦=πrad then 1◦=

( π

180

)

rad and

125 ◦= 125

(π

180

)

rad

(a) Length of minor arc,

s=rθ=( 8. 4 )( 125 )

( π

180

)

=18.3cm,

correct to 3 significant figures