Higher Engineering Mathematics, Sixth Edition

The circle and its properties 125

Area of 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 6. 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 7. Find the area of a circular metal

plate, correct to the nearest square millimetre,

having a diameter of 35.0mm.

Area of a circle=πr^2 =

πd^2

4

Whend= 35 .0 mm, area=

π( 35. 0 )^2
4
i.e. area of circular plate=962mm^2

Problem 8. Find the area of a circle having a

circumference of 60.0mm.

Circumference,c= 2 πr

from which radiusr=

c

2 π

=

60. 0

2 π

=

30. π
    Area of a circle=πr^2

i.e. area=π

(

30. 0

π

) 2

= 286 .5mm^2

Problem 9. Find the length of arc of a circle of

radius 5.5cm when the angle subtended at the

centre is 1.20rad.

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

s=( 5. 5 )( 1. 20 )=6.60cm

Problem 10. Determine the diameter and

circumference of a circle if an arc of length 4.75cm

subtends an angle of 0.91rad.

Sinces=rθthenr=

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 11. 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.

(a) Since 180◦=πrad then 1◦=

(π

180

)

rad and

125 ◦= 125

(π

180

)

rad.

Length of minor arc,

s=rθ=( 8. 4 )( 125 )

( π

180

)

=18.3cm,

correct to 3 significant figures.

(b) Length of major arc

=(circumference−minor arc)

= 2 π( 8. 4 )− 18. 3 = 34 .5cm,

correct to 3 significant figures.

(Alternatively, major arc=rθ

= 8. 4 ( 360 − 125 )(π/ 180 )=34.5cm.)

Problem 12. Determine the angle, in degrees and

minutes, subtended at the centre of a circle of

diameter 42mm by an arc of length 36mm.

Calculate also the area of the minor sector formed.

Since length of arc,s=rθthenθ=s/r

Radius,r=

diameter

2

=

42

2

=21mm

henceθ=

s

r

=

36

21

= 1 .7143rad

1 .7143rad= 1. 7143 ×( 180 /π)◦= 98. 22 ◦= 98 ◦ 13 ′=

angle subtended at centre of circle.

Area of sector

=^12 r^2 θ=^12 ( 21 )^2 ( 1. 7143 )=378mm^2.

Problem 13. A football stadium floodlight can

spread its illumination over an angle of 45◦to a

distance of 55m. Determine the maximum area that

is floodlit.