The circle and its properties 125
Area of sector
Area of a sector =
θ
360
(πr
2 ) when θ is in degrees
=
θ
2π
(πr
2
) =
1
2
r
2 θ
when θ is in radians
Problem 6. A hockey pitch has a semicircle of
radius 14.63 m around each goal net. Find the area
enclosed by the semicircle, correct to the nearest
square metre.
Area of a semicircle =
1
2
πr 2
When r = 14.63 m, area =
1
2
π(14.63) 2
i.e. area of semicircle = 336 m 2
Problem 7. Find the area of a circular metal
plate, correct to the nearest square millimetre,
having a diameter of 35.0 mm.
Area of a circle = πr 2 =
πd 2
4
When d = 35.0 mm, area =
π(35.0) 2
4
i.e. area of circular plate = 962 mm 2
Problem 8. Find the area of a circle having a
circumference of 60.0 mm.
Circumference, c = 2πr
from which radius r =
c
2π
=
60.0
2π
=
30.0
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 5.5 cm when the angle subtended at the
centre is 1.20 rad.
Length of arc, s =rθ, where θ is in radians, hence
s = (5.5)(1.20) = 6.60 cm
Problem 10. Determine the diameter and
circumference of a circle if an arc of length 4.75 cm
subtends an angle of 0.91 rad.
Since s = rθ then r =
s
θ
=
4.75
0.91
= 5.22 cm
Diameter = 2 × radius= 2 × 5.22 =10.44 cm
Circumference, c = πd = π(10.44) = 32.80 cm
Problem 11. If an angle of 125 ◦ is subtended by
an arc of a circle of radius 8.4 cm, find the length of
(a) the minor arc, and (b) the major arc, correct to
3 significant figures.
(a) Since 180 ◦ = π rad then 1 ◦ =
π
180
rad and
125 ◦ = 125
π
180
rad.
Length of minor arc,
s =rθ = (8.4)(125)
π
180
= 18.3 cm,
correct to 3 significant figures.
(b) Length of major arc
= (circumference − minor arc)
= 2π(8.4) − 18.3 = 34.5 cm,
correct to 3 significant figures.
(Alternatively, major arc =rθ
= 8.4(360 −125)(π/180) = 34.5 cm.)
Problem 12. Determine the angle, in degrees and
minutes, subtended at the centre of a circle of
diameter 42 mm by an arc of length 36 mm.
Calculate also the area of the minor sector formed.
Since length of arc, s =rθ then θ = s/r
Radius, r =
diameter
2
=
42
2
= 21 mm
hence θ =
s
r
=
36
21
= 1.7143 rad
1.7143 rad = 1.7143 × (180/π) ◦ = 98.22 ◦ = 98 ◦ 13 =
angle subtended at centre of circle.
Area of sector
=
1
2 r 2 θ =
1
2 (21) 2 (1.7143) = 378 mm 2 .
Problem 13. A football stadium floodlight can
spread its illumination over an angle of 45 ◦ to a
distance of 55 m. Determine the maximum area that
is floodlit.
Area of sector
Area of a sector =
θ
360
(πr
2 ) when θ is in degrees
=
θ
2π
(πr
2
) =
1
2
r
2 θ
when θ is in radians
Problem 6. A hockey pitch has a semicircle of
radius 14.63 m around each goal net. Find the area
enclosed by the semicircle, correct to the nearest
square metre.
Area of a semicircle =
1
2
πr 2
When r = 14.63 m, area =
1
2
π(14.63) 2
i.e. area of semicircle = 336 m 2
Problem 7. Find the area of a circular metal
plate, correct to the nearest square millimetre,
having a diameter of 35.0 mm.
Area of a circle = πr 2 =
πd 2
4
When d = 35.0 mm, area =
π(35.0) 2
4
i.e. area of circular plate = 962 mm 2
Problem 8. Find the area of a circle having a
circumference of 60.0 mm.
Circumference, c = 2πr
from which radius r =
c
2π
=
60.0
2π
=
30.0
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 5.5 cm when the angle subtended at the
centre is 1.20 rad.
Length of arc, s =rθ, where θ is in radians, hence
s = (5.5)(1.20) = 6.60 cm
Problem 10. Determine the diameter and
circumference of a circle if an arc of length 4.75 cm
subtends an angle of 0.91 rad.
Since s = rθ then r =
s
θ
=
4.75
0.91
= 5.22 cm
Diameter = 2 × radius= 2 × 5.22 =10.44 cm
Circumference, c = πd = π(10.44) = 32.80 cm
Problem 11. If an angle of 125 ◦ is subtended by
an arc of a circle of radius 8.4 cm, find the length of
(a) the minor arc, and (b) the major arc, correct to
3 significant figures.
(a) Since 180 ◦ = π rad then 1 ◦ =
π
180
rad and
125 ◦ = 125
π
180
rad.
Length of minor arc,
s =rθ = (8.4)(125)
π
180
= 18.3 cm,
correct to 3 significant figures.
(b) Length of major arc
= (circumference − minor arc)
= 2π(8.4) − 18.3 = 34.5 cm,
correct to 3 significant figures.
(Alternatively, major arc =rθ
= 8.4(360 −125)(π/180) = 34.5 cm.)
Problem 12. Determine the angle, in degrees and
minutes, subtended at the centre of a circle of
diameter 42 mm by an arc of length 36 mm.
Calculate also the area of the minor sector formed.
Since length of arc, s =rθ then θ = s/r
Radius, r =
diameter
2
=
42
2
= 21 mm
hence θ =
s
r
=
36
21
= 1.7143 rad
1.7143 rad = 1.7143 × (180/π) ◦ = 98.22 ◦ = 98 ◦ 13 =
angle subtended at centre of circle.
Area of sector
=
1
2 r 2 θ =
1
2 (21) 2 (1.7143) = 378 mm 2 .
Problem 13. A football stadium floodlight can
spread its illumination over an angle of 45 ◦ to a
distance of 55 m. Determine the maximum area that
is floodlit.