1. © T Madas

2. We can use these formulae to find areas of sectors and lengths of arcs
The area of a circle
The circumference of a circle
We can use these formulae to find areas of sectors and lengths of arcs
Arc
Sector r
© T Madas

3. We can use these formulae to find areas of sectors and lengths of arcs
The area of a circle
The circumference of a circle
We can use these formulae to find areas of sectors and lengths of arcs
Area of sector =
50°
6 cm
© T Madas

4. We can use these formulae to find areas of sectors and lengths of arcs
The area of a circle
The circumference of a circle
We can use these formulae to find areas of sectors and lengths of arcs
Length of arc =
50°
6 cm
© T Madas

5. Example
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6. Find the area of the circular segment, from the information given.
13cm
24 cm
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7. Find the area of the circular segment, from the information given.
13cm
24 cm
© T Madas

8. x Find the area of the circular segment, from the information given.
13cm
12 cm x
24 cm
5 cm
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9. Find the area of the circular segment, from the information given.
13cm
12 cm
5 cm
Area of the triangle
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10. Find the area of the circular segment, from the information given.
cm2
13cm
12 cm
5 cm
Area of the triangle
134.76°
Area of sector =
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11. © T Madas

12. Two circular sectors have radii of 5 cm and 6 cm and represent one sixth and one eighth of a circle respectively.
Calculate the area and the perimeter of each sector, correct to 1 decimal place.
area of a circle 1 6
A =
r 2 π
A = π
x 5 2
area of the sector
A = π
x 5 2 1 6
≈ 13.1 cm2
5 cm x
area of a circle
A =
r 2 π 1 8
A = π
x 6 2
area of the sector
A = π 1 8
x 6 2
≈ 14.1 cm2 x
6 cm
© T Madas

13. circumference of a circle circumference of a circle
Two circular sectors have radii of 5 cm and 6 cm and represent one sixth and one eighth of a circle respectively.
Calculate the area and the perimeter of each sector, correct to 1 decimal place.
circumference of a circle 1 6
C = 2 r π
C = 2
x π
x 5
length of the arc
5 cm
L = 2
x π
x 5 1 6
≈ 5.2 cm x
circumference of a circle
C = 2 r π 1 8
C = 2
x π
x 6
length of the arc
L = 2
x π 1 8
x 6
≈ 4.7 cm x
6 cm
© T Madas

14. © T Madas

15. 42π ⇔ ⇔ ⇔ ⇔ π π π π π π A = r 2 A = A = 18 A = r 2 A = A = 24
The diagram below shows a composite shape consisting of:
a semicircle, centre at O and a radius of 6 cm
a circular sector, centre at A corresponding to an angle of 60°.
Calculate the area of the composite shape in terms of π
Round your answer to part (1), to 3 significant figures.
area of the semicircle A = 1 2 x π
r 2 ⇔
6 cm O A = 1 2 x π
x 62 ⇔
42π A B
60° A = 18 π
area of the sector A = 1 6 x π
r 2 ⇔ A = 1 6 x π
x 122 ⇔ C A = 24 π
60° = 1 6
of a circle
© T Madas

16. The diagram below shows a composite shape consisting of:
a semicircle, centre at O and a radius of 6 cm
a circular sector, centre at A corresponding to an angle of 60°.
Calculate the area of the composite shape in terms of π
Round your answer to part (1), to 3 significant figures. A = 42 π = 42 x π ≈
132
cm2
[3 s.f.]
6 cm O
42π A B
60° C
© T Madas

17. © T Madas

18. The radii of the two sectors are r and R as shown in the diagram.
The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end.
The radii of the two sectors are r and R as shown in the diagram.
Find a formula for the area A of this pond in terms of r, R and θ.
Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° r θ R
© T Madas

19. The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end.
The radii of the two sectors are r and R as shown in the diagram.
Find a formula for the area A of this pond in terms of r, R and θ.
Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° π θ
360
R 2 x r θ R
© T Madas

20. The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end.
The radii of the two sectors are r and R as shown in the diagram.
Find a formula for the area A of this pond in terms of r, R and θ.
Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° π θ
360 θ
360
R 2 x – π
r 2 x r θ R
© T Madas

21. The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end.
The radii of the two sectors are r and R as shown in the diagram.
Find a formula for the area A of this pond in terms of r, R and θ.
Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° 2
R – r π θ
360 π θ
360
R – r 2 A =
R 2 x –
r 2 x + π r θ R
© T Madas

22. The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end.
The radii of the two sectors are r and R as shown in the diagram.
Find a formula for the area A of this pond in terms of r, R and θ.
Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° π
r 2 x θ
360 –
R 2 +
R – r 2 A = π
r 2 x θ
360 –
R 2 +
R – r 2 A =
© T Madas

23. The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end.
The radii of the two sectors are r and R as shown in the diagram.
Find a formula for the area A of this pond in terms of r, R and θ.
Find the area of a pond with this shape if r = 7 m, R = 9 m and θ = 45° π
r 2 x θ
360 –
R 2 +
R – r 2 A = c πθ
360 π 4
R – r 2 A =
R 2
– r 2 +
45π
360 π 4
9 – 7 2 c A = 92
– 72 + π 8 π 4
x 4 c A =
x 32 + 4π
+ π c A =
A = 5π
≈ 15.7 m2
© T Madas

24. © T Madas

25. area of the given sector
The prism shown has a cross section of a circular sector of radius 8 cm corresponding to a central angle of 45°.
The thickness of the prism is 3 cm.
1. Calculate the volume of the prism, correct to 3 s.f.
2. Calculate the surface area of the prism, correct to 3 s.f.
area of a circle
A =
r 2 π
45°
A = π
x 8 2
3 cm
area of the given sector
8 cm 45
360
A = π
x 8 2 x
8 cm
45°
volume of the prism 45
360
V = π
x 8 2
x 3 x
V =
24π
V ≈
75.4 cm3 [3 s.f.]
© T Madas

26. area of the given sector area of the given sector
The prism shown has a cross section of a circular sector of radius 8 cm corresponding to a central angle of 45°.
The thickness of the prism is 3 cm.
1. Calculate the volume of the prism, correct to 3 s.f.
2. Calculate the surface area of the prism, correct to 3 s.f.
area of the given sector
A = π
x 8 2 45
360 x
45°
3 cm
area of the given sector
8 cm 45
360
A = π
x 8 2 x
8 cm
45°
© T Madas

27. The prism shown has a cross section of a circular sector of radius 8 cm corresponding to a central angle of 45°.
The thickness of the prism is 3 cm.
1. Calculate the volume of the prism, correct to 3 s.f.
2. Calculate the surface area of the prism, correct to 3 s.f.
area of the given sector
x 2
A = π
x 8 2 45
360
x 2
= 16π x
45°
2πr
3 cm
area of the side rectangle
x 2
A = 3
x 8
x 2
= 48
8 cm
area of the curved surface
8 cm
45° 45
360
A =
2 x π
x 8
x 3
= 6π x
total surface area
A = 16π π
≈ 117 cm2 [3 s.f.]
© T Madas

28. © T Madas

29. area of the “larger” sector
The diagram below models the action of a single windscreen wiper rotating through an angle of 162° on a rectangular windscreen measuring 141 cm by 72 cm.
The wiper AB is 45 cm long and OA = 15 cm.
Calculate what percentage of the screen gets wiped.
141
area of a circle
A =
r 2 π
A = π
x 60 2 72
area of the “larger” sector B
162
360
162° 45
A = π
x 60 2
= 1620π A x 15 O
All measurements in cm
© T Madas

30. area of the “larger” sector area of the “inner” sector
The diagram below models the action of a single windscreen wiper rotating through an angle of 162° on a rectangular windscreen measuring 141 cm by 72 cm.
The wiper AB is 45 cm long and OA = 15 cm.
Calculate what percentage of the screen gets wiped.
141
area of a circle
A =
r 2 π
A = π
x 60 2 72
area of the “larger” sector B
162
360 45
A = π
x 60 2
= 1620π
162° A x 15 O
area of the “inner” sector
All measurements in cm
A = π
x 15 2
162
360
= π x
© T Madas

31. The diagram below models the action of a single windscreen wiper rotating through an angle of 162° on a rectangular windscreen measuring 141 cm by 72 cm.
The wiper AB is 45 cm long and OA = 15 cm.
Calculate what percentage of the screen gets wiped.
141
area of a circle
A =
r 2 π
A = π
x 60 2 72
area of the “larger” sector B
162
360 45
A = π
x 60 2
= 1620π
162° A x 15 O
area of the “inner” sector
All measurements in cm
A = π
x 15 2
162
360
= π x
area swept by the wiper is
1620π
– π
= π cm2
≈ cm2
© T Madas

32. area swept by the wiper is
The diagram below models the action of a single windscreen wiper rotating through an angle of 162° on a rectangular windscreen measuring 141 cm by 72 cm.
The wiper AB is 45 cm long and OA = 15 cm.
Calculate what percentage of the screen gets wiped.
141
%age =
area wiped
total area
x 100 c
%age = π
141 x 72
x 100 c 72
%age ≈
47% B
162° 45 A 15 O
All measurements in cm
area swept by the wiper is
1620π
– π
= π cm2
≈ cm2
© T Madas

33. Exam Question
© T Madas

34. Two circles of respective radii of 3 cm and 4 cm are overlapping each other in such a way so that their centres are 5 cm apart.
Calculate the perimeter of the compound shape, correct to 2 significant figures. C
3,4,5
= Alarm Bells !
3 cm
4 cm
∆ABC is right angled
By trig on ∆ABC : θ A B
5 cm
© T Madas

35. Simplify the diagram by showing only the relevant information
Two circles of respective radii of 3 cm and 4 cm are overlapping each other in such a way so that their centres are 5 cm apart.
Calculate the perimeter of the compound shape, correct to 2 significant figures.
Now some angle calculations:
If θ ≈ 53.13°
RCBA ≈ 36.87°
RCAD ≈ °
RCBD ≈ 73.74° C
36.87°
253.74°
3 cm
4 cm θ A B
5 cm
286.26°
reflex RCAD ≈ °
reflex RCBD ≈ °
106.26°
73.74° D
Simplify the diagram by showing only the relevant information
© T Madas

36. Two circles of respective radii of 3 cm and 4 cm are overlapping each other in such a way so that their centres are 5 cm apart.
Calculate the perimeter of the compound shape, correct to 2 significant figures.
This arc corresponds to an angle of ° on a circle with radius of 3 cm
This arc corresponds to an angle of ° on a circle with radius of 4 cm C
253.74°
3 cm
4 cm A B
286.26° D
© T Madas

37. Two circles of respective radii of 3 cm and 4 cm are overlapping each other in such a way so that their centres are 5 cm apart.
Calculate the perimeter of the compound shape, correct to 2 significant figures.
The smaller of the 2 arcs C
253.74°
3 cm
4 cm A B
286.26°
The larger of the 2 arcs D
The required perimeter
33.3 cm [2 s.f.]
© T Madas

38. © T Madas

39. In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C.
Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B
OBD is right angled
OD = 3 m
Trig to find RBOD
5 m
4 m θ C
8 m
3 m O D
5 m A
© T Madas

40. In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C.
Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B
The area of the sector
5 m θ C O D
The length of the arc
5 m A
area of the sector ≈ m2
length of the arc ≈ m
© T Madas

41. In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C.
Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B
OBC is right angled
OB = 5 m
RBOC = 53.13°
Trig to find BC
5 m
4 m θ C
3 m O D
5 m A
area of the sector ≈ m2
length of the arc ≈ m
BC = 20/3 m
© T Madas

42. In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C.
Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B
Area of OBC
5 m θ C
area of the sector ≈ m2 O D
half the shaded area ≈ 5.08 m2
5 m
shaded area ≈ m2 A
area of the sector ≈ m2
length of the arc ≈ m
BC = 20/3 m
© T Madas

43. In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C.
Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle B
length of the arc ≈ m
Length BC = 20/3 m ≈ 6.67 m
5 m θ E C
11.59
6.67
x 2
≈ 23.18
≈ 13.34 O D
5 m
Total perimeter ≈ 36.5 m A
area of the sector ≈ m2
length of the arc ≈ m
BC = 20/3 m
© T Madas

44. Exam Question
© T Madas

45. L A A circular sector has radius r and an area A .
The sector corresponds to an arc of length L .
Write an expression for A in terms of r and L . L A r
© T Madas

46. L A θ A circular sector has radius r and an area A .
The sector corresponds to an arc of length L .
Write an expression for A in terms of r and L . L A θ r
© T Madas

47. Now what? A circular sector has radius r and an area A .
The sector corresponds to an arc of length L .
Write an expression for A in terms of r and L .
Now what?
© T Madas

48. A circular sector has radius r and an area A .
The sector corresponds to an arc of length L .
Write an expression for A in terms of r and L . 2
© T Madas

49. Harder Examples
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50. curvilinear areas 1
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51. The following pattern is produced inside a square whose side length is 8 cm
Calculate the shaded area D C
P l a n
45°
The shaded region can be split into 8 congruent shapes
take one of these shapes
add the right angled triangle next to it
they make up a 45° sector
of a circle whose radius is half the square’s diagonal A B
8 cm
© T Madas

52. Calculate the shaded area
The following pattern is produced inside a square whose side length is 8 cm
Calculate the shaded area D C
Pythagoras: 4
45° 4 x 32
Area of sector:
Area of triangle: A B
8 cm
© T Madas

53. Calculate the shaded area
The following pattern is produced inside a square whose side length is 8 cm
Calculate the shaded area D C
Area of sector:
Area of triangle:
Area of sector:
Area of triangle: A B
8 cm
© T Madas

54. Calculate the shaded area
The following pattern is produced inside a square whose side length is 8 cm
Calculate the shaded area D C
Area of sector:
Area of triangle:
Shaded area: A B
8 cm
© T Madas

55. curvilinear areas 2
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56. Find the area of the shaded region
In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below.
Find the area of the shaded region
The bottom part of the shaded area consists of: D C
an equilateral triangle with side length of 8 cm
plus 2 circular segments
Look at this shaded section:
It is a circular sector, one sixth of a circle of radius 8 cm.
The area of the circular segment equals the area of this sector less the area of the equilateral triangle.
60° A
8 cm B
© T Madas

57. Find the area of the shaded region
In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below.
Find the area of the shaded region
The area of the segment D C
sector = 1/6 circle
triangle
60° A
8 cm B
© T Madas

58. Find the area of the shaded region
In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below.
Find the area of the shaded region
The area of the segment D C
32π 3
– 16 3
60° A
8 cm B
© T Madas

59. Find the area of the shaded region
In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below.
Find the area of the shaded region
Now look at one of the non - shaded regions
The area of the segment D C
The area of one of the non shaded regions equals:
One twelfth of a circle of radius 8 cm
Less the area of the segment
32π 3
– 16 3
30°
60° A
8 cm B
© T Madas

60. Find the area of the shaded region
In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below.
Find the area of the shaded region D C
32π 3
– 16 3
30°
60° A
8 cm B
© T Madas

61. Find the area of the shaded region
In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below.
Find the area of the shaded region D C
16 3
16π 3
16π 3
16 3
32π 3
– 16 3
30°
60° A
8 cm B
© T Madas

62. Find the area of the shaded region
In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below.
Find the area of the shaded region D C
The shaded region is 64 cm2 less the two non shaded regions we just found
16 3
16π 3
16π 3
16 3 A
8 cm B
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63. The Curved Surface of a Cone
STOP
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64. θ L h L A r r r
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65. © T Madas

66. © T Madas

67. © T Madas

68. © T Madas

69. h L r
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70. © T Madas