1. home and garden
2. home furnishings
3. sofas and chairs

Shape, position and movement
26
Having investigated the relationships between the radius, diameter,
circumference and area of a circle, I can apply my knowledge to solve
related problems.
MTH 4-16b
This chapter will show you how to:
• use the correct terminology to identify parts of a circle
• investigate the connection between the diameter and circumference of a circle
• calculate the circumference and area of a circle using a formula
• solve problems involving circumference and area, including compound shapes
• calculate the length of an arc and the area of a sector of a circle
• solve problems involving arcs and sectors of circles.
You should already know:
• how to calculate the perimeter and area of rectangles, triangles and common quadrilaterals
• h
ow to calculate the perimeter and area of compound 2D shapes by dividing them into rectangles and
triangles
• how to substitute values into a formula.
The circle and its parts
A circle is a set of points equidistant from a fixed point,
called the centre, designated here by O.
Circle
≠
O
You must learn all of the following terms for the
different parts of a circle.
Circumference
C
CircumferenceThe length round a circle. It is a special name for the
perimeter of a circle.
Arc
ArcOne of the two parts between two points on a
circumference.
Radius
RadiusThe distance from the centre of a circle to its
circumference. The plural of the term is ‘radii’.
DiameterThe distance across a circle through its centre. The
diameter d of a circle is twice its radius r so, d = 2r.
r
Diameter
2m
d
Chord
2
CFE Math Fourth Level_sample.indd 2
Tangent
18/05/12 6:53 PM
Properties of 2D shapes
d
andd
ChordA straight line which joins two points on the circumference
of a circle.
TangentA straight line that touches a circle at one point only on its
circumference. This point is called the point of contact.
SegmentThe region of a circle enclosed by a chord and an arc. Any
chord encloses two segments, which have different areas.
SectorA portion of a circle enclosed by two radii and one of the arcs
between them.
SemicircleOne half of a circle: either of the parts cut off by a diameter.
Diameter
Diameter
2
2m
3D objects
Chord
Chord
Tangent
Tangent
Segment
Segment
Sector
Sector
Semicircle
Semicircle
Investigation 1: Circumference of a circle
7 cm
7 cmto complete
You will need a variety of circular objects, some string or thread and a ruler
this activity.
11mm
11mm
1 Measure the circumference of each of your objects by wrapping the string round it and
marking the length when it reaches completely round your object. Unwrap
1 cmthe string
and measure this length with a ruler. Place the ruler across your object to measure the
1 cm
diameter.
14 mm
14 mm
2 Copy the following table. Complete each row using your measurements then calculate
the last column to one decimal place.
Some possible objects you could measure are given in the table.
Object
Diameter, d (cm)
Circumference,
C (cm)
3
3
6 cm
6 cm
C÷d
Drinks can
Water bottle
Crisp tube
3
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Shape, position and movement
3 a What do you notice about the last column in your table?
b How do you think the circumference is related to the diameter?
Historical approximations
In Investigation 1, you should have found that the circumference, C, of a circle is
approximately 3 times its diameter, d. The number which you have to multiply the
diameter by to get a more accurate value for the circumference is slightly more than 3,
but cannot be written down exactly as either a fraction or a decimal. We now represent
this special number using the Greek letter p (pronounced pi).
This link between the circumference and diameter has been known for thousands of
years and many different ancient civilisations used their own approximate values for
the number we now call p.
Investigation 2: Historical values for C ÷ d
Here are some of the values that have been used throughout history:
i Rhind Papyrus, Egyptian, 2000 bc
ii Babylonian, 2000 bc
iii Ptolemy, Greek, 127 ad
iv Zu Chongzhi, Chinese, 480 ad
v Aryabhata, Indian, 499 ad
256
81
25
8
3
17
120
355
113
62 832
20000
1 aUse a calculator to find the value of each one and compare it with the value given
by pressing the p key.
b Which one was the most accurate?
2The famous Greek mathematician Archimedes (287–212 bc) knew he wouldn’t be able
to find an exact value for p, but calculated that it was between
correct?
Challenge
223
71
and
22
.
7
Was he
1 Have a competition in your class to see who can memorise the most decimal
places for p. A useful way to remember a lot of decimal places is to break
the number down into blocks of five digits, as it is easier to memorise several
5-digit numbers than one long one.
2 p has now been calculated to millions of decimal places using computers. No
repeating pattern has been found and it is thought that any pattern of digits you
can think of will appear somewhere in the digits of p.
a How many decimal places of p have now been found?
b Can you find your birth date in the digits of p?
4
CFE Math Fourth Level_sample.indd 4
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Semicircle
Semicircle
3·4 m3·4 m
3 cm
3 cm
Properties
of 2D shapes and 3D objects
Calculating the circumference of a circle and
perimeter of compound shapes
≠
The formula for calculating the circumference, C, of a circle with diameter, d, is:
1·4 cm
7 cm7 cm
1·4 cm
11mm
11mm
21
mm
2
·
4
m
21 mm
2·4 m
C = p d
As the diameter is twice the radius, r, the circumference is also given by:
=1p cm
d=
1Ccm
p ë 2r = 2pr
14 mm
14 mm
2·1 m2·1 m
3·5 cm
3·5 cm
Example 26·1
·5 m
5·5tom5one
Calculate the circumference of each of the following circles. Give each answer
decimal place.
a
b
3·4 m3·4 m
6 cm6 cm
a The diameter, d = 6 cm, which gives:
C = pd = p × 6 = 18·8 cm (to 1 dp)
b The radius, r = 3·4 m, so d = 6·8 m. This gives:
C = pd = p × 6·8 = 21·4 m (to 1 dp)
Example 26·2
Calculate the perimeter of this shape. The curve is a semicircle. Give your answer to one
decimal place.
3m
1·5 m
8m
Diameter of semicircle, d = 8 - 3 = 5 m
Circumference of the semicircular part of shape is:
1
2
C = p d
1
2
(since it is half a circle)
×p×5
=
= 7·9 m (to 1 dp)
Perimeter of shape is:
P = 3 + 1·5 + 8 + 1·5 + 7·9
= 21·9 m (to 1 dp)
5
CFE Math Fourth Level_sample.indd 5
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≠
2m
2m
a a a Chord
ameter Diameter
2m
2m
5 cm
5 cm
5 cm
Shape,
position and movement
d
13
13 cm
13 cm
b b b Chord
Chord
3 cm
3 cm
3 cm
8 cm
7 cm 5 cm
7 cm
4
cm
4 πcm
In this Exercise,
take
= 3·14
or use the 5π cm
key 5oncm
your calculator. 8 cm 8 cm
4 cm
5 cm 5 5cm
cm 5 cm
5 cm 5 cm
Tangent Tangent Tangent
13 cm 13 cm
1Calculate the circumference of eachSegment
of the following circles. Give each answer to one
Segment Segment
decimal
place.
Chord
ord
a a a
bb b b
c
a
7 cm
7 cm
3 cm
7 cm
3 cm
3 cm
4 cm
6 cm
4 cm
4 cm
6 cm
6 cm
9 cm
9 cm
9 cm
1
ngent Tangent
1
1
–
–
–
2C
Sector
2C
2C
Sector
Sector
gment Segment
r
A
r
A
r
d
e
f A
Exercise
26A
8 cm
8 cm
8 cm
4
cm
4 cm 4 cm
4 cm
5 cm
4 cm
5 cm
4 cm
5 cm
Semicircle
Semicircle
Semicircle
5
cm
1
5 cm 5 cm
1 cm
1 cm–2 C
C
1–12cm
Sector
ctor
r
3·
3·4 m SHIFT
SHI
3 cm 3·4 m
3 cm
3 cm
r
4
SHIFT
4 SHI
A
A
2 Calculate the circumference of each of the following circles. Give each answer to one
4 cm
6 cm
4 cm
6 cm
4 cm
cm
decimal6place.
9 cm
9 cm
9 cm
2
cm
2 cm 2 cm
6
·
2
cm
a
b
c
5
cm
6·2 cm6·2 cm
5 cm 5 cm
3
·
4
m
37
cm
7 cm 3·4 m7 cm
3 cm
cm
1·4 cm 1·4 cm
11mm ≠ 11mm4≠ x2 21=
11mm
4 2x·24 m=
2·4 m
21 mm 21 mm 2·4 mmm
micircleSemicircle
4 cm
4 cm
4 cm
5 cm
5 cm
5 cm
1 cm
1 cm 1 cm
e
1 cm
3·5 cm
1 cm
1 cm
1 m 3·5 cm
2·1 m 3·52·cm
2
·
1
m
14
mm
·
1
4
cm
14
mm
14
mm
1·4 cm
21 mm 2·4 m
2·4 m
5· 5 m
5·5 m
d
7 cm
11mm
14 mm
1 cm
11mm
21 mm
14 mm2·1 m
Circle
O
3·4 m
6 cm
Circumference
C
2 cm
2 cm
2 cm
·2 6cm
5 cm
·2 cm
5 cm
6·26cm
5 cm
Circle
3·4 m
3
·
4
m
3
·
4
m
3 Take a 3
2p
coin
and
measure
its
diameter
to
the
nearest
millimetre.
3
·
5
cm
·
5
cm
6 cm
6 cm
6 cm
≠
2·1 m
Calculate the circumference of the coin, giving your answer to the
O
nearest millimetre.
5· 5 m
5· 5 m
≠ Big Wheel at a theme park has a diameter of 5 m. How far
4 The
would you travel in one
complete revolution of the wheel? Give
Circumference
your
answer
to
the
nearest
C metre.
3·4 m
Entrance
5 The diagram shows the dimensions of a running track at a sports centre. The bends are
semicircles.
51 m
Arc
51 m
Arc
120 m
Radius
r
Radius
r
6
Diameter
d
CFE Math Fourth Level_sample.indd 6
120 m
2m
Calculate the distance round the track. Give your answer to the nearest metre.
Diameter
2m
2m
d
7 cm
2m
13 cm
Chord
7 cm
5 cm
5 cm
18/05/12
5 cm
6:54 PM
1· 4
ce
Properties of 2D shapes and 3D objects
6 The Earth’s orbit can be taken
to be a circle with a radius of
approximately 150 million km.
Circle
Circle
Circle
O O
O
Circumference
Circumference
C
C Circumference
C
Earth’s orbit
≠
≠
Calculate
the distance the Earth
travels ≠in one orbit of the Sun.
Give your answer to the nearest
million kilometres.
Sun
Earth
7 By measuring the diameter, calculate the perimeter of this semicircular shape.
Give your answer to one decimal place.
≠
ArcArc
Arc
Circle
51 m Circle
51 m
O51 m
≠
≠
O
120120
m m
Radius
Circumference
Radius
120 m
Radius
8 In each of the following
C shapes, all the curves are semicircles.
r
r
Circumference
Calculate
the
perimeter of each shape, giving your answer to one decimal place.
2m
2m
51 m
C
r
a b c 2m
Diameter
Diameter
Diameter
d d
d
2m
Chord
Chord
Chord
2m
Tangent
Tangent
Tangent
Segment
Segment
Segment
2 m2 m
2 mm
120
Arc
51 m
Arc
51 7
m cm
7 cm
5 cm5 cm
5 cm5 cm
5 cm
5 cm
7 cm
13 cm
13 cm
5 cm1205mcm 5 cm
13 cm
Radius
120 m and those on a mountain bike
9 The wheels on a BMX
bike have a diameter of 50.8 cm
Radius
have
a diameter rof 66 cm. Which bike’s wheels make more revolutions in a 400 m
r race? How many more revolutions do they make?
2m
2m
10 The inside perimeter
7 cm of an international-standard running track is 400 m and the two
straights are 85 m long.
What is the radius
of the
curved ends?
5 cm
Diameter
2 m5 cm 5 cm
Diameter
2m
13 cm
d
d
7 cm
7 cm
Calculating the area of a circle and compound
shapes involving circular shapes
13 cm
5 cm
13 cm
5 cm
1
–
1
C
–
Chord
Chord
2C 2
The circle shown has been
1
–
split into 16 equal sectors. 2 C
r been placed A A
These
r have
Tangent
together
to
r form a shape
A
Tangent
which is roughly a rectangle.
Segment
Semicircle
Segment
Semicircle
As
the
circle
is
split
into
more
Semicircle
·4 m
1
and
more
sectors
which
are
3
cm
3·43m
–C
3 cm
2
2
placed together, the resulting shape eventually3becomes
a rectangle.
≠ ≠
4 4x2 x = =
·4 m
3 cm
The area of this rectangle will be the same as the–12 Carea of the circle. ≠
4 x2 =
r
A Sector
1
–C
2
Sector
The length of the rectangle
is half the
r
A
circumference, C, of the circle and its
r
A
width is the radius, r, of the circle.
7 cm
1cm
·4 cm
7 cm
1
·
4
11mm
11mm
Semicircle
21 mm 3·4 m2·42m·4 m
21
mm
7 cm
1·4 cm
3 cm
11mm
≠
4 x 2 = 3· 4 m
21 mm
Semicircle
2·4 m
3 cm
7
≠
4 x2
3· 4 m
3 cm
1 cm
·5 cm
1 cm
3·53cm
2m·1 m
CFE Math Fourth Level_sample.indd
7
18/05/12 6:54 PM
2
·
1
14
mm
14 mm
5 cm
Sector
Sector
Sector
=
7 cm
7 cm
5 cm
5 cm 5 cm
5 cm 5 cm
5 cm
1313
cmcm
Chord
Chord
Shape, position and movement
So, the area of the rectangle is given by:
Tangent
Tangent
Segment
Segment
1
2
1
2
A= C×r
=
= p × r × r = pr 2
×2×p×r×r
Hence, the formula for the area, A, of a circle of radius r is:
1 1
–
2 C–
2C
2
Sector
Sector
A = pr r r
AA
Calculate the area of each of the following circles. Give each answer to one
decimal place.
Example 26·3
Semicircle
Semicircle
a
b 3 cm
3 cm
A = pr2 = p × 32
1111
mm
mm
2
=
cm
2121
mm
2·4
mm 9p = 28·3
2·m
4 m(to 1 dp)
4 4 x2x2 = =
1·4
1·cm
4 cm
b The diameter, d = 3·4 m, so r = 1·7 m. This gives:
1 cm
1 cm
≠≠
a The radius, r = 3 cm, which gives:
7 cm
7 cm
3· 4
3·m
4m
A = pr2 = p × 1·72
5
3·2cm
5(to
cm1 dp)
= 2·89p = 9·13·m
2·1
2·m
1m
1414
mm
mm
5·5 ·m
5m
Example 26·4
Calculate the area of this shape. The curve is a semicircle. Give your answer to one
decimal place.
3·4
3·m
4m
6 cm
6 cm
A1
3m
A2
8m
Diameter of semicircle, d = 8 - 3 = 5 m
Radius of semicircle, r = 5 ÷ 2 = 2·5 m
Area of the semicircular part, A1 is:
1
2
A1 = πr2 (since it is half a circle)
1
2
= × π × 2·52
= 9·8 m2 (to 1 dp)
1·5 m
Area of rectangular part, A2 is:
A2 = lb
= 8 × 1·5
= 12 m2
So, total area is:
A = A1 + A2
= 12 + 9·8
= 21·8 m2 (to 1 dp)
8
CFE Math Fourth Level_sample.indd 8
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Tangent
Tangent
Tangent
Tangent
SHIFT
SHIF
SHIFT
Segment
Segment
Segment
Segment
4 cm
cm
cm
6 cm
66 cm
66 cm
cm
4 cm
44 cm
4 cmof 2D shapes
cm
Properties
3D objects
cm
9and
cm
99 cm
99 cm
In this Exercise, take π = 3·14 or use the π key on your calculator.
1
1
1
1
–
–
–
–
2C 2C 2C 2C
Sector
Sector
Sector
Sector
1Calculate the area of each of the following circles. Give each answer to one
decimal place.
r r r r
A A A A
b
c
a
cm
cm
4 cm
44 cm
5 cm
cm
44 cm
5 cm
cm
5 cm
1
cm
5
5
1
cm
1
cm
1 cm
1 cm
Semicircle
Semicircle
Semicircle
Semicircle
Exercise
26B
3·4 m
3·4 3
m·4 3m·4 m
3 cm3 cm
3 cm
3 cm
d
e
2
cm
cm
2 cm
22 cm
2 cm
cm
cm
5 cm
55 cm
55 cm
≠ ≠ ≠ 4 ≠ 4x2 4x
f
cm
cm
266cm
66··226·cm
··22 cm
7 cm7 cm
7 cm
7 cm
1·4 cm
·4 cm
·4 1cm
1· 4 1
cm
11mm
11mm
11mm
11mm
21 mm 2·4 m
21 mm
21 mm
21 mm
2·4 m
2·4 m
2· 4 m
2 Calculate the area of each of the following circles. Give each answer to one
decimal place.
a 1 cm1 cm
1 cm
1 cm
b c d 3·5 cm
·5 cm
·5 3cm
3·5 3
cm
2·1 m
2·1 m
2·1 m
2·1 m
14 mm
14 mm
14 mm
14 mm
5· 5 m
5· 5 m
5·5 m
5· 5 m
3 Calculate the area of a circular tablemat with a diameter of 21 cm. Give your answer
Circle
3·4 m
3·4 3
m·4 3m·4 m
to6 the
cm nearest square centimetre.
6 cm
6 cm
6 cm
≠
O
E
4 Measure the diameter of a 1p coin to the nearest millimetre. Calculate
the area of one face of the coin, giving your answer to the nearest square
Circumference
millimetre.
C
1
51 m
120 m
Radius
6 The minute hand of a clock has a length of 13 cm. Calculate the area
r swept by the minute hand in one hour. Give your answer to one
2m
decimal place.
11
1
2
9
3
4
Bloggs
7
d
12
10
8
Diameter
Y
5 Calculate the area of the
sports ground shown. The
bends
Arc are semicircles. Give
your answer to the nearest
square metre.
P E N
N
O N
6
5
2m
7 cm
5 cm
13 cm
Chord
Tangent
9
Segment
CFE Math Fourth Level_sample.indd 9
18/05/12 6:55 PM
5c
Shape, position and movement
4
14 0
0
60 m
11 cm
60 m
m
60
11 cm
11 cm
cm
8 Calculate the area of each of the following shapes. Give your answers to one
11
10 cm
decimal place.
10 cm
10 cm
cm
a b c 10
1 cm
10 cm
10 cm
1 cm
6 cm
cm
10 cm
10 cm
11 cm
10 cm
cm
6 cm
10 cm
cm
10
10
5 cm
6
cm
6 cm
5 cm
8 cm
cm
55 cm
8 cm
cm
88 cm
0 0 180
180
1717
00
161060 10
10
0
15150 2020
0 0 3030
180
14140 60 170
0
4 40 1
0 20 10
15
30
0
10
180 170 20
160 30
15
0
4
0 0 1010
14 0
2020
180
17
0
180 17
0 0 161
060 3030
1515
0 0 404
141 0
040
0
14 0
4
170 180
0
10
0
10
180 170 20
160
100
110
80
12
70
0
90 100
60 80
1
70 51030 90 80 110
90 70 100120
80 90
60 110 0 0 80
1006 11
0
70
100 90 80 0 11 013
12
0
7
0 1100 100 90 80 77050 01120
5 0
6
0 0 6020 1
0
1
6 120 1
13
60
3
0
0
5
513 0
12
501300
500 0
3
1
160
20
60 m
90
90
0
15
30
30
15
0
50 0
13
80
70 100
60 110
0
2
1
0
14 0
4
7 Calculate the area of one face of the
semicircular protractor shown. Give your
40 m answer to one decimal place.
40 m
40 m
m
40
9 A circular lawn has an area of 200 m2. Calculate the radius of the lawn, giving your
answer to one decimal place.
10 A circular
disc has
Length
l a circumference of 15 cm. Calculate the area of one face of the
disc, giving your answer
to one
Length
l decimal place.
10 cm
Length8 llcm
Length
10 cm
10 cm
cm
10
8 cm
Area A
8 cm
cm
8
Area A
15 cm
Area A
A
The arbelosArea
15 cm
6 cm
15 cm
cm
15
6 cm
An arbelos is the shaded area shown in the diagram. 66 cm
cm
Challenge
12 cm
15 cm
8m
9 cm
12
12 cm
cm
12
9 cm2·5 m
cm
99 cm
8m
m
88 m
2·5 m
15 cm
2·55m
mm
2·5
15
cm
16 cm
15 cm
10
cm
1 cm
7m
6 cm
16 cm
1 cm
16 cm
cm
7m
6 cm
16
1 cm
cm
m
6
cm
4
cm
1
77 m
6
cm
It is the Greek name for an ancient shoemaker’s knife and has a number
of
4 cm
interesting properties.
4
cm
4 cm
1·5 m
1 The diameter of the large semicircle is 10 cm.
1·5 m
1·5 m
m
1·5
21 m
a If the diameters of the two smaller
semicircles
are
6 cm
and
4 cm,
find:
21 m
1m
21 m
m
21
4m
i
the perimeter of the arbelos
ii the area of the arbelos1 m1 m
1m
4m
4m
m
4
b If the diameters of the two smaller semicircles are 8 cm and 2 cm, find:
6m
6m
i the perimeter
of the arbelos
ii the area of the arbelos
m
66 m
h
2 What do you notice about the area and perimeter of the arbelos? Try some
other diameters to check if you are correct.
10
5m
m
55 m
140 cm
140 cm
140 cm
cm
140
25 cm
150 cm
CFE Math Fourth Level_sample.indd 10
150 cm
25 cm
25 cm
cm
25
18/05/12 6:55 PM
π
r
O
Properties of 2D shapes and 3D
objects
d
Arcs and sectors
The arc, AB, is part of the circumference. The sector AOB is a slice
of the circle enclosed by the arc AB, and the radii OA and OB.
O
∠AOB is the angle of the sector, and is usually denoted by the
Greek letter θ (pronounced theta).
The length of the arc AB, as a fraction of the circumference, is
Arc
r
Sector
θ
A
d
π
B
a
O
θ
360
θ
So, the length of the arc AB =
× πd
360
Arc
A
Sector
Aθ
O 60°
θ
× π r2
Similarly, the area of the sector AOB =
360
d
B
B
5 cm
5
O
Example 26·5Calculate:
a the length of the arc AB
b the area of the sector AOB.
Round your answers to an appropriate degree of accuracy.
A
B
60°
a 5 cm
O
60
× p × 10 = 5·24 cm
360
60
b Area of the sector AOB =
× p × 52 = 13·1 cm2
360
a Length of the arc AB =
Exercise
26C
In this Exercise, let p = 3·142 or use the p key on your calculator.
1 For each of the circles below, calculate:
b
60°
5 ·4 m
6 cm
a
b
60°
45°
6 cm
i the length of the arc
ii the area of the sector.
r
r
O O
O O
d d
d
d
A
A
A
A
r
r
O
O
d
d
r
r
Arc Arc Arc
AArcA ArcB
Arc B B
A A
B B B
Sector
Sector
Sector
Sector
Sector
Sector
θ θ θ
θ θ θ
O O O
O O O
A A
B B B
A A
B B B
60° 60° 60°
60°60°
60° 5
cm5 cm
5 cm
5 cm5 cm
5 cm
O O O
O O O
π π πa
π π πa
d
d
Round your answers to an appropriate degree of accuracy.
a a b b
b b a
a a
b b
b
60° 60° 60°
60° 60°60°
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
d d
d d d
e e e
e e
e e 4 m4 m4 m
4 m 4 m4 m
75° 75° 75°
75° 75°75°
135°135°135°
135°135°
135°
m·4 m
5 ·4 5
5 ·4 m
5·4 m5·4 5m·4 m
CFE Math Fourth Level_sample.indd 11
c c c
c
c
45° 45° 45°
7 cm
7 cm
7 cm
45° 45°
45°
7 cm7 cm
7 cm
c
c
f f
f f
f
f
f
120°120°120°
120°120°
120°
m·5 m
2·5 2
2·5 m
2·5 m2·5 2m·5 m
12 cm
12 cm
12 cm
1220°
cm
12 cm
1220°
cm20°
20° 20°20°
7·2 7
m·2 m
7·2 m
7·2 m7·2 7m·2 m
150°150°150°
150°150°
150°
15 mm
15 mm
15 mm
15 mm
15 mm
15 mm
60° 60° 60°
60° 60°60°
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm
11
10 cm
10 cm
10 cm
10 cm
10 cm
10 cm 18/05/12
6:55 PM
O
d
Shape, position
Arc
π and movement
a
A
B
r
10 cm
b
d
60°
45°
2 A flowerbed in a park is in the shape
of a sectorB
of a circle.
A
7 cm
Sector
O
4m
10 cm
60°
θ total perimeter of the flowerbed.
a
Calculate
the
d
5 cm
75°
O
b Calculate the area of the flowerbed. O
a
d
d
c e
12120°
cm
20°
2·5 m
135°
1
5 ·4 m
Arc
B
Round
your answers to an appropriate
degree of accuracy.
d
e
f
120°
Sector
7·2 m
3 Calculate
of the shape on the right. Give your
4B m
A the total perimeter
2·5 m
θ
answer to the nearest millimetre.
15 mmc
a
b
60°
75°
135°
O
60°
45°
12 cm
150°
60°
5 cm
16 cm
30°
·
4
m
5
6 cm
π π πa a a
b b b
c
O
r
r
r
60° 60° 60°
45° 45° 45°
7 cm7 cm7 cm
O O O
10
cm
10your
cm
10 cm
4 Calculate the area of the shaded segment on the right.
Give
A
B
d nearest
d
answerdto the
square centimetre.
15 mm
60°
10 cm
135°
a
b
c
d
60°
5 cm
60°
45°
120°
12 cm
m Arc Arc
5·4Arc
O
e 10 cm
e e 30°
f
b b A A A
c c d d d 16 cm
B B 6Bcm
9 cm
120°
60° 60°
45° 45°
120°
120°
12 cm
12 cm
Sector
cm
7Sector
cm7Sector
4 m 4 m4 m
20° 20°
10 cm
5 A circular
stained
glass
window
with
a
diameter
of 2·4 m is divided
2·5 m
θ
10 cm
2·5 m
θ
2·5 m
θ
into eight equal panes. One of the panes is accidentally
smashed.
75°
75°
75°
O O O
a Calculate:
b
c
d
e
60°
45°
120°
12 cm
240°
a the area
glass that needs to be replaced
16ofcm
30°
e
f
e
f
6 cm
9 cm
120° of lead
120°
b the
length
needed to surround the pane.
2·4 m
A
B
7·2 m
7·2 m
A B
B
A
4 m4 m
3
15
·
5
m
2
·5 m to two decimal places.
Give your 2answers
15cm
mm
15 mm mm
1
60° 60° 60°
10 cm
135°
135°
135°
75° 75°
60° 60°
60°
5 cm5 cm5 cm
150°
m
5·4150°
5 ·4 m
5 ·4 m
O
O
O
For the following circles, calculate:
i the length of the arc ii the area of the sector.
A
a
75°
Challenge
Give your answers in terms of p.
15 mm
15 mm
10
10
b a a a a
b cm
b
b cm
c c c c
d d
60° 60° 60° 60° 60°
12 cm
45° 45° 45°
12 cm
12 cm
16 cm
16 cm
16 cm
30° 30°
30°
10 cm
10 cm
6 cm6 cm6 cm
135°135°
5 ·4 m
5 ·4 m
c
45° 45°
16 cm
16 cm
d d
c
12 cm
12 cm
30° 30°
d
120°120°
d
e
120°120°120°
9 cm9 cm9 cm
e e e
240°240°
9 cm9 cm
3 cm3 cm
12
CFE Math Fourth Level_sample.indd 12
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Properties of 2D shapes and 3D objects
• By working on this topic, I have learnt how to find the circumference and area of a circle or parts
of a circle.
• I can use the correct terminology to identify the different parts of a circle.
• I can explain the relationship between the circumference and diameter of a circle. Investigation 1
pages 3–4
• I can explain how different approximations for p have been used throughout history.
Investigation 2 page 4
• I can use a formula to calculate the circumference of a circle.
• I can use a formula to calculate the area of a circle.
Exercise 26A Q4
Exercise 26B Q3
• I can calculate the perimeter and area of compound shapes involving circles. Exercise 26A Q5
Exercise 26B Q5
• I can apply my knowledge of circumference and area to find the length of an arc and the area of
a sector of a circle.
Exercise 26C Q2
13
CFE Math Fourth Level_sample.indd 13
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