Transcript S5 Circles

KS4 Mathematics
S5 Circles
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Contents
S5 Circles
A S5.1 Naming circle parts
A S5.2 Angles in a circle
A S5.3 Tangents and chords
A S5.4 Circumference and arc length
A S5.5 Areas of circles and sectors
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Naming the parts of a circle
A circle is a set of points equidistant from its centre.
The distance around the
outside of a circle is called
the circumference.
radius
centre
circumference
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The radius is the
distance from the centre
of the circle to the
circumference.
The diameter is the
distance across the width
of the circle through the
centre.
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Arcs and sectors
arc
An arc is a part of the
circumference.
sector
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When an arc is bounded
by two radii a sector is
formed.
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A line moving through a circle
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Naming the parts of a circle
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Contents
S5 Circles
A S5.1 Naming circle parts
A S5.2 Angles in a circle
A S5.3 Tangents and chords
A S5.4 Circumference and arc length
A S5.5 Areas of circles and sectors
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Right angles in a semicircle
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Right angles in a semicircle
We have just seen a demonstration showing that the angle in a
semicircle is always a right angle.
We can prove this result as follows:
Draw a line from C to O. This line is a
radius of the circle.
C
x
In triangle AOC,
A
x
O
B
OA = OC
(both radii)
So, angle OAC = angle OCA
(angles at the base of an isosceles
triangle)
Let’s call these angles x.
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Right angles in a semicircle
We have just seen a demonstration showing that the angle in a
semicircle is always a right angle.
We can prove this result as follows:
In triangle BOC,
C
OB = OC
x y
A
y
x
O
(both radii)
So, angle OBC = angle OCB
B
(angles at the base of an isosceles
triangle)
Let’s call these angles y.
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Right angles in a semicircle
We have just seen a demonstration that the angle in a
semicircle is always a right angle.
We can prove this result as follows:
In triangle ABC,
C
x + x + y + y = 180° (angles in a triangle)
x y
A
y
x
O
2x + 2y = 180°
B
2(x + y) = 180°
x + y = 90°
Angle ACB = x + y = 90°
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Calculating the size of unknown angles
Calculate the size of the labelled angles in the following
diagram:
a = 37° (angles at the base of an
isosceles triangle)
37°
O
d
c = 53° (angles at the base of an
isosceles triangle)
e
c
b
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b = 90° – 37°
= 53° (angle in a semi-circle)
a
d = 180° – 2 × 53°
= 74° (angles in a triangle)
e = 180° – 74°
= 106° (angles on a line)
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The angle at the centre
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The angle at the centre
We have just seen a demonstration that shows that the angle
at the centre of a circle is twice the angle at the circumference.
We can prove this result as follows:
B
Draw a line from B, through the centre
O, and to the other side D.
x
In triangle AOB,
OA = OB
O
So, angle OAB = angle OBA
x
C
A
D
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(both radii)
(angles at the base of an isosceles
triangle)
Let’s call these angles x.
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The angle at the centre
We have just seen a demonstration that shows that the angle
at the centre of a circle is twice the angle at the circumference.
We can prove this result as follows:
In triangle BOC,
B
OB = OC
x y
(both radii)
So, angle OBC = angle OCB
O
y
x
C
(angles at the base of an isosceles
triangle)
Let’s call these angles y.
A
D
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The angle at the centre
We have just seen a demonstration that shows that the angle
at the centre of a circle is twice the angle at the circumference.
We can prove this result as follows:
angle AOD = 2x
B
angle COD = 2y
x y
(the exterior angle in a triangle is
equal to the sum of the opposite
interior angles)
O
x
2x 2y
y
C
angle AOC = 2x + 2y
= 2(x + y)
A
D
angle ABC = x + y
 angle AOC = 2 × angle ABC
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Calculating the size of unknown angles
Calculate the size of the labelled angles in the following
diagram:
a = 29° (angles at the base of an
isosceles triangle)
c
b = 180° – 2 × 29°
= 122° (angles in a triangle)
O
b
d
29°
41°
a
c = 122° ÷ 2
= 61° (angle at the centre is
twice angle on the
circumference)
d = 180° – (29° + 29° + 41° + 61°)
= 20° (angles in a triangle)
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Angles in the same segment
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Angles in the same segment
We have just seen a demonstration that shows that the angles
in the same segment are equal.
We can prove this result as follows:
C
Mark the centre of the circle O and
show angle AOB.
D
angle ADB = ½ of angle AOB
O
and angle ACB = ½ of angle AOB
B
A
(the angle at the centre of a circle is
twice the angle at the circumference)

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angle ADB = angle ACB
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Calculating the size of unknown angles
Calculate the size of the labelled angles in the following
diagram:
a = 90° – 51°
c
= 39° (angle in a semi-circle)
d
O
44°
b = 180° – (90° + 44°)
= 46° (angles in a triangle)
c = b = 46° (angles in the same
segment)
b
51° a
d = 51° (angles in the same
segment)
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Angles in a cyclic quadrilateral
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Angles in a cyclic quadrilateral
We have just seen a demonstration that shows that the
opposite angles in a cyclic quadrilateral add up to 180°.
We can prove this result as follows:
Mark the centre of the circle O and
label angles ABC and ADC x and y.
B
x
O
The angles at the centre are 2x and 2y.
2y
(the angle at the centre of a circle is
twice the angle at the circumference)
2x
C
y
D
A
2x + 2y = 360°
2(x + y) = 360°
x + y = 180°
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Angles in a cyclic quadrilateral
As a result of this theorem we can conclude that if the opposite
angles of a quadrilateral add up to 180°, a circle can be drawn
through each of its vertices.
For example, the opposite angles in this quadrilateral add up to
180°.
67°
It is a cyclic quadrilateral.
109°
113°
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71°
Remember that when two angles
add up to 180° they are often
called supplementary angles.
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Calculating the size of unknown angles
Calculate the size of the labelled angles in the following
diagram:
a = 64° (angle at the centre)
c
b
128°
d
O
e
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f
b = c = (180° – 128°) ÷ 2
= 26° (angles at the base of
an isosceles triangle)
d = 33° (angles at the base of an
isosceles triangle)
e = 180° – 2 × 33°
= 114° (angles in a triangle)
33° a
f = 180° – (e + c)
= 180° – 140°
= 40° (opposite angles in a
cyclic quadrilateral)
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Contents
S5 Circles
A S5.1 Naming circle parts
A S5.2 Angles in a circle
A S5.3 Tangents and chords
A S5.4 Circumference and arc length
A S5.5 Areas of circles and sectors
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The tangent and the radius
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Two tangents from a point
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The perpendicular from the centre to a chord
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The alternate segment theorem
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Contents
S5 Circles
A S5.1 Naming circle parts
A S5.2 Angles in a circle
A S5.3 Tangents and chords
A S5.4 Circumference and arc length
A S5.5 Areas of circles and sectors
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The value of π
For any circle the circumference is always just over
three times bigger than the diameter.
The exact number is called π (pi).
We use the symbol π because the number cannot be written
exactly.
π = 3.141592653589793238462643383279502884197169
39937510582097494459230781640628620899862803482
53421170679821480865132823066470938446095505822
31725359408128481117450284102701938521105559644
62294895493038196 (to 200 decimal places)!
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Approximations for the value of π
When we are doing calculations involving the value π
we have to use an approximation for the value.
For a rough approximation we can use 3.
Better approximations are 3.14 or 22 .
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We can also use the π button on a calculator.
Most questions will tell you which approximation to use.
When a calculation has lots of steps we write π as a symbol
throughout and evaluate it at the end, if necessary.
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The circumference of a circle
For any circle,
circumference
π=
diameter
or,
C
π=
d
We can rearrange this to make a formula to find the
circumference of a circle given its diameter.
C = πd
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The circumference of a circle
Use π = 3.14 to find the circumference of this circle.
C = πd
9.5 cm
= 3.14 × 9.5
= 29.83 cm
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Finding the circumference given the radius
The diameter of a circle is two times its radius, or
d = 2r
We can substitute this into the formula
C = πd
to give us a formula to find the circumference of a circle
given its radius.
C = 2πr
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The circumference of a circle
Use π = 3.14 to find the circumference of the following circles:
4 cm
C = πd
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C = 2πr
= 3.14 × 4
= 2 × 3.14 × 9
= 12.56 cm
= 56.52 m
C = πd
23 mm
9m
58 cm
C = 2πr
= 3.14 × 23
= 2 × 3.14 × 58
= 72.22 mm
= 364.24 cm
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Finding the radius given the circumference
Use π = 3.14 to find the radius of this circle.
12 cm
C = 2πr
How can we rearrange this to
make r the subject of the formula?
C
r= ?
2π
12
=
2 × 3.14
= 1.91 cm (to 2 d.p.)
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Finding the length of an arc
What is the length of arc AB?
A
An arc is a section of the
circumference.
6 cm
The length of arc AB is a
fraction of the length of the
circumference.
B
To work out what fraction of
the circumference it is we look
at the angle at the centre.
In this example, we have a 90°
angle at the centre.
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Finding the length of an arc
What is the length of arc AB?
A
The arc length is 14 of the
circumference of the circle.
6 cm
This is because,
90°
1
=
360°
4
B
So,
Length of arc AB =
1
4
=
1
4
× 2πr
× 2π × 6
Length of arc AB = 9.42 cm (to 2 d.p.)
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Finding the length of an arc
A
B
r
θ
For any circle with radius r and angle at the centre θ,
This is the
θ
× 2πr
Arc length AB =
circumference of
360
the circle.
2πrθ
πrθ
Arc length AB =
=
360
180
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Finding the length of an arc
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The perimeter of shapes made from arcs
Find the perimeter of these shapes on a cm square grid:
40°
The perimeter of this shape is
made from three semi-circles.
Perimeter =
40°
1
=
360°
9
1
Perimeter = 9 × π × 12 +
1
2
×π×6+
1
2
×π×4+
1
9
1
2
×π×2
3+3
= 6π cm
= 18.85 cm (to 2 d.p.)
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×π×6+
= 2π + 6
= 12.28 cm (to 2 d.p.)
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Contents
S5 Circles
A S5.1 Naming circle parts
A S5.2 Angles in a circle
A S5.3 Tangents and chords
A S5.4 Circumference and arc length
A S5.5 Areas of circles and sectors
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Formula for the area of a circle
We can find the area of a circle using the formula
Area of a circle = π × r × r
r
or
Area of a circle = πr2
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The area of a circle
Use π = 3.14 to find the area of this circle.
7 cm
A = πr2
= 3.14 × 72
= 153.86 cm2
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Finding the area given the diameter
The radius of a circle is half of its diameter, or
d
r=
2
We can substitute this into the formula
A = πr2
to give us a formula to find the area of a circle given its
diameter.
πd2
A=
4
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The area of a circle
Use π = 3.14 to find the area of the following circles:
2 cm
A = πr2
= 3.14 ×
A = πr2
22
10 m
= 12.56 cm2
A = πr2
23 mm
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= 3.14 × 52
= 78.5 m2
78 cm
A = πr2
= 3.14 × 232
= 3.14 × 392
= 1661.06 mm2
= 4775.94 cm2
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Finding the area of a sector
What is the area of this sector?
72°
6 cm
72°
× π × 62
Area of the sector =
360°
1
=
× π × 62
5
= 22.62 cm2 (to 2 d.p.)
We can use this method to find the area of any sector.
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Finding the area of a sector
A
B
r
θ
O
For any circle with radius r and angle at the centre θ,
This is the area of
θ
× πr2
Area of sector AOB =
the circle.
360
πr2θ
Area of sector AOB =
360
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Finding the area of a sector
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The area of shapes made from sectors
Find the area of these shapes on a cm square grid:
40°
40°
1
=
360°
9
Area =
1
2
× π × 32
+
1
2
× π × 12
Area =
1
9
× π × 62
–
1
2
× π × 22
–
1
9
× π × 42
= 3π cm2
=
1
9
× π × 20 cm2
= 9.42 cm2 (to 2 d.p.)
= 6.98 cm2 (to 2 d.p.)
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