Transcript Slide 1

© T Madas
The area of a circle The circumference of a circle
C = 2 r
A = r2
We can use these formulae to find
areas of sectors and lengths of arcs
Arc
Sector
r
© T Madas
The area of a circle The circumference of a circle
C = 2 r
A = r2
We can use these formulae to find
areas of sectors and lengths of arcs
50
Area of sector =  ´ 6 ´
360
2
50°
6 cm
50
=  ´ 36 ´
3601 0
= 5 » 15.7 cm2
© T Madas
The area of a circle The circumference of a circle
C = 2 r
A = r2
We can use these formulae to find
areas of sectors and lengths of arcs
50
Length of arc = 2 ´  ´ 6 ´
360
50°
6 cm
50
=  ´ 12 ´
3603 0
=
5
3
» 5.24 cm
© T Madas
© T Madas
24 cm
Find the area of the circular segment, from the
information given.
© T Madas
24 cm
Find the area of the circular segment, from the
information given.
© T Madas
Find the area of the circular segment, from the
information given.
x 2 + 122 = 132
x
5 cm
24 cm
12 cm
x 2 = 132 - 12 2
x 2 = 25
Û
Û
Û
x = 5 cm
© T Madas
Find the area of the circular segment, from the
information given.
x 2 + 122 = 132
12 cm
x 2 = 132 - 12 2
5 cm
x 2 = 25
Û
Û
Û
x = 5 cm
Area of the triangle
1´ 5
2
2
´ 24 = 60 cm
© T Madas
Find the area of the circular segment, from the
information given.
138.74 cm2
12 cm

12 = sin 
13
Û
- 1 12
sin
=
13
( )
Û
 » 67.38°
5 cm
Area of the triangle
1´ 5
2
134.76°
2
´ 24 = 60 cm
134.76
2
´

´
13
»
198
.
74
cm
Area of sector =
360
2
© T Madas
© T Madas
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.
1
6
area of a circle
A = πr 2
A = π x 52
area of the sector
A = π x 5 2 x 1 ≈ 13.1 cm2
5 cm
6
1
8
area of a circle
A = πr 2
A = π x 62
area of the sector
6 cm
A = π x 6 2 x 1 ≈ 14.1 cm2
8
© T Madas
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.
1
6
circumference of a circle
C = 2 πr
C = 2x πx 5
length of the arc
5 cm
L = 2 x πx 5
1
8
1 ≈ 5.2 cm
6
circumference of a circle
C = 2 πr
C = 2x πx 6
length of the arc
L = 2 x πx 6
6 cm
x
x
1 ≈ 4.7 cm
8
© T Madas
© T Madas
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°.
1.
Calculate the area of the composite shape in terms of π
2.
Round your answer to part (1), to 3 significant figures.
area of the semicircle
A=
A
6 cm
O
60°
B
A=
1
2
2 xπr
1
2
x
x
6
π
2
⇔
⇔
A = 18 π
area of the sector
A=
A=
C
1
60°=
of a circle
6
1
2
6 xπr
1
2
x
x
12
π
6
⇔
⇔
A = 24 π
© T Madas
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°.
1.
Calculate the area of the composite shape in terms of π
2.
Round your answer to part (1), to 3 significant figures.
A = 42 π = 42 x π ≈ 132 cm2
A
6 cm
O
60°
[3 s.f.]
B
C
© T Madas
© T Madas
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°
1.
2.
r
θ
R
© T Madas
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°
1.
2.
πR 2
x
θ
360
r
θ
R
© T Madas
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°
1.
2.
πR 2
x
θ – πr2 x θ
360
360
r
θ
R
© T Madas
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°
1.
2.
R–r
A = πR
2
x
θ – πr2 x θ + π R – r
360
360
2
2
r
θ
R
© T Madas
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.
1.
2.
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°
A = πR
2
x
θ – π r 2 x θ +Aπ=Rπ–Rr2 2x θ – π r 2 x θ + π R – r
360
360
2
360
360
2
2
© T Madas
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.
360
A = πθ R 2 – r 2 + π R – r
360
A =
45π
360
A =π
8
x
2
360
2
c
x
θ – πr2 x θ + π R – r
2
4
2
9 –
72
π
+
4
32 + π x 4
4
A = 4π + π
A = 5π ≈ 15.7 m2
9–7
2
c
A = πR
2
c
2.
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°
c
1.
© T Madas
© T Madas
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.
45°
3 cm
area of a circle
A = πr 2
A = π x 82
area of the given sector
A = π x 8 2 x 45
360
volume of the prism
45°
8 cm
V = π x 8 2 x 45 x 3
360
V = 24π
V ≈ 75.4 cm3
[3 s.f.]
© T Madas
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 x 45
45°
360
3 cm
area of the given sector
A = π x 8 2 x 45
360
45°
8 cm
© T Madas
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 x 45 x 2 = 16π
45°
3 cm
2πr
360
area of the side rectangle x 2
A = 3 x 8 x 2 = 48
area of the curved surface
A = 2 x π x 8 x 45 x 3 = 6π
360
45°
8 cm
total surface area
A = 16π + 48 + 6π ≈ 117 cm2 [3 s.f.]
© T Madas
© T Madas
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
72
area of a circle
A = πr 2
A = π x 60 2
area of the “larger” sector
162°
A
O
B
A = π x 60 2 x 162 = 1620π
360
All measurements in cm
© T Madas
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
72
area of a circle
A = πr 2
A = π x 60 2
area of the “larger” sector
162°
A
O
All measurements in cm
B
A = π x 60 2 x 162 = 1620π
360
area of the “inner” sector
A = π x 15 2 x 162 = 101.25π
360
© T Madas
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
72
area of a circle
A = πr 2
A = π x 60 2
area of the “larger” sector
162°
A
B
O
All measurements in cm
A = π x 60 2 x 162 = 1620π
360
area of the “inner” sector
A = π x 15 2 x 162 = 101.25π
360
area swept by the wiper is
1620π – 101.25π = 1518.75π cm2
≈ 4771.3 cm2
© T Madas
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.
162°
A
B
c
72
area wiped
x 100
%age =
total area
1518.75π
x 100
%age =
141 x 72
c
141
%age ≈ 47%
O
All measurements in cm
area swept by the wiper is
1620π – 101.25π = 1518.75π cm2
≈ 4771.3 cm2
© T Madas
© T Madas
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
A
θ
5 cm
3,4,5 = Alarm Bells !
B
∆ABC is right angled
By trig on ∆ABC :
tanq = 34
Û
4
q = tan-1( 3 ) Û
q » 53.13°
© T Madas
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°
C
36.87°
253.74°
A
θ
5 cm
B
73.74°
106.26°
D
286.26°
RCBA ≈ 36.87°
RCAD ≈ 106.26°
RCBD ≈ 73.74°
reflex RCAD ≈ 253.74°
reflex RCBD ≈ 286.26°
Simplify the diagram by showing only the
relevant information
© T Madas
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 253.74° on a circle
with radius of 3 cm
C
253.74°
B
A
286.26°
This arc corresponds to an
angle of 286.26° on a circle
with radius of 4 cm
D
© T Madas
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°
2´ p ´ 3 ´
B
A
D
286.26°
253.74
360
» 13.29 cm
The larger of the 2 arcs
2´ p ´ 4 ´
286.26
360
» 19.98 cm
The required perimeter
33.3 cm [2 s.f.]
© T Madas
© T Madas
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
O
8m
θ
4m
B
3m D
A
OBD is right angled
OD = 3 m
Trig to find RBOD
C
tanq = 34
Û
4
q = tan-1( 3 ) Û
q » 53.13°
© T Madas
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
p ´ 52´
θ
O
D
53.13
360
» 11.59 m2
C
The length of the arc
2´ p ´ 5 ´
53.13
360
» 4.64 m
A
area of the sector ≈ 11.59 m2
length of the arc ≈ 11.59 m
© T Madas
θ
O
B
OBC is right angled
OB = 5 m
4m
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
RBOC = 53.13°
Trig to find BC
3m D
A
area of the sector ≈ 11.59 m2
length of the arc ≈ 11.59 m
C
BC = tan q
OB
BC = tan 53.13° = 4
5
3
Û
3BC = 20
Û
Û
BC = 20
m
3
BC = 20/3 m
© T Madas
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
1 ´ 5 ´ 20 = 100 = 50 » 16.67 m2
6
3
3
2
θ
O
D
C
area of the sector ≈ 11.59 m2
half the shaded area ≈ 5.08 m2
A
shaded area ≈ 10.16 m2
area of the sector ≈ 11.59 m2
length of the arc ≈ 11.59 m
BC = 20/3 m
© T Madas
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 ≈ 11.59 m
Length BC = 20/3 m ≈ 6.67 m
θ
O
D
E
C
11.59 x 2 ≈ 23.18
6.67 x 2 ≈ 13.34
Total perimeter ≈ 36.5 m
A
area of the sector ≈ 11.59 m2
length of the arc ≈ 11.59 m
BC = 20/3 m
© T Madas
© T Madas
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
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 .
A= r ´
2
L
A
θ
L = 2 r ´

360

360
r
© T Madas
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

r
´
A=
360
L = 2 r ´

360
2
360 A =  r
r
2
r
2
360L = 2 r
2 r
2 r
Û
A=
Û L=
Û =
Û =
 r 2
360
2 r
360
360 A
r2
180L
r
Û
2
360 A =  r
Û
Û
360L = 2 r
Û
Now what?
© T Madas
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 .


=
=
360 A
r2
180L
r
Û
360 A
r
2
=
180L
r
Û
360 rA = 180 Lr 2
360 r
360 r
2
rL
A= 2
© T Madas
Harder Examples
© T Madas
© T Madas
The following pattern is produced inside a square
whose side length is 8 cm
Calculate the shaded area
D
C
45°
Plan
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
A
8 cm
B
of a circle whose radius is half the
square’s diagonal
© T Madas
The following pattern is produced inside a square
whose side length is 8 cm
Calculate the shaded area
D
C
4
x 2 = 42 + 42
x 2 = 32
x = 32
45°
4
Pythagoras:
x
Û
Û
Area of sector:
1´
8
2
p ´ ( 32) = 4p
Area of triangle:
A
8 cm
B
8 cm
© T Madas
The following pattern is produced inside a square
whose side length is 8 cm
Calculate the shaded area
D
C
Area of sector:
4p
Area of triangle: 8 cm
Area of sector:
4p
Area of triangle:
A
8 cm
B
8 cm
© T Madas
The following pattern is produced inside a square
whose side length is 8 cm
Calculate the shaded area
D
C
Area of sector:
4p
Area of triangle: 8 cm
Shaded area:
(4p
- 8 )´ 8 = 32 (p - 2)
» 36.5 cm2
A
8 cm
B
© T Madas
© T Madas
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 bottom part of the shaded
area consists of:
• an equilateral triangle with
side length of 8 cm
• plus 2 circular segments
60°
A
8 cm
B
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.
© T Madas
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 area of the segment
1´
6
p ´ 82 -
1´ 8 ´ 8 ´ sin60°
2
sector = 1/6 circle
triangle
60°
A
8 cm
B
© T Madas
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 area of the segment
1´ p
6
32p 3
1´ 8 ´ 8 ´ sin60°
´ 82 - 2
32 ´ 23 = 32p - 16 3
3
sin60° = 3
2
60°
A
8 cm
B
© T Madas
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
D non - shaded regions C
The area of the segment
1´ p
6
32p 3
1´ 8 ´ 8 ´ sin60°
´ 82 - 2
32 ´ 23 = 32p - 16 3
3
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
60°
A
8 cm
B
1 ´
12
p ´ 82 -
(323p -
)
16 3
© T Madas
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
1 ´
12
C
(323p -
p ´ 82 -
)
16 3
p - 32p + 16 3
= 64
3
12
= 163p - 323p + 16 3
=
-- 16p
3
+ 16 3
= 16 3 - 163p
60°
A
8 cm
B
1 ´
12
p ´ 82 -
(323p -
)
16 3
© T Madas
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
1 ´
12
C
p ´ 82 -
(323p -
)
16 3
p - 32p + 16 3
= 64
3
12
= 163p - 323p + 16 3
=
-- 16p
3
+ 16 3
= 16 3 - 163p
60°
A
8 cm
B
© T Madas
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
64 - 2(16 3 - 163p
A
8 cm
)
» 42 cm2
B
© T Madas
© T Madas
2´  ´ L ´
r +h = L
2
2
2
A= L ´
2
h
r
360
= 2 r
θ
360
L
L
r


A
2 r
2 r
r
© T Madas
2´  ´ L ´
r +h = L
2
2
2
A= L ´
2

360
= 2 r

360
© T Madas
r 2 + h 2 = L2
A= L ´

2
360
2´  ´ L ´

360
2 L
= 2 r
360
L
360
=r
L = 360r
= 2 r
Û
Û
Û
Û
360r
=
L
© T Madas
r 2 + h 2 = L2
A= L ´

2
360
Û
2

L
´
A=
2

L
´
A=
360r
L
360
360r
L
360
1
360r
A= L ´
360L
2
Û
Û
Û
A =  Lr
360r
=
L
© T Madas
r 2 + h 2 = L2
A= L ´

2
Û
L=
r 2 + h2
360
360r
=
L
A =  Lr
2
2
A = r r + h
© T Madas
h
L
r
A =  Lr
2
2
A = r r + h
© T Madas
© T Madas