© t madas. the area of a circle the circumference of a circle r we can use these formulae to find...
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© T Madas
© T Madas
The area of a circle The circumference of a circle
=A r 2 =C r2
r
We can use these formulae to find areas of sectors and lengths of arcs
Sector Arc
© T Madas
The area of a circle The circumference of a circle
=A r 2 =C r2
6 cm
We can use these formulae to find areas of sectors and lengths of arcs
50°
Area of sector = ´ 26 ´50
360
= ´ 36´50
36010
= 5 .» 215 7cm
© T Madas
The area of a circle The circumference of a circle
=A r 2 =C r2
6 cm
We can use these formulae to find areas of sectors and lengths of arcs
50°
Length of arc = ´ ´ 62 ´50
360
= ´ 12´50
36030
= 53
.» 5 24cm
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© T Madas
13cm
24 c
m
Find the area of the circular segment, from the information given.
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13cm
24 c
m
Find the area of the circular segment, from the information given.
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13cm
24 c
m 12 c
m
x
x 2+ 212 = 213 Ûx 2 - 212= 213 Ûx 2 = 25 Ûx = 5 cm
5 cm
Find the area of the circular segment, from the information given.
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13cm
12 c
m
x 2
5 cm
+ 212 = 213 Ûx 2 - 212= 213 Ûx 2 = 25 Ûx = 5 cm
12 ´ 5 ´ 24 = 260 cm
Area of the triangle
Find the area of the circular segment, from the information given.
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13cm
12 c
m
12
5 cm
13= sin Û
( )- 1sin 1213= Û
» °67.38
134.76°
Area of sector = ´ 213 .´
134 76
360.» 2198 74 cm
138.74 cm2
12 ´ 5 ´ 24 = 260 cm
Area of the triangle
Find the area of the circular segment, from the information given.
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© T Madas
6 cm
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.
16
18
5 cm
A = πr 2
area of a circle
A =π x 5 2
area of the sector
A = 16
π x 5 2x ≈ 13.1 cm2
A = πr 2
area of a circle
A =π x 6 2
area of the sector
A = 18
π x 6 2x ≈ 14.1 cm2
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6 cm
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.
16
18
5 cm
C = r
circumference of a circle
C = x πx 5
length of the arc
16
x ≈ 5.2 cm
2 π2
L = x πx 5 2
C = r
circumference of a circle
C = x πx 6
length of the arc
18
x ≈ 4.7 cm
2 π2
L = x πx 6 2
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© 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
π2. Round your answer to part (1), to 3 significant
figures.O
A B
C
60°
6 cm
area of the semicircle
A = 12 x π r 2 ⇔
A = 12 x πx 62 ⇔
A = π18
area of the sector
A = 16 x π r 2 ⇔
A = 16 x π x 122 ⇔
A = π2460°= 16of a circle
42π
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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
π2. Round your answer to part (1), to 3 significant
figures.O
A B
C
60°
6 cm
A = π42 = x π42
42π≈ [3 s.f.]cm2132
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© 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. Find a formula for the area A of this pond in terms
of r, R and θ.2. Find the area of a pond with this shape if r = 7 m, R
= 9 m and θ = 45°
θ
r
R
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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. Find a formula for the area A of this pond in terms
of r, R and θ.2. Find the area of a pond with this shape if r = 7 m, R
= 9 m and θ = 45°
θ
r
R
πR 2 x θ
360
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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. Find a formula for the area A of this pond in terms
of r, R and θ.2. Find the area of a pond with this shape if r = 7 m, R
= 9 m and θ = 45°
θ
r
R
πr 2 x θ
360–πR
2 x θ360
© 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. Find a formula for the area A of this pond in terms
of r, R and θ.2. Find the area of a pond with this shape if r = 7 m, R
= 9 m and θ = 45°
θ
r
R
R – r πr 2 x θ
360–πR
2 x θ360
+ π R – r2
2A =
© 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. Find a formula for the area A of this pond in terms
of r, R and θ.2. Find the area of a pond with this shape if r = 7 m, R
= 9 m and θ = 45°
πr 2 x θ
360–πR
2 x θ360
+ π R – r2
2A =πr
2 x θ360
–πR 2 x θ
360+ π R – r
2
2A =
© 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. Find a formula for the area A of this pond in terms
of r, R and θ.2. Find the area of a pond with this shape if r = 7 m, R
= 9 m and θ = 45°
πr 2 x θ
360–πR
2 x θ360
+ π R – r2
2A =
π4
+R 2– r 2
πθ360
R – r2
A =
π4
+92 – 7245π360
9 – 72
A =
π4
+x 32π8
x 4A =
+ π4πA =
A = 5π≈ 15.7 m2
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© T Madas
8 cm
3 cm45°
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.
8 cm45°
A = πr 2
area of a circle
A =π x 8 2
area of the given sector
volume of the prism
A = 45360
π x 8 2 x
V = 45360
π x 8 2 x x 3
V = 24πV ≈75.4 cm3 [3 s.f.]
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8 cm
3 cm45°
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.
8 cm45°
area of the given sector
A = 45360
π x 8 2 x
area of the given sector
A = 45360
π x 8 2 x
© T Madas
8 cm
3 cm45°
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.
8 cm45°
area of the given sector
A = 45360
π x 8 2 x
x 2
x 2
area of the side rectanglex 2
A =3x 8 x 2
area of the curved surface
A = 45360
π2 x xx 8
2πr= 16π
= 48
x 3= 6π
total surface area
A = 16π + 48 + 6π≈ 117 cm2 [3 s.f.]
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© T Madas
All measurements in cm
72
141
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.
O 15
B45A
A = πr 2
area of a circle
A =π x 60 2
area of the “larger” sector
A = 162360
π x 60 2 x = 1620π162°
© T Madas
All measurements in cm
72
141
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.
O 15
B45A162°
A = πr 2
area of a circle
A =π x 60 2
area of the “larger” sector
A = 162360
π x 60 2 x = 1620π
area of the “inner” sector
A = 162360
π x 15 2 x = 101.25π
© T Madas
All measurements in cm
72
141
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.
O 15
B45A162°
A = πr 2
area of a circle
A =π x 60 2
area of the “larger” sector
A = 162360
π x 60 2 x = 1620π
area of the “inner” sector
A = 162360
π x 15 2 x = 101.25π
area swept by the wiper is1620π – 101.25π= 1518.75π cm2
≈ 4771.3 cm2
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area wipedtotal area
All measurements in cm
72
141
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.
O 15
B45A162°
area swept by the wiper is1620π – 101.25π= 1518.75π cm2
≈ 4771.3 cm2
x 100%age =
1518.75π141 x 72
x 100%age =
47%%age ≈
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© T Madas
-1tan ( )
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.
5 cm
4 cm
A B
3,4,5
= Alarm Bells !
C
3 cm ∆ABC is right angled
By trig on ∆ABC :θ
tanq = 43 Û
q = 43 Û
53.13q » °
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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.
5 cm
4 cm
A B
C
3 cm
θ
Now some angle calculations:If θ ≈ 53.13°
CBA ≈ 36.87°CAD ≈ 106.26°CBD ≈ 73.74°
D
106.26°
36.87°
73.74° reflexCAD ≈ 253.74°reflexCBD ≈ 286.26°
253.74°
286.26°
Simplify the diagram by showing only the relevant information
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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.
4 cm
A B
C
3 cm
D
253.74°
286.26°
This arc corresponds to an angle of 253.74° on a circle with radius of 3 cm
This arc corresponds to an angle of 286.26° on a circle with radius of 4 cm
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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.
4 cm
A B
C
3 cm
D
253.74°
286.26°
The smaller of the 2 arcs2 p´ 3´ ´ 253.74
360 13.29 cm»
The larger of the 2 arcs2 p´ 4´ ´ 286.26
360 19.98 cm»
The required perimeter
33.3 cm [2 s.f.]
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© 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
A
B
C
8 m
5 m
5 m
4 m
3 m
OBD is right angledOD = 3 mTrig to find BOD
D-1tan ( )
tanq = 43 Û
q = 43 Û
53.13q » °
θ
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O
A
B
C
5 m
5 m
Dθ
The area of the sector
p 25´ ´ 53.13360
211.59 m»
The length of the arc2 p´ 5´ ´ 53.13
360 4.64 m»
area of the sector ≈ 11.59 m2length of the arc ≈ 11.59 m
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
© T Madas
O
A
B
C
5 m
5 m
Dθ
area of the sector ≈ 11.59 m2length of the arc ≈ 11.59 m
OBC is right angledOB = 5 mBOC = 53.13°Trig to find BC
BCOB = tanq Û
5BC 53.13tan °= 4
3= Û
3BC 20= Û203 mBC =
4 m
3 m
BC = 20/3 m
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
© T Madas
O
A
B
C
5 m
5 m
Dθ
area of the sector ≈ 11.59 m2length of the arc ≈ 11.59 mBC = 20/3 m
Area of OBC12 5´ 20
3´ 1006= 50
3= 216.67m»
area of the sector ≈ 11.59 m2
half the shaded area ≈ 5.08 m2
shaded area ≈ 10.16 m2
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
© T Madas
O
A
B
C
5 m
5 m
Dθ
area of the sector ≈ 11.59 m2length of the arc ≈ 11.59 mBC = 20/3 m
E
Length BC = 20/3 m ≈ 6.67 m
length of the arc ≈ 11.59 m
Total perimeter ≈ 36.5 m
11.59
6.67
x 2x 2
≈ 23.18
≈ 13.34
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
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© 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 .
r
A L
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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 .
rθ
A L
=A ´360
2r
=L ´360
2 r
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2 r2 r
2r 2r
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 ´360
2r
=L ´360
2 r
Û =A2
360
r
Û =L360
2 r
Û =360A 2r
Û =360L 2 r
Û
Û
=360A 2r
=360L 2 r
Û
Û
=2
360A
r
= 180L
r
Now what?
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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
360A
r
= 180L
r
2
360A
rÛ =180L
rÛ =360 rA 2180 Lr
360 r 360 r
=A
2
rL2
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Harder Examples
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A B
CD
The following pattern is produced inside a square whose side length is 8 cmCalculate the shaded area
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
8 cm
45°P l a n
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A B
CD
The following pattern is produced inside a square whose side length is 8 cmCalculate the shaded area
8 cm
45°
4
4Pythagoras:
x
2x = 24 24+ Û2x = 32 Û
x = 3232
Area of sector:
18´ p ( )232´ = 4p
Area of triangle:
cm8
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A B
CD
The following pattern is produced inside a square whose side length is 8 cmCalculate the shaded area
8 cm
Area of sector:
4p
Area of triangle:
cm8
Area of sector: 4pArea of triangle: cm8
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( )
A B
CD
The following pattern is produced inside a square whose side length is 8 cmCalculate the shaded area
8 cm
Area of sector: 4pArea of triangle: cm8
Shaded area:
4p 8- 8´ ( )232 p= -236.5 cm»
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© T Madas
8 cm
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
A B
CD
60°
The bottom part of the shaded area consists of:
•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.
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8 cm
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
A B
CD
60°
16´
The area of the segment
p 28´ 12- 8´ 8´ sin60´ °
trianglesector = 1/6 circle
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8 cm
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
A B
CD
60°
16´
The area of the segment
p 28´ 12- 8´ 8´ sin60´ °
323p 32- 3
2´
3sin602
°=
323 16 3p= -
32π
3 –
16 3
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8 cm
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
A B
CD
60°
16´
The area of the segment
p 28´ 12- 8´ 8´ sin60´ °
323p 32- 3
2´ 323 16 3p= -
Now look at one of the non - shaded regions
32π
3 –
16 3
30°
The area of one of the non shaded regions equals:
One twelfth of a circle of radius 8 cm
Less the area of the segment112´ p 28´ 32
3 16 3( )p- -
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8 cm
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
A B
CD
60°
32π
3 –
16 3
30°
112´ p 28´ 32
3 16 3( )p- -
112´ p 28´ 32
3 16 3( )p- -
6412p= 32
3p- 16 3+
163p= 32
3p- 16 3+
-- 163p= 16 3+
163p-16 3=
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8 cm
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
A B
CD
60°
32π
3 –
16 3
30°
112´ p 28´ 32
3 16 3( )p- -
6412p= 32
3p- 16 3+
163p= 32
3p- 16 3+
-- 163p= 16 3+
163p-16 3=
16π
3
16 3
16 316π3
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8 cm
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
A B
CD
16 316π3
The shaded region is 64 cm2 less the two non shaded regions we just found 16
32 16 3( )p- -64 242 cm»
16π
3
16 3
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STOP
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L
r
Lh
r
r 2 r
θ
r 2 +h2 = L2
2 r
´ L ´
360´ 2 = r2
A
L2´
360=A
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r 2 +h2 = L2 ´ L ´
360´ 2 = r2
L2´
360=A
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´
360= r2´ 2 ´ L
r 2 +h2 = L2
L2´
360=A
Û
L2
360= r2 Û
L
360= r Û
=L r360 Û
=L
r360
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´r360
r 2 +h2 = L2
L2´
360=A
=L
r360
Û L2´360
=A L
r360
Û
L2 ´360
1
=A L
r360
Û
L2
L360=A Û
=LrA
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r 2 +h2 = L2
L2´
360=A
=L
r360
Û =L +r h2 2
=LrA
=A r +r h2 2
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=LrA
=A r +r h2 2
Lh
r
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