Circumference and Area of Circles
The student is able to (I can):
• Develop and use formulas to find the circumference and
area of circles
• Develop and use formulas to find arc length and sector
area of circles
ππππ ((((pi)pi)pi)pi) – the ratio of the circumference to the diameter.
Since the diameter is twice the radius, this formula can also
be written as:
π is an irrational number – it never repeats and it
never ends. The symbol π is an exact number;
3.1415926… is an approximation.
C
d
π =
C d= π
2C r= π
which becomes
If you cut a circle into wedges, and arrange the wedges
into a parallelogram-shaped figure:
radius
1
2
circumference
A = bh
1
circumference radius
2
A = i
( )
1
2 radius radius
2
A = π i
2
A r= π
The more wedges, the closer the figure is to a parallelogram.
Examples
1. Find the exact circumference and area of a circle whose
diameter is 18 m.
2. Find the diameter and area of a circle whose
circumference is 22π ft.
3. Find the radius of a circle whose area is 81π cm2.
Examples
1. Find the exact circumference and area of a circle whose
diameter is 18 m.
C = πd = π(18) = 18π m
A = πr2 = π(92) = 81π m2
2. Find the diameter and area of a circle whose
circumference is 22π ft.
22π = πd
d = 22 ft
A = π(112) = 121π ft2
3. Find the radius of a circle whose area is 81π cm2.
81π = πr2
81 = r2
81 9 cmr = =
arcarcarcarc lengthlengthlengthlength – the distance along an arc. It is proportional to
the circumference of the circle.
or
360
m
L C
° 
=  
° 
where C is the circumference
(either C=πd or C=2πr).
•
mmmm°°°°
m° arc length central angle
circumference 360
360
L m
C
=
°
°
=
°
Example
Find each exact arc length.
1.
2.
120°
• 3333′′′′
72727272°°°°
8 m8 m8 m8 m
•
Example
Find each exact arc length.
1.
2.
120°
• 3333′′′′
( )
120
2 3
360
2 ft.
L
 
= π  
 
= π
( )
72
2 8
360
16
or 3.2 m
5
L
 
= π  
 
= π π
72727272°°°°
8 m8 m8 m8 m
•
3. Find the radius of the circle.
•
130130130130°°°°
rrrr
39π m
3. Find the radius of the circle.
•
130130130130°°°°
rrrr
39π m
39 130
2 360
260 14040
14040
54 m
260
r
r
r
π °
=
π °
π = π
π
= =
π
sector of asector of asector of asector of a circlecirclecirclecircle – a region bounded by a central angle. It is
proportional to the area of the circle containing the
sector.
•
R
AAAA
G
∠RAG is a central angle
RAG is a sector
Area of sector central angle
Area of circle 360
=
°
2
S
360
m
r
°
=
π °
2
360
m
S r
° 
= π  
° 
Formula:
Examples
Find the area of each sector. Leave answers in terms of π.
1.
2.
•
120° 2"2"2"2"
• 72°
10m10m10m10m
Examples
Find the area of each sector. Leave answers in terms of π.
1.
2.
•
120° 2"2"2"2"
• 72°
10m10m10m10m
( )2
2
120
2
360
4 120 480
360 360
4
in
3
S
°
= π
°
π⋅ π
= =
= π
( )2
2
72
10
360
7200
20 m
360
S
° 
= π  
° 
π
= = π

11.1 Circumference and Area of Circles

  • 1.
    Circumference and Areaof Circles The student is able to (I can): • Develop and use formulas to find the circumference and area of circles • Develop and use formulas to find arc length and sector area of circles
  • 2.
    ππππ ((((pi)pi)pi)pi) –the ratio of the circumference to the diameter. Since the diameter is twice the radius, this formula can also be written as: π is an irrational number – it never repeats and it never ends. The symbol π is an exact number; 3.1415926… is an approximation. C d π = C d= π 2C r= π which becomes
  • 3.
    If you cuta circle into wedges, and arrange the wedges into a parallelogram-shaped figure: radius 1 2 circumference A = bh 1 circumference radius 2 A = i ( ) 1 2 radius radius 2 A = π i 2 A r= π The more wedges, the closer the figure is to a parallelogram.
  • 5.
    Examples 1. Find theexact circumference and area of a circle whose diameter is 18 m. 2. Find the diameter and area of a circle whose circumference is 22π ft. 3. Find the radius of a circle whose area is 81π cm2.
  • 6.
    Examples 1. Find theexact circumference and area of a circle whose diameter is 18 m. C = πd = π(18) = 18π m A = πr2 = π(92) = 81π m2 2. Find the diameter and area of a circle whose circumference is 22π ft. 22π = πd d = 22 ft A = π(112) = 121π ft2 3. Find the radius of a circle whose area is 81π cm2. 81π = πr2 81 = r2 81 9 cmr = =
  • 7.
    arcarcarcarc lengthlengthlengthlength –the distance along an arc. It is proportional to the circumference of the circle. or 360 m L C °  =   °  where C is the circumference (either C=πd or C=2πr). • mmmm°°°° m° arc length central angle circumference 360 360 L m C = ° ° = °
  • 8.
    Example Find each exactarc length. 1. 2. 120° • 3333′′′′ 72727272°°°° 8 m8 m8 m8 m •
  • 9.
    Example Find each exactarc length. 1. 2. 120° • 3333′′′′ ( ) 120 2 3 360 2 ft. L   = π     = π ( ) 72 2 8 360 16 or 3.2 m 5 L   = π     = π π 72727272°°°° 8 m8 m8 m8 m •
  • 10.
    3. Find theradius of the circle. • 130130130130°°°° rrrr 39π m
  • 11.
    3. Find theradius of the circle. • 130130130130°°°° rrrr 39π m 39 130 2 360 260 14040 14040 54 m 260 r r r π ° = π ° π = π π = = π
  • 12.
    sector of asectorof asector of asector of a circlecirclecirclecircle – a region bounded by a central angle. It is proportional to the area of the circle containing the sector. • R AAAA G ∠RAG is a central angle RAG is a sector Area of sector central angle Area of circle 360 = ° 2 S 360 m r ° = π ° 2 360 m S r °  = π   °  Formula:
  • 13.
    Examples Find the areaof each sector. Leave answers in terms of π. 1. 2. • 120° 2"2"2"2" • 72° 10m10m10m10m
  • 14.
    Examples Find the areaof each sector. Leave answers in terms of π. 1. 2. • 120° 2"2"2"2" • 72° 10m10m10m10m ( )2 2 120 2 360 4 120 480 360 360 4 in 3 S ° = π ° π⋅ π = = = π ( )2 2 72 10 360 7200 20 m 360 S °  = π   °  π = = π