### Sectors, Segments, & Annuli - Turcotte

```Sectors,
Segments, &
Annuli
Parts of Circles
(and yes, you need to know this)
Then…
Sector
Segment
Annulus
Sector
Segment
Annulus
How do we find
the areas of
these?
We know the area of a circle
A = πr2
So…
Sector
α
Degrees: 360˚ - α
Sector
Area of the Circle:
A = πr2
α
Ratio of Sector to Circle:
(degrees) α/360
Degrees: 360˚ - α
Sector
A = πr2 (α/2π)
= πr2 (α/2π)
=(α/2) r2
α
r
Sector
A = ½ r2α
α
Degrees:
A = (α/360) πr2
r
What’s the area of this sector?
Hint: 90° is the
same as π/2
5
90°
What’s the area of this sector?
Ratio of the Sector:
R = 90°/360° or
R = (π/2)(1/2π)
Area of the Circle:
A = 52π
5
90°
Area of the Sector:
A = 52π(π/2)(1/2π)
A = 52π(90°/360°)
A = 25π/4
Segment
α
Degrees: 360˚ - α
Segment
A = ½ r2α
α
Degrees:
A = (α/360) πr2
r
Then…
Segment
Area of the Segment =
Area of the Sector –
Area of the Triangle
b
h
α
Area of the Segment:
A = ½ r2α – ½ bh (radians)
A = (α/360) πr2 – ½ bh (degrees)
r
What’s the area of this segment?
Hint: 120° is
the same as
8
120°
5
What’s the area of this segment?
Ratio of the sector:
R = 120°/360° or
R = (2π/3)(1/2π)
R = 1/3
Area of the circle:
A = 52π
8
120°
Area of the sector:
A = 25π(1/3)
A = 25π/3
5
Then…
What’s the area of this segment?
h2 = 52 – 42
h2 = 25 – 16
h2 = 9
h=3
Area of the Segment =
Area of the Sector –
Area of the Triangle
4
4
h
5
Area of the Triangle:
A = ½ bh
A = ½ (8)(3) = ½ 24 = 12
So…
What’s the area of this segment?
Area of the Sector = 25π/3
Area of the Triangle = 12
Area of the Segment:
A = As - At
A = 25π/3 – 12 = (25π/3) – (36/3)
A = (25π – 36)/3
Annulus
Area of outside circle:
A = πr12
r2
r1
Area of inside circle:
A = πr22
Annulus
Area of Annulus =
Area of Outside Circle –
Area of Inside Circle
r2
r1
Area of Annulus:
A = πr12 – πr22
What’s the Area of this annulus?
Area of outside circle:
A = 32π = 9π
Area of inside circle:
A = 22π = 4π
2
3
Area of Annulus =
Area of Outside Circle –
Area of Inside Circle
So…
What’s the Area of this annulus?
Area of Annulus :
A = Ao - Ai
A = 9π – 4π
2
3