Sectors, Segments, & Annuli - Turcotte

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Sectors,
Segments, &
Annuli
Parts of Circles
(and yes, you need to know this)
We start with a circle
Then…
Sector
Segment
Annulus
Sector
Segment
Annulus
How do we find
the areas of
these?
We know the area of a circle
r = radius
A = πr2
So…
Sector
α
r = radius
Degrees: 360˚ - α
Radians: 2π - α
Sector
Area of the Circle:
A = πr2
α
Ratio of Sector to Circle:
(degrees) α/360
(radians) α/2π
r = radius
Degrees: 360˚ - α
Radians: 2π - α
Sector
Radians:
A = πr2 (α/2π)
= πr2 (α/2π)
=(α/2) r2
α
r
Sector
Radians:
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
radians
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°)
Answer:
A = 25π/4
Segment
α
r = radius
Degrees: 360˚ - α
Radians: 2π - α
Segment
Radians:
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
2π/3 radians
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)
Answer:
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
Answer:
A = 5π
Questions?
And now for the
homework…

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