Sectors, Segments, & Annuli
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Transcript of Sectors, Segments, & Annuli
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
So…
r = radius
A = πr2
Sector
r = radiusα
Degrees: 360˚ - αRadians: 2π - α
Sector
r = radiusα
Degrees: 360˚ - αRadians: 2π - α
Area of the Circle:A = πr2
Ratio of Sector to Circle:(degrees) α/360(radians) α/2π
Sector
rα
Radians:A = πr2 (α/2π)= πr2 (α/2π)
=(α/2) r2
Sector
rα
Radians:A = ½ r2α
Degrees:A = (α/360) πr2
What’s the area of this sector?
5
90°
Hint: 90° is the same as π/2 radians
What’s the area of this sector?
5
90°
Area of the Circle:A = 52π
Ratio of the Sector:R = 90°/360° orR = (π/2)(1/2π)
Area of the Sector:A = 52π(π/2)(1/2π)A = 52π(90°/360°)
Answer:A = 25π/4
Segment
α r = radius
Degrees: 360˚ - αRadians: 2π - α
Segment
α r
Radians:A = ½ r2α
Degrees:A = (α/360) πr2
Then…
Segment
α r
Area of the Segment:A = ½ r2α – ½ bh (radians)A = (α/360) πr2 – ½ bh (degrees)
bh
Area of the Segment =Area of the Sector – Area of the Triangle
What’s the area of this segment?
120° 5
8
Hint: 120° is the same as 2π/3 radians
What’s the area of this segment?
120° 5
8
Area of the circle:A = 52π
Ratio of the sector:R = 120°/360° orR = (2π/3)(1/2π)
R = 1/3
Area of the sector:A = 25π(1/3)A = 25π/3
Then…
5
Area of the Triangle:A = ½ bhA = ½ (8)(3) = ½ 24 = 12
4h
Area of the Segment =Area of the Sector – Area of the Triangle
4
h2 = 52 – 42
h2 = 25 – 16h2 = 9h = 3
So…
What’s the area of this segment?
Area of the Segment:A = As - At
A = 25π/3 – 12 = (25π/3) – (36/3)
Area of the Sector = 25π/3Area of the Triangle = 12
What’s the area of this segment?
Answer:A = (25π – 36)/3
Annulus
r1
r2
Area of outside circle:A = πr1
2
Area of inside circle:A = πr2
2
Annulus
r1
r2
Area of Annulus =Area of Outside Circle – Area of Inside Circle
Area of Annulus:A = πr1
2 – πr22
3
2
What’s the Area of this annulus?Area of outside circle:A = 32π = 9π
Area of inside circle:A = 22π = 4π
Area of Annulus =Area of Outside Circle – Area of Inside Circle
So…
3
2
What’s the Area of this annulus?
Area of Annulus :A = Ao - Ai
A = 9π – 4π
Answer:A = 5π
Questions?
And now for the homework…