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 = radius A = πr2 So…

Sector r = radius α Degrees: 360˚ - α Radians: 2π - α

Sector Area of the Circle: A = πr2 Degrees: 360˚ - α Radians: 2π - α = radius α Degrees: 360˚ - α Radians: 2π - α Ratio of Sector to Circle: (degrees) α/360 (radians) α/2π

Sector Radians: A = πr2 (α/2π) = πr2 (α/2π) =(α/2) r2 α r

Sector Radians: A = ½ r2α α r Degrees: A = (α/360) πr2

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 r = radius α Degrees: 360˚ - α Radians: 2π - α

Segment Radians: A = ½ r2α r α Degrees: A = (α/360) πr2 Then…

Segment Area of the Segment: A = ½ r2α – ½ bh (radians) Area of the Sector – Area of the Triangle b h r α Area of the Segment: A = ½ r2α – ½ bh (radians) A = (α/360) πr2 – ½ bh (degrees)

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° 5 Area of the sector: A = 25π(1/3) A = 25π/3 Then…

What’s the area of this segment? Area of the Segment = Area of the Sector – Area of the Triangle h2 = 52 – 42 h2 = 25 – 16 h2 = 9 h = 3 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 Answer: A = (25π – 36)/3 Area of the Segment: A = As - At A = 25π/3 – 12 = (25π/3) – (36/3)

Annulus Area of outside circle: A = πr12 r2 r1 Area of inside circle:

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…