1. Sectors, Segments, & Annuli
Parts of Circles
(and yes, you need to know this)

2. We start with a circle
Then…

3. Sector

4. Segment

5. Annulus

6. Sector

7. Segment

8. Annulus

9. How do we find the areas of these?

10. We know the area of a circle
= radius
A = πr2
So…

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

12. 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

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

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

15. What’s the area of this sector?
Hint: 90° is the same as π/2 radians 5
90°

16. 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

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

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

19. 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)

20. What’s the area of this segment?
Hint: 120° is the same as 2π/3 radians 8
120° 5

21. 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…

22. 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…

23. 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)

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

25. Annulus Area of Annulus = Area of Outside Circle –
Area of Inside Circle r2 r1
Area of Annulus:
A = πr12 – πr22

26. 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…

27. What’s the Area of this annulus?
Area of Annulus :
A = Ao - Ai
A = 9π – 4π 2 3
Answer:
A = 5π

28. Questions?

29. And now for the homework…