1. Circles

2. Circumferences of Circles diameter (d) O circumference (C) The circumference (C) and the diameter (d) of a circle are related by radius (r) Since d = 2r, where r denotes the radius of the circle, we have

3. circumference of the circle = 3.14  18 cm = 56.52 cm  C =  d Do you know how to find the circumference of the circle with diameter 18 cm? Take  = 3.14,

4. Follow-up question 1 Complete the following table. 28 cm88 cm 3.5 m 22 m 7 mm14 mm Circumference of the circle d = 2r  C =  d 

5. Example 1 Solution

6. Example 2 Solution

7. Example 3 Solution

8. Example 4 Solution

9. Areas of Circles 1. Prepare a circle. 2. Divide it into 16 equal parts. 3. Cut the circle into 2 parts as shown. 4. Rearrange them to form the following figure. The formula of the area of circle can be deduced in the following way: Which kind of quadrilateral does the figure look like? parallelogram

10. Assuming the radius of the circle is r, estimate the height and the base of the figure obtained in terms of r. height base r Height  r r Base   2  r =  r The figure looks like a parallelogram. ∴ The approximate area  height  base

11. If A and r denote the area and the radius of a circle respectively, then r =  28 2 cm 2 For a circle of radius 28 cm, its area = 2464 cm 2  A =  r 2 take  =,

12. Follow-up question 2 Complete the following table. (Take  = 3.14.) (Give your answers correct to 2 decimal places if necessary.) 8 mm50.27 mm 2 5 cm 78.54 cm 2 3.57 m7.14 m d = 2r  A =  r 2 

13. Example 5 Solution

14. Example 6 Solution

15. Example 7 Solution

16. Example 8 Solution

17. Arcs and Sectors

18. To avoid confusion, the shorter arc ACB Arcs If A and B are two points on the circumference of a circle, then curve ABis called an arc, The figure shows a circle with centre O. A B O which is denoted by ‘arc AB’ or ‘ ’. and the longer arc ADB C D  AOB is the angle subtended at the centre by the arc AB. It can be simply called the angle at the centre. angle at the centre is denoted by is denoted by.

19. Lengths of Arcs Let be an arc on the circle For a circle with centre O and radius r, O r circumference = 2  r. A B and  be the angle subtended at the centre by the arc. Let’s find the lengths of arc AB for different values of ... 

20. Length of O A B (a)  = 180  = of the circumference  circumference =  2 r 2 r  r r

21.  2 r 2 r  r r = of the circumference  circumference Length of (b)  = 90  = of the circumference =  2 r 2 r  r r O A B  circumference Length of = (c)  = 45  O A B

22. =  2 r 2 r  r r of the circumference  circumference Length of (d)  = 1  = O A B What can you conclude from the above results? The ratio of the arc length to the circumference of the circle =.

23.  Therefore, O r A B For a circle with radius r and angle  subtended at the centre by an arc, length of arc = Length of arc Circumference Angle subtended at the centre Round angle = 

24. Follow-up question 3 In each of the following figures, O is the centre of the circle. O A B 10 cm O A B 9 m Solution Find the length of the arc AB in terms of . 1.2.

25. Example 9 Solution

26. Example 10 Solution

27. Example 11 Solution

28. Areas of Sectors O A B A sector is the region enclosed by an arc and two radii of a circle. In the figure, AOB is a sector enclosed by, radii OA and OB.  AOB is called the angle of the sector. angle of the sector sector

29. For a sector with radius r and angle of the sector , B O A r Therefore, area of sector =

30. Refer to the following figure. O A B 10 m Area of sector OAB 

31. Follow-up question 4 Complete the following table. 10  mm 2 3 m 8  cm 2 8 m28 m2 36   320  45 

32. Example 12 Solution

33. Example 13 Solution

34. Example 14 Solution

35. Cylinders

36. A cylinder is a solid with uniform cross-section and its two bases are circles. The solids below are all cylinders.

37. Volume of Cylinders We have learnt that Similarly, r h base For a cylinder of base radius r, height h and volume V, then volume of a cylinder = base area × height volume of a prism = base area × height V =  r 2 h

38. Refer to the following figure. 5 cm 2 cm Volume of the cylinder =   2 2  5 cm 3 = 20  cm 3 = 62.83 cm 3 (cor. to 2 d.p.)  V =  r 2 h

39. Follow-up question 5 The figure shows a cylinder of base radius r and height h. Complete the following table. r h 112  10 5 40  275  V =  r 2 h 

40. Example 15 Solution

41. Example 16 Solution

42. Example 17 Solution

43. Example 18 Solution

44. Total Surface Areas of Cylinders 1. Prepare a cylindrical can that has a piece of wrapper. 2. Cut the wrapper vertically as shown. 3. Spread out the wrapper. The lateral face of a cylinder is a curved surface. How can we find the curved surface area?

45. A B C D What is the relationship between AD and the circumference of the base of the can? AD = circumference of the base of the can What is the relationship between AB and the height of the can? AB = height of the can

46. Then, what is the curved surface area of the can? A B C D Curved surface area of the can = area of rectangle ABCD = AD × AB = the circumference of the base of the can × the height of the can

47. For a cylinder of base radius r and height h, The total surface area of the cylinder can be found by adding the areas of its two bases to its curved surface area: curved surface area of a cylinder = 2  r h base area =  r 2 r curved surface area = 2  rh base area =  r 2 total surface area of a cylinder = 2  r h + 2  r 2

48. Refer to the following figure. 3 cm 4 cm Total surface area of the cylinder  Total surface area = 2  r h + 2  r 2 = = 20  cm 2

49. Follow-up question 6 The figure shows a cylinder of base radius r and height h. Complete the following table. r h 7 Total surface area = 2  r h + 2  r 2 

50. Example 19 Solution

51. Example 20 Solution

52. Extra Teaching Example

53. Example 2 (Extra) Solution

54. Example 8 (Extra) Solution

56. Example 10 (Extra) Solution

57. Example 14 (Extra) Solution

58. Example 19 (Extra) Solution