1. How to obtain the value of  ? Method 1
12.1
How to obtain the value of  ? Method 1
radius
centre
diameter
diameter
circumference
diameter
circumference
Circumference=   diameter
Circumference = d or 2r

2. Conclusion: 3 <  < 42r r r r r r 2r 2r r r r r r r r 2r
diameter
hexagon of
perimeter
circle
nce
circumfera
square
< r 2 6
1415 . 3 8
< 3
< 3 .
1415
< 4
Conclusion: 3 <  < 4

3. How to obtain the value of  ? Method 2
As the number of sides of polygons increases ,
The ratio is getting closer to

4. How to obtain the value of  ? Method 3 12.2 – Area of circle = r2
Obtain value of . Measure number of cells. Divide it by radius2
1 cm2 2 3

5. Example 1
Find the circumference of a circle with diameter 6.4 cm. Correct your answer to 1 decimal place.  
6.4 =
Circumference = (  6.4) cm = 20.1 cm (corr. to. 1.d .p.)
Classwork 12.1
1. Diameter = 15.2 cm
2. radius = 8.71 cm
Circumference= (2    8.71) cm = 54.7 cm (corr. to. 1.d .p.)
Circumference= (  15.2) cm = 47.8 cm (corr. to. 1.d .p.)

6. Example 2
Find the radius of a circle with the circumference 15.4cm.
Let r cm be the radius of the circle.
The radius of the circle is 2.45 cm
Classwork 12.2 The circumference of a circle is 35.2cm
1. Diameter
2. The radius

7. Perimeter of the figure
Example 3
The figure is formed by two semi-circles with the same centre and two line segments. Find its perimeter correct to the nearest 0.1cm.
4 cm
8 cm A B C D
Perimeter of the figure CD AB
circle
smaller
nceof
circumfere er l
nce of
Circumfere ) ( 2 1
arg + = cm ) 2 4 1 8 ( +  = p 2 2 = 22 . 8 cm

8. Perimeter of the figure
Classwork 12.3
1. The figure is formed by two semi-circles and straight lines. Find its perimeter correct to the nearest 0.1cm.
4 cm
12 cm A B C D
Perimeter of the figure

9. Classwork 12.3
2. The figure is formed by one quarter of a circle and straight lines. Find it’s perimeter to the nearest 0.1 cm. A C B
9 cm
Perimeter of the figure

10. Example 4
The radius of a wheel is 30 cm. How far does it move in 5000 revolutions? (take  =3.14)
The distance travelled in 1 revolution
=Circumference of the coin
=(2  3.14 30) cm
=188.4 cm
The distance travelled in 5000 revolutions
= (5000  188.4) cm
= 9.42 cm
2 r
2 r
distance travelled= 5000  2 r

11. Classwork 12.4
1. The diameter of a coin is 2.5 cm. How far does it travel in 40 revolutions? (take  =3.14)
The distance travelled in 1 revolution
=Circumference of the wheel
=(3.14 2.5) cm
=7.85 cm
The distance travelled in 40 revolutions
= (7.85  40) cm
= 314 cm
2 r
2 r
distance travelled= 40  2 r

12. Classwork 12.4
2. After the wheel of a car makes 30 revolutions, the car travelled 47.1 cm. Find the diameter of the wheel. (take  =3.14)
Let a be the distance travelled in 1 revolution a  30=47.1m
a =1.57 m
The diameter of the wheel
d =1.57
d= 1.57/3.14 =1.5 m
2 r
2 r
distance travelled= 30  2 r

13. Difference of circumferences
Example 5
The difference of the diameters of two circles is 5 cm. Find the difference of their circumferences. (Express your answer in terms of  )
Let d1 and d1 be the diameters of the larger and smaller circles respectively. d2
d1 - d1 = 5 d1
Difference of circumferences
=circumference of large circle - circumference of smaller circle
=  d1 -  d2 =  (d1 -d2 )=  5 = 5  cm

14. Difference of circumferences
Questions
The sport stadium is formed by a rectangular field and two semi-circular fields at the two ends. If the track is 1. 5m wide, what is the difference in length between the inner boundary and the outer boundary of the track
Let r1 and r1 be the radius of the larger and smaller circles respectively. r2 r1
r1 - r1 = 1.5
Difference of circumferences
=circumference of large circle - circumference of smaller circle
= 2 r1 - 2 r2 = 2 (r1 -r2 )= 2   1.5 = 3  cm

15. Area= r2 How to obtain the r2 for area of circle? Lr W r r L Lr
Area=LW=r r= r2
Area= r2

16. Z H A X
Pythegoras Theorem
Z2=X2+Y2

17. Example 6
Find the areas of the following circles.(correct your answer to 0.1cm2)
(a) Radius = 4 cm
Area of the circle= (  42) cm2 =50.3 cm2 (corr. to the nearest 0.1cm2)
(b) Diameter = 12.5 cm
Radius = 6.25 cm
Area of the circle= (  6.252) cm2 =122.7 cm2 (corr. to the nearest 0.1cm2)
Example 7
Find the diameter of the circle with its area as 78.5cm2.(Take =3.14)
Let r cm be the radius of the circle.
3.14r2 = 78.5
r2 = 78.5/3.14
r2 = 25
r2 = 5
 The diameter of the circle = 2r cm = 10 cm

18. Area of smaller circle = (  1.52) cm2 = 2.25 cm2
Example 8.
Find the area of the ring formed by two concentric circles with radii 1.5 cm and 2 cm. (Express your answer in terms of )
Area of smaller circle = (  1.52) cm2 = 2.25 cm2
Area of larger circle = (  22) cm2 = 4 cm2
Area of the ring = (4-2.35) cm2 = 1.75 cm2
1.5 cm
2 cm

19. A B
16 cm A C B
5 cm
6 cm 5)
Let r be the length of minute hand

20. 6) Let d be the diameter 7)
Area of shaded area =Area of larger circle-area of smaller circle A C B
12cm

21. Area of playground=area of rectangle + 2 area of semi-circle
8 b) 9)
Area of playground=area of rectangle + 2 area of semi-circle
32 cm
164 cm

22. 10.)
11.)

23. 8 cm 4 cm 4 cm A) 10cm 16cm B) 4 cm C) F) 8 cm D) E) 4 cm 8 cm
Find the perimeter and area of the above figure which is formed by semi circles.

24. Z2=X2+Y2 BOOK 2B CHAPTER 11 H Z A X Pythegoras Theorem
ANGLE of a regular polygon =
where n is the number of sides A
E.G.

25. r A B C O
5 cm
3 cm
4 cm r B C O
4 cm A B C 2 O
4 cm
2 cm
2400

26. 360 2 x r L = p L = 2 r p 360 x L x 3600 2r r 12.3 Length of Arcs Arc
Sector
angle at the centre
360 2 x r L = p L = 2 r p
360 x  L r x
3600
2r

27. A 360 x r = p 360 x r A  = p pr2 x 3600 r A 12.3 Areas of Sectors
360 2 x r = p
Sector
angle at the centre
360 2 x r A  = p
pr2 A x
3600 r

28. 12.3 Length of Arcs and Areas of Sectors
Example 9 –find the length of arc AB L
1200
7 cm A
27
3600 B
7 cm
360 2 x r L  = p
360 2
1200 7 L  = p
= 14.7 cm

29. Example 9 –find the area of sector AOB.
72
1200
3600
7 cm B O
360 x r2 A  = p
1200 A =  p 72
360
= 51.3 cm2

30. 360 2 r L = p x x 360 8 x 3 = 8 3 360  = x x = 1350 3 8 Example 10
The length of an arc of a circle is
of its circumference. 8
Find the angle subtended by the arc.
Let x be the angle subtended by the arc.
360 2 r L = p x
arc x
360 8 x 3 = 8 3
360  = x
x = 1350

31. Example 11 X Y Z T
4 cm
XYZ= YZX = ZXY=600 (equilateral XYZ)
ZT2 + XT2 = XZ2 (Pyth.theorem)
ZT2 = XZ2 - XT2
ZT2 = 42 – 22 = 12 cm ZT
464 . 3 12 =
600
Sum of areas of 3 sectors 2
283 . 6 )
360 60 ( 3 cm =  p
Area of shaded region = Area of XYZ - Sum of areas of 3 sectors
283 . 6 )
464 3 4 2 1 ( -  = =
= cm2 corr. To 3 sig.fig.

32. Classwork 12.11 A B C D
2 cm
Shaded area =area of square – 4  area of quarter circle
=22 –12
=4 –
=0.858 cm2

33. In the figure, ABC is isosceles triangle with AB=BD=CD=2 cm,
A,B,C are centres of arc FGE, FD and ED.
Find the area of shaded area A
AB2 = BD2 + AD2 (Pyth.theorem)
AB2 = F E cm AB
243 . 4 18 =
3cm
AF=AB-BF= =1.243 m
450
450
Area of sector FAE B
3cm D
3cm C
Area of region FAED = Area of ABC- 2 area of sector FBD
Area of shaded region = area of sector FAE+ Area of region FAED
= =5.57 cm2

34. Area of shaded region= Area of circle – area of triangle
2 cm
2400 A B O
600 C
AOB= =1200
AOC=600 C
600 A O
2 cm
Area of shaded region= Area of circle – area of triangle

35. Area of shaded region =area of circle – area of triangle
CAD=600 (equilateral ) 2 2 O
600
Area of triangle= A D B 2
Area of circle =
Area of shaded region =area of circle – area of triangle
= =2.46 cm2

36. cm CT 464 . 3 12 = C AT2 + CT2 = AC2 (Pyth.theorem) CT2 = AC2 - AT2B C O
AT2 + CT2 = AC2 (Pyth.theorem)
CT2 = AC2 - AT2
CT2 = 42 – 22 = 12 cm CT
464 . 3 12 =
300
300 A
4 cm T
Area of shaded region = area of triangle – area of circle
=  = cm2

37. Areaof AOC +Area of AOB + Area of BOC= Area of ABC
10 cm
6 cm
AC2=AB2+BC2
102=62+BC2
BC2=102-62
BC2=81
BC=9 cm
Areaof AOC +Area of AOB + Area of BOC= Area of ABC
12.5r= 27
r= 2.16 cm
Area o shaded region= area of ABC-area of circle
=27- =12.34 cm2

38. A
3 cm
300 D C B
Area of shaded region = area of sector – area of triangle

39. r A B C O
5 cm
3 cm
4 cm r B C O
4 cm A B C 2 O
4 cm
2 cm
2400

40. 2cm
6 cm 1) 2) O C A
4 cm B
Find the area of the taichi figures
Find perimeter and the area of the shaded region. C O A B
4 cm 3) O A B
4 cm 4)
Find perimeter and the area of the shaded figures
Find perimeter and the area of the shaded figures

41. 1.
4 cm 2.
4 cm
4 cm
4 cm 3. 4.
4 cm
4 cm 5. 6. 7.
4 cm
4 cm
4 cm A
10.
3 cm
300
300
3 cm

42. A B C D E F
4 cm A B C D E F
4 cm 2. 1.
Find the perimeter and the area of the above shaded figures

43. 8 cm
6 cm A B C D
8 cm
2 cm
2cm O A B C
10 cm
8 cm
6 cm A B C
8 cm
10 cm A B C
Find the perimeter and the area of the above shaded figures

44. O A B
3 cm
600
2 cm
300 O A B C
10 cm
3 cm
4 cm O O
3 cm
600 D C A B

45. the area of region I is 50 cm2 and the area of region II is 32 cm2, find the area of region III in terms of .t
3 cm
1 cm  O A B C D
a) Find the perimeter of shaded part.
b)Find area of the shaded part

46. cm CT 464 . 3 12 = C AT2 + CT2 = AC2 (Pyth.theorem) CT2 = AC2 - AT2B C O
AT2 + CT2 = AC2 (Pyth.theorem)
CT2 = AC2 - AT2
CT2 = 42 – 22 = 12 cm CT
464 . 3 12 =
Area of AOC+ Area of BOC + Area of AOB = Area of ABC A
4 cm T
Area of shaded region = area of ABC – area of circle
=  = cm2

47. In ADO, a2+h2=22 (pyth. Theorem) h2=4-a2B C 2 O a h D E F
In ADO, a2+h2=22 (pyth. Theorem)
h2=4-a2 2 4 a h - =
Area AOC + Area BOC + Area AOB
= Area ABO+ Area BOC + Area AOC
a = 1 2 4 a h - = 1
h = 3
Area ABC =