Circle P6 2010 SA2 ACS P2 Q14

pi  x  30 = 94.248



180/360  =  1/2

3  -  1/2  =  2 1/2

2 1/2  x  94.248  =  235.62 cm2 (2 dp)

The shaded area is 235.62 cm2.

Circle P6 2010 SA2 SCGS P2 Q18

[2 marks for (a) and 3 marks for (b)]{Answers given: 116 cm and 364.42 cm (correct to 2 decimal places)]


43  x  2 =  86  cm

86  x  2  =  172  cm

28  x  2  =  56  cm

172  -  56  =  116  cm

a) The length of SR is  116  cm.


86  -  56  =  30  cm

pi  x  30  =   94.247 cm 

2  x  pi  x  43  =  20.177  cm 

94.247  +  20.177  =  364.42 cm (2 dp)


The perimeter is 364.42 cm.

Circle P6 2010 SA2 Nanyang P2 Q18

18. The figure below is created using the curved lines (arcs) of quadrants with radius 1 cm, 2 cm and 3 cm. Find the area of the shaded parts.
(Take pi as 3.14)

Circle of radius 3 cm  -  4 x Right-angled triangles of sides 3 cm -
[Circle of radius 1 cm  -  4 x  Right-angled triangles of sides 1 cm]
4 x Right-angled triangles  =  1/2  x  Diagonal  x  Diagonal

3.14 x 1 x 1 = 3.14

1/2 x 2 x 2 = 2

3.14 - 2 = 1.14

3.14 x 3 x 3 = 28.26

1/2 x 6 x 6 = 18

28.26 - 18 - 1.14 = 9.12 cm2

It was 9.12 cm2.

Circle P6 2010 SA2 CHIJ P2 Q13

4 x 2 + 3 x 2 = 14

1/4 x 2 x 3.14 x 4 = 6.28

1/2 x 2 x 3.14 x 3 = 9.42

14 + 6.28 + 9.42 = 29.7 cm

a) It was 29.7 cm.

1/4 x 3.14 x 4 x 4 = 12.56           A + B

1/2 x 3.14 x 3 x 3 = 14.13           C + B

14.13 - 12.56 = 1.57 cm2.

Circle P6 2010 SA2 Henry Park P2 Q18

1//2 x 2 x pi x 18 = 56.57?

3.5 x 2 x pi x 6 = 132

56.57 + 132 + 36 = 224.57? cm

a) It was 224.57 cm.   (please use pi from your calculator)

1/6 x pi x 18 x 18 = 54 pi    1/6 Big Circle (or 1/3 Semicircle)

2/3 x pi x 6 x 6 = 24 pi   2/3 Circle (1/2 + 1/6)

54 pi - 24 pi = 30 pi cm2

b) It was 30 pi cm2.

Circle P6 2010 SA2 HHK P2 Q13

The figure below shows 2 quarter circles and a rectangle. The radius of the big quarter circle is 8 cm. The radius of the small quarter circle is 4 cm. Find the difference in area between the two shaded parts X and Y. Use the calculator value of pi and give your answer correct to 1 decimal place.

1/4 x ∏ x 8 x 8 = 16 pi

1/4 x ∏ x 4 x 4 = 4 pi

16 pi - 4 pi =  12 pi  B + X

8 x 4 = 32       B + Y

12 pi  - 32 = 5.7 cm2

It was 5.7 cm2.

Please calculate ?? as I do not have a proper calculator.

Circle P6 2009 SA1 P2 RGS Q10

10. The diagram below shows a square, a semicircle and a right-angled isosceles triangle. Given that the radius of the circle is 4 cm, find the area of the shaded part. Take π as 3.14. [3]

4 x 4 = 16

1/4 x 3.14 x 4 x 4 = 12.56

16 – 12.56 = 3.44

4 x 2 = 8

1/2 x 8 x 8 = 32

32 + 3.44 = 35.44 cm²

The area of the shaded part is 35.44 cm².

Circle P6 2009 SA1 P2 RGS Q15

15. The figure is made up of 4 equal quadrants and one semicircle. AB = CD = EF = 8 cm. The total area of the unshaded parts is 33.5 cm². Take π as 3.14.

a) Find the total perimeter of the shaded parts. [1]

b) Find the total area of the shaded parts. [4]

2 x 3.14 x 8 = 50.24

50.24 + 8 + 8 = 66.24 cm

a) The total perimeter of the shaded parts is 66.24 cm.

1/2 x 3.14 x 4 x 4 = 25.12

33.5 – 25.12 = 8.38

3.14 x 8 x 8 = 200.96

200.96 – 8.38 – 8.38 = 184.2 cm²

The total area of the shaded parts is 184.2 cm².

Circle P6 2009 SA2 P2 MGS Q18

18. The shaded part below is made up of a quarter circle and a semicircle which are drawn within the rectangle ACDF. AB = 12 cm.

a) Find the area of rectangle ACDF.

b) What percentage of rectangle ACDF is shaded? Round off your answer to 2 decimal places.

12 ÷ 2 = 6

6 + 12 = 18

18 x 12 = 216 cm²

a) The area of rectangle ACDF is 216 cm².

1/4 x 22/7 x 12 x 12 = 113 1/7

1/2 x 22/7 x 6 x 6 = 56 4/7

113 1/7 – 56 4/7 = 56 4/7

                         ≈ 56.57 cm²

56 4/7 ÷ 216 x 100% = 26.19%

b) 26.19% of rectangle ACDF is shaded.

Circle P6 2009 SA2 P2 Rulang Q3

3. A piece of wire is bent to form two semicircles as shown below. Find the length of the wire used. (Take π = 3.14)

3.14 x 5 = 15.7

3.14 x 10 = 31.4

5 + 5 + 15.7 + 31.4 = 57.1 cm

Circle P6 2009 SA2 P2 Rulang Q12

12. The figure below is made up of a semicircle, 2 isosceles triangles and a square. If the area of each isosceles triangle is 32 cm², find the area of the semicircle. (Take π = 3.14)

32 x 2 = 64

64 = 8 x 8

8 x 3 = 24

24 ÷ 2 = 12

1/2 x 3.14 x 12 x 12 = 226.08 cm²

The area of the semicircle is 226.08 cm².