perimeterareaandvolume-190718073754 (1).pptx

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1. Perimeter , area and volume

2. A A A A A A Contents S8 Perimeter, area and volume S8.1 Perimeter S8.6 Area of a circle S8.2 Area S8.5 Circumference of a circle S8.3 Surface area S8.4 Volume

3. Put these shapes in order

4. Perimeter To find the perimeter of a shape we add together the length of all the sides. What is the perimeter of this shape? 1 cm 3 3 2 1 1 2 Perimeter = 3 + 3 + 2 + 1 + 1 + 2 = 12 cm Starting point

5. Perimeter of a rectangle To calculate the perimeter of a rectangle we can use a formula. length, l width, w Using l for length and w for width, Perimeter of a rectangle = l + w + l + w = 2 l + 2 w or = 2( l + w )

6. Perimeter What is the perimeter of this shape? b cm a cm 9 cm 5 cm 12 cm 4 cm The lengths of two of the sides are not given so we have to work them out before we can find the perimeter. Let’s call the lengths a and b . Sometimes we are not given the lengths of all the sides. We have to work them out using the information we are given.

7. Perimeter 9 cm 5 cm b cm 12 cm a cm 4 cm a = 12 – 5 cm = 7 cm 7 cm b = 9 – 4 cm = 5 cm 5 cm P = 9 + 5 + 4 + 7 + 5 + 12 = 42 cm Sometime we are not given the lengths of all the sides. We have to work them out from the information we are given.

8. Calculate the lengths of the missing sides to find the perimeter. P = 5 + 2 + 1.5 + 6 + 4 + 2 + 10 + 2 + 4 + 6 + 1.5 + 2 Perimeter 5 cm 2 cm 6 cm 4 cm 4 cm 2 cm 2 cm p q r s t u p = 2 cm q = r = 1.5 cm s = 6 cm t = 2 cm u = 10 cm = 46 cm

9. P = 5 + 4 + 4 + 5 + 4 + 4 Perimeter What is the perimeter of this shape? Remember, the dashes indicate the sides that are the same length. 5 cm 4 cm = 26 cm

10. Perimeter Perimeter = 4.5 + 2 + 1 + 2 + 1 + 2 + 4.5 Start by finding the lengths of all the sides. 5 m 2 m 2 m 2 m 4 m 4.5 m 4.5 m 1 m 1 m = 17 m What is the perimeter of this shape?

11. Before we can find the perimeter we must convert all the lengths to the same units. Perimeter 3 m 2.4 m 1.9 m 256 cm In this example, we can either use metres or centimetres. Using centimetres , 300 cm 240 cm 190 cm P = 256 + 190 + 240 + 300 = 986 cm What is the perimeter of this shape?

12. Equal perimeters Which shape has a different perimeter from the first shape ? A B C A B C A B C B A A

13. Contents S8 Perimeter, area and volume A A A A A A S8.2 Area S8.1 Perimeter S8.6 Area of a circle S8.5 Circumference of a circle S8.3 Surface area S8.4 Volume

14. The area of a shape is a measure of how much surface the shape takes up. Area For example, which of these rugs covers a larger surface? Rug A Rug B Rug C

15. Area of a rectangle Area is measured in square units . We can use mm 2 , cm 2 , m 2 or km 2 . The 2 tells us that there are two dimensions, length and width. We can find the area of a rectangle by multiplying the length and the width of the rectangle together. length, l width, w Area of a rectangle = length × width = lw

16. Area of a rectangle What is the area of this rectangle? 8 cm 4 cm Area of a rectangle = lw = 8 cm × 4 cm = 32 cm 2

17. Area of a right-angled triangle What proportion of this rectangle has been shaded? 8 cm 4 cm What is the shape of the shaded part? What is the area of this right-angled triangle? Area of the triangle = × 8 × 4 = 1 2 4 × 4 = 16 cm 2

18. We can use a formula to find the area of a right-angled triangle: Area of a right-angled triangle base, b height, h Area of a triangle = 1 2 × base × height = 1 2 bh

19. Area of a right-angled triangle Calculate the area of this right-angled triangle. 6 cm 8 cm 10 cm To work out the area of this triangle we only need the length of the base and the height. We can ignore the third length opposite the right angle. Area = 1 2 × base × height = × 8 × 6 1 2 = 24 cm 2

20. Area of shapes made from rectangles How can we find the area of this shape? 7 m 10 m 8 m 5 m 15 m 15 m We can think of this shape as being made up of two rectangles. Either like this … … or like this. Label the rectangles A and B. A B Area A = 10 × 7 = 70 m 2 Area B = 5 × 15 = 75 m 2 Total area = 70 + 75 = 145 m 2

21. Area of shapes made from rectangles How can we find the area of the shaded shape? We can think of this shape as being made up of one rectangle with another rectangle cut out of it. 7 cm 8 cm 3 cm 4 cm Label the rectangles A and B. A B Area A = 7 × 8 = 56 cm 2 Area B = 3 × 4 = 12 cm 2 Total area = 56 – 12 = 44 cm 2

22. E D C B A Area of an irregular shapes on a pegboard We can divide the shape into right-angled triangles and a square. Area A = ½ × 2 × 3 = 3 units 2 Area B = ½ × 2 × 4 = 4 units 2 Area C = ½ × 1 × 3 = 1.5 units 2 Area D = ½ × 1 × 2 = 1 unit 2 Area E = 1 unit 2 Total shaded area = 10.5 units 2 How can we find the area of this irregular quadrilateral constructed on a pegboard?

23. C D B A Area of an irregular shapes on a pegboard An alternative method would be to construct a rectangle that passes through each of the vertices. The area of this rectangle is 4 × 5 = 20 units 2 The area of the irregular quadrilateral is found by subtracting the area of each of these triangles. How can we find the area of this irregular quadrilateral constructed on a pegboard?

24. Area of an irregular shapes on a pegboard Area A = ½ × 2 × 3 = 3 units 2 A B C D Area B = ½ × 2 × 4 = 4 units 2 Area C = ½ × 1 × 2 = 1 units 2 Area D = ½ × 1 × 3 = 1.5 units 2 Total shaded area = 9.5 units 2 Area of irregular quadrilateral = (20 – 9.5) units 2 = 10.5 units 2 How can we find the area of this irregular quadrilateral constructed on a pegboard?

25. Area of an irregular shape on a pegboard

26. Area of a triangle What proportion of this rectangle has been shaded? 8 cm 4 cm Drawing a line here might help. What is the area of this triangle? Area of the triangle = × 8 × 4 = 1 2 4 × 4 = 16 cm 2

27. Area of a triangle

28. Area of a triangle The area of any triangle can be found using the formula: Area of a triangle = × base × perpendicular height 1 2 base perpendicular height Or using letter symbols: Area of a triangle = bh 1 2

29. Area of a triangle What is the area of this triangle? Area of a triangle = bh 1 2 7 cm 6 cm = 1 2 × 7 × 6 = 21 cm 2

30. Area of a parallelogram

31. Area of a parallelogram Area of a parallelogram = base × perpendicular height base perpendicular height The area of any parallelogram can be found using the formula: Or using letter symbols: Area of a parallelogram = bh

32. Area of a parallelogram What is the area of this parallelogram? Area of a parallelogram = bh 12 cm 7 cm = 7 × 12 = 84 cm 2 8 cm We can ignore this length

33. Area of a trapezium

34. Area of a trapezium The area of any trapezium can be found using the formula: Area of a trapezium = (sum of parallel sides) × height 1 2 Or using letter symbols: Area of a trapezium = ( a + b ) h 1 2 perpendicular height a b

35. Area of a trapezium 9 m 6 m 14 m Area of a trapezium = ( a + b ) h 1 2 = (6 + 14) × 9 1 2 = × 20 × 9 1 2 = 90 m 2 What is the area of this trapezium?

36. Area of a trapezium What is the area of this trapezium? Area of a trapezium = ( a + b ) h 1 2 = (8 + 3) × 12 1 2 = × 11 × 12 1 2 = 66 m 2 8 m 3 m 12 m

37. Area problems 7 cm 10 cm What is the area of the yellow square? We can work this out by subtracting the area of the four blue triangles from the area of the whole blue square. If the height of each blue triangle is 7 cm, then the base is 3 cm. Area of each blue triangle = ½ × 7 × 3 = ½ × 21 = 10.5 cm 2 3 cm This diagram shows a yellow square inside a blue square.

38. Area problems 7 cm 10 cm We can work this out by subtracting the area of the four blue triangles from the area of the whole blue square. There are four blue triangles so: Area of four triangles = 4 × 10.5 = 42 cm 2 Area of blue square = 10 × 10 = 100 cm 2 Area of yellow square = 100 – 42 = 58 cm 2 3 cm This diagram shows a yellow square inside a blue square. What is the area of the yellow square?

39. Area formulae of 2-D shapes You should know the following formulae: b h Area of a triangle = bh 1 2 Area of a parallelogram = bh Area of a trapezium = ( a + b ) h 1 2 b h a h b

40. Using units in formulae Remember, when using formulae we must make sure that all values are written in the same units. For example, find the area of this trapezium. 76 cm 1.24 m 518 mm Let’s write all the lengths in cm. 518 mm = 51.8 cm 1.24 m = 124 cm Area of the trapezium = ½ (76 + 124) × 51.8 = ½ × 200 × 51.8 = 5180 cm 2 Don’t forget to put the units at the end.

41. Contents S8 Perimeter, area and volume A A A A A A S8.3 Surface area S8.1 Perimeter S8.6 Area of a circle S8.2 Area S8.5 Circumference of a circle S8.4 Volume

42. To find the surface area of a shape, we calculate the total area of all of the faces. A cuboid has 6 faces. The top and the bottom of the cuboid have the same area. Surface area of a cuboid

43. To find the surface area of a shape, we calculate the total area of all of the faces. A cuboid has 6 faces. The front and the back of the cuboid have the same area. Surface area of a cuboid

44. To find the surface area of a shape, we calculate the total area of all of the faces. A cuboid has 6 faces. The left hand side and the right hand side of the cuboid have the same area. Surface area of a cuboid

45. To find the surface area of a shape, we calculate the total area of all of the faces. Can you work out the surface area of this cuboid? Surface area of a cuboid 7 cm 8 cm 5 cm The area of the top = 8 × 5 = 40 cm 2 The area of the front = 7 × 5 = 35 cm 2 The area of the side = 7 × 8 = 56 cm 2

46. To find the surface area of a shape, we calculate the total area of all of the faces. So the total surface area = Surface area of a cuboid 7 cm 8 cm 5 cm 2 × 40 cm 2 + 2 × 35 cm 2 + 2 × 56 cm 2 Top and bottom Front and back Left and right side = 80 + 70 + 112 = 262 cm 2

47. We can find the formula for the surface area of a cuboid as follows. Surface area of a cuboid = Formula for the surface area of a cuboid h l w 2 × lw Top and bottom + 2 × hw Front and back + 2 × lh Left and right side = 2 lw + 2 hw + 2 lh

48. How can we find the surface area of a cube of length x ? Surface area of a cube x All six faces of a cube have the same area. The area of each face is x × x = x 2 . Therefore: Surface area of a cube = 6 x 2

49. This cuboid is made from alternate purple and green centimetre cubes. Chequered cuboid problem What is its surface area? Surface area = 2 × 3 × 4 + 2 × 3 × 5 + 2 × 4 × 5 = 24 + 30 + 40 = 94 cm 2 How much of the surface area is green? 47 cm 2

50. What is the surface area of this L-shaped prism? Surface area of a prism 6 cm 5 cm 3 cm 4 cm 3 cm To find the surface area of this shape we need to add together the area of the two L-shapes and the area of the six rectangles that make up the surface of the shape. Total surface area = 2 × 22 + 18 + 9 + 12 + 6 + 6 + 15 = 110 cm 2

51. 5 cm 6 cm 3 cm 6 cm 3 cm 3 cm 3 cm It can be helpful to use the net of a 3-D shape to calculate its surface area. Using nets to find surface area Here is the net of a 3 cm by 5 cm by 6 cm cuboid. Write down the area of each face. 15 cm 2 15 cm 2 18 cm 2 30 cm 2 30 cm 2 18 cm 2 Then add the areas together to find the surface area. Surface Area = 126 cm 2

52. Here is the net of a regular tetrahedron. Using nets to find surface area What is its surface area? 6 cm 5.2 cm Area of each face = ½ bh = ½ × 6 × 5.2 = 15.6 cm 2 Surface area = 4 × 15.6 = 62.4 cm 2

53. Contents S8 Perimeter, area and volume A A A A A A S8.4 Volume S8.1 Perimeter S8.6 Area of a circle S8.2 Area S8.5 Circumference of a circle S8.3 Surface area

54. The following cuboid is made out of interlocking cubes. Making cuboids How many cubes does it contain?

55. We can work this out by dividing the cuboid into layers. Making cuboids The number of cubes in each layer can be found by multiplying the number of cubes along the length by the number of cubes along the width. 3 × 4 = 12 cubes in each layer. There are three layers altogether so the total number of cubes in the cuboid = 3 × 12 = 36 cubes .

56. The amount of space that a three-dimensional object takes up is called its volume . Making cuboids We can use mm 3 , cm 3 , m 3 or km 3 . The 3 tells us that there are three dimensions, length, width and height. Volume is measured in cubic units. Liquid volume or capacity is measured in ml, l, pints or gallons.

57. Volume of a cuboid We can find the volume of a cuboid by multiplying the area of the base by the height. Volume of a cuboid = length × width × height = lwh height , h length , l width , w The area of the base = length × width So:

58. Volume of a cuboid What is the volume of this cuboid? Volume of cuboid = length × width × height = 13 × 8 × 5 = 520 cm 3 5 cm 8 cm 13 cm

59. Volume and displacement

60. Volume and displacement By dropping cubes and cuboids into a measuring cylinder half filled with water we can see the connection between the volume of the shape and the volume of the water displaced. 1 ml of water has a volume of 1 cm 3 If an object is dropped into a measuring cylinder and displaces 5 ml of water then the volume of the object is 5 cm 3 . What is the volume of 1 litre of water? 1 litre of water has a volume of 1000 cm 3 .

61. What is the volume of this L-shaped prism? Volume of a prism made from cuboids 6 cm 5 cm 3 cm 4 cm 3 cm We can think of the shape as two cuboids joined together. Volume of the green cuboid = 6 × 3 × 3 = 54 cm 3 Volume of the blue cuboid = 3 × 2 × 2 = 12 cm 3 Total volume = 54 + 12 = 66 cm 3

62. Remember, a prism is a 3-D shape with the same cross-section throughout its length. Volume of a prism We can think of this prism as lots of L-shaped surfaces running along the length of the shape. Volume of a prism = area of cross-section × length If the cross-section has an area of 22 cm 2 and the length is 3 cm: Volume of L-shaped prism = 22 × 3 = 66 cm 3 3 cm

63. What is the volume of this triangular prism? Volume of a prism 5 cm 4 cm 7.2 cm Area of cross-section = ½ × 5 × 4 = 10 cm 2 Volume of prism = 10 × 7.2 = 72 cm 3

64. Volume of a prism Area of cross-section = 7 × 12 – 4 × 3 = 84 – 12 = Volume of prism = 5 × 72 = 360 m 3 3 m 4 m 12 m 7 m 5 m 72 m 2 What is the volume of this prism?

65. Contents S8 Perimeter, area and volume A A A A A A S8.5 Circumference of a circle S8.1 Perimeter S8.6 Area of a circle S8.2 Area S8.3 Surface area S8.4 Volume

66. Circle circumference and diameter

67. The value of π For any circle the circumference is always just over three times bigger than the radius. The exact number is called π (pi). We use the symbol π because the number cannot be written exactly. π = 3.141592653589793238462643383279502884197169 39937510582097494459230781640628620899862803482 53421170679821480865132823066470938446095505822 31725359408128481117450284102701938521105559644 62294895493038196 (to 200 decimal places)!

68. Approximations for the value of π When we are doing calculations involving the value π we have to use an approximation for the value. For a rough approximation we can use 3. Better approximations are 3.14 or . 22 7 We can also use the π button on a calculator. Most questions will tell you which approximation to use. When a calculation has lots of steps we write π as a symbol throughout and evaluate it at the end, if necessary.

69. The circumference of a circle For any circle: π = circumference diameter or: We can rearrange this to make an formula to find the circumference of a circle given its diameter. C = π d π = C d

70. The circumference of a circle Use π = 3.14 to find the circumference of this circle. C = π d 8 cm = 3.14 × 8 = 25.12 cm

71. Finding the circumference given the radius The diameter of a circle is two times its radius, or C = 2 π r d = 2 r We can substitute this into the formula C = π d to give us a formula to find the circumference of a circle given its radius.

72. The circumference of a circle Use π = 3.14 to find the circumference of the following circles: C = π d 4 cm = 3.14 × 4 = 12.56 cm C = 2 π r 9 m = 2 × 3.14 × 9 = 56.52 m C = π d 23 mm = 3.14 × 23 = 72.22 mm C = 2 π r 58 cm = 2 × 3.14 × 58 = 364.24 cm

73. ? Finding the radius given the circumference Use π = 3.14 to find the radius of this circle. C = 2 π r 12 cm How can we rearrange this to make r the subject of the formula? r = C 2 π 12 2 × 3.14 ≈ = 1.91 cm (to 2 d.p.)

74. Find the perimeter of this shape Use π = 3.14 to find perimeter of this shape. The perimeter of this shape is made up of the circumference of a circle of diameter 13 cm and two lines of length 6 cm. 6 cm 13 cm Perimeter = 3.14 × 13 + 6 + 6 = 52.82 cm

75. Circumference problem The diameter of a bicycle wheel is 50 cm. How many complete rotations does it make over a distance of 1 km? 50 cm The circumference of the wheel = 3.14 × 50 Using C = π d and π = 3.14, = 157 cm The number of complete rotations = 100 000 ÷ 157 ≈ 636 1 km = 100 000 cm

76. Contents S8 Perimeter, area and volume A A A A A A S8.6 Area of a circle S8.1 Perimeter S8.2 Area S8.5 Circumference of a circle S8.3 Surface area S8.4 Volume

77. Area of a circle

78. Formula for the area of a circle We can find the area of a circle using the formula. radius Area of a circle = π r 2 Area of a circle = π × r × r or

79. The circumference of a circle Use π = 3.14 to find the area of this circle. A = π r 2 4 cm ≈ 3.14 × 4 × 4 = 50.24 cm 2

80. Finding the area given the diameter The radius of a circle is half of its diameter, or We can substitute this into the formula A = π r 2 to give us a formula to find the area of a circle given its diameter. r = d 2 A = π d 2 4

81. The area of a circle Use π = 3.14 to find the area of the following circles: A = π r 2 2 cm = 3.14 × 2 2 = 12.56 cm 2 A = π r 2 10 m = 3.14 × 5 2 = 78.5 m 2 A = π r 2 23 mm = 3.14 × 23 2 = 1661.06 mm 2 A = π r 2 78 cm = 3.14 × 39 2 = 4775.94 cm 2

82. Find the area of this shape Use π = 3.14 to find area of this shape. The area of this shape is made up of the area of a circle of diameter 13 cm and the area of a rectangle of width 6 cm and length 13 cm. 6 cm 13 cm Area of circle = 3.14 × 6.5 2 = 132.665 cm 2 Area of rectangle = 6 × 13 = 78 cm 2 Total area = 132.665 + 78 = 210.665 cm 2

83. Area of a sector What is the area of this sector? 72 ° 5 cm Area of the sector = 72° 360° × π × 5 2 1 5 = × π × 5 2 = π × 5 = 15.7 cm 2 (to 1 d.p.) We can use this method to find the area of any sector.

84. Area problem Find the shaded area to 2 decimal places. Area of the square = 12 × 12 1 4 Area of sector = × π × 12 2 = 36 π 12 cm = 144 cm 2 Shaded area = 144 – 36 π = 30.96 cm 2 (to 2 d.p.)