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perimeterareaandvolume-190718073754 (2).pptx
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- 2. A A A A A A Contents S8 Perimeter, areaand 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 shapesin order
- 4. Perimeter To find theperimeter 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 arectangle 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 = 2l + 2w or = 2(l + w)
- 6. Perimeter What is theperimeter 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 bcm 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 lengthsof 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 canfind 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 shapehas a different perimeter from the first shape? A B C A B C A B C B A A
- 13. Contents S8 Perimeter, areaand 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 ofa 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 arectangle Area is measured in square units. We can use mm2 , cm2 , m2 or km2 . 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 arectangle What is the area of this rectangle? 8 cm 4 cm Area of a rectangle = lw = 8 cm × 4 cm = 32 cm2
- 17. Area of aright-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 cm2
- 18. We can usea 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 aright-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 cm2
- 20. Area of shapesmade 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 m2 Area B = 5 × 15 = 75 m2 Total area = 70 + 75 = 145 m2
- 21. Area of shapesmade 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 cm2 Area B = 3 × 4 = 12 cm2 Total area = 56 – 12 = 44 cm2
- 22. E D C B A Area of anirregular shapes on a pegboard We can divide the shape into right-angled triangles and a square. Area A = ½ × 2 × 3 = 3 units2 Area B = ½ × 2 × 4 = 4 units2 Area C = ½ × 1 × 3 = 1.5 units2 Area D = ½ × 1 × 2 = 1 unit2 Area E = 1 unit2 Total shaded area = 10.5 units2 How can we find the area of this irregular quadrilateral constructed on a pegboard?
- 23. C D B A Area ofan 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 units2 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 anirregular shapes on a pegboard Area A = ½ × 2 × 3 = 3 units2 A B C D Area B = ½ × 2 × 4 = 4 units2 Area C = ½ × 1 × 2 = 1 units2 Area D = ½ × 1 × 3 = 1.5 units2 Total shaded area = 9.5 units2 Area of irregular quadrilateral = (20 – 9.5) units2 = 10.5 units2 How can we find the area of this irregular quadrilateral constructed on a pegboard?
- 25. Area of anirregular shape on a pegboard
- 26. Area of atriangle 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 cm2
- 27. Area of atriangle
- 28. Area of atriangle 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 atriangle What is the area of this triangle? Area of a triangle = bh 1 2 7 cm 6 cm = 1 2 × 7 × 6 = 21 cm2
- 30. Area of aparallelogram
- 31. Area of aparallelogram 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 aparallelogram What is the area of this parallelogram? Area of a parallelogram = bh 12 cm 7 cm = 7 × 12 = 84 cm2 8 cm We can ignore this length
- 33. Area of atrapezium
- 34. Area of atrapezium 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 atrapezium 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 m2 What is the area of this trapezium?
- 36. Area of atrapezium 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 m2 8 m 3 m 12 m
- 37. Area problems 7 cm 10cm 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 cm2 3 cm This diagram shows a yellow square inside a blue square.
- 38. Area problems 7 cm 10cm 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 cm2 Area of blue square = 10 × 10 = 100 cm2 Area of yellow square = 100 – 42 = 58 cm2 3 cm This diagram shows a yellow square inside a blue square. What is the area of the yellow square?
- 39. Area formulae of2-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 informulae 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 cm2 Don’t forget to put the units at the end.
- 41. Contents S8 Perimeter, areaand 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 thesurface 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 thesurface 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 thesurface 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 thesurface 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 cm2 The area of the front = 7 × 5 = 35 cm2 The area of the side = 7 × 8 = 56 cm2
- 46. To find thesurface 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 cm2 + 2 × 35 cm2 + 2 × 56 cm2 Top and bottom Front and back Left and right side = 80 + 70 + 112 = 262 cm2
- 47. We can findthe 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 = 2lw + 2hw + 2lh
- 48. How can wefind 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 = x2 . Therefore: Surface area of a cube = 6x2
- 49. This cuboid ismade 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 cm2 How much of the surface area is green? 47 cm2
- 50. What is thesurface 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 cm2
- 51. 5 cm 6 cm 3cm 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 cm2 15 cm2 18 cm2 30 cm2 30 cm2 18 cm2 Then add the areas together to find the surface area. Surface Area = 126 cm2
- 52. Here is thenet of a regular tetrahedron. Using nets to find surface area What is its surface area? 6 cm 5.2 cm Area of each face = ½bh = ½ × 6 × 5.2 = 15.6 cm2 Surface area = 4 × 15.6 = 62.4 cm2
- 53. Contents S8 Perimeter, areaand 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 cuboidis made out of interlocking cubes. Making cuboids How many cubes does it contain?
- 55. We can workthis 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 ofspace that a three-dimensional object takes up is called its volume. Making cuboids We can use mm3 , cm3 , m3 or km3 . 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 acuboid 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 acuboid What is the volume of this cuboid? Volume of cuboid = length × width × height = 13 × 8 × 5 = 520 cm3 5 cm 8 cm 13 cm
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- 60. Volume and displacement Bydropping 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 cm3 If an object is dropped into a measuring cylinder and displaces 5 ml of water then the volume of the object is 5 cm3 . What is the volume of 1 litre of water? 1 litre of water has a volume of 1000 cm3 .
- 61. What is thevolume 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 cm3 Volume of the blue cuboid = 3 × 2 × 2 = 12 cm3 Total volume = 54 + 12 = 66 cm3
- 62. Remember, a prismis 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 cm2 and the length is 3 cm: Volume of L-shaped prism = 22 × 3 = 66 cm3 3 cm
- 63. What is thevolume of this triangular prism? Volume of a prism 5 cm 4 cm 7.2 cm Area of cross-section = ½ × 5 × 4 = 10 cm2 Volume of prism = 10 × 7.2 = 72 cm3
- 64. Volume of aprism Area of cross-section = 7 × 12 – 4 × 3 = 84 – 12 = Volume of prism = 5 × 72 = 360 m3 3 m 4 m 12 m 7 m 5 m 72 m2 What is the volume of this prism?
- 65. Contents S8 Perimeter, areaand 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 anddiameter
- 67. The value ofπ 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 thevalue 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 ofa 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 = πd π = C d
- 70. The circumference ofa circle Use π = 3.14 to find the circumference of this circle. C = πd 8 cm = 3.14 × 8 = 25.12 cm
- 71. Finding the circumferencegiven the radius The diameter of a circle is two times its radius, or C = 2πr d = 2r We can substitute this into the formula C = πd to give us a formula to find the circumference of a circle given its radius.
- 72. The circumference ofa circle Use π = 3.14 to find the circumference of the following circles: C = πd 4 cm = 3.14 × 4 = 12.56 cm C = 2πr 9 m = 2 × 3.14 × 9 = 56.52 m C = πd 23 mm = 3.14 × 23 = 72.22 mm C = 2πr 58 cm = 2 × 3.14 × 58 = 364.24 cm
- 73. ? Finding the radiusgiven the circumference Use π = 3.14 to find the radius of this circle. C = 2πr 12 cm How can we rearrange this to make r the subject of the formula? r = C 2π 12 2 × 3.14 ≈ = 1.91 cm (to 2 d.p.)
- 74. Find the perimeterof 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 diameterof 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 = πd and π = 3.14, = 157 cm The number of complete rotations = 100 000 ÷ 157 ≈ 636 1 km = 100 000 cm
- 76. Contents S8 Perimeter, areaand volume A A A A A AS8.6 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 acircle
- 78. Formula for thearea of a circle We can find the area of a circle using the formula. radius Area of a circle = πr2 Area of a circle = π × r × r or
- 79. The circumference ofa circle Use π = 3.14 to find the area of this circle. A = πr2 4 cm ≈ 3.14 × 4 × 4 = 50.24 cm2
- 80. Finding the areagiven the diameter The radius of a circle is half of its diameter, or We can substitute this into the formula A = πr2 to give us a formula to find the area of a circle given its diameter. r = d 2 A = πd2 4
- 81. The area ofa circle Use π = 3.14 to find the area of the following circles: A = πr2 2 cm = 3.14 × 22 = 12.56 cm2 A = πr2 10 m = 3.14 × 52 = 78.5 m2 A = πr2 23 mm = 3.14 × 232 = 1661.06 mm2 A = πr2 78 cm = 3.14 × 392 = 4775.94 cm2
- 82. Find the areaof 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 × 6.52 = 132.665 cm2 Area of rectangle = 6 × 13 = 78 cm2 Total area = 132.665 + 78 = 210.665 cm2
- 83. Area of asector What is the area of this sector? 72° 5 cm Area of the sector = 72° 360° × π × 52 1 5 = × π × 52 = π × 5 = 15.7 cm2 (to 1 d.p.) We can use this method to find the area of any sector.
- 84. Area problem Find theshaded area to 2 decimal places. Area of the square = 12 × 12 1 4 Area of sector = × π × 122 = 36π 12 cm = 144 cm2 Shaded area = 144 – 36π = 30.96 cm2 (to 2 d.p.)
Editor's Notes
- #1 The aim of this unit is to teach pupils to: Deduce and use formulae to calculate lengths, perimeters, areas and volumes in 2-D and 3-D shapes Material in this unit is linked to the Key Stage 3 Framework supplement of examples pp 234-241.
- #3 Ask pupils to explain what perimeter is and how we could work out the perimeters of the shapes shown on the board. Challenge pupils to put these shapes in order from the one that has the smallest perimeter to the one that has the largest. Establish that it is not necessary to know the exact length of the diagonal edge. Ask pupils how we know that the diagonal edges must be between 1 and 2 units long. When the activity has been completed ask, If we put these shapes in order of area from smallest to largest, would the order be the same? Establish that the order would be different because the area of a shape is not directly related to its perimeter. Ask pupils to put the shapes in order of area.
- #4 Ask pupils if they know how many dimensions measurements of perimeter have. Establish that they only have one dimension, length, even though the measurement is used for two-dimensional shapes. Tell pupils that when finding the perimeter of a shape with many sides it is a good idea to mark a starting point and then work from there adding up the lengths of all the sides.
- #5 Explain the difference between the two forms of the formula. The first formula means double the length, double the width and add the two together. The second formula means add the length and the width and double the answer.
- #6 Point out that to work out the perimeter we need to add together the lengths of every side. If we are not given some of the lengths, then we have to work them out before we can find the perimeter.
- #7 Discuss how to work out the missing sides of this shape. The side marked a cm plus the 5 cm side must be equal to 12 cm, a is therefore 7 cm. The side marked b cm plus the 4 cm side must be equal to 9 cm, b is therefore 5 cm. Can anyone see a quicker way of working out the perimeter of this shape?
- #8 Discuss how to find each missing length.
- #9 Rather than using repeated addition, pupils might suggest 2 × 5 + 4 × 4 = 26 cm.
- #10 Discuss how pupils have calculated the lengths of the missing sides.
- #12 Ask pupils to decide which shape has a different perimeter from the other three and to explain how they can tell that the other three shapes have the same perimeter. For the first set all of the shapes have a perimeter of 16 units except shape B. For the second set all of the shapes have a perimeter equal to the perimeter of the circle (3 units) except shape A. For the third set all of the shapes have a perimeter of 14 units except shape A.
- #14 Discuss how we can compare the area of the rugs by counting the squares that make up each pattern. The small squares in each pattern are all the same size. Conclude that Rug B covers the largest surface.
- #15 This formula should be revision from Key Stage 2 work.
- #16 The length and the width of the rectangle can be modified to make the arithmetic more challenging. Different units could also be used to stress that units must be the same before they are substituted into a formula.
- #17 Discuss the fact that a right-angled triangle always has half the area of the rectangle made by its height and its width. It is not necessary to know the length of the longest side opposite the right angle (the hypotenuse) to find the area.
- #18 Introduce the formula for the area of a right-angled triangle.
- #19 Talk through this example. Make sure that pupils are able to identify which length is the base and which length is the height. The lengths of the sides may be modified to make the arithmetic more difficult.
- #20 Discuss ways to divide this composite shape into rectangles. A third possibility not shown on this slide would be to take the square of area 15 m × 15 m and to subtract the area of the rectangle 10 m × 8 m. This gives us 225 m2 – 80 m2 = 145 m2.
- #21 Discuss ways to find the shaded area before revealing the solution.
- #22 Discuss how the area can be found using right-angled triangles and rectangles.
- #25 Drag the vertices to modify the shape. Ask a volunteer to divide the shape into rectangles and triangles using the pen tool set to draw straight lines. Use this to work out the area. Alternatively, draw a rectangle around the outside of the shape to find the area by subtraction.
- #26 A line is drawn on the diagram to split the shape into two rectangles each with one half shaded. Pupils should conclude from this that one half of the whole rectangle is shaded. Establish that the area of the whole rectangle is equal to the base of the shaded triangle times the height of the shaded triangle. Conclude that the area of the shaded triangle is equal to half the base times the height.
- #27 Use this activity to deduce that the area of any triangle can be found by multiplying the length of the base by the perpendicular height. Start with a right-angled triangle. Modify the triangle while keeping the length of the base and the height constant. Set the pen tool to draw straight lines and use it to draw a rectangle around the triangle. Ask pupils to use this to find the area of the triangle. Repeat for a variety of triangles.
- #28 Pupils will need to learn this formula.
- #29 Tell pupils that to work out the area of the triangle they must start by writing the formula. They can then substitute the correct values into the formula provided that they are in the same units. Emphasize that it is important to always write down the correct units at the end of the calculation. The numbers and units in the example may be modified to make the problem more challenging.
- #30 Ask pupils to suggest ways to find the area of the parallelogram. This can be done by imagining a right-angled triangle cut off from one end and moved to the other. Use the pen tool set to draw straight lines to show this if required. Drag the vertices of the parallelogram to make a rectangle. Confirm that the area of the original parallelogram and the area of the rectangle are the same. Modify the parallelogram while keeping the length of the base and the height constant. Conclude that if the length of the base and the height of the parallelogram are kept constant then the area will remain the same regardless of the slope. By looking at further examples deduce that the area of any parallelogram can be found by multiplying the length of the base by the perpendicular height. Discuss the fact that we do not need to know the diagonal length on the parallelogram; warn pupils not to fall into the trap of using this measure rather than the perpendicular height.
- #31 Pupils will need to learn this formula.
- #32 Tell pupils that to work out the area of the parallelogram they must start by writing the formula. They can then substitute the correct values into the formula provided that they are in the same units. Point out that the length of the diagonal can be ignored. Emphasize that it is important to always write down the correct units at the end of the calculation. The numbers and units in the example may be modified to make the problem more challenging.
- #33 Set the pen to draw straight lines and use it to show how to divide the trapezium into triangles and rectangles. Use this to find the areas of different trapezia. Use the pen tool to show that for any trapezium we can construct an identical trapezium rotated through 180° to make a parallelogram. If we call the lengths of the parallel sides of the trapezium a and b, then the area of the parallelogram is equal to (a + b) times the height. The area of a single trapezium is half the area of the parallelogram and so the area of any trapezium can be found using the formula ½(a + b)h.
- #34 Pupils will need to learn this formula.
- #35 Tell pupils that to work out the area of the trapezium they must start by writing the formula. They can then substitute the correct values into the formula provided that they are in the same units. Emphasize that it is important to always write down the correct units at the end of the calculation. The numbers and units in the example may be modified to make the problem more challenging.
- #36 This example shows a right trapezium. Again, the numbers and units in the example may be modified to make the problem more challenging.
- #37 Talk through this area problem. Encourage the pupils to discuss strategies before you begin. There will be a temptation to try to work out or guess the lengths of the sides of the yellow square before they work out the area. Ask what parts of the diagram the pupils can work out the area for instead. How does this help them answer the question?
- #38 As an extension, think back to the suggestions the class made at the beginning as to how to work out the area of the square. They now know that the area of the square is 58 cm²; how can they use this knowledge to work out the length of one side of the square? How can the pupils prove that the triangles are all 3 by 7? (By arguing by symmetry, or finding equal angles in the figure.)
- #39 Use this slide to summarize or review key formulae.
- #40 Emphasize that when substituting different lengths into a formula the units must be the same. Link: S7 Measures – converting units.
- #42 Discuss the meaning of surface area. The important thing to remember is that although surface area is found for three-dimensional shapes, it only has two dimensions. It is therefore measured in square units.
- #46 Emphasize the importance of working systematically when finding the surface area to ensure that no faces have been left out. We can also work out the surface area of a cuboid by drawing its net (see slide 51). This may be easier for some pupils because they would be able to see every face rather than visualizing it.
- #47 Pupils should write this formula down.
- #48 Ask pupils to use this formula to find the surface area of a cube of side length 5 cm. 6 × 52 = 6 × 25 = 150 cm2. Repeat for other numbers. As a more challenging question tell pupils that a cube has a surface area of 96 cm2. Ask them how we could work out its side length using inverse operations.
- #49 Discuss how to work out the surface area that is green. Ask pupils how we could write the proportion of the surface area that is green as a fraction, as a decimal and as a percentage.
- #50 Discuss ways to find the surface area of this solid. We could use a net of this prism to help find the area of each face.
- #51 Links: S3 3-D shapes – nets. S6 Construction and loci – constructing nets.
- #56 Link: S7 Measures – units of volume and capacity.
- #59 Each green cube represents one centimetre cubed. Ask pupils how we could use water in a measuring cylinder to find the volume of an object. Tell pupils that 1 cm3 of water will displace 1 ml of water in the beaker. Demonstrate this by dropping each cuboid into the beaker, and recording how the level of the water changes. Use this slide to demonstrate how volume is linked to capacity. Links: S7 Measures – units of volume and capacity. S7 Measures – reading scales.
- #60 Ask pupils to give the dimensions of a cube that would hold 1 litre of water. This would be a 10 cm by 10 cm by 10 cm cube. Ask pupils how many litres of water we could fit into a metre cube. (1000 litres.) A litre of water has a weight of 1 kg. A metre cube would therefore hold 1 tonne of water! Link: S7 Measures – units of volume and capacity.
- #61 Compare this with slide 50, which finds the surface area of the same shape.
- #66 Drag the circle to different sizes and ask pupils to record the length of the diameter and the circumference for each one. Challenge them to find a link between these two numbers. Tell pupils that the ratio of the diameter to the circumference of a circle is a fixed amount equal to just over 3. This means that the circumference of a circle is always just over three times the diameter. Once pupils know the link between the diameter and the circumference, hide the circumference and ask them to find it given the diameter (by multiplying the diameter by 3.14). Also hide the diameter and ask them how we can find it given the circumference (by dividing by 3.14).
- #67 Explain that pi is a number. We call it pi because it is not possible to write the number exactly. Even written to 200 decimal places, although extremely accurate, is an approximation.
- #68 It is useful to approximate pi to a value of 3 when approximating the answers of calculations.
- #69 Pupils should be asked to learn these formulae.
- #70 Tell pupils that when solving a problem like this they should always start by writing down the formula that they are using. This will minimize the risk of using the radius instead of the diameter, for example.
- #72 For each one, start by asking pupils what formula we have to use. Estimate each answer first using = 3, or use this to check the answer.
- #73 Link: A3 Formulae – changing the subject of a formula.
- #75 Explain that we can ignore any remainder when dividing 100000 by 157 because we are asked for the number of complete rotations.
- #77 This animation shows how the area of a circle can be approximated to the area of a parallelogram of base length r and height r. Watch the circle pieces rearrange into an approximate parallelogram and ask a volunteer to use the pen tool to label the length and the height in terms of r. Discuss how the diagram would look if the circle was divided into more pieces. Deduce from this that the area of a circle is r2.
- #81 Explain that rather than use the formula on the previous slide, it is usually easier to halve the diameter mentally to give the radius, before substituting it into the formula. The most common error is to forget to half the diameter to find the radius.
- #82 Compare this with slide 74, where we found the perimeter of the same shape.
- #83 Discuss how this area could be calculated before revealing the solution. The area of a sector is a fraction of the area of a full circle. We can find this fraction by dividing the angle at the centre by 360°.
- #84 Discuss how this area could be calculated before revealing the solution.