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**S8 Perimeter, area and volume**

Maths Age 11-14 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 S8 Perimeter, area and volume

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**S8 Perimeter, area and volume**

Contents S8 Perimeter, area and volume A S8.1 Perimeter A S8.2 Area

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**Put these shapes in order**

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.

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**What is the perimeter of this shape?**

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

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**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, 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. Perimeter of a rectangle = l + w + l + w = 2l + 2w or = 2(l + w)

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**What is the perimeter of this shape?**

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

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

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**Calculate the lengths of the missing sides to find the perimeter.**

5 cm p = 2 cm p 2 cm q = r = 1.5 cm q r s = 6 cm t = 2 cm s 6 cm u = 10 cm Discuss how to find each missing length. 4 cm 4 cm P = 2 cm t 2 cm u = 46 cm

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**What is the perimeter of this shape?**

Remember, the dashes indicate the sides that are the same length. 5 cm 4 cm P = = 26 cm Rather than using repeated addition, pupils might suggest 2 × × 4 = 26 cm.

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**What is the perimeter of this shape?**

Start by finding the lengths of all the sides. 4.5 m 4.5 m Perimeter = 5 m 4 m = 17 m Discuss how pupils have calculated the lengths of the missing sides. 2 m 2 m 1 m 1 m 2 m

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**What is the perimeter of this shape?**

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

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**Which shape has a different perimeter from the first shape?**

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

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**S8 Perimeter, area and volume**

Contents S8 Perimeter, area and volume A S8.1 Perimeter A S8.2 Area

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

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**Area of a rectangle 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 This formula should be revision from Key Stage 2 work. Area of a rectangle = length × width = lw

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**What is the area of this rectangle?**

Area of a rectangle What is the area of this rectangle? 4 cm 8 cm 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. Area of a rectangle = lw = 8 cm × 4 cm = 32 cm2

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**Area of a right-angled triangle**

What proportion of this rectangle has been shaded? 4 cm 8 cm What is the shape of the shaded part? 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. What is the area of this right-angled triangle? 1 2 Area of the triangle = × 8 × 4 = 4 × 4 = 16 cm2

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**Area of a right-angled triangle**

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

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**Area of a right-angled triangle**

Calculate the area of this right-angled triangle. To work out the area of this triangle we only need the length of the base and the height. 8 cm 6 cm 10 cm We can ignore the third length opposite the right angle. 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. Area = 1 2 × base × height = × 8 × 6 1 2 = 24 cm2

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**Area of shapes made from rectangles**

How can we find the area of this shape? We can think of this shape as being made up of two rectangles. 7 m Either like this … A 10 m 15 m … or like this. 8 m Label the rectangles A and B. 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. B 5 m Area A = 10 × 7 = 70 m2 15 m Area B = 5 × 15 = 75 m2 Total area = = 145 m2

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**Area of shapes made from rectangles**

How can we find the area of the shaded shape? 7 cm We can think of this shape as being made up of one rectangle with another rectangle cut out of it. A 3 cm 8 cm Label the rectangles A and B. 4 cm B Area A = 7 × 8 = 56 cm2 Discuss ways to find the shaded area before revealing the solution. Area B = 3 × 4 = 12 cm2 Total area = 56 – 12 = 44 cm2

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**Area of an irregular shapes on a pegboard**

How can we find the area of this irregular quadrilateral constructed on a pegboard? We can divide the shape into right-angled triangles and a square. A B Area A = ½ × 2 × 3 = 3 units2 Area B = ½ × 2 × 4 = 4 units2 Area C = ½ × 1 × 3 = 1.5 units2 Discuss how the area can be found using right-angled triangles and rectangles. D E Area D = ½ × 1 × 2 = 1 unit2 C Area E = 1 unit2 Total shaded area = 10.5 units2

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**Area of an irregular shapes on a pegboard**

How can we find the area of this irregular quadrilateral constructed on a pegboard? An alternative method would be to construct a rectangle that passes through each of the vertices. A B The area of this rectangle is 4 × 5 = 20 units2 C The area of the irregular quadrilateral is found by subtracting the area of each of these triangles. D

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**Area of an irregular shapes on a pegboard**

How can we find the area of this irregular quadrilateral constructed on a pegboard? Area A = ½ × 2 × 3 = 3 units2 A B 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 C D = 10.5 units2

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**Area of an irregular shape on a pegboard**

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.

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**What is the area of this triangle?**

Area of a triangle What proportion of this rectangle has been shaded? 4 cm 8 cm Drawing a line here might help. 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. What is the area of this triangle? 1 2 Area of the triangle = × 8 × 4 = 4 × 4 = 16 cm2

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Area of a triangle 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.

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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 Pupils will need to learn this formula. Or using letter symbols: Area of a triangle = bh 1 2

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**What is the area of this triangle?**

Area of a triangle What is the area of this triangle? 6 cm 7 cm Area of a triangle = bh 1 2 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. = 1 2 × 7 × 6 = 21 cm2

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**Area of a parallelogram**

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.

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**Area of a parallelogram**

The area of any parallelogram can be found using the formula: Area of a parallelogram = base × perpendicular height base perpendicular height Pupils will need to learn this formula. Or using letter symbols: Area of a parallelogram = bh

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**Area of a parallelogram**

What is the area of this parallelogram? We can ignore this length 8 cm 7 cm 12 cm 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. Area of a parallelogram = bh = 7 × 12 = 84 cm2

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Area of a trapezium 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.

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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 perpendicular height a b Pupils will need to learn this formula. Or using letter symbols: Area of a trapezium = (a + b)h 1 2

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**What is the area of this trapezium?**

Area of a trapezium What is the area of this trapezium? Area of a trapezium = (a + b)h 1 2 6 m 9 m = (6 + 14) × 9 1 2 = × 20 × 9 1 2 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. 14 m = 90 m2

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**What is the area of this trapezium?**

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 8 m 3 m = × 11 × 12 1 2 This example shows a right trapezium. Again, the numbers and units in the example may be modified to make the problem more challenging. 12 m = 66 m2

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**What is the area of the yellow square?**

Area problems This diagram shows a yellow square inside a blue square. 3 cm What is the area of the yellow square? 7 cm We can work this out by subtracting the area of the four blue triangles from the area of the whole blue square. 10 cm If the height of each blue triangle is 7 cm, then the base is 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? 3 cm. Area of each blue triangle = ½ × 7 × 3 = ½ × 21 = 10.5 cm2

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**What is the area of the yellow square?**

Area problems This diagram shows a yellow square inside a blue square. 7 cm 10 cm 3 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. There are four blue triangles so: 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.) 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

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**Area formulae of 2-D shapes**

You should know the following formulae: b h Area of a triangle = bh 1 2 b h Area of a parallelogram = bh Use this slide to summarize or review key formulae. a h b Area of a trapezium = (a + b)h 1 2

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**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 Let’s write all the lengths in cm. 518 mm = 51.8 cm 518 mm 1.24 m = 124 cm Emphasize that when substituting different lengths into a formula the units must be the same. Link: S7 Measures – converting units. 1.24 m Area of the trapezium = ½( ) × 51.8 Don’t forget to put the units at the end. = ½ × 200 × 51.8 = 5180 cm2

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