Supporting the effective use of ICT across the primary curriculum
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Mathematics Ages 10 - 11: The Sheep Pen Introduction: This lesson is about problem solving. It is an investigation into the relationship between area and perimeter. The lesson uses a spreadsheet program to perform the calculations and the graphing facility to illustrate the results. Although this is a reasonably well publicised problem for solving on a spreadsheet, the follow-up activities listed below offer a number of further challenges to the more able pupils Resources • One computer per two children for group work, or one computer and a data projector for whole class work • Spreadsheet software (e.g. Excel) • Children may need paper and pencil to help them with their calculations. Less able children may also need a calculator and squared graph paper. Previous learning Children should have the ability to work out the area and perimeter of a rectangular shape They should know how to enter numbers into a spreadsheet. Learning Objectives • To recognise that shapes with the same perimeter may have different areas • To develop a strategy for solving the problem • To enter formulae into a spreadsheet • To use a spreadsheet to draw a graph. What to do Here is the scenario: Farmer Jones has to create a new grazing pen for his sheep. He has 144 metres of fencing and all the necessary posts. However, he also has a stone wall that he can use for one side of his pen, so he only has to worry about the other three sides. He is building a rectangular pen. How should he lay out this pen in order to give the sheep the maximum area to graze? What is the maximum area in square metres that the sheep can have and what will be the dimensions of the pen? Initially the children may think that it makes no difference. They could try to complete a table like the one shown below. Width (metres) 1 2 5 10 20
Length (metres)
Perimeter (mtrs)
Area (sq metres)
This should help to illustrate that the area does change despite the fact that the perimeter is the same. The next stage is to try to produce a formula that gives the length. Some children will be able to articulate how they have worked out the length, without necessarily expressing the calculation as algebraic formulae. Lead them to the idea that the length = 144 – 2 x width. Then move them on to the spreadsheet. The children set this up in a very similar fashion to the above table. In © ictopus ltd
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Maths Ages 10 - 11: Sheep Pen 2010
effect they enter 1 into cell A1, the above formula into cell B1 and the formula for the area into cell C1. This data is then copied down from row 1 into the next 50+ rows. What do we notice about the area? At what point has it reached its maximum value? Why does the area start to get smaller again after this point? Do you notice anything particular about the shape of the pen that provides the largest area? (Teacher’s note: it is in effect two squares alongside each other.) Using the graphing facility on the spreadsheet, the children could then draw a graph of width against area. The result should produce a parabola (bell shaped curve) where the turning point of the curve represents the maximum area. Field Area
3000
2500
Area
2000
Series1
1500
1000
500
0 1
4
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10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 Width
Differentiation For less able children it might be necessary to start with perhaps 36 metres of fencing and then use centimetre graph paper (or a pin board) and a length of 36 cm string to help them create some fields. They can then work out the area by multiplying length by width or by counting squares The role of ICT The computer allows for numerous attempts in a very short period of time. Children will soon be aware that in order to find the answer to the problem without ICT they may have to do the calculation manually some 20 to 40 times. Using a spreadsheet takes the monotony out of the calculations and very quickly reveals the answer. The spreadsheet can also be used to produce a graphical interpretation of the solution. Follow-up suggestions There are a number of ways of extending this problem. a) The amount of fencing could be changed – what would happen if, for example, the farmer had 200 metres of fencing? b)
What happens if he decides against a rectangular field, but opts for a right angle triangle instead, with the stone wall being the base of the triangle (knowledge of Pythagoras’s theorem necessary here)?
c)
What if he decides to use the stone wall as a diameter and then sets out the field as a semi-circle? How much area would then be enclosed (knowledge of circumference and area of a circle is necessary here)?
d)
Assuming the farmer only wanted to create an area of 2000 square metres, what is the minimum length of fencing that he would have to buy?
e)
A similar problem relating to volume is given below. Take a sheet of card measuring 40cm by 40 cm. A small square is to be cut out from each corner (each square will be the same size). The sides are then folded up to make a box without a lid
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Maths Ages 10 - 11: Sheep Pen 2010
What is the maximum volume that the box can hold? Clue: Let the length of the side of the square be x. The length of the base is then 40 – 2x. The height of the box is x. Hence the volume is x(40 - 2x)(40 - 2x) Assessment The children should consider the difference between a random approach and an approach based around a systematic method. The children should also be encouraged to explain what strategy they adopted for solving the problem. In effect this is solution by exhaustion; that is all the possibilities are tried and then the correct answer selected manually or by drawing the graph.
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lessons2go
Maths Ages 10 - 11: Sheep Pen 2010