In this problem students measure lengths, calculate the perimeters and areas of rectangles using the formulae: perimeter = twice length plus twice width and area = length x width.
Students should know how to use a table as they investigate the mathematical question, what effect does the perimeter of a rectangle have on its area?
There are seven related problems: Peters’ String, Measurement, Level 4, Peters’ Second String, Measurement, Level 5, Peters’ Third String, Algebra, Level 6, The Old Chicken Run Problem, Algebra, Level 6 and the Polygonal String Problem, Algebra, Level 6. These show the non-link between rectangles’ areas and perimeters, going as far as showing that among all quadrilaterals with a fixed perimeter, the square has the largest area.
Peter keeps a piece of string from a parcel that came for his birthday. It is 30 cm long. He plays with it and makes different shapes. He thinks that all the rectangles he makes have the same area. His sister Miri disagrees. Who is right and why?
- Introduce the problem to the class. Allow them time to consider how they might approach the problem.
- Give each pair of students a piece of string. Have the students investigate Peter’s conjecture in any way that they want. They may prefer to draw diagrams rather than measure the sides of rectangles produced with their string.
- As they work, check on their progress, and encourage systematic recording.
- Pose the Extension Problem when appropriate. Share ideas if some students are struggling.
- Have students share solutions in a class discussion, explaining their strategies and methods of recording.
Show that Peter can make a rectangle with area 24 cm2. Miri has a piece of string too. It is only 20 cm long. Can she make a rectangle with a bigger area than than 24 cm2?
Solution to the problem
Students may take a practical approach, spreading out the string in a shape of a rectangle, measuring the length, and calculating its area.
A table such as this can show the results of the students' explorations.
It is clear that Peter’s conjecture that the areas he can make are all the same is false.
Peter is to make an area of 24 cm2 with his string. He must find L and W so that 2L + 2W = 30 and LW = 24.
An approximation method and a table are used:
We will guess L and W so that L + W = 15 (half of 30) and keep adjusting our guess until we get LW equal, or very close to, 24. (The answer we give here is sufficiently close for our purposes. Accuracy to more decimal places can be obtained if required.)
The previous table (Table 1) gives a place to start. L is clearly just under 2 and W just above 13.
L + W
Using a table similar to Table 1, we can see that Miri can produce an area of 25 cm2. (See Peter’s Third String, Level 6, for a justification of the fact that the rectangle of largest area and given perimeter is a square.) So even though Miri has a smaller string than Peter, she can produce a bigger area than a particular rectangle that he can make.