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?

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.

A second set of problems involves questions about the maximum and minimum perimeters for a given area: __Karen’s Tiles__, Measurement, Level 5 and __Karen’s Second Tiles__, Algebra, Level 6.

Copymaster of the problem (English)

squared paper or graph paper

ruler

pieces of string of various lengths

### Problem

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?

### Teaching sequence

- 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.

#### Extension problem

Show that Peter can make a rectangle with area 24 cm^{2}. 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.

length |
width |
area |

1 |
14 |
14 |

2 |
13 |
26 |

3 |
12 |
36 |

Table 1

It is clear that Peter’s conjecture that the areas he can make are all the same is false.

#### Extension Solution:

Peter is to make an area of 24 cm^{2} 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 |
L + W |
LW |

1.9 |
13.1 |
15 |
24.89 |

1.8 |
13.2 |
15 |
23.76 |

1.85 |
13.15 |
15 |
24.33 |

1.84 |
13.16 |
15 |
24.21 |

1.83 |
13.17 |
15 |
24.10 |

1.82 |
13.18 |
15 |
24.00 |

Table 2

Using a table similar to Table 1, we can see that Miri can produce an area of 25 cm^{2}. (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.