[Note: GERMAN versions of the new Week 2 tasks are available HERE. I welcome suggestions on how to improve the translations! The first 5 tasks below have become Week 5 in the book. The GERMAN versions can be seen HERE.]
Note: We have replaced the original Week 2 tasks by a completely different set, which is shown at the end of this sequence. The original Week 2 tasks, shown at the beginning of this sequence, appear as Week 5, under the heading "Similar shapes, all in a row", in the published book.
In
Week 2 we compare the lengths of paths made of curves or line segments.
The tasks can be thought of as puzzles. They involve similar shapes,
often quite simple ones such as squares, and can be used to highlight
the role played by the distributive law in multiplication.
Monday:
This task is likely to seem quite challenging at first (but also
intriguing, hopefully) since we are not given any specific lengths. The
task rests on the fact that the shapes are all similar (being squares)
and that the blue and red arrangements have the same overall width.
Some
students might resort to measuring lengths, which is fine as an initial
approach. A very practical example of this is shown below!
Or
students might try to rearrange some of the line segments, mentally or
by drawing: both sets of squares can be decomposed and the line segments
translated to make a single square whose sides are equal to the width
of the arrangements.
A more analytic approach could go something like this:
The perimeter of each square is 4 times its base. So the total perimeter of the blue squares is
4 times the base of each blue square, which is 4 times the total width of the blue squares.
Similarly, the total perimeter of the red squares is 4 times the base of each red square, which
is 4 times the total width of the red squares.
But the total widths are the same, so the total perimeters are equal.
Tuesday:
Here we simplify Monday’s task by providing specific values for some of
the lengths. This means students can directly calculate the total
length of the green lines and the red lines, and then use this to derive
the length of the blue lines.
Students
should notice that the total lengths of the green lines and the red
lines are same. In turn, they might realise that the expression for the
length of the red lines, 4×2cm + 4×7cm + 4×3cm, can be written as 4 ×
(2cm + 7cm + 3cm) by factorising, and that this gives us 4×12cm, which
is an expression for the length of the blue lines.
A
similar logic can be applied to the blue lines. Though we don’t know
the specific lengths of individual blue lines, we can imagine writing an
expression similar to the one for the red lines which, when factorised,
would give us 4×12cm. We could, of course, spell this out by using
letters for the side lengths of the individual blue squares, in this
kind of way:
4×a + 4×b + 4×c + 4×d + 4×e + 4×f = 4(a + b + c + d + e + f ) = 4×12.
Wednesday:
Here we switch from (similar) squares to (similar) equilateral
triangles. For each triangle, the sum of the lengths of the two blue
sides is equal to twice the length of the red ‘base’.
This task is really very simple -
once you realise that you don’t need to know the value of specific lengths! The
total length of all the triangles’ blue sides is the sum of twice the
length of each triangle’s red base, which is twice the total length of
the red bases, which is twice 18 cm.
2×18 cm = 36 cm.
Formally, we could write something like this, again involving the distributive law:
2×
a + 2×
b + 2×
c + 2×
d + 2×
e + 2×
f = 2(
a +
b +
c +
d +
e +
f )
= 2×18
= 36.
Thursday: This time our similar shape is curvilinear, in the form of a letter e,
rather than consisting of (straight) line segments. The same principle
applies, that the length of the (curved) line forming a given example of
the similar shape is proportional to its width.
However, we can’t
combine such shapes to make a larger version, as we could with the
squares and with the equilateral triangles. This suggests that it will
be less obvious to students that the total length of the line formed by a
row of the similar shapes is proportional to the total width of the row
and that therefore the blue and the red curved lines are the same
length.
Friday: As with the letter e
in Thursday’s task, our similar shape (the circle) is again curved, but
this time there is a known formula relating its width (or height), d, to its total length, p: p = πd.
This
week’s tasks were inspired by a SMILE poster that could be seen in many
mathematics classrooms in London, and elsewhere, in the 1980s. It asked
students to compare the length of a semicircle with the total length of
smaller semicircles placed along its diameter, as here:
The
current task is not as exciting. Its aim is to give students the
opportunity, if they have not already taken it, to formalise the
relation between the width and total length of rows of similar shapes.
For our given circles, we can write
Circumference of blue circle
= π(
a +
b +
c +
d +
e)
= π
a + π
b + π
c + π
d + π
e= sum of the circumferences
of the red circles.
THIS, below, IS THE NEW VERSION OF WEEK 2, which appears in the published book.
Week 2 looks at multiplication and at the role played by the distributive law when we multiply what might be called 'compound' numbers. A typical example would be long multiplication of two two-digit numbers, where we partition each number into tens and units before multiplying. However, rather than revisiting such a much-frequented situation, we take a fresh look by multiplying with numbers like 5½.
Monday: Here we look at the product 5½ × 5½. It is likely that most people's default position is that its value lies exactly between 5 × 5 and 6 × 6. To see beyond this is likely to take some quite careful thought (and we consider some ways of doing this in the Week's later tasks). Less thought, perhaps, is required just to perform the calculation, though the result (that 5½ × 5½ is nearer 25 than 36) may still come as a surprise!
A more general empirical argument can be applied to the task along these lines: consider the squares of consecutive numbers (or any set of equally spaced numbers). As the numbers get larger (eg 9, 10, 11) so does the gap between their squares (eg 100–81 = 19, 121–100 = 21).
We can also look at increments. What happens as we move from 5 × 5 to 5 × 5½ to 5½ × 5½? The first step produces an increase of 5 × ½, or ½ × 5. The second step produces a further increase of ½ × 5½, making a total increase of ½ × 10½. This is less that ½ × 11, which would take us halfway from 25 to 36.
It should be interesting to see what kinds of argument students bring to the task, and whether some are content to resolve the task by just calculating. But here, too, it should be interesting to see how they perform the calculation (see Wednesday's task).
Tuesday: Here we use the area model to represent the task. Such a representation can be very illuminating - once one sees it! So it should be interesting to find out whether there are students for whom the task is quite demanding. Unless students are very familiar with the model (and understand how area can be used to model multiplication), considerable thought might be needed to match the various features of the diagram to the key numerical elements.
An effective way of interpreting the diagram is to see the area B+C+D (or 2B+D) as the difference between 5½² and 5², and to see E+F+G+H+I (or 2B+3D) as the difference between 6² and 5½². These differences aren't quite the same. They differ by the amount 2D (2×½² = ½).
Wednesday: Here we present four different ways of writing 5½² as a prelude to performing the calculation. How do students respond to the different forms? Were any of them used in Monday’s task?
Given the rationale behind this Week's tasks, part c) [reinforced by part d)] is perhaps the most interesting part here, as it probably has the greatest potential to reveal the part played by 5 and by ½ in the multiplication. Of course, this can also happens with b) if it is performed as a long multiplication, but students might be too familiar with the procedure to notice!
Thursday: This task illustrates the classic misconception that occurs when students multiply 'compound' numbers, as in, say, 23×23 = 49. Part c) of Wednesday's task addresses this issue, and should help students see what is going on.
Friday: This task makes use of the difference of two squares, namely that (a + b)(a – b) = a² – b². However, the relation is left implicit. Note that we are challenging students to see the general in a specific example, rather than spot a pattern from several examples. They may not be used to doing this, though it can be very powerful in that it focusses attention on structure rather than on pattern spotting which is sometimes quite superficial.
Students will probably need to perform Fay's calculation carefully, and then carefully analyse what went on, if they are to find a quick rule for answering Question 2. It should be interesting to see whether students who manage to find an effective rule can also explain why it works.
