How many times bigger is a 14 inch pizza than a 7 inch pizza? Ask this to most non-mathematicians and they will most likely say, in a delightful mixture of confidence and incorrectness, “twice as big”.
Now, as mathematicians, we know this is not true (at least so long as we make the assumption that both pizzas have the same thickness, we ignore crusts, and so on). But I have been surprised over the years how many working teachers of mathematics, when asked to explain this, or when using it as an example to their own students, are drawn to methods such as the following:
7 inch pizza: \(Area = {\pi}r^{2} = \pi \times 3.5^{2} = 12.25\pi\) inch\(^{2}\)
14 inch pizza: \(Area = {\pi}r^{2} = \pi \times 7^{2} = 49\pi\) inch\(^{2}\)
\(\therefore\) Ratio of areas \(= \dfrac{49\pi}{12.25\pi} = 4\)
Which, perfectly correctly, establishes that the 14 inch pizza is, in fact, 4 times bigger than the 7 inch pizza.
When we happen across a correct method of doing something, it is often tempting to dust our hands and move on, confidently holding on to what we have as “the way to do it”. But I am going to argue that the above method is actually a very poor one, for two main reasons:
- It is unnecessarily longwinded, cumbersome and inaccessible to many learners.
- It bypasses a far deeper and more powerful idea regarding the scaling of area (and other measures) which can be applied to all shapes.
Consider the following “visual” argument as an alternative:
As well as being an opportunity for learners to discuss “how does this show that the 14 inch pizza is 4 times as big as the 7 inch pizza?”, I would suggest that this type of argument is far more accessible to far more people, including pizzeria owners and their customers!
Moreover, it is very clear that the circular shape of the pizza here is irrelevant to the result. Any 2D shape that is stretched by a factor of 2 in both the horizonal and vertical directions will end up with an area that is 4 times as big. It isn’t a huge step from there to generalise and say that enlargement by scale factor \(f\) results in an area that is \(f^{2}\) times as big.
Or, in other words,
for a family of similar shapes, the area of each is proportional to the square of a chosen side.
As an example of the power of this idea, consider a rectangle divided into 3 similar triangles as in the diagram below:
Each labelled side is the hypotenuse of a similar triangle, and so the area of each triangle must be equal to some constant, \(k\), times the square of that hypotenuse. It is clear from the diagram that
\(ka^{2} + kb^{2} = kc^{2}\)
Which, after cancelling the \(k\)’s, leads to what should be a very familiar result!
\(a^{2} + b^{2} = c^{2}\)
This is the simplest and, in my opinion, most satisfying proof of Pythagoras’ Theorem that I have ever come across. However, it requires a deeply embedded intuitive understanding of the scaling arguments discussed earlier for it to be both satisfying and convincing.
Often in mathematics, looking for a wider range of approaches, including simple visual arguments, can lead to ways of thinking which take us way beyond the scope of the original problem. Sometimes it is worth taking a step back from your own “favourite method”, exploring fundamentally different approaches, and encouraging learners to do the same!
By Michael Gibson