Julie's Wheels
Apply Pythagoras’ theorem to solve a problem
Devise and use problem solving strategies to explore situations mathematically (be systematic, think).
Like many problems at this Level, the main strategy is to work the problem situation around so that you can use a piece of theory that you know. In this case it is Pythagorass’ Theorem.
This is a difficult application of Pythagoras’ Theorem because it is not the one that you may first think to use. Unfortunately the centres of the three circles do not lie on a right-angled triangle. (You may like to check this using a scale diagram.) That means that you need to think a bit first
The Problem
Julie had three wheels from bikes and things that she stacked against the shed. Each wheel fitted so neatly together that Julie took this photo.
Now the radius of the largest wheel was 16cm and the radius of the middle-sized wheel was 9cm. What was the radius of Julie’s smallest wheel?
Teaching sequence
- Ask the students to write down two things that come into their minds when they first look at the problem. (circles, radius, circumference, diameter, draw diagram etc)
- List these ideas.
- Look at the list of ideas and discuss which ones they think will be useful in solving the problem.
- You may also like to talk about the skills they have for finding out unknown lengths. This may lead to someone mentioning Pythagoras’ Theorem although its relevance will probably not be apparent. At this stage avoid telling the students that Pythagoras’ Theorem will be needed and leave it as one of the suggestions.
- As the students solve the problem (in pairs) ask questions that require them to explain their reasoning. If they are stuck then you may want to more direct with the Pythagoras prompt.
- Share solutions.
Extension to the problem
Can you find a formula for the radius of the smaller wheel in terms of the radii of the other two wheels? In other words, if you are given the radii of the two larger circles as a and b, can you find c in terms of a and b?
Solution
This looks like a way to apply Pythagoras; all we’ve got to do is to find the triangles. Let the radius of the smaller wheel be c. It turns out that we now need three equations.
In triangle ABB' , AB = 16 + 9 = 25 and AB' = AD – BC = 16 – 9 = 7,
(BB' )2 = AB 2 – (AB' )2 = 252 – 72 = 576.
In triangle AEF' , AE = 16 + c and AF' = 16 – c, and
(F' E)2 = AE2 – (AF' )2 = (16 + c)2 – (16 – c)2 = 64c,
In triangle BEF, BE = 9 + c and BF = 9 – c, and
FE2 = BE2 – BF2 = (9 + c)2 – (9 – c)2 = 36c.
But BB' = F'F + EF, so
√576 = √64c + √36c
√576 = 8√c + 6<√c
14√c = 24
... √c = 12/7 ... √c = 144/49.
Extension: Using precisely the algebra in the above with 16 replaced by a and 9 replaced by b gives
c = ab/(<√a + <√b)2.
| Attachment | Size |
|---|---|
| Julie.pdf | 37.6 KB |
| JulieMaori.pdf | 52.08 KB |
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