Teaching Mathematics: What, When and Why

An in-depth examination of mathematics education, topic by topic


Difficult is done at once …

Last night a friend rang with a mathematics problem. He is selling a property with an irregular shape and wanted me to check his calculation of the area. Just in case you want to try before I proceed, the boundaries are straight with lengths (in order going around) 36.0, 21.3, 28.1 and 22.7 metres.

When I told him the problem was impossible and I could not answer, he replied that I had been studying maths for 70 years and still could not solve a simple problem. There must be a major problem with the education system, he said. I tried to explain by showing him a square with all sides 10 m and them squeezing it into a skinny rhombus. But by this time he was off on his criticism of the world and the elites that run it.

In mathematics we are rightly proud of proofs where problems that seem solvable are shown to be impossible. But in school mathematics the student never meets such problems. Perhaps a sprinkling of impossible problems might narrow the gap between the average citizen and the so-called elites.

So I’m seeking suitable problems to be included, without warning, in textbooks. Here is one, Year 11 level, where the first part suggests the second is do-able.

This could lead into a discussion of congruence, or perhaps why builders make triangles in addition to rectangles, or how surveyors record properties. Or given the sides of a quadrilateral what extra information is need to find its area? Are the edge lengths of a tetrahedron sufficient to determine its volume?

In the previous question there were too many objects that fitted the description. There are also questions where the object described does not exist. Worth a run?

More problems please?

PS: On the basis that practical minded people learn best from physical models, I made one such to solve the initial problem. Here are some pictures.

Just remembered a computation that I used as an introduction to beam theory when teaching engineering students. It’s understandable at a secondary level by students who know some statics.



14 responses to “Difficult is done at once …”

  1. […] Peachey has a new, nice post on his blog: Difficult is done at once. It is concerned with Tom’s failure to answer a […]

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  2. I have used questions like your #3 which are impossible to solve. After students have seen it a couple of times, they can recognise such problems.

    It should be possible to find the area of a triangle from the lengths of the sides. However, textbooks tend not to mention Heron’s formula for the area of a triangle. The formula also tells you if the triangle is impossible.

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    1. Indeed Terry. Am I right in saying that Heron’s formula is not in any Victorian syllabus?

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  3. Oops I had a silly error in Q1a, now fixed. Thanks Marty.

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  4. I’m relieved to see that Heron’s formula is in the 2 unit year 11 unit on trigonometry. (In NSW.)

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    1. It is?? I haven’t seen it in the current syllabus nor the new one coming in 2026.

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      1. I’m not sure how it translates to reality, but I saw it under MA-T1 in this document:

        https://www.dropbox.com/scl/fi/jucgw1kqie9q3mlqg7jyc/nsw_stage6.pdf?rlkey=1y3symfh5czhmnc12lkogkked&dl=0

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      2. The right hand column seems to be teacher’s notes whilst the left hand side are the syllabus dot points. The reference to “triangle area formula” is generally accepted in this context to mean the one involving sine. Heron’s formula has not been a requirement to learn in NSW for at least a few decades.

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  5. Similar to the area of the quadrilateral question, this question is from the new NSW syllabus lacks a bit of info. It is under their “non-routine teaching examples”.

    Find the surface area of a truncated cone that has a top of radius 3 cm and a base of radius 7 cm using similar triangles.

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    1. Isn’t the answer to this just 10pi times the slant height?

      Not every answer has to be a number…

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      1. Yes but usually the wording would be to the effect “write an expression for…”

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      2. Do we include the circular disks?

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  6. The questions seems to want that yes. So, it would be 10pi * (slant height + 1).

    The wording of the question, with reference to similar triangles, made it seem like the question writer wanted a numerical answer. That was what I was trying to get at.

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  7. Sorry it was meant to be 10pi * slant height + 58pi.

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