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Posts Tagged ‘Hero’

A surprising find

April 18, 2015 3 comments

The other day I my timehop showed me this lovely little post from last year. It includes “Heron’s Formula” for calculating the area of a triangle, as I read it I remembered thinking it was a little strange that not many people had heard of it before.

Today I was looking through a number of textbooks trying to find a decent set of questions on area, perimeter and volume for my year nines as I wanted to consolidate their learning at the start then move onto surface area. I’m not a fan of textbook misuse- ie “copy the example and try the questions” but I do sometimes use them for exercises as we have a very limited printing budget and some of them have superb exercises. For a fuller picture on.my view of textbooks, read this.

I was looking in one of my favourite textbooks:

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And I happened across this:

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There it is! Plain as day! Heron’s Formula! In a KS3 textbook!

I was disappointed that its function was described and its name wasn’t and there was no mention of why this worked. It basically reduces the question down from a geometry one to a purely algebraic substitution task and I would question the appropriateness of including it in an exercise on area, but still, I was incredibly exciting to find it there!

Are you a fan of Heron’s Formula? Had you even heard of it? Do you have a favourite textbook? I’d love to hear your views.

Perimeter, the Hero’s way

December 28, 2014 2 comments

Forgive the title, but I do love Heron’s Formula, (named after Hero of Alexandria) it’s my favourite geometric formula and this is only the second time I’ve had need to mention it in a blog post. The first was Area the Hero’s way.

This post is a look at my solution to this beautiful puzzle which was set by Ed Southall (@solvemymaths) the other day:

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This struck me as a tricky one. The fact it’s a right angled triangle meant my brain was crying out “we’ve got to use trigonometry”, but I couldn’t see an obvious was to do it. We have the area of the triangle, and there is an incircle in the picture so I figured we’d need to use the relationship that the area of a triangle is equal to the semiperimeter multiplied by the radius of the incircle. It was possibly the fact that this relationship mentions semiperimeter that put Heron’s Formula in my mind.

I sketched the problem and filled in what lengths I could deduce using circle theorems, filling gaps with variables:

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From this the semiperimeter was nice to work out, so I went down Hero’s route.

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Which led me to:

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I had one equation, but two variables. I needed another equation in the same two variables, so I used the aforementioned relationship from incircle:

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I rearranged this for x (as it was the simplest rearrangement) and subbed it into the other equation:

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I then solved this equation for r, discounting the negative, as a radius can’t be negative. Once I had r I could use the relationship area/radius = semiperimeter so twice area/radius =perimeter:

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Which rounds to 44 cm. (The question asks to the nearest cm.)

I love this question, and I’m happy with my solution. I think it’s fairly elegant and uses a nice array of mathematics, but I can’t help but think that I’ve missed something blinding obvious that would have led to a simpler solution. If you solved it a different way, I’d love to hear how you did it. (I’d also like to hear if you did it the same way!) I asked Ed how he solved it, and he used the fact that the point where the incircle meets the longest side of a triangle splits the longest side into two segments, the product of whose lengths is the area. This is a nice relationship, and not one I’d known, so I will look to explore it. This could have saved be a bit if working, as I could have jumped to tge bit where I wrote x=6.

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