Home > #MTBoS, A Level, Maths, Starters, Teaching > Perimeter, the Hero’s way

Perimeter, the Hero’s way

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:


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:


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


Which led me to:


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


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


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:


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.

  1. December 29, 2014 at 1:10 pm

    I solved this using simultaneous equations. As we are given the area of the triangle and part of one side, I came up with (12.5 + x)^2 = (12.5 + r)^2 + (r + x)^2. The area of the triangle can be expressed as (12.5 + r)(x + r) = 150. Expand these and you get:

    Pythagoras: 2r^2 + 25r + 2rx = 25x
    Area: r^2 + 12.5r + 12.5x + rx = 150

    Double the area equation to get 2r^2 + 25r + 25x + 2rx = 300

    Now subtract the Pythagoras equation from this to give 25x = 300 – 25x. 50x = 300 so x = 6.

    Substitute this into either equation to find r and then add the three sides together to find the perimeter

    • December 29, 2014 at 1:11 pm

      Aye, this seems to be the method most people used. Its a lovely solution.

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