## An irrational triangle

The sad news about Don Steward last week prompted many people to share blog posts and tweets about how he has helped and inspired them over the years. One of the people who blogged about him was Jo Morgan (@mathsjem) who wrote this piece. Jo had the pleasure of knowing Don personally and it was really nice to read the things she shared. If you missed it, do give it a read.Today I want to look at the problem she shared at the end of the post, which she says is her favourite Don Steward problem:It’s not one I remember seeing before, although I have spent so long on Don’s fantastic median website it is highly probably I have seen it. But I thought I’d have a go. Usually when working out an are of a triangle with 3 sides I will apply Heron’s formula, but I looked at this one and realised I would end up with a massive expansion of surds and quite possibly nested radicals and that would take a long time to work through by hand, so I considered other options. I thought a locigal first attempt would be to use trigonometry. I drew a sketch:

Labelled one if the angles x. I can then work out cos (x) fairly easily:

I did think I might get stuck at this point, with the limitation in the question prohibiting calculators, but actually it worked out really nicely. If I know cos (x) I can use a right angled triangle to work out sin (x):

Here I only have to square 1 (1) and the root of 26 (26) do a subtraction and take the square root of 25 (5) to complete my triangle using Pythagoras’s Theorem. All easily done without a calculator.Now I have sin x I can work out the area using area = absin(c)/2

When I started playing with the surds I wasn’t quite expecting it to fall out so nicely. Who’d have thought a triangle with 3 irrational side lengths would have such a nice integer area? I think I will explore this more when I get time. See how easy this type of triangle is to generate and if there is a rule to triangle of this sort occurring.

I said earlier that I didn’t think I’d seen this one before. But now I think about it, I think I may have. I think I remember giving it to a year 12 class maybe last year and some of them cheating and using the calculator but not getting the right answer as it had rounded the cosine and sine values, I will need to check this later on a calculator too. If this is the case, then I think this is a great discussion point when looking at rounding and why we shouldn’t round too early. I’m also left wondering how easy/difficult it would have been to get to the nice answer using Heron’s Formula, so I will give that a go in due course too.

## Incircles and semi-perimeters

Another day, another puzzle. This one also comes from the awesome Ed Southall via his Twitter account @edsouthall.

It has a circle and a triangle, two of my favourite shapes to play with, so I thought I would have a crack.

I drew out the sketch so I could mark information on and quickly spotted the radius of the circle. I drew some radii to meet the tangents as I know this gives me right angles.

I then considered this triangle, looking at it now I dont know why I chose this one. One.of the congruent ones on the base would have been in a more familiar orientation, but there you go. I know the radius, r, I know it’s a right angled triangle as it’s a tangent meeting a radius and this means I can use trig to work out the other sides.

I seem to have snapped this one before I wrote the 1 next to the y. This led me to the knowledge I have an equilateral triangle side length 2 so the area was simple.to calculate from there:

I could have used the height of course:

I could, of course, have done it the hero’s way:

All sides are length 2. So the semi perimeter is 3. So the area is (3*1*1*1)^1/2 = rt3.

While I was considering Heron’s formula I remembered that there is a link between the radius of a triangles incircle, the area of the triangle and the semi-perimeter.

The link between them is that the area of the triangle is equal to the product of its inradius and its semi-perimeter. In this case (1/rt3)*3 which equals rt3. Initially I thought this could lead to a much simpler solution, but then I realised I’d still have needed to calculate the side length to get the semi-perimeter.

*I like this puzzle. It combines shapes I love, surds, trigonometry and certainly got me thinking. I’d love to hear how you did it, especially if you did it a different way. Please let me know on social media or in the comments.*

## A surprising find

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:

And I happened across this:

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

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. *

## Area the Hero’s Way

This morning I posted this about a lovley simple triangle puzzle I had found and explored. One of the responses came from Mrs Watson (@MrsJMWatson) who tweeted this reply:

The link is to this page from Nrich and contains this puzzle:

Triangle T has sides 6,5,5 and Triangle Q has sides 8,5,5 What is the ratio Area T:Area Q?Now this is a great example if a “low barrier, high ceiling” problem. I think your first instinct are probably to go with what you are most familiar with. In today’s earlier problem the respondents who are used to working with the sine rule for area instinctively went for that, but others who are more used to working with other things went other ways.

For this problem my instinct was to use the cosine rule to find an angle, and then the sine rule for area ((1/2) absin(c)) to find the areas and simplify the ratios. My friend Steve, also a math teacher up to A level, actually worked through this, but didn’t need the sine rule for area as the ratio is easy to spot when you see the angles add up to 180 degrees (we’ll let him off for using degrees just this once).

I didn’t go down this route, I was thinking about generalising for all triangles and thought Heron’s Formula would be better for this.

I used it and found this lovely solution:

T) a=5,b=5,c=6 s=8Area: (8.3.3.2)^1/2=12 U) a=5,b=5,c=8 s=9

Area:(9.4.4.1)^1/2 = 12

So Area T: Area U = 12:12 = 1:1Discussing my method and answer with various people today I have been shocked at how many people (most it seems) haven’t heard of “Heron’s Formula”!

Heron’s FormulaHeron’s Formula is a fantastic piece of mathematics. I don’t know how I know about it. I have known about it so long it didn’t occur to me others wouldn’t. I guess I had assumed I learnt it at school, but if that was the case others would know about it. I’ve read and watched a lot about maths over the years, so I guess I must have picked it up from a book or show.

For those of you who don’t know, Heron’s Formula states:

For a triangle with side lengths a,b and cArea=(s(s-a)(s-b)(s-c))^1/2

where s is the semiperimeter (ie s=(a+b+c)/2)A truly marvellous formula. It’s named after Hero of Alexandria, who along with this formula is credited with being the first person to envisage imaginary and complex numbers.

Hero’s own proof is pretty cool, and involves cyclic quadrilaterals and properties of right angles triangles. There are lots of other proofs too, my favourite is the trigonometric proof, which I think would have been what I ended up with if I had decided to generalise this problem using trig!

Later in the discussion Mrs Watson said the phrase “Pythagorean Triples“, and I instantly saw both triangles could be cut into 2 3,4,5 triangles. I think this is the nicest solution. When I checked back to another discussion I noticed that Andy (@andycav_25) had also had this realisation. His instinct was to draw a perpendicular height. I wondered if I would have gone down this route if I’d had paper to work on instead if working mentally?!

Finally, Mrs Watson mentioned that her year 7 class do it via construction. She didn’t elaborate much on this, but I imagine she means draw both triangles accurately, measure the height, work out the areas. This is, in itself, quite nice, and shows that pupils can tackle the problem using whatever tools they have at their disposal.

This is a lovely problem, I’d love to hear how your instincts would tackle it, how your students for if you try it, and if you previously knew about Heron’s Formula!

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