## An interesting area puzzle

Regular readers will know that I have a tendency to collect puzzles and I like to have a go at them. This evening I had a crack at this while my daughter was playing before bed.

It looked interesting when I saw it on Facebook a while ago and so I thought I’d have a crack. If you haven’t yet, do it now. I want to know if you took the same approach!

I wasn’t too sure where to start, so I drew it out, labelled some stuff and came up with some equations:

I thought if I multiples two of my equations together and rearranged I could get the yellow area as the subject:

Then I needed cx:

I thought “now I need ay”, then realised I had it:

This meant I could.sub back in through the equations:

So the area:

I thought 27 was a nice answer, and I’m fairly sure it’s correct, however I have a feeling that I may have missed something blindingly obvious that would have gotten me there much quicker.

*If you did it, I’d love to hear your approach, especially if you spotted something I missed!*

## An interesting area problem

Here’s an interesting little question for you:

*Have you worked it out? How long did it take you to see it?*

It took me a few seconds at least, I had screenshotted the picture and was reaching for the pencil when the penny dropped, and that’s why I thought it was an interesting question.

The answer is, of course, 100pi. This follows easily from the information you have as the diagonal of the rectangle is clearly a radius – the top left is on the circumference and the bottom right is on the centre.

*So why didn’t I spot it immediately? *

I think the reason for me not spotting it instantly might be the misdirection in the question, the needless info that the height of the rectangle had me thinking about 6, 8, 10 triangles before I had even discovered what the question was.

I see this in students quite often at exam time, they can get confused about what they’re doing and it links to this piece I wrote earlier about analogy mistakes. The difference is I wasn’t constrained by my first instinct but all too often students can be and they can worry that it must be solved in the manner they first thought of.

Earlier today a student was working on an FP1 paper and he was struggling with a parabola question, he had done exactly this, he had assumed one thing which wasn’t the right way and got hung up in it. When he showed me the problem my instinct was the same as his, but when I hit the same dead end he had I stepped back and said “what else do we know”, then saw the right answer. I’m hoping that by seeing me do this he will realise that first instincts aren’t always correct.

I’m going to try this puzzle on all my classes tomorrow and Friday and see if they can manage it!

*How quickly did you see the answer? Do you experience this sort of thinking from your students? I’d love to hear any similar experiences.*

Cross-posted to Betterqs here.

## 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.*

## Area Puzzle – Squares and Circles

This morning I came across this puzzle from Ed Southall (@solvemymaths):

And it seemed interesting. Looking at it is seems that A is the midpoint of the arc, so equidistant from b and the lower left corner of the square. This would mean that an isosceles triangle exists in the semi-circle, which in turn implies the square has side length 4rt2 (from Pythagoras’s Theorem, as the side of the square is a diameter).

This means the square has area 32cm^2. The semi-circles have radius 2rt2 and as such have area 4pi. That leaves just the white circle. The diagonal of the square is a diameter of this, and as the side lengths are 4rt2 the diagonal must be 8 (again via Pythagoras’s Theorem). This means the circle must have a radius of 4 and hence an area of 16pi.

So the sum of the orange areas must equal the square add the 4 Semi-Circles subtract the circle add the square.

Or:

2(32) + 4(4pi) – 16pi

Which simplifies to 64cm^2.

A lovely solution, and one which shows us that the areas of the orange crescents equal the area of the square.

Towards the end of the working I realised I could have used another property of Pythagoras’s Theorem, namely that the sum of the areas of semi-circles on the two shorter sides equals the area of the semi-circle on the hypotenuse. By splitting the square into two right angled triangles I could have reasoned that the 4 smaller semi-circles provide the same area as the large circle. Which means we would get the total area to be the area of the square, add the area of the circle (from the semi-circles) subtract the area of the circle (from the circle) add the area of the square. Which again simplifies to twice the area if the square. A much more elegant solution.

## A Puzzling Heptagon

At somepoint Ed Southall (@solvemymaths) posted this heptagon puzzle:

I saved it in my phone to solve later, and forgot about it until the other day.

At first I looked at it and wasn’t sure where to start. I attacked the problem by sketching what I knew:

The heptagon is split along the base of the red shape into an isosceles trapezium and a pentagon. Because I know the interior angle of a heptagon (900/7) I know enough about this trapezium to work out the other angles (51 3/7).

This means I can deduce the angle BAE (77 1/7):

Because the heptagon is regular I know that AE = BE, and as such ABE = BAE. This means I can use the angle sum of a triangle and the sine rule to calculate each angle and each side of the triangle.

From here I could calculate the area of the triangle ABE either using absin(c)/2 or Heron’s Formula. I chose Heron’s Formula:

I know that triangles ABE and ADE are congruent so the area I’m looking for is double this area subtract the area of the overlap.

I considered the Pentagon above the line AE and how the triangles split it up into 4 triangles and a rhombus. I briefly considered the rhombus:

And quickly realised that using the rhombus properties and opposite angles I had enough information to calculate the area of the overlap:

First calculating the height, then the area:

Then I could find the area I was looking for:

*I thought this was a great puzzle, and it got me thinking about which of my students would be able to attack it. The skills needed are all skills that higher level GCSE students should have, and I think that some of my year ten class would give it a good go, but I also worry that some of my sixth formers may struggle with it. I think by exposing students to these problems early, and by sharing our own thought processes, we can start to build the resilience and mathematical thinking needed to succeed.*

*When I first attempted this problem it was late, and I made some daft errors transfering working from one line to another, and got a very wrong answer, this showed me that it’s easy to do, and we need to make sure we reiterate often the importance of checking our work to our students.*

*There was also the problem of rounding, as we were dealing with angles that didn’t give us nice exact trig ratios I had to round, through my working I rounded where it seemed sensible, but all these errors would have built up so I decided 1dp would be a good limit to round my final answer to, although now I feel nearest whole number would have been better.*

*I’m fairly sure that there is a much better and more concise way of solving this problem, but I currently can’t see it. If you do spot one, or solve it an even more long winded way, please let me know.*

## A nice area puzzle

At some point over the last few days Danny Brown (@dannytybrown) tweeted this lovely puzzle out:

While waiting in the car this afternoon I saw the screenshot I’d taken of it and had a think about it. Initially I’d made a daft mistake with the base of a triangle and got the wrong answer (5:6 if you were wondering), but after a rethink and some in head workings I got to an answer I’m now happy with. I have since written the solution down and want to write about it here.

Firstly, when attacking problems like this I like to sketch them out. This step is one many students are reluctant to take, and I try to drill it into them that sketching is always helpful in understanding problems. If I’d had a pen and paper handy when I’d attacked this problem originally I’d have sketched it and not made my daft error.

From the sketch it is easy to see the area of shape A, this was easy to visualise.

In my working I then sketched shape B:

When I visualised this I made my daft error, which was to assume the triangle had base x. From the sketch it’s easy to see that it’s not, and it will need calculating, for this I used Pythagoras’s Theorem:

To calculate the area of the triangle I’d need an angle, so I thought about what I knew, the angle at E is made from a fold along EF so the angle must be equal to DEF.

Using the tan ration of the right triangle I got tan (theta) to be rt3.

Then it was a case of calculating the height and then the area:

The working written out properly looks a lot more thorough than my phone jottings:

*This is a nice puzzle that should be accessible to higher GCSE students and definitely A Level Students, but I worry that most would give up. We need to be giving our students the tools to unlock this sort of problem. I’m not sure how we can do that explicitly. I set these tasks, give them hints and then walk them through my thinking to model how I would attack them, and this has a great effect for many. But I wonder if this is enough.*

## “Next Level” question

Back in November I wrote this post outlining the changes to the new GCSE maths curriculum and included some questions I had enjoyed from the SAMs and some that I had thought were less good.

One of my Y13 students had read the post and had had a go at some of the questions, he came to me in school and told me that he’d seen a “next level” question on my blog, that it had taken him ages to realise how to do it and that he couldn’t believe it was on the new foundation GCSE. I asked him which question it was and he said it was this one:

I considered the question. A shows a sector of a circle with an angle of pi/2 radians. *I should say 90 degrees I suppose, as it’s a GCSE question.* B shows the **same square** but this time there are 4 sectors, each with the same angle, again pi/2 radians *90 degrees*. There are no lengths marked on.

This question is relatively straightforward once you pick up the key piece of information, that the squares are the same. You don’t even need to use the angles to work out the sectors as we are dealing with as they are quarter circles. You assign a value for side length and work through each one, showing the areas are the same.

I think it’s a great question, and I love the fact that these two shading arrangements give the same area. I did, however, wonder how future foundation students would cope with it given that a Y13, *who scored a good A in maths AS and As in his other 3 ASs in Y12,* referred to it as “Next Level”.

When discussing it with said student it turned out that the maths wasn’t the issue, it was that there was no lengths marked on, and he hadn’t realised straight away that you could assign a variable to work though.

I assume this is down to the nature of the current examinations, and their tendency to give scaffolding all the way through. I like the rigour of the new qualifications, and the fact they are designed to build mathematical thinking all the way through, but I fear that for some teachers ensuring this is built in will take some getting used to.