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Symmetry and reflection

July 17, 2019 Leave a comment

This week I’ve been working on symmetry and reflection with my year 9s. As part of this we looked at creating some Rangoli style patterns that had symmetries. After discussing how to reflect shapes in mirror lines and drawing some reflections I shower them some pictures of Rangoli patterns and gave out some squares of squared paper with various mirror lines on and some coloured pencils and gave them some time to get to work.

I was interested to see some of the different approaches. Most just used the squares to create patterns in the first instance. Of these there were two basic approaches, A – complete a pattern in one quarter and reflect into the other quadrants. B – each time they coloured a square they reflected it. I would have opted for option B I think, but watching the class work it seems those who used method A made less mistakes.

There were a couple of students who didn’t just colour squares, they created some triangles, trapeziums and other shapes in their patterns. They all took approach A.

I’d be interested to hear how you would approach this.

Once we had some patterns we looked at them as a class on the visualiser, discussing their approaches, what we liked about each one finding mistakes, looking at the reflective symmetries and also discussing rotational symmetry too.

Here are some of the patterns:

This student did a quadrant at a time then shaded what was left in blue when I told them time was nearly up.

This student was doing 4 squares at a time. I like the fact each quadrant is symmetrical too. (I know technically it’s not a symmetrical pattern, but it would.have been had we had a minute or 2 more!)

This student did a bit at a time and reflected each bit in the diagonal line.

I liked all three of these, and they gave rise to a good discussion as they had all gone for more than 2 lines of symmetry and all had rotational symmetry too. More than half of the students had done this, and I wonder if that’s because the Rangoli patterns I had shown them had also done this.

Unfortunately a lot of the others opted to take their work home and I didn’t think to take photos first as there were some awesome patterns.

Categories: #MTBoS, SSM Tags: , , ,

Constructions

July 12, 2019 4 comments

One topics I have never been a fan of teaching is constructions. I think that this is due to a few factors. Firstly, there is the practical nature of the lesson, you are making sure all students in the class have, essentially, a sharp tool that could be used to stab someone. I remember when I was at school a pair of compasses being used to stab a friend of mines leg and this is something I’m always wary of.

Secondly, the skill of constructing is one that I struggled to master myself. I was terrible at art, to the point where an art teacher kept me back after class in year 8 to ask why I was spoken about in the staffroom as the top of everyone else’s class but was firmly at the bottom of his. I explained that I just couldn’t do it, although it was something I really wished I could do. He was a lovely man and a good teacher and he offered to allow me to stay back every Monday after our lesson and have some one to one sessions. I was keen and did it, this lasted all through year 8 and although my art work never improved my homework grades did, as he now knew I was genuinely trying to get better. I have always assumed the reason I am poor at art is some unknown issue with my hand to eye coordination, and I have always blamed this same unknown reason for struggling sometimes with the technical skills involved in constructions. Since coming into teaching I have worked hard to improve at these skills, and I am certainly a lot lot better than I used to be, but I still feel I have a way to go to improve.

For these reasons I chose to go to Ed Southall’s (@solvemymaths) session “Yes, but constructions” at the recent #mathsconf19. Ed had some good advice about preparation and planning, but most of that was what I would already do:

Ensure you have plenty of paper, enough equipment that is in good working order, a visualiser etc.

Plan plenty of time for students to become fluent with using a pair of compasses before moving on.

He then moved on to showing us some geometric patterns he gets students to construct while becoming familiar with using the equipment. Some of these were ones I’d not considered and he showed us good talking points to pick out and some interesting polygons that arise. The one I liked best looked like this:

This is my attempt at it, I used different coloured bic pens in order to outline some of the shapes under the visualiser.

The lesson was successful, the class can now all use a pair of compasses and we managed to have some great discussions about how we knew that the shapes we had made were regular and other facts about them.

Next week we need to move on to looking at angle bisectors, perpendicular bisectors, equilateral triangles, and the such. I hope to get them constructing circumcircles of triangles, in circles of triangles and circles inscribed by squares etc.

Here are some more of my attempts at construction:

“Constructing an incircle” – I actually did this one in Ed’s session!

“A circumcircle” – I drew the triangle too big and the circle goes off the page. Interesting to note the centre is outside the triangle for this one.

“A circle inscribed within a square” – this is difficult. Constructing a square is difficult and that is only half way there if that. This is the closest I have got so far and two sides are not quite tangent.

“A flower” – nice practice using a pair of compasses and this flower took some bisectors too.

If you have any ideas for cool things I can construct, and that I can get my students to construct, please let me know in the comments or on social media.

Saturday puzzle

December 2, 2017 2 comments

One of the first things I saw this morning when I awoke was this post from solve my maths on facebook:

That’s interesting I thought, so I thought it have a go.

The radius is given to be 2. So we have an equilateral triangle side length 2. Using my knowledge of triangle and exact trig ratios I know the height of such a triangle is root 3 and as such so is the area. 

Similarly, as the diagonal of the rectangle is 2 and the short side is 1 we can work out from Pythagoras’s Theorem that the longer side is root 3. And again it follows that so is the area.

Lastly we have the square, the diagonal is 2 and as such each side must be root 2, again this is evident from Pythagoras’s Theorem  this gives us an area of 2.

Which leaves us a nice product of the areas as 6.

I think that is correct, I’ve justvwoken up nd this post has been my working, so do about up if you spot an error. And I’d love to hear if youbsolved it a different way.

Oblongs

November 10, 2017 5 comments

Last week while we were waiting for a swimming lesson to start my daughter told me that one of her teachers had got “higgledy piggledy” about oblongs. I asked what she meant and she said that she’d accidentally called one a rectangle and had to correct herself and had informed the class that at her last school she’d had to call them rectangles but at this school had to call them oblongs and sometimes got higgledy piggledy about this. I asked my daughter why they couldn’t call them rectangles and she said that it was because squares can be rectangles too.

This set off a lengthy chain of thoughts in my head. Firstly, I was quite impressed by the fact a 5 year old could articulate all this about knowledge about shapes so well. Then I thought, does it really matter whether they call them oblongs or rectangles? Then I thought, wait a minute, why are we prohibiting the use of rectangle because it can also mean a square, but we are not prohibiting the use of oblong when it can also mean an ellipse? My chain of thought then jumped down a rabbit hole questioning whether we should actually be referring to regular or equilateral rectangular parallelograms and non – regular/equilateral parallelograms. Why are we allowing children to call a shape a triangle, when it is one possible type of triangle in a family of triangles, but not allowing them to call a shape a rectangle when it is only one possible rectangle in a family of rectangles. These thoughts stewed around in my head for a while and I thought I’d ask the twittersphere for their opinions on the matter.

These opinions fell into a couple of camps. The first cam thought that oblong was a nice enough word and they didn’t mind others using it but preferred not to themselves. The second camp felt that it was important to distinguish between an oblong and a square so important to use oblong not rectangle and the third camp thought that actually it was better to use rectangles due to the elliptical oblongs. I questioned some of the respondents from the second two groups a little further to see why they fell into these groups. Those in the second seemed unaware that the word oblong also meant ellipse and those in the third thought it was more important to excluded ellipses than squares. Stating that it was easy enough to explain away the special case that is the square.

I’ve spend rather a lot of time considering this, and am now not really sure what I think on the issue. I can’t see a problem with using a rectangle and explaining away the square as a special case. We call all triangles triangles and expand as and when required. No one bothers about calling a non-rectangular parallelogram a parallelogram, despite the fact that that could mean a rectangle. But again I’m not sure I’m massively strongly against the term oblong either. It could open up a good discussion about the term and how it could apply to ellipses, although this probably is a little too much for a year 1 classroom. I think I’m leaning towards rectangle as a preference though, as explaining away a special case is, for me, much more preferable than ignoring a whole class of oblongs.

 

If you have views on this, whichever way you lean, I’d love to hear them, either in the comments or via social media.

Categories: #MTBoS, KS2, KS3, Maths, SSM, Teaching Tags: , ,

An interesting area puzzle 

July 5, 2017 12 comments

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 dw:

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!

Is one solution more elegant?

May 14, 2016 13 comments

Earlier this week I wrote this post on mathematical elegance and whether or not it should have marks awarded to it in A level examinations, then bizarrely the next day in my GCSE class I came across a question that could be answered many ways. In fact it was answered in a few ways by my own students.

Here’s the question – it’s from the November Edexcel Non-calculator higher paper:

image

I like this question, and am going to look at the two ways students attempted it and a third way I think I would have gone for. Before you read in I’d love it if you have a think about how you would go about it and let me know.

Method 1

Before I go into this method I should state that the students weren’t working through the paper, they were completing some booklets I’d made based on questions taken from towards the end of recent exam papers q’s I wanted them to get some practice working on the harder stuff but still be coming at the quite cold (ie not “here’s a booklet on sine and cosine rule,  here’s one on vectors,” etc). As these books were mixed the students had calculators and this student hadn’t noticed it was marked up as a non calculator question.

He handed me his worked and asked to check he’d got it right.  I looked, first he’d used the equation to find points A (3,0) and D (0,6) by subbing 0 in for y and x respectively. He then used right angled triangle trigonometry to work out the angle OAD, then worked out OAP from 90 – OAD and used trig again to work out OP to be 1.5, thus getting the correct answer of 7.5. I didn’t think about the question too much and I didn’t notice that it was marked as non-calculator either. I just followed his working, saw that it was all correct and all followed itself fine and told him he’d got the correct answer.

Method 2

Literally 2 minutes later another student handed me her working for the same question and asked if it was right, I looked and it was full of algebra. As I looked I had the trigonometry based solution in my head so starter to say “No” but then saw she had the right answer so said “Hang on, maybe”.

I read the question fully then looked at her working. She had recognised D as the y intercept of the equation so written (0,6) for that point then had found A by subbing y=0 in to get (3,0). Next she had used the fact that the product of two perpendicular gradients is -1 to work put the gradient of the line through P and A is 1/2.

She then used y = x/2 + c and point A (3,0) to calculate c to be -1/2, which she recognised as the Y intercept, hence finding 5he point P (0,-1.5) it then followed that the answer was 7.5.

A lovely neat solution I thought, and it got me thinking as to which way was more elegant, and if marks for style would be awarded differently. I also thought about which way I would do it.

Method 3

I’m fairly sure that if I was looking at this for the first time I would have initially thought “Trigonometry”, then realised that I can essential bypass the trigonometry bit using similar triangles. As the axes are perpendicular and PAD is a right angle we can deduce that ODA = OAP and OPA = OAD. This gives us two similar triangles.

Using the equation as in both methods above we get the lengths OD = 6 and OA = 3. The length OD in triangle OAD corresponds to the OA in OAP, and OD on OAD corresponds to OP, this means that OP must be half of OA (as OA is half of OD) and is as such 1.5. Thus the length PD is 7.5.

Method 4

This question had me intrigued, so i considered other avenues and came up with Pythagoras’s Theorem.

Obviously AD^2 = 6^2 + 3^2 = 45 (from the top triangle). Then AP^2 = 3^2 + x^2 (where x = OP). And PD = 6 + x so we get:

(6 + x)^2 = 45 + 9 + x^2

x^2 + 12x + 36 = 54 + x^2

12x = 18

x = 1.5

Leading to a final answer of 7.5 again.

Another nice solution. I don’t know which I like best, to be honest. When I looked at the rest of the class’s work it appears that Pythagoras’s Theorem was the method that was most popular, followed by trigonometry then similar triangles. No other student had used the perpendicular gradients method.

I thought it might be interesting to check the mark scheme:

image

All three methods were there (obviously the trig method was missed due to it being a non calculator paper). I wondered if the ordering of the mark scheme suggested the preference of the exam board, and which solution they find more elegant. I love all the solutions, and although I think similar triangles is the way I’d go at it if OD not seen it, I think I prefer the perpendicular gradients method.

Did you consider this? Which way would you do the question? Which way would your students? Do you tuink one is more elegant? Do you think that matters? I’d love to know, and you can tell me in the comments or via social media!

Cross-posted to Betterqs here.

Exact Trigonometric Ratios

November 8, 2015 4 comments

This morning I read this interesting little post from Andy Lyons (@mrlyonsmaths) which looked at teaching the exact Trigonometric Ratios for certain given angles (namely 0, 30, 60, 90 and 180 degrees). The post gave a nice little info graphic linked to the unit circle to show what was going on and then focused on methods yo remember the ratios.

While reading it I thought about how I introduce these exact Trigonometric Ratios. I first like to know that my students have a thorough and in depth understanding of right angled triangles and the trigonometry involved with them (including Pythagoras’s Theorem). I feel this is imperative to learning mathematics, the Triangle is an extremely important shape in mathematics and to fully understand triangles you must first fully understand the right angled triangle. The rest follows from that.

Once these are understood then you can move on to the trigonometric graphs, showing how these can be generated from right angled triangles within the unit circle, as shown in the info graphic on Andy’s post. Once the graphs are understood then the coordinates f the x and y intercepts and the turning points give us nice exact values for angles of 0, 90 and 180 degrees. This leaves us with 30, 60 and 45 to worry about.

At this point I introduced 2 special right angled triangles. First up is the right angled isosceles triangle with unit lengths of the short sides. This obviously gives us a right angled triangle that has two 45 degree angles (as the angle sum of a triangle is 180) and a hypotenuse of rt2 (via Pythagoras’s Theorem).

image

Using our definitions of trigonometric ratios (ie sin x =opp/hyp, cos x = adj/hyp and tan x = opp/adj) we can clearly see that tan 45 = 1 and that sin 45 = cos 45 = 1/rt2. This aids the understanding more than just giving the values and allows students a method of working these values out easily if stuck.

The second triangle is an equilateral triangle of side length 2 cut in half. This gives us a right angled triangle with hypotenuse 2, short side lengths 1 and rt3 (again obtained through Pythagoras’s Theorem) and angles 30, 60 and 90.

image

Again we can use our definitions of trigonometric ratios to conclude that sin 30 = cos 60 = 1/2, sin 60 = cos 30 = rt3/2, tan 30 = 1/rt3 and tan 60 = rt3.

This is again good for deeper understanding and for seeing why sin x = cos 90 – x, and cos x = sin 90 -x. This can lead to a nice discussion around complementary angles and that the word cosine means “sine of the complementary angle”. This triangle is also a good demonstration that tan x = cot 90 – x, when you come to higher level trig.

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