Archive

Posts Tagged ‘Similarity’

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.

Rectangle Puzzle

December 27, 2014 Leave a comment

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

image

The picture shows a unit square with two congruent rectangles and a number of triangle. I looked,at the puzzle, saw no obvious answer, realised it would need a bit of thought so saved the photo until I found some time.

Yesterday I had some time, so I started by sketching the puzzle and filling in things I knew:

image

I quickly realised that the angles for all the triangle were the same, and thus they were all similar. (For some reason I missed one of the triangles on this page.) I couldn’t work out how I could find a missing length. I worked out the length if the diagonal of the rectangle using Pythagoras’s Theorem but it didn’t really help.

Then I drew some perpendiculars from each end of the diagonal and had a breakthrough.

image

If two rectangles share a diagonal, and a side length, then they must be congruent. This meant that I now knew the length of the bottom was 2x + 2y, as it’s a unit square then 2x + 2y = 1 so x+y=1/2.

I filled in these lengths:

image

Redrew my similar triangles and solved:

image

As no one likes an irrational denominator I rationalised as ended up with the satisfying solution 2-rt3.

I love this puzzle, and think it could be used as enrichment for high ability learners of many ages. After I’d solved it I looked at some solutions others had posted, and was thankful to see that other people also unnecessarily use trigonometry, and I was amazed I’d managed to avoid it, given my track record!

Similarity, and other stories

November 15, 2013 1 comment

Recently I have been looking into a variety of things. One thing is “Inquiry Maths” and another is something I found on the #matheme site through the explore the MTBoS project called “Notice and wonder”.

These got me thinking about how I could introduce some of their elements into some of my lessons. I had just introduced similarity to my year elevens and I was going to move to similar area and volume problems. So I came up with this starter:

image

I put it on the board and gave them ten minutes (I think) and let them get their teeth into it. A few were a little confused at first, but the discussions on each table enabled all pupils to make their own way to the correct answer. I didn’t know what they would notice or wonder, but I was pleasantly surprised to hear some of their comments:

“I notice that the area has gone up by four, not two. Does that mean you double the scale factor for area?”

-I loved this one, and refused to answer it, instead I asked him to enlarge the shape sf3 and see if the area was enlarged by six. I then got:

“It’s nine, not six. Why’s it nine? Stupid thing. Oh hang in, it’s squared. Oh, of course it’s squared! you times each side by it [the scale factor] and you times them together! Duh!”

Others I particularly liked were:

“I wonder if there’s a way of doing Pythagoras on triangles without right angles” (I told her that we would be meeting the cosine rule soon enough).

“the angles are the same! Wait, that’s how this SOHCAHTOA thing works isn’t it, cos it’s ratios an that.” (I said “very good, but can we use the proper name please!” then another pupil interjected with “Trigonometry”)

The lesson goes on to pose question prompts similar to those I’ve seen on inquiry maths in which we discussed similar volume and then I included a set of questions for then to attempt. I have uploaded the resource to TES:
http://www.tes.co.uk/teaching-resource/Similarity-and-other-stories-6374388/

%d bloggers like this: