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Posts Tagged ‘geometry’

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.

Area 48

July 10, 2019 Leave a comment

Today I was looking at some of Ed Southall’s (@solvemymaths) puzzles on his website. I saw this one that I had not seen before:

I thought I’d give it a crack. You should too….. go on…. Did you get an answer? Well here is how I approached it:

First I did a little sketch, as I always tell my students to do:

I labelled the points with letters as this is normally quite a good way of keeping track of things.

I then decided to let AB = 1 (I chose that bit to be 1 as I knew a unit square would lead to lots of fractions, in hindsight this also let to fractions and AB = 2 would have been better.)

This gave me a few lengths straight off the bat, and I could find BD by Pythagoras’s Theorem and hence had the area of the larger square – which I need to answer the question.

I also noticed I had a RAT (ABD) and I knew the perpendicular sides, and therefore could work out the area.

I then looked at the triangle BCH. This looked like it would be similar to ABD but I took a couple of moments to justify it to myself before moving on, just in case….

If angle ABD is x then as DAB and BCH are both 90 and the angle sums of a triangle and on a straight line are both 180 then CBH and BDA must both equal 90 – x and CHB must equal x, hence they are similar.

They are similar and the scale factor is 2 (as BC is half of AD and they are corresponding sides).

Hence the Area scale factor is 4 and the area pf BCH is a quarter of the area ABD. As Area ABD = 1 then Area BCH = ¼.

From here I took the area of the two triangles away from the area of the square ACED to get the shaded area and put it over the area of the larger square. (Well, after momentarily putting it over the area of the smaller square like a fool!).

So here I had an answer, 11/20. I clicked on the comments on Ed’s website and saw some answers that were not what I had. This had me second guessing myself, so I thought about a different approach.

I went for a coordinate geometry approach (coordinate geometry seems to have taken over from trig as my brains go to method).

I chose the origin as the common corner of the two squares and called the point where the vertex meets the horizontal point B. This mean B’s coordinates were (1,2). I called this line l1 and could spot its equation was 2x. Part of the shaded area is the area under this curve between x = 0 and x = 1 so I calculated that area to be 1.

The perpendicular through B is the other line that bounds the top of our shaded region. I know the perpendicular gradiens multiply to -1 and I know it goes through point (1,2) so I could work out the equation of this line easy enough:

Then calculate the area below it between the values x = 1 and x = 2. This gave an area of 7/4.

So I had a total shaded area of 11/4 and could divide this by the area of the large square to get 11/20 again.

I felt happier now that I had the same answer though two different methods, and I stress to my students that this is what they should be doing with any extra time in exams. Doing different methods and seeing if they get the same answer!

I hope you tried Ed’s puzzle, and if you did, please let me know how you approached it.

Nice area puzzle

April 23, 2019 2 comments

Yesterday evening I came across this lovely area puzzle on twitter:

The puzzle is from Gerry McNally (@mcnally_gerry) he says its his first, and I hope that’s “first of many”.

I reached for the nearest pen and paper and had a quick go:

As you can see, I misread the puzzle originally and thought the lower quadrilateral was a square. The large triangle is isosceles as given in the question. This allowed me to use the properties of similar triangles and the base lengths given to work out the areas of the square, both right angled triangles and the whole triangle. This then allowed me to calculate the area of the shaded quadrilateral and hence that area as a fraction of the whole.

Then I went to tweet my solution to Gerry and realised that nowhere does it say that the bottom quadrilateral is a square. I had added an assumption. This made me ponder the question some more. Instincts told me that it didn’t have to be a square, but that the solution would be the sane whether it was a square or not. But I didn’t want to leave it at that, I wanted to be sure, so I had another go.

I sketched out the triangle again:

This time I called the height of the rectangle x.

This made it trivial to find the area’s of the rectangle and the triangle GCD. Triangle HAB was easy enough to find using similar triangle properties.

and then I found the area of the whole shape again using similarity to discover the height.

This allowed me to find the shaded area:

Then when I put it as a fraction the xs cancelled and it of course reduced to the same answer.

I really like this puzzle, and would be interested to see how you approached it, please let me know in the comments or on social media.

Angle problem

December 5, 2017 2 comments

Today has been quite a geometric based day for me. I spent a couple of hours solving non-RAT trigonometry problems with year 10 and then a while with year 11 looking at various algebra angle problems. Then I went on Twitter and saw this from Ed Southall (@solvemymaths):

A couple of nice parallel lines questions that I might grow at y11 tomorrow.

Both are fairly straight forward to solve. I looked at the first one, imagines a third parallel line through the join if x and saw x must be the sum of 40 and 60 hence 100.

The second I saw an alternate angle to the 50 in the top triangle and used angle sum of a triangle is 180 to spot that x is a right angle. I glanced down at the responses and saw the vast majority had the same answers as me. That would probably have been the end of it but then I noticed this response:

The same thought process for the first one, but a significantly different approach to the second.

It made me wonder what approach others would take, and which approach my students would take. I wondered if the first problem had led this respondent into this solutions the second, and if so why it hasn’t had the same effect as me.

I don’t know if either approach is better, I just thought the differences were interesting. I’d love to hear your thoughts on it and how you would approach it.

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.

An excellent puzzle – alternate methods

July 19, 2017 2 comments

Yesterday I wrote this post looking at a nice puzzle I’d seen and how I solved it.

The puzzle again:

Lovely, isn’t It?

After I published my previous post I wondered if I may have been better using a vector approach or a coordinate geometry approach. So I gave them a try.

Coordinate Geometry

I started by sketching the figure against an axis.

I place the origin at the centre of the circle, worked out the equation if the circle and the right leg of the triangle and solved simultaneously for x. Giving x =1 and x=1/3. These x values correspond to half the base of each triangle, which shows the scale factor from the large triangle to the small one is 1/3. As the area of the large one is rt2 this gives the area of the small as rt2/9.

I like this method, probably a little better than the one prior to it.

Vectors

First I sketched it out and reasoned I could work it out easy enough with 4 vectors.

I saw that I could write AC as a sum of two others:

I knew that the length of AC was 1 so I used Pythagoras’s Theorem to calculate mu. It left me with the exact same quadratic to solve. This time mu was the fraction of DB needed so was automatically the length scale factor. The rest falls out as it did before.

As well as this, Colin Beveridge (@icecolbeveridge), maths god and general legend, tweeted a couple of 1 tweet solutions. First he used trig identities:

Trig Identities 

I assumed this was right, but checked it through to ensure I knew why was going on:

We can see beta is 2 x alpha and as such the tan value is correct. The cos value (although it is missing a negative sign that I’m sure Colin missed to test me) follows from Pythagoras’s Theorem:

This is again the scale factor as it is half the base of the small triangle and the base if large triangle is 2.

Complex Numbers

Then Colin tweeted this:

At first I wasn’t totally sure I followed so I asked for further clarification:


I had a moment of stupidity:

And then saw where Colin was going. I tried to work it through, by way of explaining here in a better manner.

I sketched it out and reasoned the direction of lines:

Then I normalised that and equated imaginary parts to get the same scale factor:

I am happy that is is valid, and that it shows Colin is right, but I’m not entirely sure this as the exact method Colin was meaning. He has promised a blog on the subject so I will add a link when it comes.

I like all these methods. I dontvthink I would have though of Colin’s methods myself though. I’d love to hear another methods you see.

A little circle problem 

June 28, 2017 Leave a comment

I’ve just seen this post from Colin Beveridge  (@icecolbeveridge) answering this question:

Naturally I had a go at it before reading Colin’s solution. When I read his I found a lovely concise solution that we slightly different to mine, so I thought I’d share mine.

I started by just drawing a right angled triangle from the centre of the circle like so.

I seem to have cut off the denominator of 6 on the angle there. I know that the hypotenuse is 12-r (where r is radius) and the side opposite the known angle is r.

This means I can use the sine ratio of pi/6 to get  

r/(12-r)  = 1/2

Which leads to:

2r = 12 – r

Then 

3r = 12

r = 4

Which is the same as Colin got. 

I’ve seen questions like this on A level papers before and I know they often throw students, so I make sure I explore lots of geometry based problems and puzzles to combat this. I’d be interested to know which way you would approach this. Colin and I used a very similar approach, just differing in the point at which we introduced the 12. Which way did you do it? 

Categories: A Level, Maths Tags: , , ,
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