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Posts Tagged ‘Starters’

An interesting area puzzle 

July 5, 2017 11 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!

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A puzzle with possibilities

December 9, 2015 Leave a comment

Brilliant’s Facebook page is a fantastic source of brain teasers, they post a nice stream of questions that can provide a mental work out and that I feel can be utilised well to build problem solving amongst our students.

Today’s puzzle was this:

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It’s a nice little question. But when I use it in class I will only use the graphic, as I feel the description gives away too much of the answer. Without the description students will need to deduce that the green area is a quarter of a circle radius 80 (so area 1600pi) with the blue semicircle radius 40 (so area 800pi)  removed, leaving a green area of 800pi.

I find the fact that the area of the blue semi circle is equal to the green area is quite nice, and in feel that with a slight rephrasing the question could really make use of this relationship. Perhaps the other blue section could be removed or coloured differently and the question instead of finding the area could be find the ratio of blue area to green area.

Another option, one I may try with my further maths class on Friday,  could be to remove the other blue section and remove the side length and ask them to prove that the areas are always equal, this would provide a great bit of practice at algebraic proof.

Can you think of any further questions that could arise from this? I’d love to hear them!

This post was cross-posted to the blog Betterqs here.

Problem solving triangles

December 7, 2015 Leave a comment

Brilliant – a lovely puzzle app and a source of many little puzzlers if you follow their Facebook page. The other day, I came across this one:

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It looked like it might be interesting so I screen shot it and thought, “I’ll have a go at that later, when I’ve got a pen. It’s bound to be nice using a bit of trigonometry and angle reasoning.”

But as I thought about it I realised I didn’t need paper. The hypotenuse of the large triangle is easy enough to find (6rt2) using Pythagoras’s Theorem. You can deduce the size of the green square is then 2rt2 as the big triangle is isosceles meaning the angles are 90, 45 and 45, as the square is only right angles then the little blue triangles in the 45 degree corners must also be isosceles. Thus the two blue and the green segments of the hypotenuse are equal.

The area of the square is then way to find (8) by squaring 2rt2. A nice easy puzzle.

My first thought had been that it would take a bit of working out, but it didn’t,  it was a very straightforward question once I got going. That got me thinking, problem solving is something that I would love my students to get better at and I’m hoping to launch a puzzle of the month in January. This sort of puzzle is ideal. It will require then to build their perseverance skills as well as their problem solving skills and will give them a mental workout. I’m going to use this as a starter this week to warm them up.

This post was cross posted to the BetterQs blog here.

Equal Products

April 23, 2015 2 comments

I come across a lot of puzzles and other maths things online and often save them for later, this evening I came across this little puzzle:

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The numbers 2,3,12,14,15,20 and 21 may be divided into two sets so that the product of the numbers in each set is equal. What is that product?

I had saved this over a year ago, and cannot remember where I got it from, but I can see why. It’s a lovely little question that I intend to use as a starter next week and see how my classes get on.

How I approached it

Before you read on have a go at it yourself. Go on, you know you want to……..

Right, good, now you can see if I went about it the same way!

My first thought was that all the fun could be taken out of this by using a calculator, typing all the numbers in and pressing square root. So when I set it I will be adding the line “and which numbers are in each set”, and this is what I set out to find.

Firstly I set out the numbers in terms of their prime factors:

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Then I tallied up the prime factors:

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From this I knew that the product must be 2x2x2x3x3x5x7 which is 2520.

This, of course, answers the original question but I wanted each set. I looked at the numbers and the first think I noticed was tgat 14 and 21 had to be in separate sets, as they had 7 as a factor. I also needed to split 15 and 20, my intuition suggested that 20 and 21 should be in separate boxes, but it was easy to spot that the 2s and 3s, wouldn’t work out so I placed them together and fit the rest on around them.

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A nice little puzzle, I wonder how my classes will find it.

Trigonometry Puzzle

February 2, 2015 Leave a comment

Chris Smith (@aap03102) produces a weekly Maths newsletter, which is fantastic. If you don’t already subscribe get in touch with him and get on the list. One of my favourite parts of the newsletter is the puzzle of the week, and this week’s was a particular nice one:

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Have you solved it yet? If not have a think before reading on….

I solved this within a few seconds, but still appreciate how nice a puzzle it is, and think I may only have solved it so fast because of a conversation I’d had Friday morning. On Friday morning I’d been teaching mechanics, and one question on resolving forces had an answer “Sin x”. When I revealed the answer one of my students said “I guess that would work as well.” Intrigued I ventured over and saw he had the answer “Cos (90 – x)”. This led to a nice class discussion of this property, a proof of it and the class attempting that question from THAT C3 exam. All this meant that this result was at the forefront of my mind, so when I saw the puzzle I immediately realised that

Sin^2(1) + sin^2(89) = Sin^2(1) + cos^2(1) = 1

And so on for 44 pairs, this just leaves sin^2(45), which is 1/2 as sin45=1/rt2.

This gives a nice final answer of 44.5, which Chris promptly confirmed. A lovely little puzzle.

Quadrilateral Puzzle

January 10, 2015 Leave a comment

Yesterday the fantastic Ed Southall (@solvemymaths) tweeted this brilliant puzzle:

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It looked fun, so I thought I’d give it a try. First I sketched it out and gave all the vertices labels. A strategy I advise my students to take.

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I then considered triangle wzg, as it was the triangle I knew most about:

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My first thought was to find the length of the hypotenuse using Pythagoras’s Theorem. This was something that I didn’t use in the end, but I had yet to really formulate at strategy, and as I tell my students, you can never have too much information. I then thought I’d find the angles, but realised that it is this defaulting to trigonometry that often leads me to overcomplicate matters, so I thought I’d leave that til later, (plus I don’t know tan 2 or tan 0.5 off the top of my head.)

I considered the area of the triangle, then sketched the next triangle I knew stuff about fyz. It was here I saw my strategy.

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The right angles meant I could use congruent triangles.

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I could work out the area of the triangle yxk, which is congruent to fyz, as half the parallelogram area is 32, which is made up of this triangle and one which has area 8.

Thus the other leg of the right angled triangle must be 6rt2, and so a=12rt2 (as y is it’s midpoint!)

From there it was a question of Pythagoras’s Theorem to find b.

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A fantastic little puzzle, one that I enjoyed solving, and one which should be accessible to higher GCSE learners. If you haven’t already do check out Ed’s website.

Rectangle Puzzle

December 27, 2014 Leave a comment

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

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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:

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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.

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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:

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Redrew my similar triangles and solved:

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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!

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