## Pythagoras type maths problem

(27 Posts)Can you help me understand ds's (Yr 5) maths homework.

It started off really easy - draw 7 squares, starting with sides of 1 cm and going up to 7 cm. Measure the diagonal.

Then it asked, what can you work out about the relationship of the length of the sides to the length of the diagonal? Can you explain this?

OK, so the lengths of the sides and diagonals are roughly as follows (rounding a bit):

1cm - 1.4cm

2cm - 2.8cm

3cm - 4.2cm

4cm - 5.6cm

5cm - 7cm

Right - we could work out the relationship between the lengths -

diagonal = length of the side + (length of the side x 0.4).

Why though? How to explain this? Can you help Oh wise ones?

brilliant conversation here, more please more. It is not sad to help each other with maths. I am an ex maths teacher and now maths author, can't say what of for obvious reasons BUT this sort of helpful support is a great thing for learning maths, thanks all

Senua, you are not sad at all. Maths is a great subject and if you did not fully appreciate that at school, now's your chance to catch up. Maths is all around us, so a good understanding of the subject is necessary to understand the world. Enjoy learning with your children. They will appreciate that.

oh hello. didn't see that you had come back.

I think that homework is great for parents- I have learned so much ~~helping DC with homework~~ second time round!

I like the graph paper answer. Get DS to draw the 1cm square (call it S) and cut it in half to make the triangle (call it T). Next get DS to draw the 2cm triangle and break it down into 1cm chunks: you can *see* that it is one S and two T. The 3cm triangle is three S and three T. The 4cm triangle is six S and four T. etc etc. You can actually *see* and *count* the little hypotenuse along the hypotenuse of the bigger triangle.

I love Maths: you can approach it so many different ways but it still comes back to the same answer!

You got full marks if you recognised the pattern. The teacher said that ideally the children should have expressed the pattern as an equation (which ds had done). He said he just wanted the kids to think why that pattern might be there. I think this is great but sometimes I think that schools overlook the fact that parents did not have homework in primary school and are therefore often slightly flummoxed by it.

At that level would the answer not be a bit more pattern related. I.e that the length of the diagonal on 3 is the same as 2 +1 and 5 is the same as 2+3or 4+1. Or that the diagonal will always be the length of the side x the diagonal of the 1cm sided triangle.

Hi Goosey. So what was the 'expected' answer? - pythagorean algebra or a simple "it's a constant ratio / linear relationship"?

LaBelle - I wasn't peeved, just wasn't sure where to go with it. As Cecily said, without just giving ds the answer, I was a bit stuck. A few more instructions might have helped. Ds loves questions like these - he loves having a chance to really think about numbers so don't be put off.

If you set it - even for your high ability group - it might be best to set it as a group exercise in school so that the parents won't just give the children the answer.

I was planning to set something similar for my mixed-age year5/6 class. No longer! I see that it is far too hard and will keep it for the highest ability group. Thank you for saving me from some rather peeved parents.

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What about drawing the squares on graph paper of 1cm squared. It will then be easy to see that the diagonal of a 2cm square is twice as much as 1cm square and 7cm square is 7 times, etc? This would help if he is a visual learner.

They're asking the Y5's to work out for themselves what Pythagorus worked out all those years ago. It is quite a tall order really because, I would assume, Pythagorus himself was was considerably older than 9 or 10 when he managed this.

Difficult one because if parents (or the combined power of mumsnet) help, we already know the answer, so the child is not doing the working out.

Noble - he is one of those kids with an intuitive grasp of numbers and once he has a concept explained to him, can apply it and understand it. He really, really loves maths. It is his thing. Sadly for me, it is not really mine.

Not sure that the teacher really expected them to provide much of an explanation, but he likes to ask challenging questions to try and make the kids "think like mathematicians" rather than just crunching the numbers. Ds loves him.

a2 + b2 = c2

Where c is the diagonal

a=b for a square

Therefore

2a2 = c2

c = a x square root of 2

c = 1.1414 a

Hope that makes sense

That is not y5 maths. Is this a school in Hong Kong?

I have to ask, is your DS exceptionally advanced at maths? I would hesitate to give that homework to a Y9 class!

Thanks noblegiraffe. I wish I had an instinctive feel for numbers. I really like the way they work but I need help with them sometimes. I just thought I might have got a little further.

Another way to think of it would be that the diagonal of the 1cm triangle is 1.4, the 2cm one is double the size so its diagonal is 2x1.4, the 3cm diagonal is 3 times as big so 3x1.4 etc.

You got this as length of side + length of side x 0.4 is the same as doing length of side x 1.4.

Which skips the need for Pythagoras and square roots, which might be easier to get.

Yay!

Am I also a teeny bit sad to get excited about helping someone become excited by Maths?

Fab and thanks very much - I think I get it now.

DS loves maths and just "gets it" in a way I have always struggled to. I refuse to allow my 9 year old's homework to be beyond me though! I am quite excited by this now (although am aware that that may be a teeny bit sad).

The tricky bit is

the square-root of (2a^2)

That is lots of multiplying together but, with multiplying, it doesn't matter in which *order* you multiply.

So you could sort out the (2 times a^2) bit, and then find the square-root of that.

Or you could consider separately **2** and **a^2**, find the sqaure-root of each and *then* multiply them together. The second version gives you the elegant answer because it highlights the point that "the square-root of a^2 is always a".

You could labour the point and say this is why we use **a**, **b**, **x**, **y** and **z** in Maths: it is precise and shows the *thinking* behind the problem in a way that merely getting out your ruler would never do.

I am struggling. I remember a2 + b2 = c2. ds had been told this (I think) at school.

It's this bit I can't quite get my head around: or (sq-root of 2) x (sq-root of a^2)

Is there a simpler way of putting it for the matematically challenged amongst us?

Call the horiontal and vetical sides of the triangle **a** and **b**, and the hypotenuse **h**

Pythagorus gives us a^2 + b^2 = h^2.

To rearrange, h is the square root of (a^2 + b^2)

However we have a special case where a=b so

h is the square root of (a^2 + a^2)

or sq-root of (2a^2)

or (sq-root of 2) x (sq-root of a^2)

the square root of a-squared will always bring you back to a

so the answer will always be **a** times the square root of 2 (which is 1.414)

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