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Angle problem

December 5, 2017 Leave a comment

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.

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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.

Dodgy Microsoft Graphics 

October 10, 2017 1 comment

So my new laptop arrived today and I quickly set about using it. It’s a Windows 10 laptop and as such has all the usual Microsoft stuff preloaded in it. I was going to set chrome as the default browser when it suggested I try Microsoft edge as it’s apparently faster and made for Windows 10. When I opened it it showed me this graphic:

Immediately I called shinanegans. The 5% difference between the green and the blue looked far too big. Initially I thought it was just down to the scale starting from 25000 and the size, but looking deeper there are also 4 extra sets if 5 notches on the blue which further add a to the illusion.

All in all a terrible diagram. Poor form Microsoft. Poor form.

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