## Ed’s Infamous Area Problem

Yesterday a colleague asked if I’d seen the maths problem that was going round and featured in the national press. I hadn’t, but was not surprised to see that it was Ed Southall (@solvemymaths) who had posed the problem that had got the world (well, the nation at least) talking maths. My initial thought was that it was great to see a positive discussion of maths in the press. Then I figured I’d need to solve it.

Here is the problem:

*What fraction of the area is shaded?*

What follows is my solution. Please attempt the problem before reading on, I’d love to see your approach.

Firstly,I did a sketch (of course I’d did. If you didn’t then why on earth not!)

I labelled the base of the rectangle 2x and the height b (it looked like a square, but I didn’t want to assume and figures if it was necessary to Ed would have told us). I realised that I was looking at 2 similar triangles (proof can be made using opposite and alternate angles), with a scale factor of 2 (the base of the bottom is double the base of the top). I know that when working with areas the scale factor is squares so using an area scale factor of 4, a for the height if the top triangle and (b – a) for the height of the lower triangle intake up with this equation:

Which solved to tell be b was 3a, thus b-a was 2a.

From here it was simple, I worked out the area of the shaded triangle and the whole rectangle put it as a fraction and simplified.

How did you do it?

## Saturday puzzle

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

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.

## Angle problem

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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