## 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.*

Similarly did this in a slightly dozy state. I’d say your method is neater and mine adds unnecessary complexity, but I got there in the end!

I got the triangle the same way as you but then, probably because I was thinking about sectors, I saw that the square’s area would be the same as that of a right isosceles triangle with two sides length r.

1/2bc(sinA) gave me an area of 2.

Same thing with the rectangle, another isoceles with two sides r but this time with with an angle of 60 between them. 1/2bc(sinA) gave me an area of sqrt(3) and therefore an overall product of 6.

Nice.