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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.

  1. December 2, 2017 at 11:22 am

    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.

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