Here is a nice little puzzle I saw from brilliant.org on Facebook.

Have you worked it out yet?

Here’s what I did:

First I drew a diagram (obviously).

And worked out the area of the triangle.

Then the area of each sector.

Leaving a subtraction to finish.

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I split the equilateral triangle down the middle making a right angled triangle. Used Pythag to find the missing side (sq root 3) – calculated the area of this and multiplied by 2, to find area of original equil. triangle.

For sectors – each has angle of 60 – place next to each other to make semicircle. Calculate area (pi/2).

Shaded pink is the difference between these (root 3 – pi/2).

Lovely method.

You seem to have taken a long way around here I reckon. Blue bits is half a unit circle – so radius is irrelevant in formula (180/360), so that’s pi/2, area of equilateral triangle is sqrt(3) (Pythag 1 + x^2 = 4) Just need height because 1/2 base is just 1.

I often do take the long way round!

Nice puzzle. I like the questions from Brilliant. A lot of them are real thinkers.

Yes, they certainly are. Their Facebook page keeps me sharp! (Although I wasn’t too sharp this morning.)

I don’t think I’ve ever considered my maths skills as anything other than amateur! I think school bashes into us all that ‘good at maths’ means elite genius who only ever gets right answers in the most efficient ways.

Haha. It certainly does in some places.