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Posts Tagged ‘Semi-Circles’

Hippocrates’s First Theorem

April 11, 2016 5 comments

Over the half term I was doing some reading for my MA and I happened across Hippocrates’s First Theorem. (Not THAT Hippocrates, THIS Hippocrates!)

Here is the mention in the book I was reading (Simmons 1993):

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It’s not a theorem I’d ever come across before, and it doesn’t seem to have any real applications, however it is still a nice theorem and it made me wonder why it worked, so I set about trying to prove it.

First I drew a diagram and assigned an arbitrary value to the hypotenuse of triangle A.

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I selected 2x, as I figured it would be easier than x later when looking at sectors.

I then decided to work out the area of half of A.

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A nice start – splitting A into two smaller right angled isosceles triangles made it nice and easy.

I then considered the area b. And that to find it I’d need to work out the area the book had shaded, I called this C.

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Then the area of B was just the area of a semi circle with the area of C subtracted from it:

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Which worked out as the area of the triangle (ie half the area of A) as required.

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This made me wonder if it worked for all triangles that are inscribed in semi circles this way – ie the areas of the semicircles on the short legs that fall outside the semicircle on the longest side equal the area of the triangle.

My first thought was that for all three vertices to sit on the edge of a semi circle in this was then the triangle must be right angled (via Thales’s Theorem).

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I called the length eg  (ie the diameter of the large semi circle and the hypotenuse of efg) x and used right angled triangle trigonometry to get expressions for the two shorter sides ef and fg. Then I found the area of the triangle:

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Then the semi circles:
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I then considered the diagram, to see where to go next:

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I could see that the shaded area needed to be found next, and that this was the area left when you subtract the triangle from the semicircle.
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I could now subtract this from the two semi circles to see if it did equal the triangle.

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Which it did. A lovely theorem that I enjoyed playing around with and proving.

I think there could be a use for this when discussing proof with classes, it’s obviously not on the curriculum, but it could add a nice bit of enrichment.

Have you come across the theorem before? Do you like it? Can you see a benefit of using it to enrich the curriculum?

Reference:

Simmons M, 1993, The Effective Teaching of Mathematics, Longman: Harlow

Area Puzzle – Squares and Circles

January 31, 2015 1 comment

This morning I came across this puzzle from Ed Southall (@solvemymaths):

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And it seemed interesting. Looking at it is seems that A is the midpoint of the arc, so equidistant from b and the lower left corner of the square. This would mean that an isosceles triangle exists in the semi-circle, which in turn implies the square has side length 4rt2 (from Pythagoras’s Theorem, as the side of the square is a diameter).

This means the square has area 32cm^2. The semi-circles have radius 2rt2 and as such have area 4pi. That leaves just the white circle. The diagonal of the square is a diameter of this, and as the side lengths are 4rt2 the diagonal must be 8 (again via Pythagoras’s Theorem). This means the circle must have a radius of 4 and hence an area of 16pi.

So the sum of the orange areas must equal the square add the 4 Semi-Circles subtract the circle add the square.

Or:

2(32) + 4(4pi) – 16pi

Which simplifies to 64cm^2.

A lovely solution, and one which shows us that the areas of the orange crescents equal the area of the square.

Towards the end of the working I realised I could have used another property of Pythagoras’s Theorem, namely that the sum of the areas of semi-circles on the two shorter sides equals the area of the semi-circle on the hypotenuse. By splitting the square into two right angled triangles I could have reasoned that the 4 smaller semi-circles provide the same area as the large circle. Which means we would get the total area to be the area of the square, add the area of the circle (from the semi-circles) subtract the area of the circle (from the circle) add the area of the square. Which again simplifies to twice the area if the square. A much more elegant solution.

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