Home > #MTBoS, A Level, Maths, Pedagogy, Resources, Starters > Area of a semi-circle – the return

Area of a semi-circle – the return

Last week I wrote this piece in which I relayed my thought process when solving this problem from Ed Southall (@solvemymaths).

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After sharing it a lot of people tweeted me their solutions and I thought I’d look at a few here. I think the variety of solutions people used, and the amount of people who engaged with this puzzle is testament both to how interesting a puzzle this is and also how fantastic maths is. If you did it a different way, do let me know!

My solution was fairly long compared to many of the others. I think this maybe down to my heavy preference to Trigonometry, and my brain’s insistence on using it if possible! Here are some other, more concise methods:

Using the Tangent Ratio

In my solution I used right angled triangle trigonometry to find the lengths of the chords, then used Pythagoras’s Theorem to find the diameter. Steve Atkinson (@Small_Ears) also used trig, spotted that as tan a = 2 realised that tan b had to be 2 and thus the remainder of the diameter must be 12. A much more efficient and concise method, but not as much fun!

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Steve’s method made me realise we are dealing with three similar triangles, and that fact alone is enough to know the ratio of the sides are equal, and hence the rest if the diameter must be 12. This idea was the one used by Rob Rolfe (@robrolfemaths).

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Drawing a line from the centre

Jo Morgan (@mathsjem) tweeted this solution:

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Which I think is brilliant. Instead of calculating the rest of the diameter, Jo went for the rest of the radius, using a line from the centre to the point the perpendicular touches the circumference. Pythagoras’s Theorem and algebraic manipulation drop the radius out quickly implying the answer in great time.

The I only need Pythagoras’s Theorem method

This was tweeted to me by Mr Draper (@mrdrapermaths):

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As you can see, he has used Pythagoras’s Theorem on all three triangles and solved the simultaneous equations to find the necessary information. Possibly not the most concise method, but certainly no less concise than my approach, and one that has a lovely feel to it.

Constructing the geometric mean

Martin Noone (@letsgetmathing) tweeted this fantastic solution:

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It is, I believe, derived from an old method for constructing square roots that I think dates back to ancient Greece. This rule is illustrated in hospital top sketch, then he’s enlarged it to ensure the one is a three. He later tweeted the proof, which he uses as an exercise in lessons and one I may start to use too!

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Ed Southall then tweeted this pic which shows the general rule and how a semi-circle can be used to construct the geometric mean of two numbers, which itself could be used to find the diameter.

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And is a great visual proof that the geometric mean can never exceed arithmetic mean.

Coordinate geometry

Stuart Price (@sxpmaths) informed me that he’d enjoyed the puzzle and left it in his board for his students to complete, and that they used a variety of methods. He mentioned coordinate geometry and equations of circles, so I thought I’d have a crack at that method:

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I drew the axes so that the origin was at the point on the diameter where the perpendicular touches it. Ie 3 cm in from the circumference. I considered the general equation of a circle:

(x-a)^2 + (y-b)^2 = r^2

Then subbed b=0 in as the centre lies on the x axis.

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I know two points on the circumference, (-3,0) and (0,6) so I subbed these in to get two simultaneous equations which I then solved.

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Once I knew a (the x coordinate of the centre) I could work out the radius, and hence the area. A brilliant solution.

Calculus

Stuart also mentioned calculus, specifically implicit differentiation. I’m not sure how this would help, to be honest, and I’d love to see this solution. I did, however, consider integration. I rearranged the equation of the circle:

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I started to integrate but realised I’d need multiple substitutions, so I ran it on Wolfram Alpha instead.

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An nice, but incredibly long winded solution!

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  1. December 8, 2014 at 10:56 pm

    The whole problem is a thing of beauty. From not only the original hypothesis started by Ed but also the number of amazing solutions to the problem. Thanks for pulling them together. As an aside I would absolutely love for an aspect of the end assessment to be of this form.

    No single solution is better than the other, a place where beauty is in the eye of the beholder but rather students would be free to solve the problem in whatever method they wanted.

    Surely this is the very essence of what we want higher level Maths to be in schools. It would reward the best students and at the very least get away from the staid and rote method of practicing methods. The main problem is that it tends to be geometry problems which lend themselves to these multiple method solutions but nevertheless it should be a style of question which is to built upon.

    I’d be really interested if @solvemymaths invisiged the amount of solutions people would come up with when designing it.

    Great work

    • December 9, 2014 at 6:08 am

      Aye, o certainly agree there. It would be great to see on tye test.

  2. January 27, 2017 at 3:05 pm

    I’m not sure I have seen the almost trivial solution using the height theorem $h^2 = pq$, which gives the answer $r = 15$ immediately.

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