Home > #MTBoS, Maths, SSM, Starters, Teaching > Area of a semi-circle puzzle

Area of a semi-circle puzzle

If you have read this blog before, you may have noticed I enjoy a good mathematical puzzle ever now and then. This week I had a bit of time to pass and I thought I’d have a crack at this one from Ed Southall (@solvemymaths):

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Well almost, I didn’t have my phone and I misremebered it slightly, so actually had a good at this:

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As you can see, it’s the same puzzle but scaled by a factor if 1/2.

The first thins I did was label the 4 points. I knew ABD was a right angled triangle, as it is a semi-circle. We are told that ABC and BCD are right angled in the question. The calculate the area we need the radius, which should be easy enough to calculate through Pythagoras’s Theorem and Right Angled Triangle Trigonometry.

I used Pythagoras’s Theorem to calculate AB.

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I could then calculate the sine of angle ABC, which I called x. I know that ABC and CBD (which I called y) are complementary, and as such Sin x = Cos y,  so Cos y = 1/rt5

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This meant I could easily work out the length BD using the fact that Cos y = adjacent/hypotenuse.

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This left me with the two short sides of the right angled triangle ABD, so the hypotenuse (the diameter of the semi-circle) was easy to calculate:

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From this the radius, then the area, follow easily.

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When I came to type this up I realised I’d solved the wrong problem, so for completeness sake, to solve the original problem I can multiply by 4 (as I need to enlarge so lengths gave increased SF 2, and I’m dealing with area) giving a final answer of: 112.5pi (or 225pi/2).

I enjoyed this puzzle, and did it without a calculator. I think had I used a calculator, it may have lost some of its appeal. I would love to see questions like this appear on the non-calculator GCSE paper. The skills/knowledge needed are circle theorems, Pythagoras’s Theorem, Right Angled Triangle Trigonometry, Surds, Area of a Semi-circle and basic number skills. All of which should be within the grasp of a decent GCSE student. There are, of course, many other solutions, some of which are explored here.

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  1. December 6, 2014 at 9:53 am

    This puzzle caught my eye, too, so I wrote it on a whiteboard and left it for my Yr12/13 students to see. So many of them had a go and between them produced 5 or 6 variations of the solution! My way was to draw the full circle and use intersecting chords. Others used Pythagoras, trigonometry (with a calculator, but hey..), coordinate geometry, similar triangles (equivalent to intersecting chords), and even implicit differentiation of the equation of a circle!

    • December 6, 2014 at 9:59 am

      That’s cool, I love the variety!

  2. December 6, 2014 at 8:25 pm

    I drew a radius to the top of the 12cm line, making a right-angled triangle with lengths 12, (r-6) and r. Pythagoras gives r=15. Nice question.

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