Problem Solving #4

I stumbled across this question in a book yesterday:



I know the problem, and in fact have solved it a few times before, but upon looking at it I couldn’t quite remember how I solved it, so I thought I’d have another stab at it. It didn’t go well…!


I knew I needed something like this diagram, because there’s just nothing else I can think of doing with the tiny amount of information I’m given.


More specifically, I just need this half (I think). But this is where I got into a pickle. I stared at the above diagram for a little while, convinced it was what I needed, but completely lost at how it helps me. Sure I have one side, and what I need is part of the triangle, but surely I need *more* information??

Now as I said, I have solved this problem a few times, and in fact, the first time it was shown to me I solved it in about a minute or two, but for whatever reason I couldn’t for the life of me get past this point yesterday. The reason why I couldn’t progress was simple – I wasn’t following my own advice on solving problems. Specifically, I’d recommend to students just finding as many things as you can until you find a way in. So after staring at that little triangle for far too long, I decided to just use Pythagoras anyway, despite “knowing” that I didn’t have enough information for it to be useful.

R^2 = r^2 + 25

So what? I surely needed to replace R with some kind of expression using r and a number… but I don’t have that information, so I have a ‘useless’ two variable equation and nothing else to work with.

I re-read the question and realised I’d been hung up with the seemingly useless application of Pythag instead of focusing on the fact that the question concerns area. Finally, I had that moment of clarity.

\pi R^2 = \pi r^2 + 25\pi

The bit I want is the shaded bit, so I can effectively ignore \pi r^2



Weird how it took me so long to ‘see’ it this time, and I spotted it so quickly a couple of years ago. I probably shouldn’t overthink that…


6 thoughts on “Problem Solving #4

  1. Note that the area of the inner circle is abitrary, so you can effectively eliminate it and treat 10 as the radius of the greater circle. Lo, 25 pi.

  2. This reminds me of the 3-D problem of a “de-cored apple” (i.e. a perfect sphere with a concentric cylinder removed) of height 10cm (when placed on the edge of the hole). What is it’s volume?

  3. One way to tackle this type of problem is to observe that the answer is the same whatever the radius of the big and small circles (as long as the tangent is 10cm long). So in particular the orange shaded area will be equal to the orange shaded area when say R=13 and r=12 (using the Pythagorean triple 5, 12, 13). In this instance the area of the large circle is Pi x 13 x 13 =169 Pi and the area of the small circle is Pi x 12 x 12 = 144 Pi so the answer is 25 Pi.

    You can even take the degenerate case when r=0 and the tangent is the diameter of the large circle. So R=5 and again the area of the orange shaded area (in this case all of the large circle since the small circle is a dot with no area) = Pi x 5^2 = 25 Pi.

    Even if you don’t like assuming that the answer is the same whatever r and R are it still may be helpful to look at some numerical examples which naturally don’t involve any algebra.

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