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From the MindYourDecisions video "Challenging problem given to 12 year olds - square in a quadrant" on YouTube:

Original question:-

enter image description here

I have a doubt in the solution given:- Make 3 additional copies and complete the circle. There will be 5 squares in the shape of a +. One of the diameters of the circle will intersect the outer corners of 2 outer squares.

I have made a very rough diagram following what has being said, I get a shape of +. After that I am confused. {The purple, yellow, and orange squares are the ones I got by symmetry}

enter image description here

Let $s = \text{side of square}$. That diameter will be the hypotenuse of a right triangle ($2$). The legs are $s$ and $3s$. So, $$10s^2 = 4\to s^2 = 2/5$$

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  • $\begingroup$ Your answer of $2/5$ matches the video's answer of $0.4$, so what's your question? If you're wondering if your approach is correct: it is. (And I consider it an improvement over the one shown in the video.) Congratulations to you! (Or, congratulations to the person whose YouTube comment included that solution.) $\endgroup$
    – Blue
    Dec 25, 2021 at 5:58

1 Answer 1

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"...I have made a very rough diagram following what has being said, I get a shape of +. After that I am confused."

So I guess this is what you are asking about:

image

Apply Pythagoras' theorem to the obvious right triangle.

$$(3s)^2+s^2=2^2$$

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  • $\begingroup$ Why do I need 10 rep to post images? :\ $\endgroup$
    – ACB
    Dec 25, 2021 at 11:48
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    $\begingroup$ To avoid spam accounts, and to not overly burden the image hosting site. $\endgroup$ Dec 25, 2021 at 11:56
  • $\begingroup$ So every circular sector would have an inscribed square. Do we have a nice formula for the side in terms of $r$ and $\theta$ ( the central angle)? $\endgroup$
    – orangeskid
    Dec 27, 2021 at 7:18
  • $\begingroup$ @orangeskid , I don't think there is a 'nice' formula. But it can be easily derived using the method described in the video above. The method in my answer (actually it is not mine) can not give a formula for every $\theta$. $\endgroup$
    – ACB
    Dec 27, 2021 at 8:59

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