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I was asked an interesting question by a friend:

You are given half a circle and a line $d$ of length $6$ is given. You build a perpendicular to the base line (the diameter of the half-circle) and complete a rectangle, such, that the top side is tangent to the circle (see an animation below). What is the area $A$ of the resultant rectangle.

rectangle construction animation

So, I have a very quick question,

if I have all the data needed to calculate the area $A$?

And, since I've been asked by another human, I assume that the answer is yes. So I conclude, that the while the area is a function $f(r, d)$, where $r$ is the radius of a circle, it should be that $r = r(d)$, and in fact it follows that $A = f(d)$. Then I pick a convenient choice of a circle and calculate the area in my head. Basically, setting $d=2r$ is one such particularly easy choice, and $$A = \frac{d^2}{2}.$$

However, I have cheated, and now I need to establish somehow, that changing the circle, have no impact on the area. (I am able to check another choice of setting an angle to $45$ degrees between a diameter and $d$).

Hence, I need to make calculations (using pen and paper). On the image below, I have added $x$ to be the length of the line to the left of the perpendicular. Then by triangle similarities we have the following:

$$ \frac{d}{2r-x} = \frac{2r}{d}$$

or, more compactly, $\frac{d^2}{2} = r\cdot (2r-x)$. Which is accidentally the area.

some additional notes

I, personally, do not like to make heavy calculation in problems, where some simple reasoning should be applied. Therefore, I am looking for a simpler, just-in-head, solution, to why the area of the rectangle on the animation below, is independent of the circle radius?

animation of the changing rectangle

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    $\begingroup$ If you are familiar with Euclid's proof of Pythagoras' theorem, your rectangle is just one half or rectangle $KCLE$. $\endgroup$ Commented Mar 18, 2022 at 18:12
  • $\begingroup$ I can't help it. +1 (also) for the use of animation to create an outstanding presentation of a Math problem. $\endgroup$ Commented Mar 18, 2022 at 18:17
  • $\begingroup$ Call $x$ the base of the rectangle (what you are calling $2r-x$). Then, the similarity you wrote becomes $d/x = 2r/d$. That is, $xr$ is constant. This is the area, because $r$ is the height of the rectangle. Not too geometric, I agree. :-) $\endgroup$ Commented Mar 18, 2022 at 18:30
  • $\begingroup$ @Intelligentipauca Could you elaborate? I don’t exactly see what’s what… $\endgroup$
    – Milten
    Commented Mar 18, 2022 at 19:55
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    $\begingroup$ @Milten, if you look at the image Intelligenti pauca referenced us, you can see that the bottom blue rectangle shares the same basis with the rectangle painted in the images of mine. However, its altitude is exactly $2r$, and not $1r$. Then in the Euclid's proof we know, that it's area is equal to the area of the square build on $d$ !! $\endgroup$
    – dEmigOd
    Commented Mar 18, 2022 at 20:22

1 Answer 1

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I will attach an image and reference the reader to the Euclid's proof of Pythagorean theorem (Euclid's proof).

This answer is credited to @Intelligenti pauca.

  1. The blue rectangle and square possess equal area (this is the Euclid's part).
  2. The green rectangle (was blue on my original drawing) and the bottom blue rectangle share the same base, while the bottom blue one is twice as high.

Some Euclid's reasoning to why

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