enter image description here

I have found this in a Facebook group :) The only thing we are given is the marked distance and we need to find the shaded area. I have tried several options but I am always missing some data. Then I tried the following: I considered two different conditions:

  1. The two side circles do not overlap; they are tangent to each other. In this case, the given distance corresponds to 3 diameters, so each is 3.33 units and the requested area is 100 sq. units.

  2. The two side circles overlap by 100%. Again the area is 100 sq. units.

It is therefore obvious that the area is independent from the amount of overlapping, i.e. from the angle of the specific line segment to the horizontal line.

Can you provide a geometrical solution? Many thanks!!

  • $\begingroup$ which group (I ask so that I could join)? $\endgroup$ – Invisible Nov 24 '20 at 18:19

Let's call $r$ the radius of a circle, and $2x$ the overlap region between two circles. Then the area is $$A=(2r)^2+4(2r)(4r-2x)=4r^2+32r^2-16rx=36r^2-16rx$$ Now let's take a look at one of the circles on the sides. Draw the perpendicular from the intersection point to the axis connecting circles. This has a length $h$. Using Pythagoras, $$h^2=r^2-(r-x)^2=2rx-x^2$$ Now looking at the given line, just take half, and we apply Pythagoras again: $$h^2+(3r-x)^2=5^2$$ Using $h^2$ from above, $$5^2=2rx-x^2+9r^2-6rx+x^2=9r^2-4rx$$ Then area is exactly four times this, so $A=4\cdot5^2=100$

enter image description here

In the above picture $DE=x$, $CD=h$, $AC=r$. Then $AD=r-x$

  • $\begingroup$ Andrei so 2x is the overlap segment, right? In other words, x is the vertical distance between the chord and the circle? The height of the circular segment? Thank you $\endgroup$ – Harry Gartner Nov 19 '20 at 15:50
  • 1
    $\begingroup$ I've added a picture to clarify things. $\endgroup$ – Andrei Nov 19 '20 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.