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Yes, Bob, it's the algebra I have difficulty with. But in addition, I don't see how you can look at what I originally described in my first post and be able to know each of the pieces that have to be solved and the formulas needed. The right triangle I can remember. But I am really very impressed with your knowing how to solve this -- and then being able to explain it in a way that a dunce like me can understand it.
I've made a diagram below that should help.
I made a square, r by r and took off a smaller amount 'a' from each side. Then I split the square into four parts.
First part is a square r - a on each side, so its area is
That's the area we need.
The second and third parts are rectangles, one is r by a, the other is a by r.
So together they have area
But if you put these three back together, to make the r by r square the rectangles overlap, so I'm counting some area twice.
Can you see that the amount I've got twice is
So to make the original square, r by r, you need
It is quite a common mistake to miss out that 2ra term.
Let's try with numbers to make sure we've got the right thing.
Can you see you don't get the right answer if you miss out the 2ra term?
So back to your question.
So here a = 0.11037
I hope that is clear.
Thanks, Bob, for all the work you put into this. With my math at a deficit, I tried reviewing some of the subjects available here to try and refresh my memory and understand each of the steps you illustrated. I also added the colored triangles in this image (i46.tinypic.com/demija.jpg) so I could better visualize what you were doing, and then I added the values of the triangle sides as they were calculated. (This was just a visual aid for me to try and avoid confusion in going between the two different views.)
Thank you so much, Bob. I printed it so I can make sure I understand it all. Not that I doubt it at all, I just want to make sure I do understand it. I'm amazed that it could be solved with Pythagoras. I thought it would be a lot more complicated than that. But it just shows how a brilliant mind can "cut through the weeds" to find only the necessary information. Thanks!
Oh good. I must have been making some sort of sense then. Brain now re-freshed. I have a plan.
Pythag on OFC
Pythag on FPC
square both sides
cancel the r squared and re-arrange
I will just check this ........
EDIT: checked. I cannot find any errors. Value seems small but fits with the original diagram where it was small.
You asked for the diameter. I'll leave that as an exercise for you.
That's exactly it! If a flat surface were to intersect with the sphere, or two spheres with one another, the intersection would be a circle. Yes?
Thanks in advance to anyone who might be able to help me. I'm not a mathematician (obviously), or much of an expert at anything worth mentioning. But I've searched all over for a solution, or a way to figure this out, and haven't found it. I ran across this site, and was fascinated with all the information available, but still didn't find an answer. Here's the problem: